Determine derivatives and equations of tangents for parametric curves.
Find the area under a parametric curve.
Use the equation for arc length of a parametric curve.
Apply the formula for surface area to a volume generated by a parametric curve.
Now that we have introduced the concept of a parameterized curve, our next step is to learn how to work with this concept in the context of calculus. For example, if we know a parameterization of a given curve, is it possible to calculate the slope of a tangent line to the curve? How about the arc length of the curve? Or the area under the curve?
Another scenario: Suppose we would like to represent the location of a baseball after the ball leaves a pitcher’s hand. If the position of the baseball is represented by the plane curve
then we should be able to use calculus to find the speed of the ball at any given time. Furthermore, we should be able to calculate just how far that ball has traveled as a function of time.
Derivatives of parametric equations
We start by asking how to calculate the slope of a line tangent to a parametric curve at a point. Consider the plane curve defined by the parametric equations
The graph of this curve appears in
[link] . It is a line segment starting at
and ending at
Graph of the line segment described by the given parametric equations.
We can eliminate the parameter by first solving the equation
for
t :
Substituting this into
we obtain
The slope of this line is given by
Next we calculate
and
This gives
and
Notice that
This is no coincidence, as outlined in the following theorem.
Derivative of parametric equations
Consider the plane curve defined by the parametric equations
and
Suppose that
and
exist, and assume that
Then the derivative
is given by
Proof
This theorem can be proven using the Chain Rule. In particular, assume that the parameter
t can be eliminated, yielding a differentiable function
Then
Differentiating both sides of this equation using the Chain Rule yields
so
But
which proves the theorem.
□
[link] can be used to calculate derivatives of plane curves, as well as critical points. Recall that a critical point of a differentiable function
is any point
such that either
or
does not exist.
[link] gives a formula for the slope of a tangent line to a curve defined parametrically regardless of whether the curve can be described by a function
or not.
Finding the derivative of a parametric curve
Calculate the derivative
for each of the following parametrically defined plane curves, and locate any critical points on their respective graphs.
This derivative is undefined when
Calculating
and
gives
and
which corresponds to the point
on the graph. The graph of this curve is a parabola opening to the right, and the point
is its vertex as shown.
Graph of the parabola described by parametric equations in part a.
which corresponds to the point
on the graph. When
we have
which corresponds to the point
on the graph. The point
is a relative minimum and the point
is a relative maximum, as seen in the following graph.
Graph of the curve described by parametric equations in part b.
This derivative is zero when
and is undefined when
This gives
as critical points for
t. Substituting each of these into
and
we obtain
0
5
0
0
5
−5
0
0
−5
5
0
These points correspond to the sides, top, and bottom of the circle that is represented by the parametric equations (
[link] ). On the left and right edges of the circle, the derivative is undefined, and on the top and bottom, the derivative equals zero.
Graph of the curve described by parametric equations in part c.
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