7.2 Calculus of Parametric Curves
Learning Objectives
 Determine derivatives and equations of tangents for parametric curves.
 Find the area under a parametric curve.
 Determine the arc length of a parametric curve.
 Apply the formula for the surface area of the surface generated by revolving a parametric curve about the xaxis or the yaxis.
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 Figure 1 below. It is a line segment starting at and ending at
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. □
The formula (*) from the previous theorem can be used to calculate the first derivative for a curve defined parametrically at the given value of , and hence the slope of the tangent line to the curve at the point corresponding to . For the purpose of sketching parametric curves, it is useful to determine, where the tangent to the curve is horizontal and where it is vertical. It has a direct analogy with considering the critical points on the graph of a curve with explicit equation , corresponding to the values of , such that or is undefined. Based on (*), we see that the tangent to the parametric curve , is horizontal where and , and the tangent is vertical where and .
Finding the Derivative for a Parametric Curve
For each of the following parametrically defined plane curves, calculate the derivative as well as determine the points, where the tangent line is horizontal and the points, where the tangent line is vertical.
Solution
 To apply (*), first calculate and
Next substitute these into (*):
Since , there are no points on the curve, where the tangent line is horizontal. Solving , we find the value of corresponding to the point on the curve, where the tangent line is vertical. Calculating and gives and yielding the point on the curve. Note that, eliminating the parameter, we can determine that this curve is a parabola opening to the right, and the point is its vertex as shown below.
 Again, we start by calculating and
Next substitute these into (*) to find :
Since , there are no points on this curve, where the tangent line is vertical. To determine the points, where the tangent line is horizontal, we solve , and find that . When , and which corresponds to the point on the curve. When ,
andwhich corresponds to the point on the curve. The following figure provides the sketch of the curve.
 Calculating and we obtain
Therefore, (*) yields
We see that when in the interval . Note that
and .
Hence, each of these values of yields a point on the curve, where the tangent line is horizontal. To find the coordinates of these points, we substitute and into and :
, , yielding the point , and
, , yielding the point .
Solving , we find within . Since is nonzero at all these values of , each of them corresponds to a point on the curve, where the tangent line is vertical. Substituting into and , we find the coordinates of these points to be , and respectively.
The above computations agree with the sketch of the parametric curve, which is a circle of radius 5 with the center at the origin.
Calculate the derivative for the curve defined by the parametric equations , and find all points on the curve, where the tangent line is horizontal or where the tangent line is vertical.
Answer
The tangent line line is horizontal at and , corresponding to and respectively. The tangent line is vertical at , corresponding to .
Slope of the Tangent Line in a Special Case
Determine the slope of the tangent line to the hypocycloid
Solution
We first calculate and
We see that , and so (*) cannot be applied to find when . However, when , , and so we can consider :
Since we deal with a indeterminate form and can apply L’Hospital’s rule.
Therefore, when , the slope of the tangent line is zero, and hence the tangent line to the hypocycloid is horizontal at the point , corresponding to , where the curve has a cusp.
Finding a Tangent Line
Find the equation of the tangent line to the parametric curve defined by the equations
Solution
We first calculate and
Next we substitute these into (*):
When so this is the slope of the tangent line. Calculating and gives and which corresponds to the point on the curve, see Figure 5 below. We now use the pointslope form of the equation of a line to find the equation of the tangent line at this point:
Find the equation of the tangent line to the curve defined by the equations
Answer
The equation of the tangent line is
SecondOrder Derivatives
Our next goal is to see how to take the second derivative of a function defined parametrically. The second derivative of a function is defined to be the derivative of the first derivative; that is,
Since we can replace the on both sides of this equation with This gives us
If we know as a function of t, then this formula is straightforward to apply.
Calculate the second derivative for the plane curve defined by the equations
Answer
From before, we know that the second derivative is “responsible” for concavity of the curve with an explicit equation : the curve is concave upward where , and it is concave downward where Since, locally, a parametric curve usually admits eliminating the parameter and obtaining an explicit equation, we can still look at the sign of , to determine where the curve is concave upward and where it is concave downward. Because, in practice, we won't be finding an explicit equation of the curve, but we will be using (**) to find as a function of , it is the intervals in terms of that we will be referring to when discussing concavity of parametric curves.
Examining Concavity of a Parametric Curve
Determine where the parametric curve , is concave upward and where it is concave downward.
Solution
Applying (*), we find that Using (**) together with the quotient rule, we obtain
We now need to determine for which values of , is positive, and for which values of it is negative. Factoring the numerator and denominator, we rewrite :
The numerator has zeros and , while the denominator has a zero of multiplicity 3. Using sample points or any other appropriate method, we find that , and hence the parametric curve is concave upward, when and , and , implying that the curve is concave downward, when and .
Determine where the parametric curve , is concave upward.
Answer
The curve is concave upward when .
Integrals Involving Parametric Equations
Now that we have seen how to calculate the derivative of a plane curve, the next question is this: How do we find the area under a curve defined parametrically? Recall the cycloid defined by the equations Suppose we want to find the area of the shaded region in the following graph.
To derive a formula for the area under the curve defined by the functions
we assume that is increasing and differentiable and start with an equal partition of the interval Suppose and consider the following graph.
We use rectangles to approximate the area under the curve. The height of a typical rectangle in this parametrization is for some value in the ith subinterval, and the width can be calculated as . It follows that the area of the ith rectangle is given by
Then a Riemann sum for the area is
Multiplying and dividing each area by gives
Taking the limit as approaches infinity, we obtain
Note that if is decreasing, that is, the curve is traced from left to right, everything in the above derivation stays the same except that the width of a typical rectangle becomes , which results in the formula
This leads to the following theorem.
Area under a Parametric Curve
Consider the plane curve defined by the parametric equations
and assume that is differentiable.
 If is increasing, then the area under this curve is given by

If is decreasing, then the area under this curve is given by
Finding the Area under a Parametric Curve
Find the area under one arc of the cycloid defined by the equations
Solution
To determine whether is increasing or decreasing we look at the sign of . We have that , and hence is increasing. Applying the above theorem, we have
Find the area under the upper half of the hypocycloid defined by the equations
Answer
Hint
Use the above theorem, along with the identities and Note that is decreasing.
Arc Length of a Parametric Curve
The same way we did for a regular curve with explicit equation or , to derive a formula for the arc length of a parametric curve, we approximate it by a union of line segments as shown in the following figure.
Given a plane curve defined by the parametric equations we start by partitioning the interval into n equal subintervals: The width of each subinterval is The length of the th line segment can be found as follows:
Adding those from , we obatin an approximation of the arc length s of the parametric curve:
If we assume that and are differentiable functions of t, then the Mean Value Theorem applies, so in each subinterval there exist and such that
With this, becomes
This is a Riemann sum that approximates the arc length over a partition of the interval If we further assume that the derivatives are continuous and let the number of points in the partition increase without bound, the approximation approaches the exact arc length. This gives
When taking the limit, the values of and are both contained within the same evershrinking interval of width so they must converge to the same value.
We can summarize this method in the following theorem.
Arc Length of a Parametric Curve
Consider the plane curve defined by the parametric equations
and assume that and are smooth, that is, their derivatives and are continuous. Then the arc length of this curve is given by
Now suppose that the parameter can be eliminated, leading to a function We are going to show that the above formula agrees with the formula for the arc length of a regular curve derived in Section 2.4. We have and the Chain Rule gives Substituting this into the above formula gives
Here we have assumed that and the case when is analogous (the extra minus is going to disappear when the limits of integration are interchanged). Using a substitution , we have that and letting and we obtain the formula
which is exactly the one we had before.
Finding the Arc Length of a Parametric Curve
Find the arc length of the semicircle defined by the equations
Solution
The parametric curve is shown in Figure 9 below. To determine its length, we use the formula:
Note that the formula for the arc length of a semicircle is and the radius of this circle is 3. This is a great example of using calculus to derive a known geometric formula.
Find the arc length of the curve defined by the equations
Answer
We now return to the problem posed at the beginning of the section about a baseball leaving a pitcher’s hand. Ignoring the effect of air resistance (unless it is a curve ball!), the ball travels in a parabolic path. Assuming the pitcher’s hand is at the origin and the ball travels left to right in the direction of the positive xaxis, the parametric equations for this curve can be written as
where t represents time. We first calculate the distance the ball travels as a function of time. This distance is represented by the arc length. We can modify the arc length formula slightly. First rewrite the functions and using v as an independent variable, so as to eliminate any confusion with the parameter t:
Then we write the arc length formula as follows:
The variable v acts as a dummy variable that disappears after integration, leaving the arc length as a function of time t. To integrate this expression, one needs to make a trigonometric substitution , which will lead to a constant multiple of an integral of . After some technical computations, this will result in
This function represents the distance traveled by the ball as a function of time. To calculate the speed, take the derivative of this function with respect to t. While this may seem like a daunting task, it is possible to obtain the answer directly from the Fundamental Theorem of Calculus:
Therefore,
One third of a second after the ball leaves the pitcher’s hand, the distance it travels is equal to
This value is just over three quarters of the way to home plate. The speed of the ball is
This speed translates to approximately 95 mph—a majorleague fastball.
Surface Area Generated by a Parametric Curve
Recall the problem of finding the surface area of a surface of revolution. In Section 2.4, we derived a formula for the surface area of a surface generated by revolving the curve from to around the xaxis:
We now consider a surface of revolution generated by revolving a parametrically defined curve around the xaxis as shown in the following figure.
The formula for its surface area is
provided that is nonnegative on
Finding Surface Area
Find the surface area of a sphere of radius r centered at the origin.
Solution
We start by parametrizing the upper semicircle with center at the origin and radius :
When this curve is revolved around the xaxis, it generates a sphere of radius r. To calculate the surface area of the sphere, we use the above formula:
This agrees with the geometric you might have seen before.
Find the area of the surface generated by revolving the plane curve defined by the equations
around the xaxis.
Answer
Hint
When evaluating the integral, use a usubstitution.
Key Concepts
 The derivative of the parametrically defined curve and can be calculated using the formula Using the derivative, we can find the equation of a tangent line to a parametric curve.
 If , the area under the parametric curve can be determined by using the formula where the choice of sign depends on whether is increasing or decreasing over .
 The arc length of a parametric curve can be calculated by using the formula
 The area of a surface obtained by revolving a parametric curve around the xaxis is given by provided when . If the curve is revolved around the yaxis, then the formula is
provided when .
Key Equations
 Derivative of parametric equations
 Secondorder derivative of parametric equations
 Area under a parametric curve
, where the sign depends on the sign of  Arc length of a parametric curve
 Surface area generated by a parametric curve about a coordinate axis
(revolving about xaxis)
(revolving about yaxis)
Exercises
For the following exercises, find as a function of the parameter t.
1.
Answer
2.
3.
Answer
4.
For the following exercises, each set of parametric equations represents a line. Without eliminating the parameter, find the slope of each line.
5.
Answer
6.
7.
Answer
0
For the following exercises, determine the slope of the tangent line at the point corresponding to the given value of the parameter.
8.
9.
Answer
10.
11.
Answer
Slope is undefined.
12.
For the following exercises, find all points on the parametric curve where the tangent line has the given slope.
13.
Answer
, where is integer, corresponding to the points and .
14.
15.
Answer
corresponding to the point (note that is not in the domain of )
16.
For the following exercises, write an equation of the tangent line to the given parametric curve at the point that corresponds to the specified value of the parameter t.
17.
Answer
18.
19.
Answer
20. Consider the parametric curve Find all values of the parameter that correpsond to the points on the curve where the tangent line is horizontal.
21. Consider the parametric curve Find all values of the parameter that correpsond to the points on the curve where the tangent line is vertical.
Answer
.
For the following exercises, find all points on the given parametric curve where the tangent line is horizontal or vertical.
22.
23.
Answer
No horizontal tangents. Vertical tangents at
24.
25.
Answer
Horizontal tangent at vertical tangents at
For the following exercises, find
26.
27.
Answer
28.
29.
Answer
For the following exercises, find at the specified value of the parameter.
30.
31.
Answer
4
For the following exercises, find t intervals on which the given parametric curve is concave up and t intervals on which it is concave down.
32.
33.
Answer
Concave up on
34.
35.
Answer
Concave up on and concave down on .
36. Sketch and find the area under one arch of the cycloid Here is a fixed real number and is a parameter.
37. Find the area below the curve and above the xaxis.
Answer
2
38. Find the area enclosed by the ellipse
39. Find the area of the region below the curve and above the xaxis over the interval
Answer
For the following exercises, find the total area of the regions between the parametric curves and the xaxis. In exercises 4143 is a fixed real number.
40.
41.* .
Answer
42. (the “hourglass”)
43.[T] (the “teardrop”)
Answer
For the following exercises, find the arc length of the given parametric curve.
44.
45.
Answer
46.
47.
Answer
48.
49. (the hypocycloid)
Answer
50. Find the length of one arch of the cycloid
51. Find the distance traveled by a particle with position as t varies in the given time interval:
Answer
52. Find the length of the curve
For the following exercises, set up but do not evaluate the integral that represents the area of the surface obtained by rotating the given parametric curve about the xaxis.
53.
Answer
54.
55.
Answer
56.
For the following exercises, find the area of the surface obtained by rotating the given parametric curve about the xaxis.
57.
Answer
58.
For the following exercises, set up but do not evaluate the integral that represents the area of the surface obtained by rotating the given parametric curve about the yaxis.
59.
Answer
60.
61. Find the area of the surface generated by revolving about the yaxis.