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.