How to evaluate vector-valued functions

 Goal

 Evaluate vector valued functions

Definition

A vector-valued function is a function of the form:

The functions f, g, h are real-valued functions of the parameter t. Vector-valued functions are also written in the form:

The first form defines a two-dimensional vector-valued function. The second form describes a tri-dimensional vector-valued function. 

Example

For each of the vector-valued functions, evaluate r(0), r(π/2), r(2π/3). Do any of these functions have domain restrictions?

Solution

a. Let's substitute each of the value of t in the function:

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

b. Let's do the same thing for the second function:

 

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. 

Practice

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?

Arc length of a parametric curve

 Let's consider the plane curve defined by the following parametric equations:

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:

This video gives an idea of where this formula originates:

 

Example

Find the arc length of the semicircle defined by the equations:

Solution

Here is the graph of the semicircle:

Let's use the formula to find the arc length:

Practice

Find the arc length of the curve defined by the equations:

Area under a parametric curve

 Goal: Finding the area under the curve of a parametric equation

Area under parametric curve

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

We know that the area is given by:

We assume that the curve is given by the parametric equations: x = x(t) y = y(t). The formula above becomes:

The formula that allows to find the area under a parametric curve is then given by:

Example

Find the area under curve of the cycloid define by the equations:

x(t) = t-sint   y(t) = 1-cost  0  ≤ t ≤ 2π

Solution

Using the formula we have:

Practice

Find the area under the curve of the hypocycloid defined by the following equations:

x(t) = 3cost + cos3t   y(t) = 3sint - sin3t  0 ≤  t ≤ π 

Second Order Derivatives

 The second order derivative of a function y = f(x) is the derivative of the first derivative of the function.

The first derivative of the function f is dy/dx.

The derivative of the first derivative is :d/dx[dy/dx] = d²y/dx²

The relation equality being commutative, we can write: d²y/dx² = d/dx[dy/dx].

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.

The function here is dy/dx. Let's calculate its derivative:

 d²y/dx² = d/dt[dy/dx]./dx/dt.

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.

Examples

Calculate the second derivative  d²y/dx² for the plane curve defined by the parametric equations:

 Solution

We have :

 d²y/dx² = d/dt[dy/dx]./dx/dt.

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.

dy/dx = y'(t)/x'(t) = 2/2t = 1/t

Let's calculate dx/dt also:

dx/dt = 2t

The expression of d²y/dx² becomes:

d²y/dx² = d/dt (1/t)/2t = -1/t²/2t = (-1/t²)(1/2t) = -1/2t³

Practice

Calculate the second derivative  d²y/dx² for the plane curve defined by the parametric equations:

Locate any critical points on its curve.

 

Derivative of parametric equations

 Objective: to find the derivative of parametric curves

Derivatives of parametric equations

Let's consider the plane curve defined by the parametric equations:

x(t) = 2t + 3  y(t) = 3t - 4       -2≤ x≤3

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.

The curve of the parametric equations is a line starting at (-1, -10) and ending at (9, 5). 

Let's find the explicit form of the parametric equations. To do this we start by eliminating t in x(t):

Let's substitute t in y(t):

The slope of the tangent line is 3/2. The derivative is dy/dx = 3/2

But dy/dx = dy/dt/dx/dt = y'(t)/x'(t)

x'(t) = 2 y'(t) = 3

dy/dx = 3/2

In both cases dy/dt = 3/2. The expression dy/dx = y'(t)/x'(t) is the derivative of the function.

Theorem. Derivative of Parametric equations

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:

Examples

Calculate the derivative dy/dx for each of the parametrically defined curves and locate any critical points on their respective graph. 

Solutions

a) Let's calculate dy/dx: dy/dx = y'(t)/x'(t)

From the parametric equations in example a) we have: y'(t) = 2 and x'(t) = 2. Therefore:

dy/dx = 2/2t = 1/t

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.

b)  I provided a guided solution for examples b) and c) .We have for example b)

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.

The point (-1, 6) is a relative maximum and the point (3, 2) is a relative minimum.

c) We have : dy/dx = -cost/sint

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).

When sint = 0, the corresponding values of t are : t = 0, t = π, t = 2π. The critical points are respectively:

(5. 0), (-5, 0) and (5. 0).

Here is the graph of the parametric equations defined in part example c)

Practice

Calculate the derivative dy/dx for the plane curve defined by the following equations and locate any critical points on its graph:

 

Parameterization of a curve

 Objective:

Transform the function of a curve into parametric equations.

Parameterization of a curve

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.

Example

Find two different pairs of parametric equations to represent the graph of y = 2x² - 3

Solution

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.

The first set of pair of parametric equations is x(t) = t y = 2t²-3

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.

For the second pair of equations let's choose x(t) = 3t-2

Let's substitute x in the function:

 

Therefore a second parameterization of the function is represented by:

Practice

Find two different sets of parametric equations to represent the graph of y = x² + 2x

Converting the parametric equations of a curve into an explicit form

 Objective: Convert the parametric equations of a curve into the form y = f(x)

Conversion of the parametric equations of a curve into an explicit form

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. 

Examples:

a) let's solve example a)

Let's find t in the first equation. We can also find it in the second equation.

Let's substitute t in the second equation:

This is the equation of a parabola opening upward. The limits of the parameter lead to a domain restriction of the function.

The interval for t is mentioned in the problem: -6≤t≤-2

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].

b) I am not going to solve completely the example 2. I'll give some steps. 

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.

Practice

Eliminate the parameter for the plane curve defined by the following parametric equations and describe the resulting graph:

x(t) = 2 + 3/t  y(t) = t-1  2≤t≤6

What are parametric equations and how to graph their curve

 Objectives:

1. Define parametric equations

2. Plot their curve

Considerations

Let's consider the orbit of the earth around the sun. 

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.

Let's superimpose the ellipse on a system of coordinates (x,y):

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.

Definition

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 parameter. 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 parametric curve or plane curve, and is denoted by C

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). 

Examples

Solution

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.

Example c.

In this case, use multiples of π/6 for t and create another table of values:

t          x(t)                  y(t)              t             x(t)                     y(t)

0          4                      0                 7π/6        -2⎷3 ≃ -3.5       2

π/6       2⎷3≃ 3.5        2                4π/3           -2                   -2⎷ 3≃ -3.5

π/3        2                  2⎷3≃ 3.5    3π/2            0                     -4

π/2        0                   4                 5π/3            2                      -2⎷≃ -3.5

2π/3       -2                 2⎷ ≃ 3.5     11π/6         2⎷≃ 3.5            2

5π/6        2⎷3≃ -3.5      2              2π               4                       0

π              -4                 0                  

The graph of this plane curve appears in the following graph:

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).

Practice. Solve example b) 

Evaluating non-elementary integrals

 Objective:

Use Taylor series to evaluate non-elementary integrals

Example

Solution

Practice

Solving differential equations using power series

 Objective: Solve differential equations using power series.

Let's consider the differential equation y'(x) = y. Recall that this is a first order separable equation and its solution is y = Ce^x. For most differential equations there are not any analytical tools to solve them. Power series are an extremely useful tool to solve them. The technique we use is to look for a solution of the form 

Then we determine the coefficients in order to solve the differential equation.

Example

Solution

Using the uniqueness of power series representations, we know that these series can only be equal if their coefficients are equal. Therefore:

Practice