tag:blogger.com,1999:blog-69953538317331681872024-03-16T12:47:27.069-04:00ALTEREDZINEThe purpose of this blog is to teach Math subjects : Calculus, Pre-calculus, Algebra and Basic Math. Now it focuses on Higher Math. You can also find posts on learning theories, study skills and web resources for learning. Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.comBlogger155125tag:blogger.com,1999:blog-6995353831733168187.post-83638872508955841852024-03-15T23:43:00.009-04:002024-03-15T23:51:12.300-04:00Functions of two variables<p> Goal: recognize a function of two variables and identify its domain and range</p><p><b>Function of two variables</b></p><p>The definition of a function with two variables is similar to that of the function with one variable. The only difference is that we map a pair of two variables to another variable instead of mapping one variable to another variable.</p><p><b>Definition</b></p><p>A function of two variables z = f(x,y) maps each ordered pair (x,y) in a subset D of the real plane R² to a unique real number z. The set D is called the domain of the function. The range of f is the set of all real numbers z that has at least one ordered pair (x,y) 𝜺 D such that f(x,y) = z as shown in the following figure:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgstrInU5ofstztPtW80_mRoDqwXBgktenNXSClgz7WEdWUtxD3-ycM2_G4lOb787LMmBgjURkTkTLOcBappm81zmcf_OedETiQ-SDJ-DVqfFSlBWTB3jHP6be1d5pQB2hHT3R_6RUauVMsX_yCMmdIHb-Qh_fY2nUQ7KD-SBe0xhCBfnuTq463wpWHP74" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="210" data-original-width="489" height="137" src="https://blogger.googleusercontent.com/img/a/AVvXsEgstrInU5ofstztPtW80_mRoDqwXBgktenNXSClgz7WEdWUtxD3-ycM2_G4lOb787LMmBgjURkTkTLOcBappm81zmcf_OedETiQ-SDJ-DVqfFSlBWTB3jHP6be1d5pQB2hHT3R_6RUauVMsX_yCMmdIHb-Qh_fY2nUQ7KD-SBe0xhCBfnuTq463wpWHP74" width="320" /></a></div><br /><br /><p></p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><b style="text-align: left;"><br /></b></b></div><div class="separator" style="clear: both; text-align: center;"><b><b style="text-align: left;">Examples </b></b></div><div class="separator" style="clear: both; text-align: center;"><b><b style="text-align: left;"><br /></b></b></div><div class="separator" style="clear: both; text-align: center;"><b><b style="text-align: left;"><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgZV8OUBqQWUML-kdq-LRHhXUxXwIEipGFGGVnAa5Ce1PfSkfylPJDiPzVgrkG_o7QN0UTo0pRir09yAReiQ-XvBajB3lZrnz9YmDmdABNg9bqRFGvaCLwMaTY8ZiRYmq-gv1M2BHJU0Kvl_J9ZIa9tvk6GH5i6xokhjnWn5rdmFbkn5qdk3xpFfZf5QhM" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="98" data-original-width="447" height="70" src="https://blogger.googleusercontent.com/img/a/AVvXsEgZV8OUBqQWUML-kdq-LRHhXUxXwIEipGFGGVnAa5Ce1PfSkfylPJDiPzVgrkG_o7QN0UTo0pRir09yAReiQ-XvBajB3lZrnz9YmDmdABNg9bqRFGvaCLwMaTY8ZiRYmq-gv1M2BHJU0Kvl_J9ZIa9tvk6GH5i6xokhjnWn5rdmFbkn5qdk3xpFfZf5QhM" width="320" /></a></div><br /> </b></b></div><div class="separator" style="clear: both; text-align: center;"><b><b style="text-align: left;"><br /></b></b></div><p></p><p><b>Solution</b></p><p>a. This is an example of a linear function with two variables. There is no pair of variables (x, y) for which the function is not defined. Therefore the domain of the function is R².</p><p>To determine the range of the function, let's determine the set of reals z for which f(x,y) = z. We have 3x +5y + 2 = z. Let's solve this equation by choosing x = 0. We have 5y + 2 = z. y = z-2/5. The pair (0, z-2/5) is a solution of the equation 3x + 5y + 2 = z for any value of z. The range of the function is R.</p><p>b. For the function g to have a real value we need 9 - x² - y²≥ 0 or - x² - y²≥ -9 x² + y² ≤ 9 </p><p>The domain D is defined as follow: D = {(x,y)ε R²/ x² + y² ≤ 9}.The graph of this set of points is described as a disk of radius 3. The graph includes the boundary as shown below.</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgbxNXw_WwWT-laKrpQ5ImuEzMYhY4_-8gioAm6CpcP-Idv5oxq1E4j64IADTixm8XT9zvRloyMNUf8FggEoKUjWay3VxR-5dV33tA9SEnthIgNDtr0vfNGFLUgoYQOBylzOUY1JVsioSgR7iY-tMYqW4F2GnZc1e1NF37HiydDyFq0WDAMaosIVXHFbaU" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="422" data-original-width="417" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEgbxNXw_WwWT-laKrpQ5ImuEzMYhY4_-8gioAm6CpcP-Idv5oxq1E4j64IADTixm8XT9zvRloyMNUf8FggEoKUjWay3VxR-5dV33tA9SEnthIgNDtr0vfNGFLUgoYQOBylzOUY1JVsioSgR7iY-tMYqW4F2GnZc1e1NF37HiydDyFq0WDAMaosIVXHFbaU" width="237" /></a></div><br /><br /><p></p><p> To determine the range of the function, we have to find out the set of reals for which g(x,y) = z. The domain is made of circles starting from (0, 0) and ending at the boundary circle defined by x² + y² = 9. Let's find z for (0,0) i.e a point of the domain starting at the origin. We have g(0,0) = z. </p><p>g(0,0) = ⎷9-(0)²-(0)² = ⎷9 = 3. Let's take a point of the boundary circle i.e (0,3). We have g(0,3) = ⎷9-(0)² - (3)⁰ = 0. The range is [0,3].</p><p><b>Practice</b></p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhQdRgfeQx12TsOjNd1Id7gMwGY-aKBOPbCNFMZw5aenNh7DcOg2c-3vdlzN3MtYBA8FvqEUuw2fqZhtMxrVnbpJ9xkymMploGsvExQAENw966GsqSyg0OfEpqh_eN8T48SPOk3M5q6S-Cd5RGvhQV9jJmXYHu4NT0W9Ln7tiOSYf8xMxhuoeeXIQN7hUs" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="49" data-original-width="549" height="29" src="https://blogger.googleusercontent.com/img/a/AVvXsEhQdRgfeQx12TsOjNd1Id7gMwGY-aKBOPbCNFMZw5aenNh7DcOg2c-3vdlzN3MtYBA8FvqEUuw2fqZhtMxrVnbpJ9xkymMploGsvExQAENw966GsqSyg0OfEpqh_eN8T48SPOk3M5q6S-Cd5RGvhQV9jJmXYHu4NT0W9Ln7tiOSYf8xMxhuoeeXIQN7hUs" width="320" /></a></div><br /><br /><p></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p> </p><p><b><br /></b></p>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-70146711267643961472024-03-02T14:46:00.000-05:002024-03-02T14:46:14.345-05:00Arc length in polar curves (continued)<p> In the last post, we set the formula for the arc length of a polar curve. Now let's do an example. </p><p><b>Example</b></p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgqor5ABNL3by3rZb7cKFVhFAGZUFyZj5VW32Nvr0zmNlj05fOFKhGrMgTDPKVFuBdoTCdmoMwG1sh-NxdceFEVkouFCdTeZUijeOwpWlQUPFix1MyN-8h9I6EByMAzo3tTgMkXasU5g5oYSl5zsAS1hrBy1lstAioyGrW6-2HvYPC2SWvHFFxLkUB6G2Y" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="86" data-original-width="466" height="59" src="https://blogger.googleusercontent.com/img/a/AVvXsEgqor5ABNL3by3rZb7cKFVhFAGZUFyZj5VW32Nvr0zmNlj05fOFKhGrMgTDPKVFuBdoTCdmoMwG1sh-NxdceFEVkouFCdTeZUijeOwpWlQUPFix1MyN-8h9I6EByMAzo3tTgMkXasU5g5oYSl5zsAS1hrBy1lstAioyGrW6-2HvYPC2SWvHFFxLkUB6G2Y" width="320" /></a></b></div><div class="separator" style="clear: both; text-align: center;"><b><br /></b></div><div class="separator" style="clear: both; text-align: center;"><b>Solution</b></div><div class="separator" style="clear: both; text-align: center;"><b><br /></b></div><div class="separator" style="clear: both; text-align: center;"><b> </b>When θ<b> = 0, r = 2 + 2cos0 = 2 + 2 = 4. </b>As θ goes from 0 to 2ℼ, the cardioid is traced exactly once. Therefore 0 and 2ℼ represent the limits of integration. Using f(θ) = <b>2 + 2cos0 , ɑ</b> = 0 and β = 2π<b> , </b>the formula for the arc length becomes: </div><div class="separator" style="clear: both; text-align: center;"> </div><div class="separator" style="clear: both; text-align: center;"><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgqTbUbbnNR_8mCTXAefIoBf6HKXXXobsnitocshZNT5bfGT7we4TbvQR1mENo0rtagl5ON7j__XiJoWzL0s3M9YyHC2iqjxwb_3nndAXaX7MFwuU6KQ6U5tdR_XFU6PYH-k6VjoZ7IW9kHYcjmuL0Z8EK1nSqFVIFXLcyPrZo16G_7K2t3St7TveMwk-k" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="360" data-original-width="397" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEgqTbUbbnNR_8mCTXAefIoBf6HKXXXobsnitocshZNT5bfGT7we4TbvQR1mENo0rtagl5ON7j__XiJoWzL0s3M9YyHC2iqjxwb_3nndAXaX7MFwuU6KQ6U5tdR_XFU6PYH-k6VjoZ7IW9kHYcjmuL0Z8EK1nSqFVIFXLcyPrZo16G_7K2t3St7TveMwk-k" width="265" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div>We have 1 + cosθ = 2cos²θ/2. Multiplying by 2: 2 + 2cosθ = 4cos²θ/2. </div><div class="separator" style="clear: both; text-align: center;"> </div><div class="separator" style="clear: both; text-align: center;"> <a href="https://blogger.googleusercontent.com/img/a/AVvXsEjG6M5VcWwoaqAVGZUL6jDLXY4In-zlb-zIuIFPorWhbQF-8U0h4Y80gWQ9ee_5CJUZ3MCZIRnuR-9jcT9zAzhX11f3txJHkd1LdIZXDCPBp1u5uTgC2XsiN1tI5zfa-e2Jw7ygFyHiBv3oVMLv5l3CFgfRoKiSTtSMvIf4A2q77excBWrinnU1ecdMsh0" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="197" data-original-width="250" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEjG6M5VcWwoaqAVGZUL6jDLXY4In-zlb-zIuIFPorWhbQF-8U0h4Y80gWQ9ee_5CJUZ3MCZIRnuR-9jcT9zAzhX11f3txJHkd1LdIZXDCPBp1u5uTgC2XsiN1tI5zfa-e2Jw7ygFyHiBv3oVMLv5l3CFgfRoKiSTtSMvIf4A2q77excBWrinnU1ecdMsh0" width="305" /></a></div><div class="separator" style="clear: both; text-align: center;"><div class="separator" style="clear: both; text-align: center;"><br />The absolute value is necessary because cosine is negative for some values of its domain. To resolve this issue, change the limit from 0 to ℼ and double the result. This strategy works because cosine is positive between 0 and ℼ/2 . </div><div class="separator" style="clear: both; text-align: center;"> </div><div class="separator" style="clear: both; text-align: center;"><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgbCGgIkCR8tiwdLx6Yfm8dDt70zwwv0n2_CeVmbugFqLjDfXbVJTzPQQ8n-ImLQxls64FBtKSaiq44yeoYwGOs91m_-tr5IWBBZ1vtUi9fVb5P7YgrogalC5oyWlcpNDGcU6MsSnFfIf7lX9ynM8Kkq0myeSvwfiAf2gVsaFvR-99vNF7l9mDGTuUmjUQ" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="180" data-original-width="216" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEgbCGgIkCR8tiwdLx6Yfm8dDt70zwwv0n2_CeVmbugFqLjDfXbVJTzPQQ8n-ImLQxls64FBtKSaiq44yeoYwGOs91m_-tr5IWBBZ1vtUi9fVb5P7YgrogalC5oyWlcpNDGcU6MsSnFfIf7lX9ynM8Kkq0myeSvwfiAf2gVsaFvR-99vNF7l9mDGTuUmjUQ" width="288" /></a></div><br /><b>Practice</b> </div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;">Find the arc length of r = 3sinθ </div><br /> :</div><div class="separator" style="clear: both; text-align: center;"><b> <div class="separator" style="clear: both; text-align: center;"><br /></div><br /> </b></div><div class="separator" style="clear: both; text-align: center;"><span style="text-align: left;"> </span></div><div><br /><p></p></div>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-39238893120679458022024-02-24T14:38:00.000-05:002024-02-24T14:38:11.334-05:00Arc length in polar curves<p><b>Goal : </b>find<b> </b>a formula<b> </b>for the<b> </b>arc length of a curve in polar coordinates</p><p><b>Arc length of a curve in polar coordinates</b></p><p>To find the formula for the arc length of a curve in polar coordinates, let's start from the formula of the arc length of a parametrized curve (x(t), y(t)) for a≤ t ≤b in rectangular coordinates.</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEg1xT4W01vXCv1oi_6zb9Uv6kx7O9p8WIUAn1ZQ_lI8GSryb1agU_VZiDswTGJHPXg5JMEg4tQeIPgx_4hAs4fBjUZ14l99IfRjS3A3kGN11g3c7zWC4PInOQ_ASiJ0sHKq0cegtJl8zShsrpXN-TAzU1MlnFRy5SFmaksB3C78Scw8o0wFIUzP15JANF8" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="70" data-original-width="272" height="82" src="https://blogger.googleusercontent.com/img/a/AVvXsEg1xT4W01vXCv1oi_6zb9Uv6kx7O9p8WIUAn1ZQ_lI8GSryb1agU_VZiDswTGJHPXg5JMEg4tQeIPgx_4hAs4fBjUZ14l99IfRjS3A3kGN11g3c7zWC4PInOQ_ASiJ0sHKq0cegtJl8zShsrpXN-TAzU1MlnFRy5SFmaksB3C78Scw8o0wFIUzP15JANF8" width="320" /></a></div><br /><br /><p></p><p></p><div class="separator" style="clear: both; text-align: center;"><br /></div>In polar coordinates the curve is defined by r = f(θ) and we also have:<div>x = rcosθ = f(θ)cosθ and y = rsinθ =f(θ)sinθ . Let's calculate dx/dθ and dy/dθ:</div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhhoZ3TD6njuVSbPNJbrxhDZxxVi4MmEV6B-Ry11FYxGjwOruGFtu8006IGxRa_OXeW5dqyqE0L-hEa4AWSv6JXxBhCNYkEHWwv8Hs2d0aiIVp23ngZdkyDGSqB2MJno1ZxOpcVRr8fJuksnySVmlRMvZ794qNfb8RLHVtDlOAOPZ4hjMSdxfh-BRhoMXg" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="75" data-original-width="254" height="94" src="https://blogger.googleusercontent.com/img/a/AVvXsEhhoZ3TD6njuVSbPNJbrxhDZxxVi4MmEV6B-Ry11FYxGjwOruGFtu8006IGxRa_OXeW5dqyqE0L-hEa4AWSv6JXxBhCNYkEHWwv8Hs2d0aiIVp23ngZdkyDGSqB2MJno1ZxOpcVRr8fJuksnySVmlRMvZ794qNfb8RLHVtDlOAOPZ4hjMSdxfh-BRhoMXg" width="320" /></a></div><br /><br /></div><div>Let's replace dt by dθ and a and b by ɑ and β, which define the limits of integration of the curve in polar coordinates, in the formula for the arc length above:</div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEiCGBuqvNQIz9To6PXseACkYX8W6NtwBNyudvwBqMMJrgqOgefi2ogoXiW3flxgJ6Ui4kVy3MwWNJ5PScOnBLqMkJ5UkcfPdGjUtCnJZU-ReInz3eYZQLoTJHy9co_4v7pLzQZXQSGsOFjVqM7TxE7RNO3-_W7T_p0QT_VPeFJkaGMse18Th2Wo8bZzAYI" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="380" data-original-width="594" height="205" src="https://blogger.googleusercontent.com/img/a/AVvXsEiCGBuqvNQIz9To6PXseACkYX8W6NtwBNyudvwBqMMJrgqOgefi2ogoXiW3flxgJ6Ui4kVy3MwWNJ5PScOnBLqMkJ5UkcfPdGjUtCnJZU-ReInz3eYZQLoTJHy9co_4v7pLzQZXQSGsOFjVqM7TxE7RNO3-_W7T_p0QT_VPeFJkaGMse18Th2Wo8bZzAYI" width="320" /></a></div><br />This leads to the following theorem:</div><div><br /></div><div><b>Theorem</b></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEilcHfaKgMWaZg24KlWjfd7j72v9hl6ujRxaGEMcTuQjviGB1mbguFHqglB2ZmbCp2xsKMGeb58DQ3kkfubnT4wKSS0d6PUcw2VzrvM0PewWCTl2jADreRJ2aPo6Rdxa6SY5CJbOIwt32hi8YGu4yoiabVUSTHLVFpo0ZAfH2IASKXReyqGkYmRv5pd84o" style="margin-left: 1em; margin-right: 1em;"><img data-original-height="102" data-original-width="754" height="86" src="https://blogger.googleusercontent.com/img/a/AVvXsEilcHfaKgMWaZg24KlWjfd7j72v9hl6ujRxaGEMcTuQjviGB1mbguFHqglB2ZmbCp2xsKMGeb58DQ3kkfubnT4wKSS0d6PUcw2VzrvM0PewWCTl2jADreRJ2aPo6Rdxa6SY5CJbOIwt32hi8YGu4yoiabVUSTHLVFpo0ZAfH2IASKXReyqGkYmRv5pd84o=w640-h86" width="640" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEiy8D7qfU24jO1hFQJfN-UzGWrKwMT89jKbdttUijo6-2M9CoAaOVGSyzbT4uob8WxDZwFHbRmeFw_UU8VyhRywu00Tvbx50I8FNopf36he78S_oQu8E0oW4-rFXraXgrCeaZ6Hie1g7IzlKZTK8uMpL-3NooWtcyptJGJnj8yf1AcVjb2-2IErQaW8-3Y" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="75" data-original-width="467" height="51" src="https://blogger.googleusercontent.com/img/a/AVvXsEiy8D7qfU24jO1hFQJfN-UzGWrKwMT89jKbdttUijo6-2M9CoAaOVGSyzbT4uob8WxDZwFHbRmeFw_UU8VyhRywu00Tvbx50I8FNopf36he78S_oQu8E0oW4-rFXraXgrCeaZ6Hie1g7IzlKZTK8uMpL-3NooWtcyptJGJnj8yf1AcVjb2-2IErQaW8-3Y" width="320" /></a></div><br /><br /><br /></div><div><br /><p></p><p> </p></div>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-76218685479693238542024-02-23T22:51:00.000-05:002024-02-23T22:51:14.921-05:00Area between two polar curves<p> <b>Goal</b>: find the area between 2 polar curves</p><p><b>Area between 2 polar curves</b></p><p>The procedure to find the area between 2 polar curves is similar to that of the area of 2 curves in the system of coordinates in the cartesian plane, We find the points of intersection between the 2 curves and identify the functions that define the outer curve and the inner curve respectively.</p><p><b>Example</b></p><p>Find the area outside of the cardioid r = 2 + 2 sinθ and inside the circle r = 6 sinθ</p><p><b>Solution</b></p><p> First, draw a graph containing both curves</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhmrgRElHJgJH6HEi9aBvssQz86ABuYAX6cAGpJGSr-J0SgtU3hj8gYqgU-Kp5hl5gaIgBP6UQLiHH4gYJiBhLKV0zPwdIXqmeNnWMqJ5JuEgWn9pRz8tAlTR9by592ZVlEz8zxxTNxhs0l2-wXobLUhbbDP2PwzZNZi2cZl8-Rcp0QxhmRSRvvWmANTF0" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="356" data-original-width="354" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEhmrgRElHJgJH6HEi9aBvssQz86ABuYAX6cAGpJGSr-J0SgtU3hj8gYqgU-Kp5hl5gaIgBP6UQLiHH4gYJiBhLKV0zPwdIXqmeNnWMqJ5JuEgWn9pRz8tAlTR9by592ZVlEz8zxxTNxhs0l2-wXobLUhbbDP2PwzZNZi2cZl8-Rcp0QxhmRSRvvWmANTF0" width="239" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div>To find the limits of integration, let's find the points of intersection by setting the 2 functions equal to each other and solving for θ<p></p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjOID0sV_j35QdIT_2NlP0PjPjVwK5AyutBWWdfzgKGP4gJJsSWduuwRRrtDGRRQ5G-GewcsITA_NZt1VYIlqRcVJEMa1rBDIJV70YFwUO8bt2fFGk5j_lYcVEAALlwRSL9_RZ_hBIosfC-ja41TLjS8FXJJ2WWFV2f9JvJBs7HY8pbfbCWGa-Kb7oX3KU" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="99" data-original-width="192" height="165" src="https://blogger.googleusercontent.com/img/a/AVvXsEjOID0sV_j35QdIT_2NlP0PjPjVwK5AyutBWWdfzgKGP4gJJsSWduuwRRrtDGRRQ5G-GewcsITA_NZt1VYIlqRcVJEMa1rBDIJV70YFwUO8bt2fFGk5j_lYcVEAALlwRSL9_RZ_hBIosfC-ja41TLjS8FXJJ2WWFV2f9JvJBs7HY8pbfbCWGa-Kb7oX3KU" width="320" /></a></div><br /><br />The solutions of this equation are θ = ℼ/6 and θ = 5ℼ/6, which are the limits of integration. The graph of the circle, in red.. is the outer curve. The graph of the cardioid, in blue, is the inner curve. To find the area between the 2 curves, let's subtract the area of the cardioid from that of the circle.<p></p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEi54OuuTtuljDbY6ipnkmDE41Mg8TCpZdaqvxPNRjcrgtuqNrHicKap2MsLEgzbHvJWHNjn-wY1b_yliSjMesEdmQehD3q_xc8UEj1avEovbswMYwLcfqJoaZlAqEXDp1b2At5l1LjeX-e3znlprFoNmM3PWmEyjQKlcrf8y4D50r33qwVhh_W8sCEYlo8" style="margin-left: 1em; margin-right: 1em;"><img data-original-height="382" data-original-width="594" height="258" src="https://blogger.googleusercontent.com/img/a/AVvXsEi54OuuTtuljDbY6ipnkmDE41Mg8TCpZdaqvxPNRjcrgtuqNrHicKap2MsLEgzbHvJWHNjn-wY1b_yliSjMesEdmQehD3q_xc8UEj1avEovbswMYwLcfqJoaZlAqEXDp1b2At5l1LjeX-e3znlprFoNmM3PWmEyjQKlcrf8y4D50r33qwVhh_W8sCEYlo8=w400-h258" width="400" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><b>Practice</b><p></p><p>Find the area inside the circle r = 4 cosθ and outside the circle r = 2.<br /><br /></p>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-31760652038961581572024-02-20T22:37:00.000-05:002024-02-20T22:37:20.835-05:00Area of a region bounded by a polar curve (continued)<p> In the previous post, we set the formula to find the area of a region bounded by a polar curve. Let's do an application.</p><p><b>Example. </b>Find the area of one petal of the rose defined by the equation r = 3sin (2θ)</p><p>Solution</p><p>Here is the graph of the of the petal of the rose</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEh7pGneJQkITlqAZz5VJ5uaIBvbsJtplXIcSizk6NCkQsMUVtkl9WGzq3QNcxxTHJsWBuGD80RwUE_5XAFo9bLjX05hGr73UC7xB-p5KuP4ZBBgVpY8jntod6GQOc-XrH3d1sKfkOYgXJ5NhxoPZbbtFNJaBH62g6fju3dy70NwSD0N1MXAGFhL1kL3vyc" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="354" data-original-width="354" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEh7pGneJQkITlqAZz5VJ5uaIBvbsJtplXIcSizk6NCkQsMUVtkl9WGzq3QNcxxTHJsWBuGD80RwUE_5XAFo9bLjX05hGr73UC7xB-p5KuP4ZBBgVpY8jntod6GQOc-XrH3d1sKfkOYgXJ5NhxoPZbbtFNJaBH62g6fju3dy70NwSD0N1MXAGFhL1kL3vyc" width="240" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><br />The first petal of the rose is traced out from the polar coordinates (0, 0) and (0, 𝝅/2). To find the area inside the petal, let's use the formula of the area of the region bounded by a polar curve. In this formula we substitute 𝛼 by 0 and 𝛽 by 𝝅/2. <p></p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhA__Xpcn-yHQcio-6i7voGdOufTR6YsU427YOGh57VQfGh3ZGghQjQGARDtjHUOd6DxUrzZrBhDIeg0LBPwu4O3dHaYIurT0jTTm1BTxSAkdOIjGFWazYrC729MvgMVCvc9fsw2Zf0xNGQ3p7m0S71UfZ7b-yEKPHtpVbhQYQGTBXTa8PB-P3wjHnzT1w" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="114" data-original-width="216" height="169" src="https://blogger.googleusercontent.com/img/a/AVvXsEhA__Xpcn-yHQcio-6i7voGdOufTR6YsU427YOGh57VQfGh3ZGghQjQGARDtjHUOd6DxUrzZrBhDIeg0LBPwu4O3dHaYIurT0jTTm1BTxSAkdOIjGFWazYrC729MvgMVCvc9fsw2Zf0xNGQ3p7m0S71UfZ7b-yEKPHtpVbhQYQGTBXTa8PB-P3wjHnzT1w" width="320" /></a></div><br /><br /><p></p><p>To evaluate this integral, let's use the formula sin²𝛂 = 1 - cos(2𝛂) with 𝛂 = 2𝜃</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEiI3f749m-UqZM5MlizawXIeC-tI5GM2eUAsDe7hP7hbvoQlmIJmfl3e9WRny0erG8G3t4-WLt5ZTv5H9Ho1chB0emZHtkoSiQvL8SHr1rvnEMgDkPP-xiQSXD0Va28YoSbh4w0aMD3waA6YDTbINVIQeTj5dfa5AKOTrcZi3cbY9rWN8-NH4dPQ1Q0ECc" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="280" data-original-width="290" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEiI3f749m-UqZM5MlizawXIeC-tI5GM2eUAsDe7hP7hbvoQlmIJmfl3e9WRny0erG8G3t4-WLt5ZTv5H9Ho1chB0emZHtkoSiQvL8SHr1rvnEMgDkPP-xiQSXD0Va28YoSbh4w0aMD3waA6YDTbINVIQeTj5dfa5AKOTrcZi3cbY9rWN8-NH4dPQ1Q0ECc" width="249" /></a></div><p><b>Practice</b></p><p>Find the area inside the cardioid defined by the equation r = 1-cos𝜃</p><p></p>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-17451102219289464032024-02-17T16:53:00.001-05:002024-02-17T23:56:06.174-05:00Areas of regions bounded by polar curves<p> To find the area of a region bounded by a curve in rectangular coordinates, we use the Riemann sum to approximate the area under the curve by using rectangles. In polar coordinates, we are going to use the Riemann sum also to find the area bounded by a curve but instead of using rectangles we will use sectors of a circle. Let's consider the curve defined by the function r = f(θ) where α ≤ θ ⩽ 𝛃. Our goal is to find the area bounded by the curve and the 2 radial lines θ = α and θ =𝛃. </p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgdLOTI2sK6PpH1E3COPiLl8w_3p2gES65zg_hto3V79BpbdlloJB0wtL-ie-Xi_8GZCVLrV7T240ie2jm69TF7qTR-cz-Tt_rqivue_OJP4VgcYCgHfT4_DjtYzULtRcsryjeHcd0whbvKfvxNeaJcYhATZlNNlY1kLtaA9IF-41A3ddRXtfSvByC2MDI" style="margin-left: 1em; margin-right: 1em;"><img data-original-height="669" data-original-width="668" height="400" src="https://blogger.googleusercontent.com/img/a/AVvXsEgdLOTI2sK6PpH1E3COPiLl8w_3p2gES65zg_hto3V79BpbdlloJB0wtL-ie-Xi_8GZCVLrV7T240ie2jm69TF7qTR-cz-Tt_rqivue_OJP4VgcYCgHfT4_DjtYzULtRcsryjeHcd0whbvKfvxNeaJcYhATZlNNlY1kLtaA9IF-41A3ddRXtfSvByC2MDI=w400-h400" width="400" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div>Let's start by dividing the area into sectors of equal width. We name the width Δθ and it's calculated by using this formula: Δθ = 𝛃 - α/n. Let's find the area of the sectors. They have equal area since their measurement is equal. The area of each sector is used to approximate the area between line segments. We sum the area of the sectors to approximate the total area. Let's find the formula for the area of a sector.<div><br /></div><div>The area of a circle is given by A = 𝝅r². The length of a circle is 360 degrees or 2π. The surface for one radian is A = 𝝅r²/2π = r²/2. The area for a sector of Δθ radians is A = Δθ r²/2. This represents the area of any sector. Let's call it Aᵢ and substitute r by f(θ). Aᵢ = 1/2 [f(θ)]².</div><div><br /></div><div>Let's add the areas of all the sectors to approximate the area bounded by the polar curve and the radial lines :</div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEilILIuP0XhJ_8qCRif9Vo8THsKsiGETG0LRwmMN6YYlU-854lnyDLy6u_e-yPk2gSjSZKSxsY5G9UnhqbJPzYaXEhHMUHyxLwQxb7iZysNNptRXTvi5p-gD8c7kHgM4jOwFFe0gpyvx3UYmVepTo4WtEvXK71nzUTGMXjn7fbfXlWFEXAbyv-mc7P1vck" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="74" data-original-width="286" height="83" src="https://blogger.googleusercontent.com/img/a/AVvXsEilILIuP0XhJ_8qCRif9Vo8THsKsiGETG0LRwmMN6YYlU-854lnyDLy6u_e-yPk2gSjSZKSxsY5G9UnhqbJPzYaXEhHMUHyxLwQxb7iZysNNptRXTvi5p-gD8c7kHgM4jOwFFe0gpyvx3UYmVepTo4WtEvXK71nzUTGMXjn7fbfXlWFEXAbyv-mc7P1vck" width="320" /></a></div><br /><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div>Let's divide the sector in as many subintervals as possible. At some point we approach infinity. The area of the sector is then given by:<p></p></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEi1Lm2LBppjE4PxNeQgLyZDYjWesFzqu57QxyY_61B9U0slCcbK1WHeAtdige0EVB2nRUrRtNkgNqvLR851H18fEGwPUAHcccfhPmitGT2d5Rdu7OD8dAX9ND0dpcenYvixgaGT47JTVxvPJh3zzZyVx5i2hnPLGIApGt0z5crX6VDbDYDG7xzymZKPUgg" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="67" data-original-width="269" height="80" src="https://blogger.googleusercontent.com/img/a/AVvXsEi1Lm2LBppjE4PxNeQgLyZDYjWesFzqu57QxyY_61B9U0slCcbK1WHeAtdige0EVB2nRUrRtNkgNqvLR851H18fEGwPUAHcccfhPmitGT2d5Rdu7OD8dAX9ND0dpcenYvixgaGT47JTVxvPJh3zzZyVx5i2hnPLGIApGt0z5crX6VDbDYDG7xzymZKPUgg" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><b>Theorem</b> </div><div class="separator" style="clear: both; text-align: center;"> </div><div class="separator" style="clear: both; text-align: center;">Suppose f is continuous and non negative on the interval <span style="text-align: left;"> </span><span style="text-align: left;">α ≤ θ ⩽ 𝛃 with 0 ≤</span> <span style="text-align: left;">α - </span><span style="text-align: left;">𝛃 ≤ 2𝝅. The area of the</span> region bounded by the graph r = f(θ) between the radial lines θ = <span style="text-align: left;">α and </span>θ = <span style="text-align: left;">𝛃</span> <span style="text-align: left;"> </span> is:</div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjMNUpJ358_xkASinM7k7HS_zh-1knmbZb77x_29kyeWk0cgwIWjNQcQkrirRPtKHD_piNm-H8Zi3Eilav3pVxGzkoGYWae5Z4sX-P6O0vr_ZLJQaDYBOv9CSuT2ItU0TgoxfpLtBdD4HrLPngmy8HfS1WGEXEixk4UBsSgbgOJpiiPVouY7bZc8Dh0pTI" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="75" data-original-width="292" height="82" src="https://blogger.googleusercontent.com/img/a/AVvXsEjMNUpJ358_xkASinM7k7HS_zh-1knmbZb77x_29kyeWk0cgwIWjNQcQkrirRPtKHD_piNm-H8Zi3Eilav3pVxGzkoGYWae5Z4sX-P6O0vr_ZLJQaDYBOv9CSuT2ItU0TgoxfpLtBdD4HrLPngmy8HfS1WGEXEixk4UBsSgbgOJpiiPVouY7bZc8Dh0pTI" width="320" /></a></div><br /><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"> </div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><br /><br /></div><div><br /></div>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-36171876102576395862024-02-12T22:49:00.000-05:002024-02-12T22:49:59.528-05:00Transforming polar equations into rectangular coordinates<p> Transforming polar equations into rectangular coordinates leads to find a relation between x and y. To do so, we use the formulas that allow to convert points between coordinates.</p><p><b>Example</b></p><p>Rewrite each of the following equations into rectangular coordinates and identify the graph.</p><p>a. θ = π/3</p><p>b. r = 3</p><p>c. r = 6cosθ-8sinθ</p><p><b>Solution</b></p><p>a,Let's take the tangent of of both sides:</p><p>tanӨ = tan𝝅/3 = ⎷3</p><p>Let's substitute tanӨ by y/x:</p><p>y/x = ⎷3 y = x⎷3. This is the equation of a straight line passing through the origin and of slope ⎷3. In general any polar equation of the form θ = K represents a straight line passing through the pole and with slope tanK.</p><p>b. Let's use the equation x² + y² = r². Let's substitute r by 3: x² + y² = 9. This is the equation of a circle centered at the origin. In general, any polar equation of the form r = k where k is a positive constant represents a circle centered at the origin and with radius k.</p><p>c. Let's multiply both sides by r:</p><p>r² = 6rcosθ-8rsinθ.</p><p>Let's substitute rcosθ by x and rsinθ by y.</p><p>r² = 6x - 8y.</p><p>Let's use the equation x² + y² = r² and substitute r² :</p><p>x² + y² = 6x - 8y</p><p>x² - 6x + y² + 8y = 0</p><p>x² - 6x +9-9 + y² + 8y + 16-16 = 0</p><p>(x² - 6x +9) +(y² + 8y + 16) -25= 0</p><p>(x-3)² + (y + 4)² = 25</p><p>This the equation of a circle centered at the point (3, -4) with radius r = 5.</p><p><b>Practice</b></p><p>Rewrite the equation r = secθtanθ in rectangular coordinates and identify its graph.</p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><br /></div><b><br /><br /></b><p></p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><br /></b></div><b><br /><br /></b><p></p><p></p><div class="separator" style="clear: both; text-align: center;"><br /></div><br /><br /><p></p><p></p><div class="separator" style="clear: both; text-align: center;"><br /></div><br /><br /><p></p>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-20484292653912124052024-02-10T00:06:00.006-05:002024-02-10T00:09:31.770-05:00Polar curves<p> Goal: graph a curve in the polar coordinate system</p><p><b>Polar curve</b></p><p>The same way we graph a function y = f(x) in the cartesian plane, we can generate a curve for the function r = f(θ) in the polar coordinate system. We start by creating a list of values for the independent values θ. Then we create another list for the dependent variable r. This process generates a list of ordered pairs that can be plotted in the polar coordinate system.</p><p><b>Problem-solving strategy</b></p><p><b>1.</b> Create a table with 2 columns: one for θ and the other for r.</p><p>2. Create a list of values for θ</p><p>3. Calculate the corresponding values r for each value of θ.</p><p>4. Plot each pair (r,θ) in the coordinate plane</p><p>5. Connect the points and look for a pattern.</p><p><b>Example</b></p><p>Graph the curve defined by the function r = 4 sinθ. Identify the curve and rewrite the equation in rectangular coordinates.</p><p><b>Solution</b></p><p>The function is a multiple of the sine function. Since sine is periodic the given function is periodic. The period of sine is 2ℼ. The period of the function is 2ℼ.. We choose the values of θ between 0 and 2ℼ. Here is the table of values:</p><p></p><div class="separator" style="clear: both; text-align: center;"><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjwOW6fk7L416wgDeZqiU4cEQHmdVfSBokfdKWTVoNxVmbCwcMaCBZM_USdILo9q9dr5UOskeDTPSW-9lIEkkXyNd1Pqgfuosz7_K6Wx8yDxrxZ3BWvUd1DbLaFzwX02OVUpDUrzAIjGxElrSF42sDd4X-mY4_16fhn1kUZVrpfN7D-QWlHmvRK9KAWSW8" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="444" data-original-width="765" height="373" src="https://blogger.googleusercontent.com/img/a/AVvXsEjwOW6fk7L416wgDeZqiU4cEQHmdVfSBokfdKWTVoNxVmbCwcMaCBZM_USdILo9q9dr5UOskeDTPSW-9lIEkkXyNd1Pqgfuosz7_K6Wx8yDxrxZ3BWvUd1DbLaFzwX02OVUpDUrzAIjGxElrSF42sDd4X-mY4_16fhn1kUZVrpfN7D-QWlHmvRK9KAWSW8=w640-h373" width="640" /></a></div><div class="separator" style="clear: both; text-align: center;">Here is the graph of the function: </div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEg-dIQXRBWtDm-Q4UXgjyi7Zo6tBaTPYQE69zF9uB-KeuS55ekg6wgubakmqqMfBKJzg37MSvhQUBOgu8nAWn-d0dqVATqwYWRsNGleWirlRz8WXefKbNm4cbCuZlRMM73k-2TdH6rFFUl0-pzsV3GmFKKQ78nSLojS1FayVAf_R9CYUGibfoTDwSwQlaM" style="margin-left: 1em; margin-right: 1em;"><img data-original-height="417" data-original-width="417" height="400" src="https://blogger.googleusercontent.com/img/a/AVvXsEg-dIQXRBWtDm-Q4UXgjyi7Zo6tBaTPYQE69zF9uB-KeuS55ekg6wgubakmqqMfBKJzg37MSvhQUBOgu8nAWn-d0dqVATqwYWRsNGleWirlRz8WXefKbNm4cbCuZlRMM73k-2TdH6rFFUl0-pzsV3GmFKKQ78nSLojS1FayVAf_R9CYUGibfoTDwSwQlaM=w400-h400" width="400" /></a></div><br /><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div> Rewrite the given equation in rectangular coordinates means writing in function of x and y. To do so, let's use the equation x² + y² = r². Let's multiply the equation r = 4sinθ by r, This gives r² = 4rsinθ. We know that y = 4sinθ. Therefore we have r² = 4y. . Let's substitute r² in the initial equation. We have </div><div class="separator" style="clear: both; text-align: center;">x² + y² = 4y. Let's put it in standard form. </div><div class="separator" style="clear: both; text-align: center;"> </div><div class="separator" style="clear: both; text-align: center;"><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEggsXc-24MnJHx4jeGB6tC6Y1WrZHqejO2DGACbZNlbo5HioHKFXr6J0VSDqBYXUfIvbDCbbb7AdRNrrkfHdJLRLdizhqHIaPAt_glwtd8pR7qIJ72WQD583_QiBjK9wPeeWJOiqlAa1JUjLUS0EAbu4HdRPSbeTMstvbxVVqlhPTV8cM7AmraAe7iDSXs" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="138" data-original-width="250" height="177" src="https://blogger.googleusercontent.com/img/a/AVvXsEggsXc-24MnJHx4jeGB6tC6Y1WrZHqejO2DGACbZNlbo5HioHKFXr6J0VSDqBYXUfIvbDCbbb7AdRNrrkfHdJLRLdizhqHIaPAt_glwtd8pR7qIJ72WQD583_QiBjK9wPeeWJOiqlAa1JUjLUS0EAbu4HdRPSbeTMstvbxVVqlhPTV8cM7AmraAe7iDSXs" width="320" /></a></div><br /><br /></div><div class="separator" style="clear: both; text-align: center;">This is the equation of a circle of radius 2 and center (0,2) in the rectangular coordinate system'</div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><b>Practice</b> </div><div class="separator" style="clear: both; text-align: center;"> </div><div class="separator" style="clear: both; text-align: center;"><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjVblk0L6f6fp4YOZ7Fs-xXsd_-657zq44ElQRcIIB4HP1CR4p8hRzmzFvuBeZ1fz6pfGelEDWvJWwr08vOYLKm23V1G8Q6J4CuFvsD7qA2SqkaAWsiSSHMdYAYAr6d342BbNjZTOZ5zZ2bziPVBU6vM-2eRha2AELH0jjB4_QLxOEF4vPXnAewfW4hNlk" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="41" data-original-width="511" height="51" src="https://blogger.googleusercontent.com/img/a/AVvXsEjVblk0L6f6fp4YOZ7Fs-xXsd_-657zq44ElQRcIIB4HP1CR4p8hRzmzFvuBeZ1fz6pfGelEDWvJWwr08vOYLKm23V1G8Q6J4CuFvsD7qA2SqkaAWsiSSHMdYAYAr6d342BbNjZTOZ5zZ2bziPVBU6vM-2eRha2AELH0jjB4_QLxOEF4vPXnAewfW4hNlk=w640-h51" width="640" /></a></div><br /> </div><div class="separator" style="clear: both; text-align: center;"> <br /><br /></div><br /><br /><p></p><p></p><div class="separator" style="clear: both; text-align: center;"><br /></div><br /><br /><p></p>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-77358383903683503352024-02-09T22:11:00.008-05:002024-02-09T22:51:42.254-05:00Plotting points in the polar plane<p> Goal: plot points in the polar plane</p><p><b>Plotting a point in the polar plane</b></p><p>The polar representation of a point is not unique. For example, the polar coordinates (2, π/3) and (2, 7π/3) both represent the point (1, ⎷3). Also the value of r can be negative. Aa a result the polar coordinate (-2. 2𝛑/3) represents also the point (1, ⎷3) in the cartesian plane as demonstrated here:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgSi8IyCYP8rnxe_GCH_h1tZZ-DfzTjbMKeViP9JYE-AqSpaZO6SHFa26JvVE_lTzhnaNC5_AL5mIdwkbWX-qEVmZUOgnGrjTw9a0j2uS9Ja0e0OWgIfMvNNxM6NK60v1mPEJ8Sr38TOm8bCLugkom5WJl-qek5vUdGEP8nIjhDocgm3y1EnVw3T9fKffQ" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="98" data-original-width="513" height="76" src="https://blogger.googleusercontent.com/img/a/AVvXsEgSi8IyCYP8rnxe_GCH_h1tZZ-DfzTjbMKeViP9JYE-AqSpaZO6SHFa26JvVE_lTzhnaNC5_AL5mIdwkbWX-qEVmZUOgnGrjTw9a0j2uS9Ja0e0OWgIfMvNNxM6NK60v1mPEJ8Sr38TOm8bCLugkom5WJl-qek5vUdGEP8nIjhDocgm3y1EnVw3T9fKffQ=w400-h76" width="400" /></a></div><br /><br /><p></p><p></p><div class="separator" style="clear: both; text-align: center;"><br /></div><p>Every point in the plane has an infinite number of representations in polar coordinates. The polar representation of a point in the plane has a visual interpretation. First, we have r that represents the distance of the segment that joins the origin to the point. Then Ө is the angle made of that segment and the positive direction of the x-axis. Positive angles are measured in counterclockwise direction and negative angles are measured in clockwise direction. Here is a representation of the polar coordinate system.</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjzTwoholVA6jVLWaZis5h7xO378trR1Oytk5zADXYTDIKWtv2YjHcPDCU9h63UzalCNyvt9KV4p3wg1rjPmEJ0RKKFIY1y4EHAs5pTdxwj6SKK4lLAh8FD2CoEZox2CupUWFHofzja6UxKX_rO8xiJ50P7X0cKzJnmy8UqNjnTPLHbpOLhnHTpfRWcbS0" style="margin-left: 1em; margin-right: 1em;"><img data-original-height="499" data-original-width="576" height="347" src="https://blogger.googleusercontent.com/img/a/AVvXsEjzTwoholVA6jVLWaZis5h7xO378trR1Oytk5zADXYTDIKWtv2YjHcPDCU9h63UzalCNyvt9KV4p3wg1rjPmEJ0RKKFIY1y4EHAs5pTdxwj6SKK4lLAh8FD2CoEZox2CupUWFHofzja6UxKX_rO8xiJ50P7X0cKzJnmy8UqNjnTPLHbpOLhnHTpfRWcbS0=w400-h347" width="400" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;">The positive x-axis in the cartesian plane is the polar axis, The origin of the coordinate system is the pole and corresponds to r = 0. The innermost circle of the figure above contains all the points with r = 1 meaning all the points of which the distance from the pole is 1. Then the figure progresses from r = 2 and so on. To plot a point in the coordinate system start from the angle. If the angle is positive, move in counterclockwise direction from the polar axis. If the angle is negative move in clockwise direction of the polar axis. If the value of r is positive consider the ray that is terminal to the angle. If r is negative, consider the opposite ray terminal to the angle. </div><div class="separator" style="clear: both; text-align: center;"> </div><div class="separator" style="clear: both; text-align: center;"><b>Example </b></div><div class="separator" style="clear: both; text-align: center;"><b> </b></div><div class="separator" style="clear: both; text-align: center;">Plot each of the following points in the polar plane </div><div class="separator" style="clear: both; text-align: center;"> <div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEiwL2sUmdaABLrJYb-mmILJto5lat_Rt5i2LPRY592GYxtEwwByn6XcFaS8-q_0avdGjptf4pPcDJTLMEp3fgZolvAfqqTx7uDOm3vnA6nK8dQCcT3d-zBqTQyPvsH-UwOrxR6w8hjckz5lEJ1jTex5yPcvYYatb6a_8vGgnpIZ-TDNUIVU3b6hcaqqy_M" style="margin-left: 1em; margin-right: 1em;"><img data-original-height="96" data-original-width="105" height="183" src="https://blogger.googleusercontent.com/img/a/AVvXsEiwL2sUmdaABLrJYb-mmILJto5lat_Rt5i2LPRY592GYxtEwwByn6XcFaS8-q_0avdGjptf4pPcDJTLMEp3fgZolvAfqqTx7uDOm3vnA6nK8dQCcT3d-zBqTQyPvsH-UwOrxR6w8hjckz5lEJ1jTex5yPcvYYatb6a_8vGgnpIZ-TDNUIVU3b6hcaqqy_M=w200-h183" width="200" /></a></div><br /> .</div><div class="separator" style="clear: both; text-align: center;"><b> </b></div><div class="separator" style="clear: both; text-align: center;"><b>Solution </b></div><div class="separator" style="clear: both; text-align: center;"><b><br /></b></div><div class="separator" style="clear: both; text-align: center;"><b><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhr6ldjX3N6V_cHVfWAppZcEGvQbI6ypOEAy5E9YKfFoFvFdX1jcZytYBDPAalgM4nQ17tlIbcP57u63qavWmyP2NIVGlSJQBS18uWgpgfvZHk4ktep0LvSXNcfby-regCunF3VHSguq9tQ-LjmmkLkkgZoyCCvTURMxX_kAgyDbbdM6Smj-FALj69RdBo" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="417" data-original-width="417" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEhr6ldjX3N6V_cHVfWAppZcEGvQbI6ypOEAy5E9YKfFoFvFdX1jcZytYBDPAalgM4nQ17tlIbcP57u63qavWmyP2NIVGlSJQBS18uWgpgfvZHk4ktep0LvSXNcfby-regCunF3VHSguq9tQ-LjmmkLkkgZoyCCvTURMxX_kAgyDbbdM6Smj-FALj69RdBo" width="240" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;">Practice </div><div class="separator" style="clear: both; text-align: center;"> <b> </b></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEiZLdsngE_K70LuQaalaqtfa25fwKr9aHVATM38UF-PdDfD8WE8RS9ZWju1Qh_IRxmReJwe5wMQAgsbsMODsuNmdXvPvsc9DIL3S56gO9cNoaZvwnLVv6NIf6lb10bTqxMjZ5Pwx_1WNgD-9gNNvRtDi7uAVi8l1Tka7fSDCCcyGQjQxkeK1szvZBUeqvE" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="47" data-original-width="362" height="42" src="https://blogger.googleusercontent.com/img/a/AVvXsEiZLdsngE_K70LuQaalaqtfa25fwKr9aHVATM38UF-PdDfD8WE8RS9ZWju1Qh_IRxmReJwe5wMQAgsbsMODsuNmdXvPvsc9DIL3S56gO9cNoaZvwnLVv6NIf6lb10bTqxMjZ5Pwx_1WNgD-9gNNvRtDi7uAVi8l1Tka7fSDCCcyGQjQxkeK1szvZBUeqvE" width="320" /></a></div><br /></b></div><div class="separator" style="clear: both; text-align: center;"> </div><div class="separator" style="clear: both; text-align: center;"> </div><div class="separator" style="clear: both; text-align: center;"> </div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><br /><br /><p></p>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-9628204536199504962024-02-02T22:07:00.001-05:002024-02-02T22:07:37.565-05:00What are Polar Coordinates? How to convert points between rectangular and polar coordinates?<p> <b>Goals:</b></p><p>1) Define polar coordinates</p><p>2) Convert points between rectangular and polar coordinates</p><p><b><span style="font-size: medium;">Polar Coordinates</span></b></p><p><b>Definition</b></p><p>The rectangular coordinate system provides a way to map points to ordered pairs. The polar coordinate system provides an alternative method for mapping points to ordered pairs.</p><p>Let's consider the following figure:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEimj8dN2yuNPDhDEqHRnPuhgkZ72jgy2YxKNp8wyP9b6VioQJlcQHnwzxlSld202XkVycaefbHan6mMzboSqStgmUwTeVqb5vUOJmzfuw8XJzWRO5wnl-gvNhEsBLArkTBeNH1xO_dc5rcUjqDrhtjN6u-XU6eBdFIRO4Jz2mpcOfTb7qULfYGEc5WbHIc" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="309" data-original-width="304" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEimj8dN2yuNPDhDEqHRnPuhgkZ72jgy2YxKNp8wyP9b6VioQJlcQHnwzxlSld202XkVycaefbHan6mMzboSqStgmUwTeVqb5vUOJmzfuw8XJzWRO5wnl-gvNhEsBLArkTBeNH1xO_dc5rcUjqDrhtjN6u-XU6eBdFIRO4Jz2mpcOfTb7qULfYGEc5WbHIc" width="236" /></a></div><br /><br /><p></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p>The point P has has (x,y) for coordinates in the rectangular coordinate system. The line segment that joins the origin to the point P measures the distance between these two points. This distance is designated by the letter r.. The angle between this segment and the positive part of the x-axis is designated by Ө. Now the point P can not only be referred by its cartesian coordinate x and y but by r and Ө. These 2 letters represent the polar coordinates of the point P.</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjWxFxOSOw0oUhyx1--r24TY9598OfR1eHSGwXaRqqWF9AenP-5pLvYPAakMD4lsS6OsMVUM-SGsTyQDeS9BAO7EYKGqsZf4Yt589zVPe6sSrhdJoM4BGCmnVfar6k5rHP1_W9Yu0m9XyvxzQg1kVRxKxvADAsdPVb0Z1EPja3d_1qoq1okqtU3QT3NnFk" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="249" data-original-width="594" height="269" src="https://blogger.googleusercontent.com/img/a/AVvXsEjWxFxOSOw0oUhyx1--r24TY9598OfR1eHSGwXaRqqWF9AenP-5pLvYPAakMD4lsS6OsMVUM-SGsTyQDeS9BAO7EYKGqsZf4Yt589zVPe6sSrhdJoM4BGCmnVfar6k5rHP1_W9Yu0m9XyvxzQg1kVRxKxvADAsdPVb0Z1EPja3d_1qoq1okqtU3QT3NnFk=w640-h269" width="640" /></a></div><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p>Every point (x,y) in the cartesian coordinate system can be represented in the polar coordinate system by an ordered pair (r, Ө), The first coordinate is called radial coordinate. The second coordinate is called angular coordinate.<p></p><p><b>Converting points between coordinate systems</b></p><p><b>Theorem</b></p><p>Given a point P in the plane represented by its cartesian coordinates (x,y) and its polar coordinates (r,θ), the following relations hold true:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhjf0BZFf4-Z_QVNHjm3G1oz2hnoCZCYy_lUMnhZlkQyPSOMzddy216vhiBtt5-Gde66y0_UXqS3IdX6j7C6Pxdna0L8SUDg97A-jfjyzu7EPv-WSNWGGomLAhlbUVG60ac5RynBTpugSFd8DvJUUalDd6LOVgQhpE6TmIHTMd7wIDnDOovKaCpGnZQcVY" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="114" data-original-width="542" height="134" src="https://blogger.googleusercontent.com/img/a/AVvXsEhjf0BZFf4-Z_QVNHjm3G1oz2hnoCZCYy_lUMnhZlkQyPSOMzddy216vhiBtt5-Gde66y0_UXqS3IdX6j7C6Pxdna0L8SUDg97A-jfjyzu7EPv-WSNWGGomLAhlbUVG60ac5RynBTpugSFd8DvJUUalDd6LOVgQhpE6TmIHTMd7wIDnDOovKaCpGnZQcVY=w640-h134" width="640" /></a></div><br /><br /><p></p><p></p><div class="separator" style="clear: both; text-align: center;"><br /></div><br />These formulas can be used to convert from cartesian coordinates to polar coordinates and vice-versa.,<p></p><p><b>Examples</b> </p><p><b>Converting between rectangular and polar coordinates</b></p><p>1. Convert each of the following points into the polar coordinate system:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjq55FHQvyIy7LJr6jv9ZTy2y-V4_vSl07urf7wQNyPSnagSvBaBTjPtz6cyNeayeo7Chu8UllLVqoAGhQJjYXFGJpjXGrVu-6wX3XblfG302Jbw8ra9bL5K8Yh1DoTiaCkdSmeDK391xWEapF37z-d6Koi0LRMl7dte1PipDhGwfId1ns8AsqQOAbx-PY" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="54" data-original-width="85" height="127" src="https://blogger.googleusercontent.com/img/a/AVvXsEjq55FHQvyIy7LJr6jv9ZTy2y-V4_vSl07urf7wQNyPSnagSvBaBTjPtz6cyNeayeo7Chu8UllLVqoAGhQJjYXFGJpjXGrVu-6wX3XblfG302Jbw8ra9bL5K8Yh1DoTiaCkdSmeDK391xWEapF37z-d6Koi0LRMl7dte1PipDhGwfId1ns8AsqQOAbx-PY=w200-h127" width="200" /></a></div><br /> <p></p><p><br /></p><p><br /></p><p><br /></p><p>2. Convert each of the following points into the rectangular coordinate system:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgEzuqFMHbBacuuzeipDJ8z9WE670JwSKLa20kgpcpvXF1mz-vrbneQNmWWhDHoYKr_9PEdH5-GT1JCgyxgvox5n0_nI_ArOZ7BHtONtN-DXa6BxE6psfXJoPjiVpi-AY5cgVAVpCwAe4gKQnjoGlcx3wif0WfUCm6uMI1nRCfqoqKiqBHUNH3eQ_-kBW8" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="49" data-original-width="96" height="102" src="https://blogger.googleusercontent.com/img/a/AVvXsEgEzuqFMHbBacuuzeipDJ8z9WE670JwSKLa20kgpcpvXF1mz-vrbneQNmWWhDHoYKr_9PEdH5-GT1JCgyxgvox5n0_nI_ArOZ7BHtONtN-DXa6BxE6psfXJoPjiVpi-AY5cgVAVpCwAe4gKQnjoGlcx3wif0WfUCm6uMI1nRCfqoqKiqBHUNH3eQ_-kBW8=w200-h102" width="200" /></a></div><br /><br /><p></p><p><br /></p><p><br /></p><p><b>Solution</b></p><p><b>1.</b></p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjEpN-kcQyOnqn62OIPYGGF2O9BAFUfOBtdJVgPZDktR7ywaa8xPlywnn3bw4DN4sOamG3nWze7dVSLueztQ1aIRlkkJLzzOfx3Kbsy75Gd-mXtea_bPp868ZlSCdEacpLCBIcUV1zGSDdc1i-YM7ehkbiRBHmLcjni7X9uSDE-q3BQAMhCMNA7XWYEfas" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="377" data-original-width="624" height="386" src="https://blogger.googleusercontent.com/img/a/AVvXsEjEpN-kcQyOnqn62OIPYGGF2O9BAFUfOBtdJVgPZDktR7ywaa8xPlywnn3bw4DN4sOamG3nWze7dVSLueztQ1aIRlkkJLzzOfx3Kbsy75Gd-mXtea_bPp868ZlSCdEacpLCBIcUV1zGSDdc1i-YM7ehkbiRBHmLcjni7X9uSDE-q3BQAMhCMNA7XWYEfas=w640-h386" width="640" /></a></b></div><p><b><b><br /></b></b></p><p><b><b><br /></b></b></p><p><b><b><br /></b></b></p><p><b><b><br /></b></b></p><p><b><b><br /></b></b></p><p><b><b><br /></b></b></p><p><b><b><br /></b></b></p><p><b><b><br /></b></b></p><p><b><b><br /></b></b></p><p><b><b><br /></b></b></p><p><b><b><br /></b></b></p><p><b><b><br /></b></b></p><b>2.</b><p></p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjWIBzx5GXi_notuEWtoC0oiWW_X4Tg7pjk74w6k2CI8sGUZ6SP6u9HvYlVdakhGP4DsgrDkTG3GGWdJ0SlaAY9iJ0bsYeZc1bJslbC_3-7kfxOAqfMpA9JQJnabcYNqWtX4Zih9KsEnUwl7hQLsH5_ZIqAJX6ghePf97kQzb_F2VWFuG4fm1dDu1k0jnE" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="345" data-original-width="609" height="362" src="https://blogger.googleusercontent.com/img/a/AVvXsEjWIBzx5GXi_notuEWtoC0oiWW_X4Tg7pjk74w6k2CI8sGUZ6SP6u9HvYlVdakhGP4DsgrDkTG3GGWdJ0SlaAY9iJ0bsYeZc1bJslbC_3-7kfxOAqfMpA9JQJnabcYNqWtX4Zih9KsEnUwl7hQLsH5_ZIqAJX6ghePf97kQzb_F2VWFuG4fm1dDu1k0jnE=w640-h362" width="640" /></a></b></div><b><br /><br /><br /></b><p></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p><b>Practice</b></p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><a href="https://blogger.googleusercontent.com/img/a/AVvXsEiLSs8ErHa1Bb8VlF_uDjilXDflvj4wNsC9HJBudWMhmJtnc4lKobCkVBdzC-8wjy5XWNfWTV-ThWwosrBjzYAm3Q-V4NtAVLtVXUOymJyrB-rKVwDaMihWf0gAIDYTqEVjaRNmEEQkQ8IGKEmeoDPe-cyl8B4_-rfoTPhTA3m-zH0fht9aHElogRkZ7ZA" style="margin-left: 1em; margin-right: 1em;"><img data-original-height="45" data-original-width="600" height="48" src="https://blogger.googleusercontent.com/img/a/AVvXsEiLSs8ErHa1Bb8VlF_uDjilXDflvj4wNsC9HJBudWMhmJtnc4lKobCkVBdzC-8wjy5XWNfWTV-ThWwosrBjzYAm3Q-V4NtAVLtVXUOymJyrB-rKVwDaMihWf0gAIDYTqEVjaRNmEEQkQ8IGKEmeoDPe-cyl8B4_-rfoTPhTA3m-zH0fht9aHElogRkZ7ZA=w640-h48" width="640" /></a></b></div><b><br /><br /></b><p></p>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-75476453222758950852024-01-26T23:22:00.002-05:002024-01-26T23:22:43.706-05:00Integrals of Vector-valued Functions<p> <b>Goal: Calculate the integral of vector-valued functions</b></p><p><b>Integrals of vector-valued functions</b></p><p>The antiderivative and definite integral of vector-valued functions can be calculated by taking the antiderivative and definite integral of the real-valued functions. This is the same way we have been doing with the derivative.</p><p>The antiderivative of a vector-valued function appears in applications. For example, if a vector-valued function represents the velocity of an object at a time t, then its antiderivative represents its position. Or, if the function represents the acceleration of the object at a given time, then the antiderivative represents its velocity.</p><p><b>Definition</b></p><p>Let f, g, h be real-valued functions integrable over the closed interval [a, b].</p><p>1. The indefinite integral of a vector-valued function r(t) = f(t)i + g(t)j is given by:</p><p>∫[f(t)i + g(t)j]dt = [∫f(t)dt]i + [∫g(t)]j</p><p>The definite integral of a vector-valued function is:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgpTMo38Y1tNB3BKGeGf7XN4Lnbwc9uPcXiNru4Qi5NXMECnj9-gEYAquxrbjV2U4uiyftKNzJ7uDq_1HjuuXVXoGdH73FkLNwogjCNVbWPJphqZgiBbTPonKjmsZibwkgFDuiv-1i5RAWqnVVdSborBsmJlYWg3baH8nGMAZzry0WKXvjtPqhiYBMQAEA" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="64" data-original-width="486" height="42" src="https://blogger.googleusercontent.com/img/a/AVvXsEgpTMo38Y1tNB3BKGeGf7XN4Lnbwc9uPcXiNru4Qi5NXMECnj9-gEYAquxrbjV2U4uiyftKNzJ7uDq_1HjuuXVXoGdH73FkLNwogjCNVbWPJphqZgiBbTPonKjmsZibwkgFDuiv-1i5RAWqnVVdSborBsmJlYWg3baH8nGMAZzry0WKXvjtPqhiYBMQAEA" width="320" /></a></div><br /><br /><p></p><p><br /></p><p>2. The indefinite integral of a vector-valued function r(t) = f(t)i + g(t)j + h(t)k is given by:</p><p>∫[f(t)i + g(t)j + h(t)j] = [∫f(t)]i + [∫g(t)]j + [∫h(t)]k</p><p>The definite integral of a vector-valued function is:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEh1nFDvq2hCXXtgj3o8jQYYjmyjDizsO9nrn0Rq650arFx9hLwZUHQHrk6llE-Y2R1z-3Ep_ifNOsGFcNqwGKkxl0m4QUYjNXEAdC2tN6xXK1UsELBPKKJz_4Zb33zX1kClKO-Bd1LKAc2LgSyRcPEl1PwXwRyRpmm3xES9q-hnpD06b37c6wOkJ3A_EYk" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="69" data-original-width="700" height="32" src="https://blogger.googleusercontent.com/img/a/AVvXsEh1nFDvq2hCXXtgj3o8jQYYjmyjDizsO9nrn0Rq650arFx9hLwZUHQHrk6llE-Y2R1z-3Ep_ifNOsGFcNqwGKkxl0m4QUYjNXEAdC2tN6xXK1UsELBPKKJz_4Zb33zX1kClKO-Bd1LKAc2LgSyRcPEl1PwXwRyRpmm3xES9q-hnpD06b37c6wOkJ3A_EYk" width="320" /></a></div><p><br /></p><p><br /></p><b>Examples</b><p></p><p>Calculate each of the following integrals:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgCf_GFCLvzhWuoxUXkhTZJl62xGZRIHX4VqbH79rhsw4lbXb3FVvWWsvHcrmn5b2AyOBe1mtfFAW3L-M32YOJSukPSZUk_HdtrIyrasKDjq-UrR5Cr2EJYbEA8XE0-mSnDLPqrLXsRrmJhhq_xncwxnHSAjV-w5rOianEwOuQaJf1oUMjv1Fc7UGiKTEc" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="99" data-original-width="456" height="69" src="https://blogger.googleusercontent.com/img/a/AVvXsEgCf_GFCLvzhWuoxUXkhTZJl62xGZRIHX4VqbH79rhsw4lbXb3FVvWWsvHcrmn5b2AyOBe1mtfFAW3L-M32YOJSukPSZUk_HdtrIyrasKDjq-UrR5Cr2EJYbEA8XE0-mSnDLPqrLXsRrmJhhq_xncwxnHSAjV-w5rOianEwOuQaJf1oUMjv1Fc7UGiKTEc" width="320" /></a></div><br /><br /><p></p><p><br /></p><p><b>Solution</b><br /><br /></p><p>a. Let's use the first part of the definition of the integral of a space curve:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEguVXW-l38uWjnSu_SgdOKzV2I3ALAu8uspn-A5hsSi9fpuYgN4DITWM-ZBCtiqFJ_4lTAwW6pwfxXNN771hyxmzpbgDEut7e8_D3T_PKjKtvYhqXMzuWoEJ7atudqD4RHBzki5PM0HwYgWbbAReJd7o1mKFfoZ_cMEbCSm_br1GUMkX1Hg8hb84280YJI" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="54" data-original-width="426" height="41" src="https://blogger.googleusercontent.com/img/a/AVvXsEguVXW-l38uWjnSu_SgdOKzV2I3ALAu8uspn-A5hsSi9fpuYgN4DITWM-ZBCtiqFJ_4lTAwW6pwfxXNN771hyxmzpbgDEut7e8_D3T_PKjKtvYhqXMzuWoEJ7atudqD4RHBzki5PM0HwYgWbbAReJd7o1mKFfoZ_cMEbCSm_br1GUMkX1Hg8hb84280YJI" width="320" /></a></div><br /><br /><p></p><p><br /></p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEiBzHz7oPG0D0FFz5L0hyLtufrPDWLIFWRiloGP2eSraAYlBouln72_28mWLaJrrkI4oGOlOTMXw2DKHdYlcp44vM-Vbk0XGQF_J4_NwXG3I3APpDEMpnM-dGVujObovaXb7Xg6Sr3Ipk_t7ohpY1RlEtRjBM6iqUVRxTJn8-bOH8_pNPqv_YhW7AwFUT8" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="92" data-original-width="519" height="57" src="https://blogger.googleusercontent.com/img/a/AVvXsEiBzHz7oPG0D0FFz5L0hyLtufrPDWLIFWRiloGP2eSraAYlBouln72_28mWLaJrrkI4oGOlOTMXw2DKHdYlcp44vM-Vbk0XGQF_J4_NwXG3I3APpDEMpnM-dGVujObovaXb7Xg6Sr3Ipk_t7ohpY1RlEtRjBM6iqUVRxTJn8-bOH8_pNPqv_YhW7AwFUT8" width="320" /></a></div><br /><br /><p></p><p><br /></p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEh2MISgpb4cC2veQwgo8rFQgtmc0-opphsP3EOah_q0mq2UjEfLJz_fSHQVaKt_eK1muZ_MwtvEGIr82aSsC7Us8-Ci7_1DC3-AQ5OJTOBpiBJB0oEBu7vnODtj0vEs2jb6ZbAImouhMVmUEYuSknB5yIQfz0YiIZ1X2dhFDYqsyLNY49bIEFkzn8IFX-E" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="371" data-original-width="697" height="341" src="https://blogger.googleusercontent.com/img/a/AVvXsEh2MISgpb4cC2veQwgo8rFQgtmc0-opphsP3EOah_q0mq2UjEfLJz_fSHQVaKt_eK1muZ_MwtvEGIr82aSsC7Us8-Ci7_1DC3-AQ5OJTOBpiBJB0oEBu7vnODtj0vEs2jb6ZbAImouhMVmUEYuSknB5yIQfz0YiIZ1X2dhFDYqsyLNY49bIEFkzn8IFX-E=w640-h341" width="640" /></a></div><br /><br /><p></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><b>Practice</b></p><p>Calculate the following integral:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEik-rWDyPJQs0tCYMaMkfq-AVH_tigZ_9aR5MI34QX4i6_ZuqvHn8yqH_H71BnqzEcOIX9sxSrzVYnOaBrngtRNZhxXQVyOR1rN3fhbGPYEkg4WxbI75CAjwmXJ-SdGGk_Po4Q7QOhHHFNz3M4YE1F5MT5zl2y9upK5K-NaB60YzyA37KuQi57lqtXGV-0" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="63" data-original-width="285" height="71" src="https://blogger.googleusercontent.com/img/a/AVvXsEik-rWDyPJQs0tCYMaMkfq-AVH_tigZ_9aR5MI34QX4i6_ZuqvHn8yqH_H71BnqzEcOIX9sxSrzVYnOaBrngtRNZhxXQVyOR1rN3fhbGPYEkg4WxbI75CAjwmXJ-SdGGk_Po4Q7QOhHHFNz3M4YE1F5MT5zl2y9upK5K-NaB60YzyA37KuQi57lqtXGV-0" width="320" /></a></div><br /><br /><p></p>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-68726192534383267582024-01-26T23:22:00.001-05:002024-01-26T23:22:15.133-05:00Tangent vector and unit tangent vector<p> <b>Goal</b>:</p><p>1. Define tangent vector and unit tangent vector</p><p><b>Tangent vector and unit tangent vector</b></p><p>Let's remind that the derivative of a function at a point is the slope of the tangent line to the graph of this function at this point. In the case of a vector-valued function the derivative provides a tangent vector to the graph of this function.</p><p>Let's consider a vector-valued function and see what a tangent vector looks like on its graph. We have the vector-valued function r(t) = sint<b>i </b>+ cost<b>j . </b>Its derivative is r'(t) = costi-sintj. Let's substitute t = π/6 in both functions: </p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhv_ZJXaYYZs9OPGzshMWC91yiaTlI5m_9QCFn8AJ0YBJzznD05sgdxPiV25fXgYhy1KrfDeZcClz-G-nv6j6FlBVLlCUO7bPw6fl1nZNWdBjekC9CpTkloUeEzGGvSSI-PZkxXPM5zDVBNGvSz5X1zyhuWdCuLvp6kyE3qPiRzxhQoK5FebM-C5eJd4dI" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="68" data-original-width="432" height="50" src="https://blogger.googleusercontent.com/img/a/AVvXsEhv_ZJXaYYZs9OPGzshMWC91yiaTlI5m_9QCFn8AJ0YBJzznD05sgdxPiV25fXgYhy1KrfDeZcClz-G-nv6j6FlBVLlCUO7bPw6fl1nZNWdBjekC9CpTkloUeEzGGvSSI-PZkxXPM5zDVBNGvSz5X1zyhuWdCuLvp6kyE3qPiRzxhQoK5FebM-C5eJd4dI" width="320" /></a></div><br /><br /><p></p><p><br /></p><p>Here is the graph of the function with the vectors r(π/6) r'(π/6):</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhCZqfd88FwK1Z8z_Ub-_5qOGNIqsPKXQxh9RmyVcAAlaacvL-n6JWUAFrbm5J5sESGvNG5wivhbDRZPvcNbJ-HEpIH6hTWnT0qcDZ4HzFL7b7iKdGME-oTc7LZkCTG-inXoQcZz4dEFbyYDicGxGAk7ubSadxWI9ggA76VVb-aM30lv_IpMDx8Tt-Vxz4" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="353" data-original-width="394" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEhCZqfd88FwK1Z8z_Ub-_5qOGNIqsPKXQxh9RmyVcAAlaacvL-n6JWUAFrbm5J5sESGvNG5wivhbDRZPvcNbJ-HEpIH6hTWnT0qcDZ4HzFL7b7iKdGME-oTc7LZkCTG-inXoQcZz4dEFbyYDicGxGAk7ubSadxWI9ggA76VVb-aM30lv_IpMDx8Tt-Vxz4" width="268" /></a></div><br /><br /><p></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p>Notice that the the tangent vector r'(π/6) is tangent to the circle at the corresponding point t = π/6. This is an example of the tangent vector to the plane curve defined by the function r(t) = sint<b>i </b>+ cost<b>j</b></p><p><b>Definition</b></p><p>Let C be a curve defined by a vector-valued function r, and assume that r'(t) exists when t = t₀. A tangent vector v at t = t₀.is any vector such that, when the tail of the vector is placed at point r(t₀) on the graph, vector v is tangent to curve C. Vector r'(t₀) is an example of a tangent vector at point t = t₀.. Furthermore, assume that r'(t) ≠ 0. The principal unit tangent vector at t is defined to be:</p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgAGeA_r5_wNmt9J-UYRdgOHTjWL9LZxRG9Qo6cuaRZUIe1fCTr1JnQFK3EAvSJ0v2whAuhClKEU-Gk8SNBQfSPfk-2YkZqKwL1lxg-EPM9RxirA0uOsmTyKz1y1LZ5Ryabhxi0fIgl30Shnx8cdxsE6ge01YdVIxsGCGTObBMsEC0h_EtBUltWTwQ4M7A" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="108" data-original-width="439" height="79" src="https://blogger.googleusercontent.com/img/a/AVvXsEgAGeA_r5_wNmt9J-UYRdgOHTjWL9LZxRG9Qo6cuaRZUIe1fCTr1JnQFK3EAvSJ0v2whAuhClKEU-Gk8SNBQfSPfk-2YkZqKwL1lxg-EPM9RxirA0uOsmTyKz1y1LZ5Ryabhxi0fIgl30Shnx8cdxsE6ge01YdVIxsGCGTObBMsEC0h_EtBUltWTwQ4M7A" width="320" /></a></b></div><b><br /><br /></b><p></p><p> </p><p><br /></p><p>The unit tangent vector is exactly what it sounds like: a unit vector that is tangent to the curve. To calculate a unit tangent vector, find the derivative r'(t). Second, calculate the magnitude of the derivative. The third step is to divide the derivative by its magnitude.</p><p><b>Example</b></p><p></p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><a href="https://blogger.googleusercontent.com/img/a/AVvXsEj_UHJxtZz_WJxiwbV3XSWRb1qBPBh9rbRULTk-lzjo7ym-TGzsygm0IlgCbIcYI-rfhpfX8k1NY-12MM4PiHI-JjcOPUE2wZWTvcPDIXL_sJtMxc_c0EsyZyIM11guLIiDMzwdXQfY5tfustpuwJeQR5dxq9yvDjRqWIU6aHRxdMOob-E77yprqJSCv4w" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="106" data-original-width="554" height="76" src="https://blogger.googleusercontent.com/img/a/AVvXsEj_UHJxtZz_WJxiwbV3XSWRb1qBPBh9rbRULTk-lzjo7ym-TGzsygm0IlgCbIcYI-rfhpfX8k1NY-12MM4PiHI-JjcOPUE2wZWTvcPDIXL_sJtMxc_c0EsyZyIM11guLIiDMzwdXQfY5tfustpuwJeQR5dxq9yvDjRqWIU6aHRxdMOob-E77yprqJSCv4w=w400-h76" width="400" /></a></b></div><b><br /><br /></b><br /><p></p><p><br /></p><p><b>Solution</b></p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><a href="https://blogger.googleusercontent.com/img/a/AVvXsEigdeBkq6DQr7QrgEVnMVzLOQ-A3CYb8B4gSEGX7R3OB8pc59DDpkfEY0oqoacnJW0HPUc-ExiY5A-5eqM7OZFme6xBudsEXP06TxX1I60zYQcV7844c_wRVJXZUWd2lQbI7NnafsgsznYrsbMUY3JW3A5BabSf35Jb2WrsPFzm_qVJOnt_3og0QkKHxpY" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="389" data-original-width="699" height="357" src="https://blogger.googleusercontent.com/img/a/AVvXsEigdeBkq6DQr7QrgEVnMVzLOQ-A3CYb8B4gSEGX7R3OB8pc59DDpkfEY0oqoacnJW0HPUc-ExiY5A-5eqM7OZFme6xBudsEXP06TxX1I60zYQcV7844c_wRVJXZUWd2lQbI7NnafsgsznYrsbMUY3JW3A5BabSf35Jb2WrsPFzm_qVJOnt_3og0QkKHxpY=w640-h357" width="640" /></a></b></div><b><br /><br /></b><p></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><b>Practice</b></p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhnB0VGeFRHt3fokkS3mwfv5wIIEQTZZ2tolkabllFzz7BgV3dE3fGosSzFauQ31bDNsZqm61YJozKBMwpvYEvquacrJSRijrsrL0BXhthxCqN59847NbSOWcBFxqt5f15lkH_DJlMd88iObfIE5QOc2xFvizX-TvDKdV2Lh-WkTXoicfX_s3GGYgcxtng" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="87" data-original-width="546" height="64" src="https://blogger.googleusercontent.com/img/a/AVvXsEhnB0VGeFRHt3fokkS3mwfv5wIIEQTZZ2tolkabllFzz7BgV3dE3fGosSzFauQ31bDNsZqm61YJozKBMwpvYEvquacrJSRijrsrL0BXhthxCqN59847NbSOWcBFxqt5f15lkH_DJlMd88iObfIE5QOc2xFvizX-TvDKdV2Lh-WkTXoicfX_s3GGYgcxtng=w400-h64" width="400" /></a></b></div><b><br /><br /></b><p></p><p><br /></p>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-57957785179314110832024-01-22T15:07:00.005-05:002024-01-22T15:07:33.985-05:00Properties of the derivatives of vector-valued functions<p> <b>Goal: Calculate the derivatives of vector-valued functions using the properties of derivatives</b></p><p><b>Properties of the derivatives of vector-valued functions</b></p><p>The constant multiple rule, the sum and difference rule, the product rule and the chain rule used in real valued functions are also applicable for vector-valued functions.</p><p><b>Theorem</b></p><p>Let r and u be differentiable vector-valued functions of t. Let f be a differentiable real-valued function of t and c a scalar.</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjn83pCluKJLcdy8VocP1IlSRee3jOrObEdt3muSCSfkfY-rX_wgF1tjI-KY25-yWCqmUqH55aHvGFJ8KynfQV4I0ig04B4Rkqlt6lhA_Nyhyd2wLc1O9NC_S7BbFIRXiBbt-1lRYRv7qaFVqvpx_BCwRbfKmIjeVuVg6nL8vOsIVxy-iV04WatS1Af9TY" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="249" data-original-width="487" src="https://blogger.googleusercontent.com/img/a/AVvXsEjn83pCluKJLcdy8VocP1IlSRee3jOrObEdt3muSCSfkfY-rX_wgF1tjI-KY25-yWCqmUqH55aHvGFJ8KynfQV4I0ig04B4Rkqlt6lhA_Nyhyd2wLc1O9NC_S7BbFIRXiBbt-1lRYRv7qaFVqvpx_BCwRbfKmIjeVuVg6nL8vOsIVxy-iV04WatS1Af9TY=s16000" /></a></div><br /><br /><p></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p>In property IV, we use dot product and in property V cross product.</p><p><b>Example:</b></p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhkYEqslItLiiy2xV_prnMSBXYW__zcZA3AKfGU9J2cs4pjkKttYQODkCrcmivPqnXBh_5SQdd_e_TaOPTkHgg3ldntuhtOjG7KoprXGoITHgnbiFDH50I7LHtyYdx0cs6OIfWvC4PYL0PbAW6mBLIf7Uzb2HAtzVV--vakn4bKp785JX5SWtFT4fe5Ko4" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="170" data-original-width="548" src="https://blogger.googleusercontent.com/img/a/AVvXsEhkYEqslItLiiy2xV_prnMSBXYW__zcZA3AKfGU9J2cs4pjkKttYQODkCrcmivPqnXBh_5SQdd_e_TaOPTkHgg3ldntuhtOjG7KoprXGoITHgnbiFDH50I7LHtyYdx0cs6OIfWvC4PYL0PbAW6mBLIf7Uzb2HAtzVV--vakn4bKp785JX5SWtFT4fe5Ko4=s16000" /></a></b></div><b><br /><br /></b><p></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p>Calculate each of the derivatives using the properties of the derivatives of vector-valued functions.</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjcnb21z0m_YNuyT_q32M5Vs-NbhdD-M6cVwl5tF2s2fHfDpGkTLmo-JOwsil_ToWYAdZ4Ym2PTF-ItdStY-OKhXNlXl5YAX4jPSCG_bgKH-uyEFB5mxalNH9tBYIHDQ73eIze9qigVLEaKy2SJi0TsSTKUCJ9mvW8snuq2CFiMRVWT6zG1buyQNVCnLlw" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="68" data-original-width="166" height="131" src="https://blogger.googleusercontent.com/img/a/AVvXsEjcnb21z0m_YNuyT_q32M5Vs-NbhdD-M6cVwl5tF2s2fHfDpGkTLmo-JOwsil_ToWYAdZ4Ym2PTF-ItdStY-OKhXNlXl5YAX4jPSCG_bgKH-uyEFB5mxalNH9tBYIHDQ73eIze9qigVLEaKy2SJi0TsSTKUCJ9mvW8snuq2CFiMRVWT6zG1buyQNVCnLlw" width="320" /></a></div><br /><br /><p></p><p><br /></p><p><br /></p><p><br /></p><p><b>Solution</b></p><p><b>a.</b></p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><a href="https://blogger.googleusercontent.com/img/a/AVvXsEiYIW-TXvYzEP5hByyMmvrk2pvjUpzH3I25S1u-5-5IZZusy2V3uaKw6eU5ZuOaYB3EH6p0aDmVDpB6nOSv1q8FLk3br7ERFLG2gOp5IgxGCbWc6VbGfbtkFjKubK4hgFB7hsljEKr8urS2dB639FY-J6qL1K1YxMj1H6nR6N1HRqC-UJEBsc6-yXWpQM4" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="40" data-original-width="582" src="https://blogger.googleusercontent.com/img/a/AVvXsEiYIW-TXvYzEP5hByyMmvrk2pvjUpzH3I25S1u-5-5IZZusy2V3uaKw6eU5ZuOaYB3EH6p0aDmVDpB6nOSv1q8FLk3br7ERFLG2gOp5IgxGCbWc6VbGfbtkFjKubK4hgFB7hsljEKr8urS2dB639FY-J6qL1K1YxMj1H6nR6N1HRqC-UJEBsc6-yXWpQM4=s16000" /></a></b></div><b><br /><br /></b><p></p><p><b><br /></b></p><p>According to property IV (dot product), we have:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgu6o1_ySeEOao9rqcgR8_jxvsrH_Sa2E4z-sQvWVuW6P-9JKVQcKtlsjKigz2n4Xp1Od0zk4FX96Vd-TJoFB9ZrBkvsLV3K-jVOFdZlreexyrdhk0E2lPNrVlpdV2NOM9wPGh38CqHPuKzkKUJ-s50hIrvXK9Yja0LgtKIBheNOreANG3XtQuDCXooY80" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="47" data-original-width="369" height="41" src="https://blogger.googleusercontent.com/img/a/AVvXsEgu6o1_ySeEOao9rqcgR8_jxvsrH_Sa2E4z-sQvWVuW6P-9JKVQcKtlsjKigz2n4Xp1Od0zk4FX96Vd-TJoFB9ZrBkvsLV3K-jVOFdZlreexyrdhk0E2lPNrVlpdV2NOM9wPGh38CqHPuKzkKUJ-s50hIrvXK9Yja0LgtKIBheNOreANG3XtQuDCXooY80" width="320" /></a></div><br /><br /><p></p><p><br /></p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEil1rLxJkjvB6-RnDBnFPzCbt_Z97OrvmuhOE7Rr8miU6EqHJIcneiTwyYowv-4FG87gtnwN1UWOqPm-_aq4BFXk6ZRYqdYc1KU38W7jA1gYD5PdIAm1IN2W40UbKJPQ8sAAsdDHu8zUTq28vPnq5_VHUmLGi8GCpfn952DuVkfpMIuRv5o4-hE0qR5C2I" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="155" data-original-width="557" src="https://blogger.googleusercontent.com/img/a/AVvXsEil1rLxJkjvB6-RnDBnFPzCbt_Z97OrvmuhOE7Rr8miU6EqHJIcneiTwyYowv-4FG87gtnwN1UWOqPm-_aq4BFXk6ZRYqdYc1KU38W7jA1gYD5PdIAm1IN2W40UbKJPQ8sAAsdDHu8zUTq28vPnq5_VHUmLGi8GCpfn952DuVkfpMIuRv5o4-hE0qR5C2I=s16000" /></a></div><br /><br /><p></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p>b. To solve this problem, we need to adapt it to property V (cross product). Property V is stated as follow:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgMBuDJrVnDBJSZgGpbMWrmJD4-zCks5qeH2c5QfvSBvzljg3tnnTU1c_Oc0xoj3yFZI3rviJP-UE7OiRdEyivHgUZM8WqWB_TKuEJT__qjtl3U7bqM2AefIA50E6M-1CbnqzH4zFjhMVei7jy2VdrlXDpV9837A37q-L8K6yNt3Lj0m3niAIb_iia4oPk" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="68" data-original-width="435" height="50" src="https://blogger.googleusercontent.com/img/a/AVvXsEgMBuDJrVnDBJSZgGpbMWrmJD4-zCks5qeH2c5QfvSBvzljg3tnnTU1c_Oc0xoj3yFZI3rviJP-UE7OiRdEyivHgUZM8WqWB_TKuEJT__qjtl3U7bqM2AefIA50E6M-1CbnqzH4zFjhMVei7jy2VdrlXDpV9837A37q-L8K6yNt3Lj0m3niAIb_iia4oPk" width="320" /></a></div><br /><br /><p></p><p><br /></p><p>Recall that the cross product of any vector with itself is zero. Let's calculate the second derivative of u(t):</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEguq9bkSjLm3TECFQiaY_qewJvOOH3umetAeExg_6c4Kpz5iwg6v3aT9ekCf0-GES4heQooKDcdUQ9aHq8cQO7nk1zwWsQBA8gi0akESs01DeoEJKm9ij9vt5-3uqfnHCYwtKV1fylPcvo31RYO3K59lBYSYjqBwDodgnwI1-b2YQ458eQ2_nUSmdshKR8" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="56" data-original-width="519" height="35" src="https://blogger.googleusercontent.com/img/a/AVvXsEguq9bkSjLm3TECFQiaY_qewJvOOH3umetAeExg_6c4Kpz5iwg6v3aT9ekCf0-GES4heQooKDcdUQ9aHq8cQO7nk1zwWsQBA8gi0akESs01DeoEJKm9ij9vt5-3uqfnHCYwtKV1fylPcvo31RYO3K59lBYSYjqBwDodgnwI1-b2YQ458eQ2_nUSmdshKR8" width="320" /></a></div><br /><br /><p></p><p>Finally we have:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgttrXcGK2UnYiubgivuLUKSgw8qcuAqubrfaS-55KPJqO_KZvCr2-EFjXZnZ5OjXjM7tqUv-3rP5i4ZgcaN4liJRgfWXsRRqbYIZyJTd0_i5vfniWMwHIAt83orjaQtbFyrhKCxIietysZvA2JKKGT1ucEZKu-Ppc_fFg-w1uxWYTsnSCK4aDpi7X780k" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="203" data-original-width="645" height="126" src="https://blogger.googleusercontent.com/img/a/AVvXsEgttrXcGK2UnYiubgivuLUKSgw8qcuAqubrfaS-55KPJqO_KZvCr2-EFjXZnZ5OjXjM7tqUv-3rP5i4ZgcaN4liJRgfWXsRRqbYIZyJTd0_i5vfniWMwHIAt83orjaQtbFyrhKCxIietysZvA2JKKGT1ucEZKu-Ppc_fFg-w1uxWYTsnSCK4aDpi7X780k=w400-h126" width="400" /></a></div><br /><br /><p></p><p><br /></p><p><br /></p><p><br /></p><p><b>Practice</b></p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjeK16D_8ob1PShQwU_81iZuwfONwAtlvaUM-nDgGVJsKh9srzcl6io7fS7sSviEWgvP-5K9q7huVxMHJispy_1raTtXT0KYN1He6hEQ3rU36N2cpdJluzrUEw_EewjMB0glSOKFageJi4Khjs7pj1X48FO2guSq46oz3tObqAkcEZx5WRTO5o6kyczCOs" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="64" data-original-width="738" height="56" src="https://blogger.googleusercontent.com/img/a/AVvXsEjeK16D_8ob1PShQwU_81iZuwfONwAtlvaUM-nDgGVJsKh9srzcl6io7fS7sSviEWgvP-5K9q7huVxMHJispy_1raTtXT0KYN1He6hEQ3rU36N2cpdJluzrUEw_EewjMB0glSOKFageJi4Khjs7pj1X48FO2guSq46oz3tObqAkcEZx5WRTO5o6kyczCOs=w640-h56" width="640" /></a></b></div><b><br /><br /></b><p></p><p><br /></p><p><b><br /></b></p><div class="separator" style="clear: both; text-align: center;"><b><br /></b></div><div class="separator" style="clear: both; text-align: center;"><b><br /></b></div><div class="separator" style="clear: both; text-align: center;"><b><br /></b></div><div class="separator" style="clear: both; text-align: center;"><b><br /></b></div><b><br /><br /><div class="separator" style="clear: both; text-align: center;"><br /></div><br /></b><p></p><p><b><br /></b></p><p><b><br /></b></p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><br /></b></div><b><br /><br /></b><p></p><scribe-shadow id="crxjs-ext" style="height: 0px; left: 0px; overflow: visible; position: fixed; top: 0px; width: 0px; z-index: 2147483647;"></scribe-shadow>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-73097816230670421052024-01-12T21:28:00.003-05:002024-01-12T21:50:54.254-05:00Differentiation of vector-valued functions<p><b>Goal</b>: Find the derivative of vector-valued functions by calculating the derivative of the real-valued functions of the vector-valued function.</p><p><b>Differentiation of vector-valued functions</b></p><p>Let's consider the vector-valued function r(t) = f(t)i + g(t)j or r(t) = f(t)i + g(t)j + h(t)k, we can find the derivative of these vector-valued functions by calculating the derivative of the real-valued functions . The following theorem allows us to do that.</p><p><b>Theorem</b></p><p>Let f, g, h be differentiable functions of t:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEia-yhKnmOi1CTt2VweQQy0lowG6BOYKOmNYKJVNcpaMaaR0Eck1ls5bPzN4osvWMBjKLKHLUdlksMrqQMspYyaVIDXKiVBpG5B9V4u4B5ewTvKlvh41g2lTlqcdTaxMKTuTPzoYcEbeM2a0RLjL9T-e4vjAXKxF-Jsv_uZ-7dPRIUCcAiaxRfnQzTQ-p8" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="58" data-original-width="618" src="https://blogger.googleusercontent.com/img/a/AVvXsEia-yhKnmOi1CTt2VweQQy0lowG6BOYKOmNYKJVNcpaMaaR0Eck1ls5bPzN4osvWMBjKLKHLUdlksMrqQMspYyaVIDXKiVBpG5B9V4u4B5ewTvKlvh41g2lTlqcdTaxMKTuTPzoYcEbeM2a0RLjL9T-e4vjAXKxF-Jsv_uZ-7dPRIUCcAiaxRfnQzTQ-p8=s16000" /></a></div><br /><br /><p></p><p><br /></p><p><b>Example</b></p><p>Use differentiation to find the derivative of the following vector-valued functions:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjDY3Mvu4p9Z1a8DTx_9F4TwnkM4rdPzCCIyNkxpmXsixy3rOXOSfL3leGOF8WeruEeGAq1VRMZy5OTJpGevXgvKMqc7LxZK0VCkxLMohShA-n24vhERPCH42yl2o4cMuacVFTwvyYIGZIZ_s4dV7GgPsz_nAkbGGtGI-5AQy_pYBRYh_J495Akwx2xXN4" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="90" data-original-width="323" height="89" src="https://blogger.googleusercontent.com/img/a/AVvXsEjDY3Mvu4p9Z1a8DTx_9F4TwnkM4rdPzCCIyNkxpmXsixy3rOXOSfL3leGOF8WeruEeGAq1VRMZy5OTJpGevXgvKMqc7LxZK0VCkxLMohShA-n24vhERPCH42yl2o4cMuacVFTwvyYIGZIZ_s4dV7GgPsz_nAkbGGtGI-5AQy_pYBRYh_J495Akwx2xXN4" width="320" /></a></div><br /><br /><p></p><p><br /></p><p><br /></p><p><b>Solution</b></p><p>Let's use the formula of the differentiation of vector-valued functions and the formulas of the differentiation of real-valued functions</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhYSTzRJf4-g2awyviygOvJp8yxV1wXz6flwxRw2KnSbX5c83U6TrbKtdV8iFpPZPFePKe41sIV24uH75t7vbALMtlrreMxR1O3HEaCQwq0sF4TvosAMoeZ0wQpo1lns4IJ_ML3RokTQAzuYlBpc-NQRrQLcbZCRV2dt_69gIvA4_kZ2OpM__JX_Q7st7I" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="229" data-original-width="714" height="205" src="https://blogger.googleusercontent.com/img/a/AVvXsEhYSTzRJf4-g2awyviygOvJp8yxV1wXz6flwxRw2KnSbX5c83U6TrbKtdV8iFpPZPFePKe41sIV24uH75t7vbALMtlrreMxR1O3HEaCQwq0sF4TvosAMoeZ0wQpo1lns4IJ_ML3RokTQAzuYlBpc-NQRrQLcbZCRV2dt_69gIvA4_kZ2OpM__JX_Q7st7I=w640-h205" width="640" /></a></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><b>Practice</b><div><br /></div><div>Calculate the derivative of the following function:</div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjshZaWjqpVlZOQPjqqiUPUFBbcGGqbobEwvSYTBqmiXfMbwJPInIEaaVV_xUIIjRRldb762rjc-iXgRzI5ecJ5lyJmaH3rs-p-vyTqqsJhve5MCLg_svn6UGjAQ9nnJD7bkOSsUnEMe9KRby_mfjdWYNKLrbpY5isD0Fk1rkt_dMbZQQg8iT2cDxti8uw" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="34" data-original-width="368" height="30" src="https://blogger.googleusercontent.com/img/a/AVvXsEjshZaWjqpVlZOQPjqqiUPUFBbcGGqbobEwvSYTBqmiXfMbwJPInIEaaVV_xUIIjRRldb762rjc-iXgRzI5ecJ5lyJmaH3rs-p-vyTqqsJhve5MCLg_svn6UGjAQ9nnJD7bkOSsUnEMe9KRby_mfjdWYNKLrbpY5isD0Fk1rkt_dMbZQQg8iT2cDxti8uw" width="320" /></a></div><br /><br /></div><div><br /></div><div><br /><br /><p></p><p></p><div class="separator" style="clear: both; text-align: center;"><span style="text-align: left;"><br /></span></div><div class="separator" style="clear: both; text-align: center;"><span style="text-align: left;"><br /></span></div><div class="separator" style="clear: both; text-align: center;"><span style="text-align: left;"><br /></span></div><div class="separator" style="clear: both; text-align: center;"><span style="text-align: left;"><br /></span></div><div class="separator" style="clear: both; text-align: center;"><span style="text-align: left;"><br /></span></div><div class="separator" style="clear: both; text-align: center;"><span style="text-align: left;"><br /></span></div><div class="separator" style="clear: both; text-align: center;"><span style="text-align: left;"><br /></span></div><div class="separator" style="clear: both; text-align: center;"><span style="text-align: left;"><br /></span></div><div class="separator" style="clear: both; text-align: center;"><br /></div><br /><p></p><p><br /></p></div>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-67390372916786690112024-01-12T20:21:00.001-05:002024-01-12T21:34:57.091-05:00Derivative of vector-valued functions<p> O<b>bjective</b>: Define the derivative of a vector-valued function</p><p>Since the limit of a vector-valued function has been defined, we can now define its derivative. The definition of the derivative of a vector-valued is similar to that of real-valued functions. The only difference is in the range of the derivative of vector-valued functions. Since the range of a vector-valued function is made of vectors, the range of the derivative of the vector-valued function consists also of vectors. </p><p><b>Definition</b></p><p>The derivative of a vector-valued is defined by:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEicQwxPStScesdWDvSDk73l5obm3R8dQGQOxE5SIWkz46y_RHHayXn3K_g_JKB5bSwuxuRODBgfq6xC2X3z6Umu70cpwT27ili1EGi2Q1AMpCdlBXIkDMcnNxhAaxjKYD9hUevxeE3G9hM_9OYYnJ6NqdxqUYMoEvva8vMWJzW3-ZwvDIARmyzWapCMnYA" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="74" data-original-width="262" height="90" src="https://blogger.googleusercontent.com/img/a/AVvXsEicQwxPStScesdWDvSDk73l5obm3R8dQGQOxE5SIWkz46y_RHHayXn3K_g_JKB5bSwuxuRODBgfq6xC2X3z6Umu70cpwT27ili1EGi2Q1AMpCdlBXIkDMcnNxhAaxjKYD9hUevxeE3G9hM_9OYYnJ6NqdxqUYMoEvva8vMWJzW3-ZwvDIARmyzWapCMnYA" width="320" /></a></div><br /><br /><p></p><p><br /></p><p><br /></p><p>provided that the limit exists. If r'(t) exists, then r is differentiable at t. If r is differentiable for all values of t in an open interval (a, b), then r is differentiable in this interval. The following two limits' must exist as well for the function r to be differentiable in a closed interval [a, b]. </p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjgFlxuoHYpBkNNSuiNb7tqMfYmFUnjcepA5fDmd9wN_uWAjAinYXjBkZczwREBWIP8h_fjp1bpXoxb_1Zcyti6KYxSsp24wr1ZoEBGwfWf7PsYzJLNpF6Y7hAUntehpRm1SoOwarpc10K-BVAPl2Atl44Lfyg3CY1OqT5rICMeWx36hZlf6d2NI-z37yM" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="68" data-original-width="271" height="80" src="https://blogger.googleusercontent.com/img/a/AVvXsEjgFlxuoHYpBkNNSuiNb7tqMfYmFUnjcepA5fDmd9wN_uWAjAinYXjBkZczwREBWIP8h_fjp1bpXoxb_1Zcyti6KYxSsp24wr1ZoEBGwfWf7PsYzJLNpF6Y7hAUntehpRm1SoOwarpc10K-BVAPl2Atl44Lfyg3CY1OqT5rICMeWx36hZlf6d2NI-z37yM" width="320" /></a></div><br />and<p></p><p><br /></p><p><br /></p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEiz5WwSvFKI6VsTkcTo1kVasEBOqIY3S_5q6s17nhcOl0UQyTmk1VPFfesEBpJAv3dmQH4O50eu1k1RX6HXM3zrafJG3M92VzoqnEblv3iPze6BcwRqleeZmEY7LbREZ8jkfUlBJbVUCKSAC3WzzyzosMl1Ed7eI8VMHvCRmbUtXy7BjnjOrQ0Epv7TGZ8" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="61" data-original-width="272" height="72" src="https://blogger.googleusercontent.com/img/a/AVvXsEiz5WwSvFKI6VsTkcTo1kVasEBOqIY3S_5q6s17nhcOl0UQyTmk1VPFfesEBpJAv3dmQH4O50eu1k1RX6HXM3zrafJG3M92VzoqnEblv3iPze6BcwRqleeZmEY7LbREZ8jkfUlBJbVUCKSAC3WzzyzosMl1Ed7eI8VMHvCRmbUtXy7BjnjOrQ0Epv7TGZ8" width="320" /></a></div><br /><br /><p></p><div><br /></div><div><br /></div><div><b>Example</b></div><div><b><br /></b></div><div><b><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhtIIB3c8HvoYbUcYZDp0DINKsjOJGSEewxf5vpU5eRHkipw8vuX9gn5IS7MOZoG_W12qj8npX2MXtnJ4b84-I_0CKHtKvVCQaOwQoA0MYCBwpzBv2dIIsGEVMzwJLa-aTZu60qf3oIixgc114dm7ygJwJ5ttKdtwG6WqIIVrMshxlAy4b0HjiXQoOq0Ro" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="46" data-original-width="299" height="49" src="https://blogger.googleusercontent.com/img/a/AVvXsEhtIIB3c8HvoYbUcYZDp0DINKsjOJGSEewxf5vpU5eRHkipw8vuX9gn5IS7MOZoG_W12qj8npX2MXtnJ4b84-I_0CKHtKvVCQaOwQoA0MYCBwpzBv2dIIsGEVMzwJLa-aTZu60qf3oIixgc114dm7ygJwJ5ttKdtwG6WqIIVrMshxlAy4b0HjiXQoOq0Ro" width="320" /></a></div><br /><br /></b></div><div><b><br /></b></div><div><b><br /></b></div><div><b>Solution</b></div><div><b><br /></b></div><div>Let's apply the formula:</div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEj2QqfkimNegq-S3O-wBWQdWZMrm3LFGvnSY9RrwS4Tjttw4f1ez6tdl7XDaLiusEshzcndnPUAje14NA64SZtbxIP1fLVdgOaCV4SIiBKN2gQD4afXZTcTYpd4ak2hh-Nbv9Z9YYFPo8OkyU11fVuty0HyfLW59LeD59GntGCmnfSeH_eCYZwoFj-Y8W4" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="285" data-original-width="549" src="https://blogger.googleusercontent.com/img/a/AVvXsEj2QqfkimNegq-S3O-wBWQdWZMrm3LFGvnSY9RrwS4Tjttw4f1ez6tdl7XDaLiusEshzcndnPUAje14NA64SZtbxIP1fLVdgOaCV4SIiBKN2gQD4afXZTcTYpd4ak2hh-Nbv9Z9YYFPo8OkyU11fVuty0HyfLW59LeD59GntGCmnfSeH_eCYZwoFj-Y8W4=s16000" /></a></div><br /><br /></div><div><b><br /></b></div><div><b><br /></b></div><div><b><br /></b></div><div><b><br /></b></div><div><b><br /></b></div><div><b><br /></b></div><div><b><br /></b></div><div><b><br /></b></div><div><b><br /></b></div><div><b><br /></b></div><div><b><br /></b></div><div><b><br /></b></div><div><b><br /></b></div><div><b><br /></b></div><div><b><br /></b></div><div><b>Practice</b></div><div><b><br /></b></div><div>Use the definition to find the derivative of the vector-valued function:</div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEg3NOa5HE_admK_tSfs56LKmQrt17F_6Otc7GxNA8dxHN3BRx_FAF4EhUD-21bB0HNyw6C-hN6MJWHaOrFAv1Btw0zsFiVOozBCGOHSxadMQekpzrT729A8rnfpx2B-DZPIywLbXnxmMNHwKHIjUu0r7wDFqUtqj8Cdxoz6VwXwb_8gRfhPlZnifj10bp4" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="33" data-original-width="258" height="41" src="https://blogger.googleusercontent.com/img/a/AVvXsEg3NOa5HE_admK_tSfs56LKmQrt17F_6Otc7GxNA8dxHN3BRx_FAF4EhUD-21bB0HNyw6C-hN6MJWHaOrFAv1Btw0zsFiVOozBCGOHSxadMQekpzrT729A8rnfpx2B-DZPIywLbXnxmMNHwKHIjUu0r7wDFqUtqj8Cdxoz6VwXwb_8gRfhPlZnifj10bp4" width="320" /></a></div><br /><br /></div><div><b><br /></b></div>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-65995322409926477392024-01-05T23:28:00.001-05:002024-01-05T23:30:52.698-05:00Limit of a vector-valued function<p> Goal: Determine the limit of a vector-valued function</p><p><b>Definition</b></p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><a href="https://blogger.googleusercontent.com/img/a/AVvXsEiR5aWdzBiEqb_91fjTdfJYHriXudTWkK5EVKhE8ufV4IcOwn2VuHyEkg1XVhsIlFJgFmeWr8tRkgu2hlMoXLp4udZuNmFX_GiYfrj67ZvUo3Myzbfq6ScCAhISUsXeKlTzs9GB7VrOcXdyu-ONcY2eM2GF86YdNAkmFbj-phgeWeVhnmrPYNoPd2xniVo" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="165" data-original-width="560" height="118" src="https://blogger.googleusercontent.com/img/a/AVvXsEiR5aWdzBiEqb_91fjTdfJYHriXudTWkK5EVKhE8ufV4IcOwn2VuHyEkg1XVhsIlFJgFmeWr8tRkgu2hlMoXLp4udZuNmFX_GiYfrj67ZvUo3Myzbfq6ScCAhISUsXeKlTzs9GB7VrOcXdyu-ONcY2eM2GF86YdNAkmFbj-phgeWeVhnmrPYNoPd2xniVo=w400-h118" width="400" /></a></b></div><b><br /><br /></b><p></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p>This is a rigorous definition. In practice we apply the following theorem:</p><p><b>Theorem</b></p><p><b>Limit of a vector-valued function</b></p><p>Let f, g, h be functions of t and r(t) = f(t)i + g(t)j, the limit of r as t approaches a is defined by:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEh7DOoKYWYTxKvIZGZelpNKAhWhUu5UgaWDLoEgvvoSkMhnX94h-mSUl1XW9WWNF5daWajMkNuBJVRR25El1ZV-nKnm1uZI_mlHvuNcRsvTusiGgyiU4cCs_aWfArEiz5dmrmMWmXdkINYutpG6rLUnL2UgQQWGENOa9qSUVd-yM8szi3VZVbWKr-s6b2U" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="54" data-original-width="306" height="56" src="https://blogger.googleusercontent.com/img/a/AVvXsEh7DOoKYWYTxKvIZGZelpNKAhWhUu5UgaWDLoEgvvoSkMhnX94h-mSUl1XW9WWNF5daWajMkNuBJVRR25El1ZV-nKnm1uZI_mlHvuNcRsvTusiGgyiU4cCs_aWfArEiz5dmrmMWmXdkINYutpG6rLUnL2UgQQWGENOa9qSUVd-yM8szi3VZVbWKr-s6b2U" width="320" /></a></div><br /><br /><p></p><p><br /></p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgWBsZZkmlos-isRR1uVXEF75HbKqf2MkbCYEzMYGrpvfcXZW48LhgsKN8TgtR_BUPzT7vy9K44ojyF6Zd7B4yx1PnG0uYUnHtSdpiAARdbC6ngfHAbc6pQl5IbUgB7ec3KHvLUKUELHC6vlxmlSewdg3u8Gw2vOjZqRVwJ7aBhmWEBLTjCKdXXQrNEQlM" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="48" data-original-width="352" height="44" src="https://blogger.googleusercontent.com/img/a/AVvXsEgWBsZZkmlos-isRR1uVXEF75HbKqf2MkbCYEzMYGrpvfcXZW48LhgsKN8TgtR_BUPzT7vy9K44ojyF6Zd7B4yx1PnG0uYUnHtSdpiAARdbC6ngfHAbc6pQl5IbUgB7ec3KHvLUKUELHC6vlxmlSewdg3u8Gw2vOjZqRVwJ7aBhmWEBLTjCKdXXQrNEQlM" width="320" /></a></div><p><br /></p><p><br /></p>Similarly, the limit of the vector-valued function r(t) = f(t)i + g(t)j + h(t)k as t approaches a is defined by:<p></p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEiJ4qDlL3qfFEI8Fl5sa78ptB2VKw5DsEuRZNm64quSWr2mmyTP4EvgGHPDspXytH8yrM2OlCXlj-yeigaNm26EVj6f11RP37gqbrWfIB2SeYqT1Hn4Ga5Zr9ibfxpcgIxpPUQGwYr39Rq9VzBX7f5waTK2OCQCV8Nje1pKWEfsIxBijlhY5AUsmha9sI4" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="95" data-original-width="584" height="65" src="https://blogger.googleusercontent.com/img/a/AVvXsEiJ4qDlL3qfFEI8Fl5sa78ptB2VKw5DsEuRZNm64quSWr2mmyTP4EvgGHPDspXytH8yrM2OlCXlj-yeigaNm26EVj6f11RP37gqbrWfIB2SeYqT1Hn4Ga5Zr9ibfxpcgIxpPUQGwYr39Rq9VzBX7f5waTK2OCQCV8Nje1pKWEfsIxBijlhY5AUsmha9sI4=w400-h65" width="400" /></a></div><br /><br /><br /><p></p><p><br /></p><p><b>Examples</b></p><p>For each of the vector-valued functions calculate lim r(t) when t approaches 3:</p><p>a. r(t) = (t²- 3t + 4)i + (4t + 3)j</p><p>b. r(t) = (2t-4/t+1)i + (t/t²+1)j + (4t - 3)k</p><p><b>Solution</b></p><p><b>a. </b>Using the definition of limit above and substituting t, we have:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEho6QHPJSKJjet0ho5nIiZBzWVReAggte8HOxh3teKW3-bkng0Lx5YRR2Em6HJqkH4Tpq6aVxsrNxa2Ffc6lPs42cn5JmZ2ZnwDA31VcARpftlzV2_BRqpDa_j-1UQkotUcW6UPofxCK06BrgwTnB18pA2NHM-O77iz-MbHSEATIOinFYUexXuAc8cVM9M" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="113" data-original-width="431" height="84" src="https://blogger.googleusercontent.com/img/a/AVvXsEho6QHPJSKJjet0ho5nIiZBzWVReAggte8HOxh3teKW3-bkng0Lx5YRR2Em6HJqkH4Tpq6aVxsrNxa2Ffc6lPs42cn5JmZ2ZnwDA31VcARpftlzV2_BRqpDa_j-1UQkotUcW6UPofxCK06BrgwTnB18pA2NHM-O77iz-MbHSEATIOinFYUexXuAc8cVM9M" width="320" /></a></div><br /><br /><p></p><p> </p><p><br /></p><p>b. Doing the same:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEiBgCCEI38d7MSdq7XWacl6Ou3-SP1Vi0bgFgURgURbBGUIvRgCBSJfB1vw0fN8UZh6W3rBonZyh6qEtBDbhjiBKRjVohpR1ujrbMS0p6SnqQBxGwvxbmPVYTQMvsjOUTKnvK1L1jinenBz35315RwXM7ZuDJM7FQDfXyHfp76N2ajoRNhgoOTrtcHSJPk" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="125" data-original-width="533" height="94" src="https://blogger.googleusercontent.com/img/a/AVvXsEiBgCCEI38d7MSdq7XWacl6Ou3-SP1Vi0bgFgURgURbBGUIvRgCBSJfB1vw0fN8UZh6W3rBonZyh6qEtBDbhjiBKRjVohpR1ujrbMS0p6SnqQBxGwvxbmPVYTQMvsjOUTKnvK1L1jinenBz35315RwXM7ZuDJM7FQDfXyHfp76N2ajoRNhgoOTrtcHSJPk=w400-h94" width="400" /></a></div><br /><br /><p></p><p><br /></p><p><br /></p><p><b>Practice</b></p><p>Calculate lim<b>r(t) </b>when t approaches 2 for:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgYZA_FkzCJZ1uKs2VXtELb5DhsFykRWgsAhoLQuhCaHEk9_nxaSkzrubb2C9pB_cuuVzYz0yA-bDk5Lhm3WK8fCufnlxcXH-vMSDPQJV2daE2iQSb4ShbdEmPVGWiTu8Ze6zDQwF_TWONSnf9tkokz4DNR6ntFBW8vgtTGmpgVYHD4sAuES8_aySEyork" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="36" data-original-width="394" height="29" src="https://blogger.googleusercontent.com/img/a/AVvXsEgYZA_FkzCJZ1uKs2VXtELb5DhsFykRWgsAhoLQuhCaHEk9_nxaSkzrubb2C9pB_cuuVzYz0yA-bDk5Lhm3WK8fCufnlxcXH-vMSDPQJV2daE2iQSb4ShbdEmPVGWiTu8Ze6zDQwF_TWONSnf9tkokz4DNR6ntFBW8vgtTGmpgVYHD4sAuES8_aySEyork" width="320" /></a></div><br /><br /><p></p><p><br /></p><p><br /></p><p></p><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><br /><br /><p></p><p></p><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhiJYC35ewKWHDZPiL9z29VU9Hx_UN9u0rDlkDiN0OyZpAoLDu_UAP4BfIA9AjvRL6TfIsG7y-IVlYGCZqZ2KUiOjd4fORPIBmEgavRmULVrl-dZH4qpk8w9m_QpiYPt21HSfveiZTXM_L6svma5yjeTbQZPeWrJatPVsDa8jo9BGKW14cVTh0qKgLqJ7Q" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><br /></a>
</div>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-29415080186390231352024-01-05T22:00:00.000-05:002024-01-05T22:00:02.003-05:00Graph of a vector-valued function<p><b>Goal: Graph of a vector-valued function</b></p><p><b>Considerations</b></p><p>A vector is defined by two quantities: magnitude and direction. If we consider an initial point and move in a certain direction according to a certain distance, we arrive at a second point. The second point is the terminal point. To determine the coordinates of a vector, we subtract the coordinates of the original point from those of the terminal point.</p><p>A vector is said to be in standard position if if its original point is located in the origin. In order to maintain the uniqueness of the graph of a vector-valued function, we start by graphing the vectors in the domain of the function.. The graph of a vector-valued function of the form r(t) = f(t)i + g(t)j is the sets of all (t, r(t)) and the path it traces is called a plane curve. The graph of a vector-valued function of the form r(t) = f(t)i + g(t)j + h(t)k and the path it traces is called a space curve. Any representation of a plane curve or space curve using a vector-valued function is called a vector parameterization of the curve. </p><p><b>Examples</b></p><p>Create<b> </b> a<b> </b>graph of each of the vector-valued functions:</p><p>a. The plane curve represented by r(t) = 4costi + 3sintj 0 ≤ t ≤ 2ℼ</p><p>b. The plane curve represented by r(t) = 4costi + 3sintj 0 ≤ t ≤ π</p><p><b>Solutions</b></p><p>a. We start by creating a table of values made of values of t in the domain of the vector valued-function and corresponding values of r(t). We graph each of the vectors of the second column in standard form and connect the terminal point of each vector to form a curve. It turns out that the curve is an ellipse centered at the origin </p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEiZq9iPyZfuuROiFgYnVjCHqChuBL5fqJXxP-TOfyojd_1tpsIazZJoLyvJrzXIH9FE_deNajpe-UMC0TqkPxqyaFJfnho27sGU2P5tj56AYAtgL-YTRDXWoaiH26LfZyfepvIJyTBdTDs3gAb_4124dwwtZyrCptOvVqXBuI8F4owfgetkqDZ2hRsufoM" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="302" data-original-width="576" height="210" src="https://blogger.googleusercontent.com/img/a/AVvXsEiZq9iPyZfuuROiFgYnVjCHqChuBL5fqJXxP-TOfyojd_1tpsIazZJoLyvJrzXIH9FE_deNajpe-UMC0TqkPxqyaFJfnho27sGU2P5tj56AYAtgL-YTRDXWoaiH26LfZyfepvIJyTBdTDs3gAb_4124dwwtZyrCptOvVqXBuI8F4owfgetkqDZ2hRsufoM=w400-h210" width="400" /></a></div><br /><br /><p></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><b>Graph</b></p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><a href="https://blogger.googleusercontent.com/img/a/AVvXsEidxEykDnfI4XfrXEB2yTx6snTLevuTeu99IJW-KTk74JwGluDcKYSo5XpXihoX1h2pioCzWJToxP_YGQ00VikO2BrQtRK9ZSpVZnWrXYQyWhtZPyDepzDLQWh7UBrrGwjvAYLikUoCilEca-NBeskGjazX5ZZtuNiShSxYjAvPXhoD59iniL8NV9f6ssM" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="422" data-original-width="417" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEidxEykDnfI4XfrXEB2yTx6snTLevuTeu99IJW-KTk74JwGluDcKYSo5XpXihoX1h2pioCzWJToxP_YGQ00VikO2BrQtRK9ZSpVZnWrXYQyWhtZPyDepzDLQWh7UBrrGwjvAYLikUoCilEca-NBeskGjazX5ZZtuNiShSxYjAvPXhoD59iniL8NV9f6ssM" width="237" /></a></b></div><b><br /><br /></b><p></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p>b. Let's do the same for the same for the second example.</p><p><b>Table of values</b></p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhQ9nsIBV-Zc5iULpRyuA8qnJIlH9qTlv8NM5be0bs6Io_DAQ4jZl0t-3bz0LWU6leFYwAdntxV4M3Hi4_jItO7GYtraNqczCTLgMJ6csYYXSd61EEbhwjWKmNCdz4KfbNAIfA_t1CXxA0AyZzHgJ1Ujoki0-x8-Pecd2fgor1XSlxz_QdIaXow8prt1Ss" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="319" data-original-width="593" height="215" src="https://blogger.googleusercontent.com/img/a/AVvXsEhQ9nsIBV-Zc5iULpRyuA8qnJIlH9qTlv8NM5be0bs6Io_DAQ4jZl0t-3bz0LWU6leFYwAdntxV4M3Hi4_jItO7GYtraNqczCTLgMJ6csYYXSd61EEbhwjWKmNCdz4KfbNAIfA_t1CXxA0AyZzHgJ1Ujoki0-x8-Pecd2fgor1XSlxz_QdIaXow8prt1Ss=w400-h215" width="400" /></a></b></div><b><br /><br /></b><p></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><b>Graph</b></p><p>The graph of this curve is an also an ellipse centered at the origin.</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEi0YHonVse8_FA_6awLwdaZV5m7w-sQrJXz9SX0obM1JBRbRHKVz-VF-e0-5Gpb-PU9QTdT4Kth3MNFGaJy28IRdBgqH0X5q3Wu5v_QdXoo0R2gp74auGTLpK-R3gK3HOl1QE2BORulu085J_IMh3LHI83IaUkkFeV3LJ7dmNuRohsILN90w3_j-vWuhk8" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="422" data-original-width="425" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEi0YHonVse8_FA_6awLwdaZV5m7w-sQrJXz9SX0obM1JBRbRHKVz-VF-e0-5Gpb-PU9QTdT4Kth3MNFGaJy28IRdBgqH0X5q3Wu5v_QdXoo0R2gp74auGTLpK-R3gK3HOl1QE2BORulu085J_IMh3LHI83IaUkkFeV3LJ7dmNuRohsILN90w3_j-vWuhk8" width="242" /></a></div><br /><br /><p></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><b>Practice</b></p><p>Create of the vector-valued function:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjawyh3w_vWT4x5hmUTKG4gmY9l4gU8fWIOrf_tKk1caBjIqow_fcmbm_SXp8AQX_m-nmlXIRZywWa5iffSigTSDH6TvSX1x8X0OUGDNGcmsNRwY4UvLcganTc5-9LGXIdOtMTgZOaT1YLzVxM0SGJVPHYna52RNZ2H3VI1fTfIABjMpR-FO1n0mWc2dUs" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="39" data-original-width="339" height="37" src="https://blogger.googleusercontent.com/img/a/AVvXsEjawyh3w_vWT4x5hmUTKG4gmY9l4gU8fWIOrf_tKk1caBjIqow_fcmbm_SXp8AQX_m-nmlXIRZywWa5iffSigTSDH6TvSX1x8X0OUGDNGcmsNRwY4UvLcganTc5-9LGXIdOtMTgZOaT1YLzVxM0SGJVPHYna52RNZ2H3VI1fTfIABjMpR-FO1n0mWc2dUs" width="320" /></a></div><br /><br /><p></p>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-70140738444084835512023-12-23T21:11:00.000-05:002023-12-23T21:11:06.313-05:00How to evaluate vector-valued functions<p><b> Goal</b></p><p> Evaluate vector valued functions</p><p><b>Definition</b></p><p>A vector-valued function is a function of the form:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhk7TOQOyf-oUs9q8PtQgjaPfkYGWtrKCUsg8PpckQ368S1zVdBLM1-15JePgCTEK6eLwi22Gdu2rjg6wILbapLaKkwjWyCZ6erFOQ2uerjQjqBjNy4lPs0epOwc6kCW5Nb7QewlPVZGN18sbHTYsw3EUTw8faWS889h9V5Z9e_Ia5n2SDt68yYMoIUVlc" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="38" data-original-width="493" height="25" src="https://blogger.googleusercontent.com/img/a/AVvXsEhk7TOQOyf-oUs9q8PtQgjaPfkYGWtrKCUsg8PpckQ368S1zVdBLM1-15JePgCTEK6eLwi22Gdu2rjg6wILbapLaKkwjWyCZ6erFOQ2uerjQjqBjNy4lPs0epOwc6kCW5Nb7QewlPVZGN18sbHTYsw3EUTw8faWS889h9V5Z9e_Ia5n2SDt68yYMoIUVlc" width="320" /></a></div><br /><br /><p></p><p>The functions f, g, h are real-valued functions of the parameter t. Vector-valued functions are also written in the form:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEh3mKYkJ1SKRUaD_IzdF7eZ07_DxVyZmzyO8VSSTfcvhqISCFP23GZu3rqdgeSLnDqrqyM5zJzapz76kx7XqiyhBYADbqLc__Rj-b6vrHsHjJ1c5d0HGq3eZ1KBWzqk_hi8PabjB8O_fzLgL1DOdaNupm_SAlyfYi9wsw7q1VgLuPwB9A4DRtptdU9rkDE" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="30" data-original-width="428" height="22" src="https://blogger.googleusercontent.com/img/a/AVvXsEh3mKYkJ1SKRUaD_IzdF7eZ07_DxVyZmzyO8VSSTfcvhqISCFP23GZu3rqdgeSLnDqrqyM5zJzapz76kx7XqiyhBYADbqLc__Rj-b6vrHsHjJ1c5d0HGq3eZ1KBWzqk_hi8PabjB8O_fzLgL1DOdaNupm_SAlyfYi9wsw7q1VgLuPwB9A4DRtptdU9rkDE" width="320" /></a></div><div><br /></div><div><br /></div><div><br /></div><div>The first form defines a two-dimensional vector-valued function. The second form describes a tri-dimensional vector-valued function. </div><div><br /></div><div><b>Example</b></div><div><br /></div><div>For each of the vector-valued functions, evaluate r(0), r(π/2), r(2π/3). Do any of these functions have domain restrictions?</div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEirTC3KunDZxVRsDfxpDsTn3WEdAofmz23GWKC1sE1ZcqvMlULZUScuFqF0Zvk29We6jXLQzv94iVeqIWim7TukhLsl6syBR2xMElvN5_mlj8_xRpjkAB2NSljaLLxadhngR1GDJ28ZfE9IGVqRxzLZkEbybzh7Vw4TDm4-Bj-Puwi7-AC87L9EdLYqdjc" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="55" data-original-width="278" height="63" src="https://blogger.googleusercontent.com/img/a/AVvXsEirTC3KunDZxVRsDfxpDsTn3WEdAofmz23GWKC1sE1ZcqvMlULZUScuFqF0Zvk29We6jXLQzv94iVeqIWim7TukhLsl6syBR2xMElvN5_mlj8_xRpjkAB2NSljaLLxadhngR1GDJ28ZfE9IGVqRxzLZkEbybzh7Vw4TDm4-Bj-Puwi7-AC87L9EdLYqdjc" width="320" /></a></div><br /><br /></div><div><br /></div><div><br /></div><div><br /></div><div><b>Solution</b></div><div><b><br /></b></div><div><b>a. </b>Let's substitute each of the value of t in the function:</div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgXp9PS31ALS48HGxWBu4AFs8Lcru4Chrt_59b1s7xqpzh1zdSA9BcwaATCI2VnG4OL2obX3Exy-yvF1LsKtJLDC5H-0mwkc1DGDvQD0HTlPpmG4xN43apT8SA0r0_MC22P3fLFFCaOowW4d-OSoy_UfA0uvJJpzCMM5giEdYiUg9OMreiiu753E_5pvfQ" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="219" data-original-width="394" height="178" src="https://blogger.googleusercontent.com/img/a/AVvXsEgXp9PS31ALS48HGxWBu4AFs8Lcru4Chrt_59b1s7xqpzh1zdSA9BcwaATCI2VnG4OL2obX3Exy-yvF1LsKtJLDC5H-0mwkc1DGDvQD0HTlPpmG4xN43apT8SA0r0_MC22P3fLFFCaOowW4d-OSoy_UfA0uvJJpzCMM5giEdYiUg9OMreiiu753E_5pvfQ" width="320" /></a></div><br /><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div>To determine a domain restriction let's consider each component function separately. The function cost is defined for all values of t. The function sint is also defined for all values of t. Therefore the function r(t) is defined for all values of t</div><div><br /></div><div>b. Let's do the same thing for the second function:</div><div><br /></div><div> <div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhZn_5U_mysWeREHGgpwLP44XvMavvwE0KlmF1XxElgvtwJQQ9jOzwCuoJxwoXcuFfl8sH7QBSmC5OUhncNOVTiaIlW4isxzVpKnMRxwcWNdcPBCVz5rfueinqEANoWnIAOn1vmLgVVV5bBo42t7Z7bcay2CSN6HjARRQhmHFXuy7lgMFVsWm3m9WD0PUY" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="202" data-original-width="559" height="145" src="https://blogger.googleusercontent.com/img/a/AVvXsEhZn_5U_mysWeREHGgpwLP44XvMavvwE0KlmF1XxElgvtwJQQ9jOzwCuoJxwoXcuFfl8sH7QBSmC5OUhncNOVTiaIlW4isxzVpKnMRxwcWNdcPBCVz5rfueinqEANoWnIAOn1vmLgVVV5bBo42t7Z7bcay2CSN6HjARRQhmHFXuy7lgMFVsWm3m9WD0PUY=w400-h145" width="400" /></a></div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div>The component functions tant and sect are not defined for odd multiples of π/2. Therefore the vector valued-function r(t) is not defined for odd multiples of π/2. </div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhPEvPRXrrXYiWCD243NRalo8Z5P8fKmMmwk1mvkVzVz0doNWs0NeVhkQpNVhVkWrCpve6NPC_rHYegfKi5DBrEJnQ3cHj3lyMh5vQGtbLVU1qSpPa8AEFwUvVdzOZT0UkSs4lBb09Gbvl04wA-vJPZyRyPQ2G7_05PIfPDVxi-oB26D6eFCKmBuRuvEKA" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="50" data-original-width="492" height="41" src="https://blogger.googleusercontent.com/img/a/AVvXsEhPEvPRXrrXYiWCD243NRalo8Z5P8fKmMmwk1mvkVzVz0doNWs0NeVhkQpNVhVkWrCpve6NPC_rHYegfKi5DBrEJnQ3cHj3lyMh5vQGtbLVU1qSpPa8AEFwUvVdzOZT0UkSs4lBb09Gbvl04wA-vJPZyRyPQ2G7_05PIfPDVxi-oB26D6eFCKmBuRuvEKA=w400-h41" width="400" /></a></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><b>Practice</b><br /><br /></div><div>For the vector-valued function r(t) = (t²-3t)i + (4t + 1)j, evaluate r(0), r(1), r(-4). Does this function have any domain restrictions?</div><div><br /></div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><br /></div><br /><br /></div><div><br /></div><div><br /></div><div><br /></div>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-19588282816211338492023-12-16T00:05:00.005-05:002023-12-16T00:07:28.788-05:00Arc length of a parametric curve<p> Let's consider the plane curve defined by the following parametric equations:</p><p>x = x(t) y = y(t) t₁≤t ≤t₂ and let's assume that x(t) and y(t) are differentiable functions of t, then the arc length of the parametric curve is given by:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEj5jL3T40KzHKi0fCoSdT1z_vWiMvH9ol6-wLLJ5Gz1lqGJkNCfmnI3164PcRnjlrx6JOqrOHMq2ET8ChXJhX2SDCJkGXEhU6ovbKQuBndsEp_VxL_2od-VWObiIIj7mAP9O6C3gQirdu6JJiyg_Pu0AwKqIqnoTI-QyE7UNQrp48QxN_I0Jaep5b7YxGU" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="98" data-original-width="282" height="111" src="https://blogger.googleusercontent.com/img/a/AVvXsEj5jL3T40KzHKi0fCoSdT1z_vWiMvH9ol6-wLLJ5Gz1lqGJkNCfmnI3164PcRnjlrx6JOqrOHMq2ET8ChXJhX2SDCJkGXEhU6ovbKQuBndsEp_VxL_2od-VWObiIIj7mAP9O6C3gQirdu6JJiyg_Pu0AwKqIqnoTI-QyE7UNQrp48QxN_I0Jaep5b7YxGU" width="320" /></a></div><br /><br /><p></p><p><br /></p><p><br /></p><p><br /></p><p>This video gives an idea of where this formula originates:</p><p> </p>
<iframe allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" allowfullscreen="" frameborder="0" height="315" src="https://www.youtube.com/embed/DXHM7LYGcbI?si=eTBBckxWam_M58Rj" title="YouTube video player" width="560"></iframe><div><br /></div><div><b>Example</b></div><div><br /></div><div>Find the arc length of the semicircle defined by the equations:</div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEg6hjLVEe5kklb-KdCpjbnkH5g7Nxp9xVNleaTwoCCxW-T9IodHcnnL3YvTo52fSPnNRgkvuB-FEPmKTKmeg2CJGB-K95q9MWXYpOFba_MvYwnvTe1BjFTJMOAO5qiyjFqWaEYa60udYnOc1ufY7ZFo34IbtB0HhNfmPFdbCACLYH6SkOzbMDhumTx2L84" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="46" data-original-width="368" height="40" src="https://blogger.googleusercontent.com/img/a/AVvXsEg6hjLVEe5kklb-KdCpjbnkH5g7Nxp9xVNleaTwoCCxW-T9IodHcnnL3YvTo52fSPnNRgkvuB-FEPmKTKmeg2CJGB-K95q9MWXYpOFba_MvYwnvTe1BjFTJMOAO5qiyjFqWaEYa60udYnOc1ufY7ZFo34IbtB0HhNfmPFdbCACLYH6SkOzbMDhumTx2L84" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><b>Solution</b></div><div><b><br /></b></div><div>Here is the graph of the semicircle:<br /><br /></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhPUU-vP2Vb73OQXdTsKlK53GabLQL5Yc077YfKrb8d3w6j0zkl0KRQVM0dHjiptPLP5-vIaTA7yHrVXmS3YMq8XqMLpxrnkbgY_6i8W4lFS_li51ub8ZyncH2KEeueQaRo_GyCdNilJCorVgv0XZ0NiOSx5wz6nVaqBND5ssdOz9jQrU-KdOrwygW2y8s" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="347" data-original-width="339" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEhPUU-vP2Vb73OQXdTsKlK53GabLQL5Yc077YfKrb8d3w6j0zkl0KRQVM0dHjiptPLP5-vIaTA7yHrVXmS3YMq8XqMLpxrnkbgY_6i8W4lFS_li51ub8ZyncH2KEeueQaRo_GyCdNilJCorVgv0XZ0NiOSx5wz6nVaqBND5ssdOz9jQrU-KdOrwygW2y8s" width="234" /></a></div><br />Let's use the formula to find the arc length:</div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjtkJvNApJZv0OEMc9pRLx7B7iDLcACVA116g7xEbpEEGy4vcspMNF3A3zXwktx6KhWrgePdlorjAbgNYMQNCWRw9OueQZJ7dS9cEInht3PzGATen_vNiVdKs4ehzDDYlp7uh9PT5aUt9VH3fqWFFZtxUQHlqNx29jLWXNSFeOCRw_XyJMPSB3M0mz6vfA" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="305" data-original-width="310" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEjtkJvNApJZv0OEMc9pRLx7B7iDLcACVA116g7xEbpEEGy4vcspMNF3A3zXwktx6KhWrgePdlorjAbgNYMQNCWRw9OueQZJ7dS9cEInht3PzGATen_vNiVdKs4ehzDDYlp7uh9PT5aUt9VH3fqWFFZtxUQHlqNx29jLWXNSFeOCRw_XyJMPSB3M0mz6vfA" width="244" /></a></div><br /><b>Practice</b></div><div><b><br /></b></div><div>Find the arc length of the curve defined by the equations:</div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEijpHlut2M1vqU00MCyTNh4VQsafzvsHybkyiQLnoWH2Ek7HZpvgszjFy5sFh8i3VTStARHU3Z2JiOQo_Hbp3U3PyYuXVey0NnYbEZBi_3496L7GYwjr5R3BqlcaPrpku7q6PexBYAM-xuoAcQxKJctoe1MLNHsGrLQcI-GE-SKRc2tAay7uNpyhhg_IZo" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="37" data-original-width="312" height="38" src="https://blogger.googleusercontent.com/img/a/AVvXsEijpHlut2M1vqU00MCyTNh4VQsafzvsHybkyiQLnoWH2Ek7HZpvgszjFy5sFh8i3VTStARHU3Z2JiOQo_Hbp3U3PyYuXVey0NnYbEZBi_3496L7GYwjr5R3BqlcaPrpku7q6PexBYAM-xuoAcQxKJctoe1MLNHsGrLQcI-GE-SKRc2tAay7uNpyhhg_IZo" width="320" /></a></div><br /><br /></div>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-2631831551079314762023-12-09T00:38:00.000-05:002023-12-09T00:38:38.152-05:00Area under a parametric curve<p> Goal: Finding the area under the curve of a parametric equation</p><p><b>Area under parametric curve</b></p><p>Let's consider the the area of the curve bounded by the curve y = f(x), the x-axis and the verticals x = a and x = b</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiHS5WwlBpLEnBGWPF_oTmypPRU9bD92HLV3ZLiu3zteKVbWFgVaDUx4X2lvPq8u1o_bcpRd0LZT7QLnXNh3meUqsv9tUo8-kW6V3DNHB9yYCZYht4SUHpP5lyESy8YLFoxdMewSxa4BmEI6pQprUMmDJRkVF44lOLp0SZfofZcCWYBqItPf0rnAta5kwI/s1944/IMG_20231208_225704.jpg" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" data-original-height="1840" data-original-width="1944" height="303" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiHS5WwlBpLEnBGWPF_oTmypPRU9bD92HLV3ZLiu3zteKVbWFgVaDUx4X2lvPq8u1o_bcpRd0LZT7QLnXNh3meUqsv9tUo8-kW6V3DNHB9yYCZYht4SUHpP5lyESy8YLFoxdMewSxa4BmEI6pQprUMmDJRkVF44lOLp0SZfofZcCWYBqItPf0rnAta5kwI/s320/IMG_20231208_225704.jpg" width="320" /></a></div><br /><b><br /></b><p></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p>We know that the area is given by:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEigy1oWywHDC_4zg6_WQpwgeewxPjFCFgbNQXT2qYytsIpgGQueV1ySA8O8OfAlTxcvTyOXxXJmNr8dmCfkollkWXjXBMw882Yuz9rf9128vAHIXTNWAM6GQWx2eXxpGzKO0BZzj8Lpw7onHsBQflQbM7JyiAyUQPfoLpCDNwMxbHMBx1AyY83TKr0j4ro" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="45" data-original-width="114" height="79" src="https://blogger.googleusercontent.com/img/a/AVvXsEigy1oWywHDC_4zg6_WQpwgeewxPjFCFgbNQXT2qYytsIpgGQueV1ySA8O8OfAlTxcvTyOXxXJmNr8dmCfkollkWXjXBMw882Yuz9rf9128vAHIXTNWAM6GQWx2eXxpGzKO0BZzj8Lpw7onHsBQflQbM7JyiAyUQPfoLpCDNwMxbHMBx1AyY83TKr0j4ro=w200-h79" width="200" /></a></div><br /><br /><p></p><p></p><div class="separator" style="clear: both; text-align: center;"><br /></div>We assume that the curve is given by the parametric equations: x = x(t) y = y(t). The formula above becomes:<p></p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEiBDi8-gr6ZNTWFyhTk_jrhHFzKJq0BPbvAGASqfznB_PsFGxObMJcJweZ3IWApGvXY3BnjzbNg0QIUe1dMI2bI7kCP8dIPMSYT77KRMCFTOmjbzop7VgFS8RJCj50jK25iSaUwCRXo4ENYZmnTJAWVhKhJ40Qn0KIykDF9w5T-kzTw4TaUCwvJ_kGIZrY" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="57" data-original-width="370" height="49" src="https://blogger.googleusercontent.com/img/a/AVvXsEiBDi8-gr6ZNTWFyhTk_jrhHFzKJq0BPbvAGASqfznB_PsFGxObMJcJweZ3IWApGvXY3BnjzbNg0QIUe1dMI2bI7kCP8dIPMSYT77KRMCFTOmjbzop7VgFS8RJCj50jK25iSaUwCRXo4ENYZmnTJAWVhKhJ40Qn0KIykDF9w5T-kzTw4TaUCwvJ_kGIZrY" width="320" /></a></div><br /><br /><p></p><p><br /></p><p>The formula that allows to find the area under a parametric curve is then given by:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEi_l1vnaNSg6_n7Qgmzo1rj0_CZo86R4aXEAKD5QTflXHoBSjKGm65IoWlVqtC4kEh0eksMVKawv6vefHxojM-sP_VtEzkq-ahTuFoh-sU3aj8eVmfKFp6j65VpfOcZXCr5UZNXcFjOnJb-XUyPH0E1fxjrxNQikj_EjmOtWaFp2McQJ9FFTQE3OWqGSmw" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="64" data-original-width="122" height="105" src="https://blogger.googleusercontent.com/img/a/AVvXsEi_l1vnaNSg6_n7Qgmzo1rj0_CZo86R4aXEAKD5QTflXHoBSjKGm65IoWlVqtC4kEh0eksMVKawv6vefHxojM-sP_VtEzkq-ahTuFoh-sU3aj8eVmfKFp6j65VpfOcZXCr5UZNXcFjOnJb-XUyPH0E1fxjrxNQikj_EjmOtWaFp2McQJ9FFTQE3OWqGSmw=w200-h105" width="200" /></a></div><br /><br /><p></p><p><b><br /></b></p><p><b><br /></b></p><p><br /></p><p><b>Example</b></p><p>Find the area under curve of the cycloid define by the equations:</p><p>x(t) = t-sint y(t) = 1-cost 0 ≤ t ≤ 2π</p><p><b>Solution</b></p><p>Using the formula we have:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgpKltpy34soPPV0GtYeBmNh6rl-k1DNxSpV540dF_jc_xuDbvm9a7nVMkcHIwf-CvU165wwe72a2b-OKrc51ORtB1oXO_QHv0NrCQmBicgXav3uSRGK7uCI_LxTqme45Zh-vxlLBfgNMTuxDfoncVUIagywDE1CKmMviIp5KHG2iuDUqEXW_09YYn--48" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="369" data-original-width="338" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEgpKltpy34soPPV0GtYeBmNh6rl-k1DNxSpV540dF_jc_xuDbvm9a7nVMkcHIwf-CvU165wwe72a2b-OKrc51ORtB1oXO_QHv0NrCQmBicgXav3uSRGK7uCI_LxTqme45Zh-vxlLBfgNMTuxDfoncVUIagywDE1CKmMviIp5KHG2iuDUqEXW_09YYn--48" width="220" /></a></div><br /><br /><p></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p><b><br /></b></p><p><b>Practice</b></p><p>Find the area under the curve of the hypocycloid defined by the following equations:</p><p>x(t) = 3cost + cos3t y(t) = 3sint - sin3t 0 ≤ t ≤ π </p>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-56941355706479206612023-12-01T23:47:00.003-05:002023-12-01T23:47:30.376-05:00Second Order Derivatives<p> The second order derivative of a function y = f(x) is the derivative of the first derivative of the function.</p><p>The first derivative of the function f is dy/dx.</p><p>The derivative of the first derivative is :d/dx[dy/dx] = d²y/dx²</p><p>The relation equality being commutative, we can write: d²y/dx² = d/dx[dy/dx].</p><p>Let's apply the formula of the first derivative. According to this formula, the derivative of y = f(x) is the derivative of the function y with respect to t divided by the derivative of x with respect to t.</p><p>The function here is dy/dx. Let's calculate its derivative:</p><p> d²y/dx² = d/dt[dy/dx]./dx/dt.</p><p>The second order derivative is the derivative of the derivative of the first derivative with respect to t divided by the derivative of x with respect to t.</p><p><b>Examples</b></p><p>Calculate the second derivative d²y/dx² for the plane curve defined by the parametric equations:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEh8DyBTQi2IhXzkwuAJ3r89aIIeE8VMVbU9QQbOriYgpvm62nVelhJwBNK4_HWRY_nJ9OGdF4n5B-vxliCoahRT-hbS8HveCbyQuJ4OO7gac-PgSXHGbm6ns-4kCet8Y3S1EpnH_Lu0CujyQ9FitAWHtMiyawqLPy8H_7cxNh3qnc0hz1BESc6sK6PDu6w" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="38" data-original-width="346" height="35" src="https://blogger.googleusercontent.com/img/a/AVvXsEh8DyBTQi2IhXzkwuAJ3r89aIIeE8VMVbU9QQbOriYgpvm62nVelhJwBNK4_HWRY_nJ9OGdF4n5B-vxliCoahRT-hbS8HveCbyQuJ4OO7gac-PgSXHGbm6ns-4kCet8Y3S1EpnH_Lu0CujyQ9FitAWHtMiyawqLPy8H_7cxNh3qnc0hz1BESc6sK6PDu6w" width="320" /></a></div><br /><br /><p></p><p> <b>Solution</b></p><p>We have :</p><p> d²y/dx² = d/dt[dy/dx]./dx/dt.</p><p>Let's calculate dy/dx. According to the formula of the derivative, we know that the derivative is equal to the derivative of y with respect to t divided by the derivative of x with respect to t.</p><p>dy/dx = y'(t)/x'(t) = 2/2t = 1/t</p><p>Let's calculate dx/dt also:</p><p>dx/dt = 2t</p><p>The expression of d²y/dx² becomes:</p><p>d²y/dx² = d/dt (1/t)/2t = -1/t²/2t = (-1/t²)(1/2t) = -1/2t³</p><p><b>Practice</b></p><p>Calculate the second derivative d²y/dx² for the plane curve defined by the parametric equations:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjA1icVPXOpTLx_38v1fMMT36YYEs8Z9q057MD57txhdauCBBFv1Ehz9TwEI1b2qdonDus3n4Om3tln4WOIfIszAxSiMa0sqVnLc6g0z0TitGVp8rGlYXXXq6ozEbBOjlBUw3rktEZvmGgnANlVyAT-hti3uxp2qKh1qYiE0k9lO5T_H6Flgjqj4Y78e6g" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="33" data-original-width="404" height="26" src="https://blogger.googleusercontent.com/img/a/AVvXsEjA1icVPXOpTLx_38v1fMMT36YYEs8Z9q057MD57txhdauCBBFv1Ehz9TwEI1b2qdonDus3n4Om3tln4WOIfIszAxSiMa0sqVnLc6g0z0TitGVp8rGlYXXXq6ozEbBOjlBUw3rktEZvmGgnANlVyAT-hti3uxp2qKh1qYiE0k9lO5T_H6Flgjqj4Y78e6g" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div>Locate any critical points on its curve.<br /><br /><p></p><p><b><br /></b></p><p><br /></p><p> </p><p><br /></p><p><br /></p><p><b><br /></b></p>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-18037759190160165312023-11-10T16:58:00.001-05:002023-11-10T17:09:56.624-05:00Derivative of parametric equations<p> <b>Objective:</b> to find the derivative of parametric curves</p><p><b>Derivatives of parametric equations</b></p><p>Let's consider the plane curve defined by the parametric equations:</p><p>x(t) = 2t + 3 y(t) = 3t - 4 -2≤ x≤3</p><p>Our objective is to determine the derivative of the parametric equations. Let's note that the derivative of a function is a function that represents the slopes of all the tangent lines at different points of the curve. If we know the slope of a tangent line at a point we can know the slopes of all the tangent lines at different points of the curve of the function. The slope of a tangent line at a point is the derivative of the function at a point. The function that allows to find the derivative at a point allows to determine the derivatives of all points of the function. Determining the slope of a tangent line to the parametric curve will allow us to determine the derivative of the parametric equations.</p><p>The curve of the parametric equations is a line starting at (-1, -10) and ending at (9, 5). </p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgp6-YtWTOIk6FX3OZ3LqpGnj7rwo7iFI77lHXXvD_sM26PctSw3C2seGyb0JHOmRLLUCdfk7g1zPaoJ-yCVTbbT4NuR1YI7seWqJm2mFaPDxz9dGuzW6-J1H1dm1N74iLnY05J-qGAs7m9gBJ0DNiEIz63ZDWdPBH91stRxqtyNJgV0r6ahnsP_zYaVL0" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="647" data-original-width="489" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEgp6-YtWTOIk6FX3OZ3LqpGnj7rwo7iFI77lHXXvD_sM26PctSw3C2seGyb0JHOmRLLUCdfk7g1zPaoJ-yCVTbbT4NuR1YI7seWqJm2mFaPDxz9dGuzW6-J1H1dm1N74iLnY05J-qGAs7m9gBJ0DNiEIz63ZDWdPBH91stRxqtyNJgV0r6ahnsP_zYaVL0" width="181" /></a></div><br />Let's find the explicit form of the parametric equations. To do this we start by eliminating t in x(t):<p></p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEg5QtcnNNoBxzEtqijviTHIymFBYnN5sF2sFCjxfPgclBJnwss91zwRwMeLup1AYaqHo3MWjS6qLNzPZsujYtqZ0Ed8sPKoc3TqJPmVzcK1IbztOQ6uM17ZFWh2RTdKTMnHSmblw4ml2PeazUfAgH4oBoCqZjMCQNGPDhonmLvAEyB2jG4W_yiYWvXMr1g" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="96" data-original-width="153" height="201" src="https://blogger.googleusercontent.com/img/a/AVvXsEg5QtcnNNoBxzEtqijviTHIymFBYnN5sF2sFCjxfPgclBJnwss91zwRwMeLup1AYaqHo3MWjS6qLNzPZsujYtqZ0Ed8sPKoc3TqJPmVzcK1IbztOQ6uM17ZFWh2RTdKTMnHSmblw4ml2PeazUfAgH4oBoCqZjMCQNGPDhonmLvAEyB2jG4W_yiYWvXMr1g" width="320" /></a></div><br /><br /><p></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p>Let's substitute t in y(t):</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgPe7-0lpm9JAs9LtSuDp4abe1HDJA4ehtg53q_se0uueO75-vuGQwC7l3iU2zD-4r3J-7mv1FOQgtqLP5g2DWigYBUcXebPmumfwOat_-Wc1AIDG9D1YmK5NrapPBJ2M0_gWwElNEiUHpCfzPtXQoGrXfmaThSkl3AFErWDchRysrxaeztPyreB8zO98k" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="137" data-original-width="203" height="216" src="https://blogger.googleusercontent.com/img/a/AVvXsEgPe7-0lpm9JAs9LtSuDp4abe1HDJA4ehtg53q_se0uueO75-vuGQwC7l3iU2zD-4r3J-7mv1FOQgtqLP5g2DWigYBUcXebPmumfwOat_-Wc1AIDG9D1YmK5NrapPBJ2M0_gWwElNEiUHpCfzPtXQoGrXfmaThSkl3AFErWDchRysrxaeztPyreB8zO98k" width="320" /></a></div><br />The slope of the tangent line is 3/2. The derivative is dy/dx = 3/2<p></p><p>But dy/dx = dy/dt/dx/dt = y'(t)/x'(t)</p><p>x'(t) = 2 y'(t) = 3</p><p>dy/dx = 3/2</p><p>In both cases dy/dt = 3/2. The expression dy/dx = y'(t)/x'(t) is the derivative of the function.</p><p><b>Theorem. Derivative of Parametric equations</b></p><p>Consider the plane curve defined by x = x(t) and y = y(t). Suppose that x'(t) and y'(t) exist, and then assume x'(t) ≠ 0. Then the derivative dy/dx is given by:</p><p><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjd45VHuKUHuqiEg2i9iYjzs6L6lffuRQ9gHNcSl3gkhkyRm8nzTCQxWltnXFwmLEPhc-KaQVBqsAuxBsL1uFCqNYothArKAQsNFdFSEKTOnZO3tpj1j0zW-CeLAoRsTauSb9fWA6Xyyb9VDklegmK4_SC65iVsXvL4_Bs45uQ1LyJRBci4pC5AdSc_or8" style="clear: left; margin-bottom: 1em; margin-right: 1em; text-align: center;"><img alt="" data-original-height="59" data-original-width="191" height="99" src="https://blogger.googleusercontent.com/img/a/AVvXsEjd45VHuKUHuqiEg2i9iYjzs6L6lffuRQ9gHNcSl3gkhkyRm8nzTCQxWltnXFwmLEPhc-KaQVBqsAuxBsL1uFCqNYothArKAQsNFdFSEKTOnZO3tpj1j0zW-CeLAoRsTauSb9fWA6Xyyb9VDklegmK4_SC65iVsXvL4_Bs45uQ1LyJRBci4pC5AdSc_or8" width="320" /></a></p><p><b>Examples</b></p><p>Calculate the derivative dy/dx for each of the parametrically defined curves and locate any critical points on their respective graph. </p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgWjcGqbz9Xz4JZAMBIrxMjagMue72KyDleK8bZMUiiP8AbXcvNNeDr-Sw_KnllI8nMe3y_bfkVEJ70nUDVYCc9juQ0qksUqn7iitK25Ixxszzh0Go6OQBouAX0nqq1F7n6d0SUHHcFx5RSSJdzjc4e21i2Xll3fBsp3zZSV1iArDk70eu1KcOT9Epm5Os" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="83" data-original-width="445" height="60" src="https://blogger.googleusercontent.com/img/a/AVvXsEgWjcGqbz9Xz4JZAMBIrxMjagMue72KyDleK8bZMUiiP8AbXcvNNeDr-Sw_KnllI8nMe3y_bfkVEJ70nUDVYCc9juQ0qksUqn7iitK25Ixxszzh0Go6OQBouAX0nqq1F7n6d0SUHHcFx5RSSJdzjc4e21i2Xll3fBsp3zZSV1iArDk70eu1KcOT9Epm5Os" width="320" /></a></div><p><br /></p><p><br /></p><br /><b>Solutions</b><p></p><p>a) Let's calculate dy/dx: dy/dx = y'(t)/x'(t)</p><p>From the parametric equations in example a) we have: y'(t) = 2 and x'(t) = 2. Therefore:</p><p>dy/dx = 2/2t = 1/t</p><p>A critical point is a point where the derivative is equal to zero or undefined. The derivative is undefined for t = 0. Let's determine the critical point. Substituting t in the parametric equations, we find x = -3 and y = -1. (-3, -1) represents the critical point. The graph of the curve is a parabola opening on the right and the point (-3, -1) is a vertex of this parabola.</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEiQL0nAsYRvmqatVrlc6PpAndauTn8dOSX6ZjZRmQZp7mbjdWxTq_4SL7tF6xVKTKTJlaIJlxER7eB2SA2MgL7bYJoqZqS_s2CP6NDDnE0vX74zW5kyVe9DbDCadn0jWal-rPArC9AdhSlHVIHGHhASrBs6jDcpVCMd2XMyx2lEK78d8okLCFbeGset2MY" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="347" data-original-width="417" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEiQL0nAsYRvmqatVrlc6PpAndauTn8dOSX6ZjZRmQZp7mbjdWxTq_4SL7tF6xVKTKTJlaIJlxER7eB2SA2MgL7bYJoqZqS_s2CP6NDDnE0vX74zW5kyVe9DbDCadn0jWal-rPArC9AdhSlHVIHGHhASrBs6jDcpVCMd2XMyx2lEK78d8okLCFbeGset2MY" width="288" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><br />b) I provided a guided solution for examples b) and c) .We have for example b)<p></p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjEy3xTEqwsDK_I5__xISmysIheSg4K-anw-0L4VV_rpfpIhlLsEFS-dm59nrs5GSN1GRhp5j6QTgBQP158gwGbIGkKJmL9cZ7PnjrVhdwsibN92pfFoOx1RBcEh2G4T-A36jWo40xrzjJOO8lk9IKefUzaSGML0dtTYdAfmOUCtCfFjua0v2qsmdwBYaI" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="42" data-original-width="102" height="132" src="https://blogger.googleusercontent.com/img/a/AVvXsEjEy3xTEqwsDK_I5__xISmysIheSg4K-anw-0L4VV_rpfpIhlLsEFS-dm59nrs5GSN1GRhp5j6QTgBQP158gwGbIGkKJmL9cZ7PnjrVhdwsibN92pfFoOx1RBcEh2G4T-A36jWo40xrzjJOO8lk9IKefUzaSGML0dtTYdAfmOUCtCfFjua0v2qsmdwBYaI" width="320" /></a></div><br /><br /><p></p><p><br /></p><p><br /></p><p><br /></p><p>Since the denominator is different of zero, the only critical points are those for which the numerator is equal to zero. That happens when t = -1 or t = 1 for which the critical points are: (-1, 6) and (3, 2) respectively. Here is the graph of the parametric curve.</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjyiTPdlF1a5CZoPKgx5M5FdLCTdoUF1j5niKqkp98pAHeZPnvQ2gAHdwF_w-rPpyrDsysfJ7-bma233YFc3j0zRBA_4yiwYC13S__6SmwoDx8TVZBSbJwn75n865MEDwxwIj6QSWiksS45oHpc9PQrxnK3oMHTbkEIp3_bffVt01fCuuY23esT4rdgC0c" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="347" data-original-width="379" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEjyiTPdlF1a5CZoPKgx5M5FdLCTdoUF1j5niKqkp98pAHeZPnvQ2gAHdwF_w-rPpyrDsysfJ7-bma233YFc3j0zRBA_4yiwYC13S__6SmwoDx8TVZBSbJwn75n865MEDwxwIj6QSWiksS45oHpc9PQrxnK3oMHTbkEIp3_bffVt01fCuuY23esT4rdgC0c" width="262" /></a></div><br />The point (-1, 6) is a relative maximum and the point (3, 2) is a relative minimum.<p></p><p>c) We have : dy/dx = -cost/sint</p><p>The derivative is 0 for cost = 0. That happens when t = π/2 or t = 3𝛑/2 to which correspond respectively the critical points (0,5) and (0, -5).</p><p>When sint = 0, the corresponding values of t are : t = 0, t = π, t = 2π. The critical points are respectively:</p><p>(5. 0), (-5, 0) and (5. 0).</p><p>Here is the graph of the parametric equations defined in part example c)</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhRYT-h1h8FE_FKDaW83nrfeBDkvVlHyvAKpabhNG_1OdaPSz1sQ23egyFzW6kCPbUP188CJcSuyqhhaBvs47Vgwhk1ISk1DiFYa9GFXYBamGDfDNyH1Ix-ic3cETXGyq61w-SQlp7SD8ostQxjSiIifTnQ6-mbojNcmm3aTLEwjQhVxjbr2sLtvKdwoeU" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="497" data-original-width="490" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEhRYT-h1h8FE_FKDaW83nrfeBDkvVlHyvAKpabhNG_1OdaPSz1sQ23egyFzW6kCPbUP188CJcSuyqhhaBvs47Vgwhk1ISk1DiFYa9GFXYBamGDfDNyH1Ix-ic3cETXGyq61w-SQlp7SD8ostQxjSiIifTnQ6-mbojNcmm3aTLEwjQhVxjbr2sLtvKdwoeU" width="237" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><br /><b>Practice</b><p></p><p>Calculate the derivative dy/dx for the plane curve defined by the following equations and locate any critical points on its graph:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgSUNZrB5E9ukcgBv80QMeJQqgwvImgiA3twP947BQsWmWqULeYglOaDQo3JX4_8Uf0FqSn6Lg4nm-C6xU-tUyc5bNPu5AS7h2Am66g3NEHLzzm_yttDe_6LF9WcblMe4TpPNofVv9BzUKs59ItDKDmuoSdy6fGPAc3b7yanv5Su7NHayZ7MfkV-on9jVA" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="35" data-original-width="408" height="27" src="https://blogger.googleusercontent.com/img/a/AVvXsEgSUNZrB5E9ukcgBv80QMeJQqgwvImgiA3twP947BQsWmWqULeYglOaDQo3JX4_8Uf0FqSn6Lg4nm-C6xU-tUyc5bNPu5AS7h2Am66g3NEHLzzm_yttDe_6LF9WcblMe4TpPNofVv9BzUKs59ItDKDmuoSdy6fGPAc3b7yanv5Su7NHayZ7MfkV-on9jVA" width="320" /></a></div><br /><br /><p></p><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><b> </b></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><br /><br /><p></p><p><br /></p><p><b><br /></b></p>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-55559969119199950582023-10-28T00:39:00.001-04:002023-10-28T00:47:28.632-04:00Parameterization of a curve<p> <b>Objective:</b></p><p>Transform the function of a curve into parametric equations.</p><p><b>Parameterization of a curve</b></p><p>In the previous lesson we learn how to eliminate the parameter in the parametric equations of a curve. In this post we do the reverse meaning transforming a function of a curve into parametric equations.</p><p><b>Example</b></p><p>Find two different pairs of parametric equations to represent the graph of y = 2x² - 3</p><p><b>Solution</b></p><p>One of the easiest way to do this is to write x(t) = t and substitute x in the function. The result is y = 2 t²-3.</p><p>The first set of pair of parametric equations is x(t) = t y = 2t²-3</p><p>Since there is no restriction on the domain of the function, we can have a variety of expressions of x in function of t and then substitute x in the function.</p><p>For the second pair of equations let's choose x(t) = 3t-2</p><p>Let's substitute x in the function:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEispinApWe_N4hMIcnjth1Xdhc7rb_85v6h2lAqNjGdeTdF1RxriPXNIWKwnkyNFo5Mej_ry-bZY9uQ5oo78tdAwp3VrK2hDPP8CifYBJ2xPmSjZTirY-YE7HnTt3mh4Z6z1VXfPzskk-1PAsOSlNjiLja0oEsmLonQGCPJ_iCjgDUvhnaXRmLnIQ4HTx4" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="133" data-original-width="255" height="167" src="https://blogger.googleusercontent.com/img/a/AVvXsEispinApWe_N4hMIcnjth1Xdhc7rb_85v6h2lAqNjGdeTdF1RxriPXNIWKwnkyNFo5Mej_ry-bZY9uQ5oo78tdAwp3VrK2hDPP8CifYBJ2xPmSjZTirY-YE7HnTt3mh4Z6z1VXfPzskk-1PAsOSlNjiLja0oEsmLonQGCPJ_iCjgDUvhnaXRmLnIQ4HTx4" width="320" /></a></div><br /><br /><p></p><p> </p><p><br /></p><p><br /></p><p><br /></p><p>Therefore a second parameterization of the function is represented by:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhsdxWFRbf_8hznaCo-1YCTAB1MqMZzDy287Axxz3d0H9nYlwcIZiKQvKGmuLDw8UW7XbLPZRzgaJQZIYM6hLwj8ARrGgVk6pzXuA91a4ULRr_Z9qIl4Ks3bctKJy42dDyGFpmfM4kKOLXNb97ccZTVWsAxONcYr8yAU1AZxxFmS1opYmLZUehMwEj-Ih4" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="37" data-original-width="342" height="35" src="https://blogger.googleusercontent.com/img/a/AVvXsEhsdxWFRbf_8hznaCo-1YCTAB1MqMZzDy287Axxz3d0H9nYlwcIZiKQvKGmuLDw8UW7XbLPZRzgaJQZIYM6hLwj8ARrGgVk6pzXuA91a4ULRr_Z9qIl4Ks3bctKJy42dDyGFpmfM4kKOLXNb97ccZTVWsAxONcYr8yAU1AZxxFmS1opYmLZUehMwEj-Ih4" width="320" /></a></div><br /><br /><br /><b>Practice</b><p></p><p>Find two different sets of parametric equations to represent the graph of y = x² + 2x<br /></p><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><br /><p></p>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-33918941494149023862023-10-20T20:49:00.004-04:002023-10-20T20:49:45.924-04:00Converting the parametric equations of a curve into an explicit form<p> <b>Objective</b>: Convert the parametric equations of a curve into the form y = f(x)</p><p><b>Conversion of the parametric equations of a curve into an explicit form</b></p><p>Usually we are more accustomed in graphing the curve of a function represented explicitly meaning representing a relation between y and x. It is possible to convert the parametric equations of a curve into its explicit form. The process is very simple. It consist in finding the expression of t in one of the parametric equations and then substituting it in the other. </p><p><b>Examples:</b></p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><a href="https://blogger.googleusercontent.com/img/a/AVvXsEiFy8dh0rWYYPHnMbU-vGZWjozMpcL8D6VGQ8Is6Wte1XQiTKdfryKQhpZwGZcCYilKmNfHMLON6kunUfp_txDlIm14Bemp6FnecOHfh84-VoWi--nhoX81dvMVGzajNSS4CZmznOes2yIcqyFyDodN7Ro8FxMXj-POZe5JCNbM6eJDCXghHODMGUiheQI" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="61" data-original-width="438" height="45" src="https://blogger.googleusercontent.com/img/a/AVvXsEiFy8dh0rWYYPHnMbU-vGZWjozMpcL8D6VGQ8Is6Wte1XQiTKdfryKQhpZwGZcCYilKmNfHMLON6kunUfp_txDlIm14Bemp6FnecOHfh84-VoWi--nhoX81dvMVGzajNSS4CZmznOes2yIcqyFyDodN7Ro8FxMXj-POZe5JCNbM6eJDCXghHODMGUiheQI" width="320" /></a></b></div><b><br /><br /></b><p></p><p><b><br /></b></p><p><b>a) </b>let's solve example a)</p><p>Let's find t in the first equation. We can also find it in the second equation.</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEj4IAx8a5zRp02h_PBJMrnGJntSFNfEeQc5I3uCK6wkOfudDh8u5pNniG_lpKDuTcBjhqKJx85FGeQoj4GYDT0duR82fI5I-lyne78c-Ls3rCPjshlNOpSd01va5bqGotKk8DPTfAdVU-QcFfyFD_5tWbQ1sVFmMT-iDJs_haccYyPmCq_CC-3FaR95O0Q" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="135" data-original-width="176" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEj4IAx8a5zRp02h_PBJMrnGJntSFNfEeQc5I3uCK6wkOfudDh8u5pNniG_lpKDuTcBjhqKJx85FGeQoj4GYDT0duR82fI5I-lyne78c-Ls3rCPjshlNOpSd01va5bqGotKk8DPTfAdVU-QcFfyFD_5tWbQ1sVFmMT-iDJs_haccYyPmCq_CC-3FaR95O0Q" width="313" /></a></div><br /><br /><p></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p>Let's substitute t in the second equation:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhP6E9efAR0xEi-zmo3X24JC_ENdfHCoXaAh1FP9iVqCXcXgO7NiwmfmELvoSjestAnFGp-0ylDk27jnH6RPy68TeLqtYpI5SV8Waaec_0LZCKq5nyGlFIxch59aJko3TixK8wz6pd6kB8jNIoeB-cGP8z0z6_cOQN4-urSP7RvZr8iT41RG0sqPao0PsI" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="139" data-original-width="202" height="220" src="https://blogger.googleusercontent.com/img/a/AVvXsEhP6E9efAR0xEi-zmo3X24JC_ENdfHCoXaAh1FP9iVqCXcXgO7NiwmfmELvoSjestAnFGp-0ylDk27jnH6RPy68TeLqtYpI5SV8Waaec_0LZCKq5nyGlFIxch59aJko3TixK8wz6pd6kB8jNIoeB-cGP8z0z6_cOQN4-urSP7RvZr8iT41RG0sqPao0PsI" width="320" /></a></div><br /><br /><p></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p><br /></p><p>This is the equation of a parabola opening upward. The limits of the parameter lead to a domain restriction of the function.</p><p>The interval for t is mentioned in the problem: -6≤t≤-2</p><p>Let's substitute t = -6 in the expression of x, we find x = 0. Let's substitute t = -2 in the expression of x, we find x = 4. The domain of the function is D = [0,4].</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgY70nlFO1QyuZotdteHn-XrFpMYwB1FAVNCOzD7V77Oob8Wob4Ibgfn6BHxuoEZtn6qRmIUzQlcwz1_gv78SzPjIotN1hVumvqB3tVRvvElgepdZnEcJbxgGHXsofBZoFsPSwBDCWSa0z-vV_G5hez9fgxcnONWbA_VuzfvL6HOc5JSvVZXvbuq-UU4aE" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="385" data-original-width="267" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEgY70nlFO1QyuZotdteHn-XrFpMYwB1FAVNCOzD7V77Oob8Wob4Ibgfn6BHxuoEZtn6qRmIUzQlcwz1_gv78SzPjIotN1hVumvqB3tVRvvElgepdZnEcJbxgGHXsofBZoFsPSwBDCWSa0z-vV_G5hez9fgxcnONWbA_VuzfvL6HOc5JSvVZXvbuq-UU4aE" width="166" /></a></div><br /><br /><p></p><p></p><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><br />b) I am not going to solve completely the example 2. I'll give some steps. <p></p><p>Instead of solving for t find sint and cost in both equations. Substitute sint and cost in the identity sin²t + cos²t = 1. The equation found is the equation of an ellipse centered in the origin.</p><p><b>Practice</b></p><p>Eliminate the parameter for the plane curve defined by the following parametric equations and describe the resulting graph:</p><p>x(t) = 2 + 3/t y(t) = t-1 2≤t≤6</p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><br /></b></div><b><br /><br /></b><p></p><p><b><br /></b></p>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0tag:blogger.com,1999:blog-6995353831733168187.post-29551296485403611802023-10-10T21:21:00.008-04:002023-10-13T12:44:54.851-04:00What are parametric equations and how to graph their curve<p><b> Objectives:</b></p><p>1. Define parametric equations</p><p>2. Plot their curve</p><p><b>Considerations</b></p><p>Let's consider the orbit of the earth around the sun. </p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhrJQqr9RfWaagSchaTajYvps3dF6J7LMnGc_fOoDOhgcWuMcOAp5n38ZUI4JuDfRCjScXL648VHJZPEEmhhuK70xbmtBTOQWommZj9roW2Iqn7g4Nf25Mo4NWm4k5hRNKyHtP_UkUq-jk0usupcheeGXoM9MDUtu0HV665N6D3Wf8w9bkhpt3bQj-2Jxk" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img alt="" data-original-height="346" data-original-width="458" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEhrJQqr9RfWaagSchaTajYvps3dF6J7LMnGc_fOoDOhgcWuMcOAp5n38ZUI4JuDfRCjScXL648VHJZPEEmhhuK70xbmtBTOQWommZj9roW2Iqn7g4Nf25Mo4NWm4k5hRNKyHtP_UkUq-jk0usupcheeGXoM9MDUtu0HV665N6D3Wf8w9bkhpt3bQj-2Jxk" width="318" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div>The earth revolves around the sun in 365.25 days. In this case, we consider 365 days. Each number in the figure represents the position of the earth in respect to the sun. The letter t represents the number of the Day. For example t = 1 means Day 1 that corresponds to January 1st, t = 274 means Day 274 and corresponds to October 1, and so on. According to Kepler's laws of planetary movement, the orbit of the earth around the sun is an ellipse with the earth at one of the foci.<p></p><p>Let's superimpose the ellipse on a system of coordinates (x,y):</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgla-0GwGfBWothLVQZPEm5zdH_h4bJPYLyOVqofSslRLEZ1T1TuwD0wRWKQbnSLBg2u6LL_MKFaJH-c0aEzNu0SMp0OzI9GbrhldsFQMQKZMCneEdwiJUb_qCeIgueDD7wlib3Sa_zyQC7OLECOvWwpwSptmdsYjgSb8p7g-ygOXowdW8DgMBbMWzQ7Jk" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="422" data-original-width="492" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEgla-0GwGfBWothLVQZPEm5zdH_h4bJPYLyOVqofSslRLEZ1T1TuwD0wRWKQbnSLBg2u6LL_MKFaJH-c0aEzNu0SMp0OzI9GbrhldsFQMQKZMCneEdwiJUb_qCeIgueDD7wlib3Sa_zyQC7OLECOvWwpwSptmdsYjgSb8p7g-ygOXowdW8DgMBbMWzQ7Jk" width="280" /></a></div><br /><br />Now that the orbit is placed in a system of coordinates, each point is defined as a pair (x,y) where x and y are functions of t. Each point of the ellipse is defined by its coordinates (x(t), y(t)). This pair defines the parametric equations of the ellipse where t is the parameter.<p></p><p></p><div class="separator" style="clear: both; text-align: center;"><br /></div><b>Definition</b><div><br /></div><div>If x and y are continuous functions of t on an interval I, then the equations x = x(t) and y = y(t) are called parametric equations and t is called the <b>parameter</b>. The set of points (x,y) obtained as t varies over the interval I is called the graph of the parametric equations. The graph of parametric equations is called a <b>parametric curve</b> or plane curve, and is denoted by C<div><p>Notice in this definition that x and y are used in two ways. The first is as functions of the independent variable t. As t varies over the interval I, the functions x(t) and y(t) generate a set of pairs (x,y). This set of ordered pairs generates the graph of the parametric equations. In this second usage, to designate the ordered pairs, x and y are variables. It is important to distinguish the variables x and y from the functions x(t) and y(t). </p><div class="separator" style="clear: both; text-align: center;"><br /></div><b>Examples</b><div><b><br /></b></div><div><b><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjIzWF6AzIkLm1lQZxuXJaC4E4h9amP-TXS_z5TPbBtub6VEXUddBbq-IIOAMEujhQY1G3jkZ1TM5qKaNuMBvbkVobomBlt8XlCPps6FAQkxxUus7NaL3QvOwFiif6aC3sGe8VWhDBdxnPBa4Pm-QrUMTLyeabdHDBEc1bv9OB9rPbuGUg_WBcRpVDxeZY" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="162" data-original-width="494" height="210" src="https://blogger.googleusercontent.com/img/a/AVvXsEjIzWF6AzIkLm1lQZxuXJaC4E4h9amP-TXS_z5TPbBtub6VEXUddBbq-IIOAMEujhQY1G3jkZ1TM5qKaNuMBvbkVobomBlt8XlCPps6FAQkxxUus7NaL3QvOwFiif6aC3sGe8VWhDBdxnPBa4Pm-QrUMTLyeabdHDBEc1bv9OB9rPbuGUg_WBcRpVDxeZY=w640-h210" width="640" /></a></div><br /><br /></b></div><div><b><br /></b><div><b><br /></b></div><div><b><div class="separator" style="clear: both; text-align: center;"><br /></div>Solution</b></div><div><b><br /></b></div><div><b><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEiDTmC7INi8U9RP1bir3eBzeP5TAZ895by-NbkyI83KVK35F-wN4lT-o8R3seLY3iuH0BSFfNfcp5jCyf5rf6vNfGuNQSD2GOKcxcRCxCtxohudKYabvdU4_xnLJJOsBYjjjrJ90GFxg-JO-4T9R5x7xESSQDvmdcwfYszWkEHMVv-nWPu6P3c1PeLkBj8" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img data-original-height="365" data-original-width="725" height="322" src="https://blogger.googleusercontent.com/img/a/AVvXsEiDTmC7INi8U9RP1bir3eBzeP5TAZ895by-NbkyI83KVK35F-wN4lT-o8R3seLY3iuH0BSFfNfcp5jCyf5rf6vNfGuNQSD2GOKcxcRCxCtxohudKYabvdU4_xnLJJOsBYjjjrJ90GFxg-JO-4T9R5x7xESSQDvmdcwfYszWkEHMVv-nWPu6P3c1PeLkBj8=w640-h322" width="640" /></a></div><br /><br /><br /></b><div><b><br /></b></div><div><div class="separator" style="clear: both; font-weight: bold; text-align: center;"><br /></div>Plotting the points on the second and third columns allows to draw the graph of the parametric equations. The arrows on the graph indicate the orientation of the graph that is the direction of a point on the graph as t varies from -3 to 2.</div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><br /></div><br /><div class="separator" style="clear: both; text-align: center;"><br /></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgvIWuwEjdj_LZt8Iew8grjwEBvcVRlKlFBAVrIt9Z8RaTdFcXLd7G8WA_xOSeikzYwktfu578AjDqt4tF_rfpE91fIjRtscrce34Hk6zKxevHeG_OkAk-dCWUo0YN7wFdsc9aTWqSCh9gCwd1G7aShvZeh3CSH6ztZICAjogykaxQmHvRkCq8uBXEZXkU" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="497" data-original-width="488" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEgvIWuwEjdj_LZt8Iew8grjwEBvcVRlKlFBAVrIt9Z8RaTdFcXLd7G8WA_xOSeikzYwktfu578AjDqt4tF_rfpE91fIjRtscrce34Hk6zKxevHeG_OkAk-dCWUo0YN7wFdsc9aTWqSCh9gCwd1G7aShvZeh3CSH6ztZICAjogykaxQmHvRkCq8uBXEZXkU" width="236" /></a></div><br /><b>Example c.</b></div><div><b><br /></b></div><div>In this case, use multiples of π/6 for t and create another table of values:</div><div><br /></div><div>t x(t) y(t) t x(t) y(t)</div><div><br /></div><div>0 4 0 7π/6 -2⎷3 ≃ -3.5 2</div><div><br /></div><div>π/6 2⎷3≃ 3.5 2 4π/3 -2 -2⎷ 3≃ -3.5</div><div><br /></div><div>π/3 2 2⎷3≃ 3.5 3π/2 0 -4</div><div><br /></div><div>π/2 0 4 5π/3 2 -2⎷≃ -3.5</div><div><br /></div><div>2π/3 -2 2⎷ ≃ 3.5 11π/6 2⎷≃ 3.5 2</div><div><br /></div><div>5π/6 2⎷3≃ -3.5 2 2π 4 0</div><div><br /></div><div>π -4 0 <br /><p></p><p><b><br /><br /></b>The graph of this plane curve appears in the following graph:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjpO4D1YnC5wMuS3gypKYxGJyBu-qI9AXcKcmVAdf9_etmXlUwOfkViDHirrP2xRGvxu7yECKd1Amw-SJBYDXrPyegYYDGPsgYzQ8s6o7I5tsqCGSBiVlaDbSekU9fadnkFLLNB-5oJgiPqQ0FP4PLZsJ0InEt5wli5f4Pg79OaYNSYKPNyR3KTy7wDGRQ" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="423" data-original-width="417" height="240" src="https://blogger.googleusercontent.com/img/a/AVvXsEjpO4D1YnC5wMuS3gypKYxGJyBu-qI9AXcKcmVAdf9_etmXlUwOfkViDHirrP2xRGvxu7yECKd1Amw-SJBYDXrPyegYYDGPsgYzQ8s6o7I5tsqCGSBiVlaDbSekU9fadnkFLLNB-5oJgiPqQ0FP4PLZsJ0InEt5wli5f4Pg79OaYNSYKPNyR3KTy7wDGRQ" width="237" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div>This is the graph of a circle with radius 4 centered at the origin, with a counterclockwise orientation. The starting point and the ending point of the curve both have coordinates (4,0).<p></p><p><b>Practice</b>. Solve example b) <br /><br /></p></div></div></div></div></div>Yves Simonhttp://www.blogger.com/profile/07146463382649045709noreply@blogger.com0