Download Calculus :Polar Coordinates Parametric Equations and more Lecture notes Vector Analysis in PDF only on Docsity!
Polar Coordinates,
Parametric Equations
10.1 Polar Coordinates
Coordinate systems are tools that let us use algebraic methods to understand geometry.
While the rectangular (also called Cartesian) coordinates that we have been using are
the most common, some problems are easier to analyze in alternate coordinate systems.
A coordinate system is a scheme that allows us to identify any point in the plane or
in three-dimensional space by a set of numbers. In rectangular coordinates these numbers
are interpreted, roughly speaking, as the lengths of the sides of a rectangle. In polar
coordinates a point in the plane is identified by a pair of numbers (r, θ). The number θ
measures the angle between the positive x-axis and a ray that goes through the point, as
shown in figure 10.1.1; the number r measures the distance from the origin to the point.
Figure 10.1.1 shows the point with rectangular coordinates (1,
3) and polar coordinates
(2, π/3), 2 units from the origin and π/3 radians from the positive x-axis.
..
..
...
..
..
...
..
...
..
...
..
..
...
..
...
..
..
...
..
...
..
..
...
..
...
..
...
..
..
...
..
..
...
..
...
..
...
..
...
..
..
...
..
..
...
..
...
..
...
..
..
...
..
...
..
..
...
..
...
..
..
...
..
...
..
...
..
..
...
..
..
...
..
...
..
...
..
...
..
..
...
..
..
...
..
...
..
...
.... .. ... .. .. .
... ... ..... .
..
..
..
..
..
...
...
...
Figure 10.1.1 Polar coordinates of the point (1,
240 Chapter 10 Polar Coordinates, Parametric Equations
Just as we describe curves in the plane using equations involving x and y, so can we
describe curves using equations involving r and θ. Most common are equations of the form
r = f (θ).
EXAMPLE 10.1.1 Graph the curve given by r = 2. All points with r = 2 are at
distance 2 from the origin, so r = 2 describes the circle of radius 2 with center at the
origin.
EXAMPLE 10.1.2 Graph the curve given by r = 1 + cos θ. We first consider y =
1 + cos x, as in figure 10.1.2. As θ goes through the values in [0, 2 π], the value of r tracks
the value of y, forming the “cardioid” shape of figure 10.1.2. For example, when θ = π/2,
r = 1 + cos(π/2) = 1, so we graph the point at distance 1 from the origin along the
positive y-axis, which is at an angle of π/2 from the positive x-axis. When θ = 7π/4,
r = 1 + cos(7π/4) = 1 +
2 / 2 ≈ 1 .71, and the corresponding point appears in the fourth
quadrant. This illustrates one of the potential benefits of using polar coordinates: the
equation for this curve in rectangular coordinates would be quite complicated.
.................. ....... ...... ..... ..... .... ..... ... .... .... .... ... .... ... ... ... .... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .... ... ... .... ... ... .... .... ... ..... .... .... ..... ..... ...... ........ ............ ..........
.............
........
......
.....
.....
....
....
....
....
...
....
...
....
...
....
...
...
...
...
....
...
..
....
...
...
...
...
...
...
...
..
....
..
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
....
...
...
....
...
...
....
...
...
....
....
....
....
.....
....
......
......
.......
.................
.
..
..
..
..
..
..
..
...
..
..
..
..
..
..
...
..
..
..
...
..
..
...
..
...
..
...
..
...
...
..
...
...
...
....
...
...
....
....
....
.....
.....
.....
.........
.............
........... .............. ........ ..... ..... ..... ... .... ... ... ... .... .. ... ... .. .. ... .. .. ... .. .. ... .. .. .. .. .. .. .. .. ... .. .. .. .. .. ... .. .. ... ... ... ..... ...... .... ... .. ... .. .. ... .. .. .. .. .. .. ... .. .. .. .. .. .. .. ... .. .. .. ... .. ... .. ... .. ... ... ... ... .... ... .... .... ..... ...... ...... ........... .....................
...........
......
.......
....
.....
....
....
...
....
...
...
...
...
...
...
...
..
...
..
...
..
...
..
...
..
..
...
..
..
..
..
..
...
..
..
..
..
..
..
..
..
...
..
... .. ... ... ... .
.. ... ... ... ..
.... .. ... ... .
.. ... ... ... ..
. ... ... ... .. .
.
.....
.. ....
.
..
..
...
..
..
Figure 10.1.2 A cardioid: y = 1 + cos x on the left, r = 1 + cos θ on the right.
Each point in the plane is associated with exactly one pair of numbers in the rect-
angular coordinate system; each point is associated with an infinite number of pairs in
polar coordinates. In the cardioid example, we considered only the range 0 ≤ θ ≤ 2 π,
and already there was a duplicate: (2, 0) and (2, 2 π) are the same point. Indeed, every
value of θ outside the interval [0, 2 π) duplicates a point on the curve r = 1 + cos θ when
0 ≤ θ < 2 π. We can even make sense of polar coordinates like (− 2 , π/4): go to the direc-
tion π/4 and then move a distance 2 in the opposite direction; see figure 10.1.3. As usual,
a negative angle θ means an angle measured clockwise from the positive x-axis. The point
in figure 10.1.3 also has coordinates (2, 5 π/4) and (2, − 3 π/4).
The relationship between rectangular and polar coordinates is quite easy to under-
stand. The point with polar coordinates (r, θ) has rectangular coordinates x = r cos θ
and y = r sin θ; this follows immediately from the definition of the sine and cosine func-
tions. Using figure 10.1.3 as an example, the point shown has rectangular coordinates
242 Chapter 10 Polar Coordinates, Parametric Equations
EXAMPLE 10.1.6 Graph r = 2 sin θ. Because the sine is periodic, we know that we
will get the entire curve for values of θ in [0, 2 π). As θ runs from 0 to π/2, r increases
from 0 to 2. Then as θ continues to π, r decreases again to 0. When θ runs from π to
2 π, r is negative, and it is not hard to see that the first part of the curve is simply traced
out again, so in fact we get the whole curve for values of θ in [0, π). Thus, the curve looks
something like figure 10.1.5. Now, this suggests that the curve could possibly be a circle,
and if it is, it would have to be the circle x
+ (y − 1)
= 1. Having made this guess, we
can easily check it. First we substitute for x and y to get (r cos θ)
+ (r sin θ − 1)
expanding and simplifying does indeed turn this into r = 2 sin θ.
.
..
..
...
..
..
..
..
..
..
..
..
...
..
..
..
...
..
..
..
...
..
...
..
...
..
...
...
..
...
...
...
...
....
...
....
....
....
....
......
......
.......
..............
.............. ............... ....... ...... ..... ..... .... .... ... .... ... ... ... ... ... ... ... .. ... .. ... .. ... .. .. ... .. .. .. .. ... .. .. .. .. .. .. ... .. .. .. .. .. .. .. .. ... .. .. .. .. .. .. .. .. ... .. .. .. ... .. .. ... .. .. ... .. ... .. ... ... ... .. .... ... ... ... .... ... .... ..... .... ...... ...... ........ .....................................
........
......
......
....
.....
....
...
....
...
...
...
....
..
...
...
...
..
...
..
...
..
..
...
..
..
...
..
..
..
...
..
..
..
..
..
..
..
..
...
..
..
.
Figure 10.1.5 Graph of r = 2 sin θ.
Exercises 10.1.
1. Plot these polar coordinate points on one graph: (2, π/3), (− 3 , π/2), (− 2 , −π/4), (1/ 2 , π),
Find an equation in polar coordinates that has the same graph as the given equation in
rectangular coordinates.
2. y = 3x ⇒ 3. y = − 4 ⇒
4. xy
2
= 1 ⇒ 5. x
2
+ y
2
6. y = x
3
⇒ 7. y = sin x ⇒
8. y = 5x + 2 ⇒ 9. x = 2 ⇒
10. y = x
2
+ 1 ⇒ 11. y = 3x
2
− 2 x ⇒
12. y = x
2
+ y
2
Sketch the curve.
13. r = cos θ 14. r = sin(θ + π/4)
15. r = − sec θ 16. r = θ/2, θ ≥ 0
17. r = 1 + θ
1
2
18. r = cot θ csc θ
19. r =
sin θ + cos θ
20. r
2
= −2 sec θ csc θ
10.2 Slopes in polar coordinates 243
In the exercises below, find an equation in rectangular coordinates that has the same graph as
the given equation in polar coordinates.
21. r = sin(3θ) ⇒ 22. r = sin
2
23. r = sec θ csc θ ⇒ 24. r = tan θ ⇒
10.2 Slopes in polar oordinates
When we describe a curve using polar coordinates, it is still a curve in the x-y plane. We
would like to be able to compute slopes and areas for these curves using polar coordinates.
We have seen that x = r cos θ and y = r sin θ describe the relationship between polar
and rectangular coordinates. If in turn we are interested in a curve given by r = f (θ),
then we can write x = f (θ) cos θ and y = f (θ) sin θ, describing x and y in terms of θ alone.
The first of these equations describes θ implicitly in terms of x, so using the chain rule we
may compute
dy
dx
dy
dθ
dθ
dx
Since dθ/dx = 1/(dx/dθ), we can instead compute
dy
dx
dy/dθ
dx/dθ
f (θ) cos θ + f
(θ) sin θ
−f (θ) sin θ + f
(θ) cos θ
EXAMPLE 10.2.1 Find the points at which the curve given by r = 1 + cos θ has a
vertical or horizontal tangent line. Since this function has period 2π, we may restrict our
attention to the interval [0, 2 π) or (−π, π], as convenience dictates. First, we compute the
slope:
dy
dx
(1 + cos θ) cos θ − sin θ sin θ
−(1 + cos θ) sin θ − sin θ cos θ
cos θ + cos
θ − sin
− sin θ − 2 sin θ cos θ
This fraction is zero when the numerator is zero (and the denominator is not zero). The
numerator is 2 cos
θ + cos θ − 1 so by the quadratic formula
cos θ =
= − 1 or
This means θ is π or ±π/3. However, when θ = π, the denominator is also 0, so we cannot
conclude that the tangent line is horizontal.
Setting the denominator to zero we get
− sin θ − 2 sin θ cos θ = 0
sin θ(1 + 2 cos θ) = 0,
so either sin θ = 0 or cos θ = − 1 /2. The first is true when θ is 0 or π, the second when θ
is 2π/3 or 4π/3. However, as above, when θ = π, the numerator is also 0, so we cannot
10.3 Areas in polar coordinates 245
Sketch the curves over the interval [0, 2 π] unless otherwise stated.
7. r = sin θ + cos θ 8. r = 2 + 2 sin θ
9. r =
+ sin θ 10. r = 2 + cos θ
11. r =
+ cos θ 12. r = cos(θ/2), 0 ≤ θ ≤ 4 π
13. r = sin(θ/3), 0 ≤ θ ≤ 6 π 14. r = sin
2
15. r = 1 + cos
2
(2θ) 16. r = sin
2
17. r = tan θ 18. r = sec(θ/2), 0 ≤ θ ≤ 4 π
19. r = 1 + sec θ 20. r =
1 − cos θ
21. r =
1 + sin θ
22. r = cot(2θ)
23. r = π/θ, 0 ≤ θ ≤ ∞ 24. r = 1 + π/θ, 0 ≤ θ ≤ ∞
25. r =
10.3 Areas in polar oordinates
We can use the equation of a curve in polar coordinates to compute some areas bounded
by such curves. The basic approach is the same as with any application of integration: find
an approximation that approaches the true value. For areas in rectangular coordinates, we
approximated the region using rectangles; in polar coordinates, we use sectors of circles,
as depicted in figure 10.3.1. Recall that the area of a sector of a circle is αr
/2, where
α is the angle subtended by the sector. If the curve is given by r = f (θ), and the angle
subtended by a small sector is ∆θ, the area is (∆θ)(f (θ))
/2. Thus we approximate the
total area as
n− 1
i=
f (θ
i
In the limit this becomes
b
a
f (θ)
dθ.
EXAMPLE 10.3.1 We find the area inside the cardioid r = 1 + cos θ.
(1+cos θ)
dθ =
1+2 cos θ+cos
θ dθ =
(θ + 2 sin θ +
sin 2θ
EXAMPLE 10.3.2 We find the area between the circles r = 2 and r = 4 sin θ, as shown
in figure 10.3.2. The two curves intersect where 2 = 4 sin θ, or sin θ = 1/2, so θ = π/6 or
246 Chapter 10 Polar Coordinates, Parametric Equations
..
..
..
...
..
..
..
...
..
..
..
...
..
..
...
..
...
..
...
..
...
..
...
..
...
...
..
...
...
...
...
...
...
...
...
...
....
...
....
....
....
...
.....
....
.....
.....
.....
......
........
.......
...........
.......................................... ........... ....... ........ ..... ..... ..... .... ...
...
..
..
...
..
...
..
...
..
..
...
..
..
...
..
...
..
...
..
..
...
..
...
..
..
...
..
..
...
..
...
..
...
..
..
...
..
...
..
..
...
..
...
..
..
...
..
...
..
...
..
..
...
..
..
...
..
...
..
..
...
..
...
..
...
..
..
...
..
..
...
..
...
...... .... ... .... .... ... .... ... ... .... ... ... ... ..
...
...
...
...
...
...
...
...
...
...
...
...
..
....
...
..
....
..
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
..
....
..
...
....
..
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
..
....
..
...
....
..
...
...
...
...
...
...
...
...
.... ... ... ... ... .. ... ... .. ... ... .. ... .. ... .. ... .. ... .. .
....
.....
....
....
....
....
....
.....
....
....
....
....
....
.....
....
....
....
....
....
.....
....
....
....
....
....
.....
....
....
....
....
....
.....
....
....
.....
....
....
....
....
....
.....
....
....
....
....
....
.....
....
....
....
....
....
........ .. ... .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. ... .. .. .. .. .
........
........
........
........
........
........
........
.........
.......
........
.........
.......
........
.........
........
.......
.........
........
........
........
........
........
........
........
........
.........
.......
........
.........
Figure 10.3.1 Approximating area by sectors of circles.
5 π/6. The area we want is then
16 sin
θ − 4 dθ =
Figure 10.3.2 An area between curves.
This example makes the process appear more straightforward than it is. Because
points have many different representations in polar coordinates, it is not always so easy to
identify points of intersection.
EXAMPLE 10.3.3 We find the shaded area in the first graph of figure 10.3.3 as the
difference of the other two shaded areas. The cardioid is r = 1 + sin θ and the circle is
r = 3 sin θ. We attempt to find points of intersection:
1 + sin θ = 3 sin θ
1 = 2 sin θ
1 /2 = sin θ.
This has solutions θ = π/6 and 5π/6; π/6 corresponds to the intersection in the first quad-
rant that we need. Note that no solution of this equation corresponds to the intersection
248 Chapter 10 Polar Coordinates, Parametric Equations
14. Find the area inside r = 2 cos θ and outside r = 1. ⇒
15. Find the area inside r = 2 sin θ and above the line r = (3/2) csc θ. ⇒
16. Find the area inside r = θ, 0 ≤ θ ≤ 2 π. ⇒
17. Find the area inside r =
18. Find the area inside both r =
3 cos θ and r = sin θ. ⇒
19. Find the area inside both r = 1 − cos θ and r = cos θ. ⇒
20. The center of a circle of radius 1 is on the circumference of a circle of radius 2. Find the area
of the region inside both circles. ⇒
21. Find the shaded area in figure 10.3.4. The curve is r = θ, 0 ≤ θ ≤ 3 π. ⇒
Figure 10.3.4 An area bounded by the spiral of Archimedes.
10.4 Parametri Equations
When we computed the derivative dy/dx using polar coordinates, we used the expressions
x = f (θ) cos θ and y = f (θ) sin θ. These two equations completely specify the curve,
though the form r = f (θ) is simpler. The expanded form has the virtue that it can easily
be generalized to describe a wider range of curves than can be specified in rectangular or
polar coordinates.
Suppose f (t) and g(t) are functions. Then the equations x = f (t) and y = g(t)
describe a curve in the plane. In the case of the polar coordinates equations, the variable
t is replaced by θ which has a natural geometric interpretation. But t in general is simply
an arbitrary variable, often called in this case a parameter, and this method of specifying
a curve is known as parametric equations. One important interpretation of t is time.
In this interpretation, the equations x = f (t) and y = g(t) give the position of an object
at time t.
10.4 Parametric Equations 249
EXAMPLE 10.4.1 Describe the path of an object that moves so that its position at
time t is given by x = cos t, y = cos
t. We see immediately that y = x
, so the path lies
on this parabola. The path is not the entire parabola, however, since x = cos t is always
between −1 and 1. It is now easy to see that the object oscillates back and forth on the
parabola between the endpoints (1, 1) and (− 1 , 1), and is at point (1, 1) at time t = 0.
It is sometimes quite easy to describe a complicated path in parametric equations
when rectangular and polar coordinate expressions are difficult or impossible to devise.
EXAMPLE 10.4.2 A wheel of radius 1 rolls along a straight line, say the x-axis. A
point on the rim of the wheel will trace out a curve, called a cycloid. Assume the point
starts at the origin; find parametric equations for the curve.
Figure 10.4.1 illustrates the generation of the curve (click on the AP link to see an
animation). The wheel is shown at its starting point, and again after it has rolled through
about 490 degrees. We take as our parameter t the angle through which the wheel has
turned, measured as shown clockwise from the line connecting the center of the wheel
to the ground. Because the radius is 1, the center of the wheel has coordinates (t, 1).
We seek to write the coordinates of the point on the rim as (t + ∆x, 1 + ∆y), where
∆x and ∆y are as shown in figure 10.4.2. These values are nearly the sine and cosine
of the angle t, from the unit circle definition of sine and cosine. However, some care is
required because we are measuring t from a nonstandard starting line and in a clockwise
direction, as opposed to the usual counterclockwise direction. A bit of thought reveals
that ∆x = − sin t and ∆y = − cos t. Thus the parametric equations for the cycloid are
x = t − sin t, y = 1 − cos t.
...
.
...
..
..
..
...
..
..
...
..
...
..
..
...
...
..
...
...
...
...
...
...
....
...
...
....
....
....
....
....
.....
.....
.....
.....
.......
.......
........
.............
............................................... ............. ........ ....... ....... ..... ..... ..... ..... .... .... .... .... .... ... ... .... ... ... ... ... ... ... .. ... ... .. .. ... .. ... .. .. ... .. .. .. ... .. ...
..
...
..
..
..
..
...
..
...
..
..
...
..
...
...
..
...
...
...
...
...
...
....
...
...
....
....
....
....
....
....
......
.....
.....
.......
.......
........
.............
............................................... ............. ........ ....... ....... ..... ..... ..... ..... .... .... .... .... .... ... ... .... ... ... ... ... .. ... ... ... .. ... .. ... .. ... .. .. ... .. .. .. ... .. ...
..
..
...
..
..
..
...
..
..
...
..
...
..
...
...
..
...
...
...
...
...
...
....
...
...
....
....
...
.....
....
....
......
.....
.....
.......
.......
........
.............
............................................... ............. ........ ....... ....... ..... ..... ..... ..... .... .... .... .... .... ... ... .... ... ... ... ... .. ... ... ... .. ... .. ... .. ... .. .. ... .. .. .. ... . ..
. ..
.....
...
..
..
..
...
..
..
...
.......
.. ...... ... ... .. .. ... .. .. .. .. ... ... ... .................
...
....
..
..
...
..
..
..
..
..
...
..
.
t
.... ... .... ... ... .... ... .... ... .... ... ..... .. .. .. .. .. .. .. .. .. .. .. .. .. .. ... .. .. .. .. ............ ..
......
.....
....
...
...
...
...
..
...
..
...
..
..
..
..
..
...
..
..
..
..
..
..
..
..
...
..
..
...
..
...
..
...
...
....
...
.....
......
........................ ...... .... .... .... ... .. ... ... .. ... .. .. .. .. ... .. .. .. .. .. .. .. .. ... .. .. .. ... .. .. ... ... .. .... ... .... ..... ...... ........... ............
......
.....
....
...
...
...
...
..
...
..
...
..
..
..
..
..
...
..
..
..
..
..
..
..
..
...
..
..
...
..
...
..
...
...
....
...
.....
......
........................ ...... .... .... .... ... .. ... ... .. ... .. .. .. .. ... .. .. .. .. .. .. .. .. ... .. .. .. ... .. .. ... ... .. .... ... .... ..... ...... ...........
Figure 10.4.1 A cycloid. (AP)
∆x
∆y
... ... .... ... .... ... .... ... .... ... .... ... .... ... ... .... .... ... .... ... ... .... ... .... .... .... .. .. .. .. .. .. .. .. .. .. ... .. .. .. .. .. .. .. .. .. .. .. .. .. .. ... .. .. .. .. .. .. .. .. .. .. .. .. .. .. ... .. ..............
..........
........
.....
.....
.....
....
....
....
...
...
...
...
...
...
...
..
...
...
..
...
..
..
...
..
..
...
..
..
..
...
..
..
..
..
..
..
..
...
..
..
..
..
..
..
..
..
...
..
..
..
..
..
..
..
..
...
..
..
...
..
..
...
..
..
...
..
...
...
..
...
...
...
..
....
...
...
....
....
...
.....
.....
.....
........
.........
.............................. ......... ....... ...... .... ..... .... .... .... ... ... ... ... ... ... ... .. ... .. ... .. ... .. ... .. .. ... .. .. .. .. .. ... .. .. .. .. .. .. .. .. ... .. .. .. .. .. .. .. .. ... .. .. .. .. .. ... .. .. .. ... .. .. ... .. ... .. ... .. ... ... ... .. ... .... ... ... .... .... .... .... ..... ...... ....... ........... ............
Figure 10.4.2 The wheel.
10.5 Calculus with Parametric Equations 251
.......... ....... ..... ... .... ... ... ... ... .. .. ... .. .. .. .. ... .. .. .. .. .. .. .. .. ... .. .. ... .. .. ... .. ... ... .... ... ..... ..... ........ ............
........
.....
.....
...
...
....
..
...
...
..
..
...
..
..
..
..
..
...
..
..
..
..
..
..
..
..
...
..
..
...
..
...
...
..
...
....
....
.....
.......
.........
...
.....
...
...
...
...
..
...
..
..
..
...
..
..
..
..
..
..
..
...
..
..
..
..
..
..
..
..
...
..
..
..
..
...
..
..
..
...
..
...
..
...
..
...
...
..
....
..
...
....
...
...
....
...
....
.....
....
.....
.....
......
.......
..........
....................
....... .................... .......... ........ ...... ...... ..... ..... ..... .... .... .... .... .... ... .... ... .... ... ... ... ... ... ... .. ... ... ... .. ... ... .. ... .. ... .. ... .. ... .. .. ... .. .. ... .. .. ... .. .. ... .. .. .. .. .. ... .. .. .. .. .. ... .. .. .. .. .. ... .. .. .. .. .. .. .. .. ... .. .. .. .. .. .. .. .. .. .. .. ... .. .. .. .. .. .. .. .. ... .. .. .. .. .. .. .. ... .. .. .. .. .. ... .. .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. ... .. ... .. ... ..
.. .....
.. ....
.... .. .. ....
.
.
....
.
.
...
..
.
...
..
....
..
...
...
...
...
..
....
..
...
.
..
...
.
.
...
..
.
...
..
....
..
...
...
...
...
..
...
.
..
...
.
..
...
.
..
.....
.
.....
.
.....
.
.....
.
.....
.
.....
.
.....
.
.....
..
.
Figure 10.4.4 An involute of a circle.
10.5 Cal ulus with Parametri Equations
We have already seen how to compute slopes of curves given by parametric equations—it
is how we computed slopes in polar coordinates.
EXAMPLE 10.5.1 Find the slope of the cycloid x = t−sin t, y = 1−cos t. We compute
x
= 1 − cos t, y
= sin t, so
dy
dx
sin t
1 − cos t
Note that when t is an odd multiple of π, like π or 3π, this is (0/2) = 0, so there is
a horizontal tangent line, in agreement with figure 10.4.1. At even multiples of π, the
fraction is 0/0, which is undefined. The figure shows that there is no tangent line at such
points.
Areas can be a bit trickier with parametric equations, depending on the curve and the
area desired. We can potentially compute areas between the curve and the x-axis quite
easily.
EXAMPLE 10.5.2 Find the area under one arch of the cycloid x = t−sin t, y = 1−cos t.
We would like to compute
y dx,
but we do not know y in terms of x. However, the parametric equations allow us to make
a substitution: use y = 1 − cos t to replace y, and compute dx = (1 − cos t) dt. Then the
integral becomes
(1 − cos t)(1 − cos t) dt = 3π.
252 Chapter 10 Polar Coordinates, Parametric Equations
Note that we need to convert the original x limits to t limits using x = t − sin t. When
x = 0, t = sin t, which happens only when t = 0. Likewise, when x = 2π, t − 2 π = sin t
and t = 2π. Alternately, because we understand how the cycloid is produced, we can see
directly that one arch is generated by 0 ≤ t ≤ 2 π. In general, of course, the t limits will
be different than the x limits.
This technique will allow us to compute some quite interesting areas, as illustrated by
the exercises.
As a final example, we see how to compute the length of a curve given by parametric
equations. Section 9.9 investigates arc length for functions given as y in terms of x, and
develops the formula for length:
b
a
dy
dx
dx.
Using some properties of derivatives, including the chain rule, we can convert this to use
parametric equations x = f (t), y = g(t):
b
a
dy
dx
dx =
b
a
dx
dt
dx
dt
dy
dx
dt
dx
dx
v
u
dx
dt
dy
dt
dt
v
u
(f
(t))
+ (g
(t))
dt.
Here u and v are the t limits corresponding to the x limits a and b.
EXAMPLE 10.5.3 Find the length of one arch of the cycloid. From x = t − sin t,
y = 1 − cos t, we get the derivatives f
= 1 − cos t and g
= sin t, so the length is
(1 − cos t)
+ sin
t dt =
2 − 2 cos t dt.
Now we use the formula sin
(t/2) = (1 − cos(t))/2 or 4 sin
(t/2) = 2 − 2 cos t to get
4 sin
(t/2) dt.
Since 0 ≤ t ≤ 2 π, sin(t/2) ≥ 0, so we can rewrite this as
2 sin(t/2) dt = 8.
