where and are functions defined on some interval . From our definition of a parametric curve, it should be clear that you can always associate a parametric curve with a vector-valued function by just considering the curve traced out by the head of the vector.

It is clear that this formula doesn't make sense if at some particular value of . If at that same value of , then it turns out the graph has a vertical tangent at that point.

While the concept of arc length is very useful for the theory of parametric curves, it turns out to be very difficult to compute in all but the simplest cases.

>f:=t->[2*cos(t),2*sin(t)];You can evaluate this function at any value of t in the usual way.

>f(0);This is how to access a single component. You would use f(t)[2] to get the second component.

>f(t)[1]One way to graph the function is with the

>VPlot(f(t),t=-Pi..Pi);Notice how the derivative command works.

>diff(f(t),t);The output shows the derivative of each

- Enter the parametric function
.
- A)
- Plot the parametric equation with a domain of .
- B)
- Find the value of the function points at the following five t-values , , , , and .
- C)
- Plot the function and the five points.

- A)
- By looking at the graph, which points will have a slope of zero, a positive slope, a negative slope, or an undefined slope.
- B)
- Find the derivative of the points with a positive or negative slope.
- C)
- What do the tangent lines look like at the points with undefined slopes? The equation of the derivatives will have what value in the denominator? Looking at the formula for the derivative of a parametric equation, where does the denominator come from? Evaluate the denominator at these points.

- Find the arclength of the inner curve of the function and the outer curve of the function.

2010-02-04