In this post we are going to solve another example of level curve and contour map.
Example
Solution
Let's substitute c by 0 in the function:
Let's square both sides and multiply by -1
Let's put the x terms together, the y terms together and 8 to both sides:
Let's group the pairs of terms containing the same variable in parentheses and put 4 in factor
from the first pair
Let's complete the square of the expressions in parentheses and add the correct terms to the
hand right side
Let's factor the left hand side and simplify the right hand:
Let's divide both sides by 16:
This equation equation describes an ellipse centered at (-1,2). Here is its graph:
We can continue to find level curves for c=1, c= 2, c= 3, c= 4.
For example for c= 1, the equation of the ellipse is:
(x - 1)²/4 + (y + 2)²/4 = 11/4. We have a different equation but still an ellipse centered at (1, -2)
For c = 4, the level curve is the point (-1, 2)
The domain is the set of (x,y) such as the expression under the radical is greater than or equal to 0 meaning that 8+8x-4y-4x²-y≥ 0. Solving this inequality leads to (x - 1)²/4 + (y + 2)²/4 ≤ 4. All pairs (x,y) that satisfy this inequality are found inside the ellipse described by the expression
(x - 1)²/4 + (y + 2)²/4 .
The domain is the set of pairs (x,y) that are located inside the ellipse. The range is the the set of values for c. These values are 0, 1, 3, 3. 4. They constitute the range of the function.
Practice
Find and graph the level curve of the following function: g(x,y) = x² + y² - 6x +2y corresponding
to c = 15
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