Section 9.3 Calculus and Parametric Equations
Key Idea 9.3.2. Finding dydx with Parametric Equations.
Let x=f(t) and y=g(t), where f and g are differentiable on some open interval I and fβ²(t)β 0 on I. Then
Definition 9.3.3. Tangent and Normal Lines.
Let a curve C be parametrized by x=f(t) and y=g(t), where f and g are differentiable functions on some interval I containing t=t0. The tangent line to C at t=t0 is the line through (f(t0),g(t0)) with slope m=gβ²(t0)/fβ²(t0), provided fβ²(t0)β 0.
The normal line to C at t=t0 is the line through (f(t0),g(t0)) with slope m=βfβ²(t0)/gβ²(t0), provided gβ²(t0)β 0.
If the tangent line at t=t0 has a slope of 0, the normal line to C at t=t0 is the line x=f(t0).
If the normal line at t=t0 has a slope of 0, the tangent line to C at t=t0 is the line x=f(t0).
Example 9.3.4. Tangent and Normal Lines to Curves.
Let x=5t2β6t+4 and y=t2+6tβ1, and let C be the curve defined by these equations.
Find the equations of the tangent and normal lines to C at t=3.
Find where C has vertical and horizontal tangent lines.
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We start by computing \(\fp(t) = 10t-6\) and \(g'(t) =2t+6\text{.}\) Thus
\begin{equation*} \frac{dy}{dx} = \frac{2t+6}{10t-6}\text{.} \end{equation*}Make note of something that might seem unusual: \(\frac{dy}{dx}\) is a function of \(t\text{,}\) not \(x\text{.}\) Just as points on the curve are found in terms of \(t\text{,}\) so are the slopes of the tangent lines. The point on \(C\) at \(t=3\) is \((31,26)\text{.}\) The slope of the tangent line is \(m=1/2\) and the slope of the normal line is \(m=-2\text{.}\) Thus,
the equation of the tangent line is \(\ds y=\frac12(x-31)+26\text{,}\) and
the equation of the normal line is \(\ds y=-2(x-31)+26\text{.}\)
This is illustrated in Figure 9.3.5.
Figure 9.3.5. Graphing tangent and normal lines in Example 9.3.4 -
To find where \(C\) has a horizontal tangent line, we set \(\frac{dy}{dx}=0\) and solve for \(t\text{.}\) In this case, this amounts to setting \(g'(t)=0\) and solving for \(t\) (and making sure that \(\fp(t)\neq 0\)).
\begin{equation*} g'(t)=0 \Rightarrow 2t+6=0 \Rightarrow t=-3\text{.} \end{equation*}The point on \(C\) corresponding to \(t=-3\) is \((67,-10)\text{;}\) the tangent line at that point is horizontal (hence with equation \(y=-10\)). To find where \(C\) has a vertical tangent line, we find where it has a horizontal normal line, and set \(-\frac{\fp(t)}{g'(t)}=0\text{.}\) This amounts to setting \(\fp(t)=0\) and solving for \(t\) (and making sure that \(g'(t)\neq 0\)).
\begin{equation*} \fp(t)=0 \Rightarrow 10t-6=0 \Rightarrow t=0.6\text{.} \end{equation*}The point on \(C\) corresponding to \(t=0.6\) is \((2.2,2.96)\text{.}\) The tangent line at that point is \(x=2.2\text{.}\) The points where the tangent lines are vertical and horizontal are indicated on the graph in Figure 9.3.5.
Example 9.3.6. Tangent and Normal Lines to a Circle.
Find where the unit circle, defined by x=cos(t) and y=sin(t) on [0,2Ο], has vertical and horizontal tangent lines.
Find the equation of the normal line at t=t0.
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We compute the derivative following Key Idea 9.3.2:
\begin{equation*} \frac{dy}{dx} = \frac{g'(t)}{\fp(t)} = -\frac{\cos(t) }{\sin(t) }\text{.} \end{equation*}The derivative is \(0\) when \(\cos(t) = 0\text{;}\) that is, when \(t=\pi/2,\, 3\pi/2\text{.}\) These are the points \((0,1)\) and \((0,-1)\) on the circle. The normal line is horizontal (and hence, the tangent line is vertical) when \(\sin(t) =0\text{;}\) that is, when \(t= 0,\,\pi,\,2\pi\text{,}\) corresponding to the points \((-1,0)\) and \((0,1)\) on the circle. These results should make intuitive sense.
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The slope of the normal line at \(t=t_0\) is \(\ds m=\frac{\sin(t_0) }{\cos(t_0) } = \tan(t_0)\text{.}\) This normal line goes through the point \((\cos(t_0) ,\sin(t_0) )\text{,}\) giving the line
\begin{align*} y \amp =\frac{\sin(t_0) }{\cos(t_0) }(x-\cos(t_0) ) + \sin(t_0)\\ \amp = (\tan(t_0) )x\text{,} \end{align*}as long as \(\cos(t_0) \neq 0\text{.}\) It is an important fact to recognize that the normal lines to a circle pass through its center, as illustrated in Figure 9.3.7. Stated in another way, any line that passes through the center of a circle intersects the circle at right angles.
Figure 9.3.7. Illustrating how a circle's normal lines pass through its center
Example 9.3.8. Tangent lines when dydx is not defined.
Find the equation of the tangent line to the astroid x=cos3(t), y=sin3(t) at t=0, shown in Figure 9.3.9.
We start by finding \(x'(t)\) and \(y'(t)\text{:}\)
Note that both of these are 0 at \(t=0\text{;}\) the curve is not smooth at \(t=0\) forming a cusp on the graph. Evaluating \(\frac{dy}{dx}\) at this point returns the indeterminate form of β0/0β.
We can, however, examine the slopes of tangent lines near \(t=0\text{,}\) and take the limit as \(t\to 0\text{.}\)
We have accomplished something significant. When the derivative \(\frac{dy}{dx}\) returns an indeterminate form at \(t=t_0\text{,}\) we can define its value by setting it to be \(\lim\limits_{t\to t_0}\)\(\frac{dy}{dx}\text{,}\) if that limit exists. This allows us to find slopes of tangent lines at cusps, which can be very beneficial.
We found the slope of the tangent line at \(t=0\) to be 0; therefore the tangent line is \(y=0\text{,}\) the \(x\)-axis.
Subsection 9.3.1 Concavity
We continue to analyze curves in the plane by considering their concavity; that is, we are interested in d2ydx2, βthe second derivative of y with respect to x.β To find this, we need to find the derivative of dydx with respect to x; that is,Key Idea 9.3.10. Finding d2ydx2 with Parametric Equations.
Let x=f(t) and y=g(t) be twice differentiable functions on an open interval I, where fβ²(t)β 0 on I. Then
Example 9.3.11. Concavity of Plane Curves.
Let x=5t2β6t+4 and y=t2+6tβ1 as in Example 9.3.4. Determine the t-intervals on which the graph is concave up/down.
Concavity is determined by the second derivative of \(y\) with respect to \(x\text{,}\) \(\frac{d^2y}{dx^2}\text{,}\) so we compute that here following Key Idea 9.3.10.
In Example 9.3.4, we found \(\ds\frac{dy}{dx} = \frac{2t+6}{10t-6}\) and \(\fp(t) = 10t-6\text{.}\) So:
The graph of the parametric functions is concave up when \(\frac{d^2y}{dx^2} \gt 0\) and concave down when \(\frac{d^2y}{dx^2} \lt 0\text{.}\) We determine the intervals when the second derivative is greater/less than 0 by first finding when it is 0 or undefined.
As the numerator of \(\ds -\frac{9}{(5t-3)^3}\) is never 0, \(\frac{d^2y}{dx^2} \neq 0\) for all \(t\text{.}\) It is undefined when \(5t-3=0\text{;}\) that is, when \(t= 3/5\text{.}\) Following the work established in Section 3.4, we look at values of \(t\) greater/less than \(3/5\) on a number line:
Reviewing Example 9.3.4, we see that when \(t=3/5=0.6\text{,}\) the graph of the parametric equations has a vertical tangent line. This point is also a point of inflection for the graph, illustrated in Figure 9.3.12.
The video in Figure 9.3.13 shows how this information can be used to sketch the curve by hand.
Example 9.3.14. Concavity of Plane Curves.
Find the points of inflection of the graph of the parametric equations x=βt, y=sin(t), for 0β€tβ€16.
We need to compute \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\text{.}\)
The points of inflection are found by setting \(\frac{d^2y}{dx^2}=0\text{.}\) This is not trivial, as equations that mix polynomials and trigonometric functions generally do not have βniceβ solutions.
In Figure 9.3.15.(a) we see a plot of the second derivative. It shows that it has zeros at approximately \(t=0.5,\,3.5,\,6.5,\,9.5,\,12.5\) and \(16\text{.}\) These approximations are not very good, made only by looking at the graph. Newton's Method provides more accurate approximations. Accurate to 2 decimal places, we have:
The corresponding points have been plotted on the graph of the parametric equations in Figure 9.3.15.(b). Note how most occur near the \(x\)-axis, but not exactly on the axis.
Subsection 9.3.2 Arc Length
We continue our study of the features of the graphs of parametric equations by computing their arc length.Factor out the fβ²(t)2:
=β«baβfβ²(t)2+gβ²(t)2β 1fβ²(t)dxβ=dt=β«t2t1βfβ²(t)2+gβ²(t)2dt.Theorem 9.3.17. Arc Length of Parametric Curves.
Let x=f(t) and y=g(t) be parametric equations with fβ² and gβ² continuous on [t1,t2], on which the graph traces itself only once. The arc length of the graph, from t=t1 to t=t2, is
Example 9.3.18. Arc Length of a Circle.
Find the arc length of the circle parametrized by x=3cos(t), y=3sin(t) on [0,3Ο/2].
By direct application of Theorem 9.3.17, we have
Apply the Pythagorean Theorem.
\begin{align*} \amp = \int_0^{3\pi/2} 3 \, dt\\ \amp = 3t\Big|_0^{3\pi/2} = 9\pi/2\text{.} \end{align*}This should make sense; we know from geometry that the circumference of a circle with radius 3 is \(6\pi\text{;}\) since we are finding the arc length of \(3/4\) of a circle, the arc length is \(3/4\cdot 6\pi = 9\pi/2\text{.}\)
Example 9.3.19. Arc Length of a Parametric Curve.
The graph of the parametric equations x=t(t2β1), y=t2β1 crosses itself as shown in Figure 9.3.20, forming a βteardrop.β Find the arc length of the teardrop.
We can see by the parametrizations of \(x\) and \(y\) that when \(t=\pm 1\text{,}\) \(x=0\) and \(y=0\text{.}\) This means we'll integrate from \(t=-1\) to \(t=1\text{.}\) Applying Theorem 9.3.17, we have
Unfortunately, the integrand does not have an antiderivative expressible by elementary functions. We turn to numerical integration to approximate its value. Using 4 subintervals, Simpson's Rule approximates the value of the integral as \(2.65051\text{.}\) Using a computer, more subintervals are easy to employ, and \(n=20\) gives a value of \(2.71559\text{.}\) Increasing \(n\) shows that this value is stable and a good approximation of the actual value.
Subsection 9.3.3 Surface Area of a Solid of Revolution
Related to the formula for finding arc length is the formula for finding surface area. We can adapt the formula found in Theorem 7.4.13 from Section 7.4 in a similar way as done to produce the formula for arc length done before.Theorem 9.3.21. Surface Area of a Solid of Revolution.
Consider the graph of the parametric equations x=f(t) and y=g(t), where fβ² and gβ² are continuous on an open interval I containing t1 and t2 on which the graph does not cross itself.
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The surface area of the solid formed by revolving the graph about the x-axis is (where g(t)β₯0 on [t1,t2]):
Surface Area =2Οβ«t2t1g(t)βfβ²(t)2+gβ²(t)2dt. -
The surface area of the solid formed by revolving the graph about the y-axis is (where f(t)β₯0 on [t1,t2]):
Surface Area =2Οβ«t2t1f(t)βfβ²(t)2+gβ²(t)2dt.
Example 9.3.22. Surface Area of a Solid of Revolution.
Consider the teardrop shape formed by the parametric equations x=t(t2β1), y=t2β1 as seen in Example 9.3.19. Find the surface area if this shape is rotated about the x-axis, as shown in Figure 9.3.23.
The teardrop shape is formed between \(t=-1\) and \(t=1\text{.}\) Using Theorem 9.3.21, we see we need for \(g(t)\geq 0\) on \([-1,1]\text{,}\) and this is not the case. To fix this, we simplify replace \(g(t)\) with \(-g(t)\text{,}\) which flips the whole graph about the \(x\)-axis (and does not change the surface area of the resulting solid). The surface area is:
Once again we arrive at an integral that we cannot compute in terms of elementary functions. Using Simpson's Rule with \(n=20\text{,}\) we find the area to be \(S=9.44\text{.}\) Using larger values of \(n\) shows this is accurate to 2 places after the decimal.
Exercises 9.3.4 Exercises
Terms and Concepts
1.
True or False? Given parametric equations x=f(t) and y=g(t), dydx=fβ²(t)/gβ²(t), as long as gβ²(t)β 0.
True
False
2.
Given parametric equations x=f(t) and y=g(t), the derivative dydx as given in Key Idea 9.3.2 is a function of ?
3.
True or False? Given parametric equations x=f(t) and y=g(t), to find d2ydx2, one simply computes ddt(dydx).
True
False
4.
True or False? If dydx=0 at t=t0, then the normal line to the curve at t=t0 is a vertical line.
True
False
In the following exercises, parametric equations for a curve are given.
Find dydx.
Find the equations of the tangent and normal line(s) at the point(s) given.
Sketch the graph of the parametric functions along with the found tangent and normal lines.
5.
x=t, y=t2;t=1
6.
x=βt, y=5t+2;t=4
7.
x=t2βt, y=t2+t;t=1
8.
x=t2β1, y=t3βt;t=0 and t=1
9.
x=sec(t), y=tan(t) on (βΟ/2,Ο/2);t=Ο/4
10.
x=cos(t), y=sin(2t) on [0,2Ο];t=Ο/4
11.
x=cos(t)sin(2t), y=sin(t)sin(2t) on [0,2Ο]; t=3Ο/4
12.
x=et/10cos(t), y=et/10sin(t); t=Ο/2
In the following exercises, find t-values where the curve defined by the given parametric equations has a horizontal tangent line. Note: these are the same equations as in Exercises 9.3.4.5 β Exercise 9.3.4.12.
13.
x=t, y=t2
14.
x=βt, y=5t+2
15.
Find the values of t where the curve defined by x=t2βt, y=t2+t has a horizontal tangent line. (If there is more than one, use commas to separate them.)
tβ{}
At what point(s) in the plane are the horizontal tangent lines? (If there is more than one, use commas to separate them.)
16.
x=t2β1, y=t3βt
17.
x=sec(t), y=tan(t) on (βΟ/2,Ο/2)
18.
Find the values of t in [0,2Ο) where the curve defined by x=cos(t), y=sin(2t) has a horizontal tangent line. (If there is more than one, use commas to separate them.)
tβ{}
At what point(s) in the plane are the horizontal tangent lines? (If there is more than one, use commas to separate them.)
19.
x=cos(t)sin(2t), y=sin(t)sin(2t) on [0,2Ο]
20.
x=et/10cos(t), y=et/10sin(t)
In the following exercises, find t=t0 where the graph of the given parametric equations is not smooth, then find limtβt0dydx.
21.
Find t=t0 where the graph of x=1t2+1, y=t3 is not smooth, then find limtβt0dydx.
t0=
limtβt0dydx=
22.
Find t=t0 where the graph of x=βt3+7t2β16t+13, y=t3β5t2+8tβ2 is not smooth, then find limtβt0dydx.
t0=
limtβt0dydx=
23.
x=t3β3t2+3tβ1,y=t2β2t+1
24.
x=cos2(t),y=1βsin2(t)
In the following exercises, parametric equations for a curve are given. Find d2ydx2, then determine the intervals on which the graph of the curve is concave up/down. Note: these are the same equations as in Exercises 9.3.4.5 β Exercise 9.3.4.12.
25.
x=t,y=t2
26.
x=βt,y=5t+2
27.
For the curve defined by x=t2βt, y=t2+t, find d2ydx2 (as a function of t). Then determine the intervals on which the graph of the curve is concave up/down.
d2ydx2=
The curve is concave up for t in the interval(s) . (If there is more than one interval, use commas to separate them.)
The curve is concave down for t in the interval(s) . (If there is more than one interval, use commas to separate them.)
28.
x=t2β1,y=t3βt
29.
x=sec(t),y=tan(t) on (βΟ/2,Ο/2)
30.
For the curve defined by x=cos(t), y=sin(2t), find d2ydx2 (as a function of t). Then determine the intervals (within [0,2Ο)) on which the graph of the curve is concave up/down.
d2ydx2=
The curve is concave up for t in the interval(s) . (If there is more than one interval, use commas to separate them.)
The curve is concave down for t in the interval(s) . (If there is more than one interval, use commas to separate them.)
31.
x=cos(t)sin(2t),y=sin(t)sin(2t) on [βΟ/2,Ο/2]
32.
x=et/10cos(t),y=et/10sin(t)
In the following exercises, find the arc length of the graph of the parametric equations on the given interval(s).
33.
Find the arc length of the graph defined by x=β3sin(2t), y=3cos(2t) on [0,Ο].
34.
Find the arc length of the graph defined by x=et/10cos(t), y=et/10sin(t) on [0,2Ο] and [2Ο,4Ο].
On [0,2Ο], the length is .
On [2Ο,4Ο], the length is .
35.
Find the arc length of the graph defined by x=5t+2, y=1β3t on [β1,1].
36.
x=2t3/2,y=3t on [0,1]
In the following exercises, numerically approximate the given arc length.
37.
Approximate the arc length of one petal of the rose curve x=cos(t)cos(2t),y=sin(t)cos(2t) using Simpson's Rule and n=4.
38.
Approximate the arc length of the βbow tie curveβ x=cos(t),y=sin(2t) using Simpson's Rule and n=6.
39.
Approximate the arc length of the parabola x=t2βt,y=t2+t on [β1,1] using Simpson's Rule and n=4.
40.
A common approximate of the circumference of an ellipse given by x=acos(t),y=bsin(t) is Cβ2Οβa2+b22. Use this formula to approximate the circumference of x=5cos(t), y=3sin(t) and compare this to the approximation given by Simpson's Rule and n=6.
In the following exercises, a solid of revolution is described. Find or approximate its surface area as specified.
41.
Find the surface area of the sphere formed by rotating the circle x=2cos(t),y=2sin(t) about:
the x-axis and
the y-axis.
42.
Find the surface area of the torus (or βdonutβ) formed by rotating the circle x=cos(t)+2,y=sin(t) about the y-axis.
43.
Approximate the surface area of the solid formed by rotating the βupper right halfβ of the bow tie curve x=cos(t),y=sin(2t) on [0,Ο/2] about the x-axis, using Simpson's Rule and n=4.
44.
Approximate the surface area of the solid formed by rotating the one petal of the rose curve x=cos(t)cos(2t),y=sin(t)cos(2t) on [0,Ο/4] about the x-axis, using Simpson's Rule and n=4.