calculus with parametric curves

Section 9.2 Calculus with Parametric Curves

Calculus with Parametric Curves

Suppose f and g are differentiable functions and we want to find the tangent line at a point on theparametric curve x = f(t), y = g(t) where y is also a differentiable function of x. Then the Chain Rulegives

dy

dt=

dy

dx· dx

dt

If dx/dt 6= 0, we can solve for dy/dx:

dy

dx=

dy

dtdx

dt

ifdx

dt6= 0 (1)

It can be seen from (1) that the curve has a horizontal tangent when dy/dt = 0 (provided that dx/dt 6= 0)and it has a vertical tangent when dx/dt = 0 (provided that dy/dt 6= 0).

EXAMPLE 1: A curve C is defined by the parametric equations x = t2, y = t3 − 3t.

(a) Show that C has two tangents at the point (3, 0) and find their equations.

(b) Find the points on C where the tangent is horizontal or vertical.

(c) Determine where the curve is concave upward or downward.

(d) Sketch the curve.

1


EXAMPLE 1: A curve C is defined by the parametric equations x = t2, y = t3 − 3t.

(a) Show that C has two tangents at the point (3, 0) and find their equations.

(b) Find the points on C where the tangent is horizontal or vertical.

(c) Determine where the curve is concave upward or downward.

(d) Sketch the curve.

Solution:

(a) Notice that x = t2 = 3 when t = ±√

3. Therefore, the point (3, 0) on C arises from two values of theparameter, t =

√3 and t = −

√3. This indicates that the curve C crosses itself at (3, 0). Since

dy

dx=

dy/dt

dx/dt=

3t2 − 3

2t=

3

2

(

t − 1

t

)

the slope of the tangent when t = ±√

3 is dy/dx = ±6/(2√

3) = ±√

3, so the equations of the tangents at(3, 0) are

y =√

3(x − 3) and y = −√

3(x − 3)

(b) C has a horizontal tangent when dy/dx = 0, that is, when dy/dt = 0 and dx/dt 6= 0. Since dy/dt =3t2 − 3, this happens when t2 = 1, that is, t = ±1. The corresponding points on C are (1,−2) and(1, 2). C has a vertical tangent when dx/dt = 2t = 0, that is, t = 0. (Note that dy/dt 6= 0 there.) Thecorresponding point on C is (0, 0).

(c) To determine concavity we calculate the second derivative:

d2y

dx2=

d

dt

(

dy

dx

)

dx

dt

=

3

2

(

1 +1

t2

)

2t=

3(t2 + 1)

4t3

Thus the curve is concave upward when t > 0 and concave downward when t < 0.

(d) Using the information from parts (b) and (c), we sketch C:

EXAMPLE 2:

(a) Find the tangent to the cycloid x = r(θ − sin θ), y = r(1 − cos θ) at the point where θ = π/3.

(b) At what points is the tangent horizontal? When is it vertical?

2


EXAMPLE 2:

(a) Find the tangent to the cycloid x = r(θ − sin θ), y = r(1 − cos θ) at the point where θ = π/3.

(b) At what points is the tangent horizontal? When is it vertical?

Solution:

(a) The slope of the tangent line is

dy

dx=

dy/dθ

dx/dθ=

r sin θ

r(1 − cos θ)=

sin θ

1 − cos θ

When θ = π/3, we have

x = r(π

3− sin

π

3

)

= r

(

π

3−

√3

2

)

, y = r(

1 − cosπ

3

)

=r

2

anddy

dx=

sin(π/3)

1 − cos(π/3)=

√3/2

1 − 1

2

=√

3

Therefore the slope of the tangent is√

3 and its equation is

y − r

2=

√3

(

x − rπ

3+

r√

3

2

)

or√

3x − y = r

(

π√3− 2

)

(b) The tangent is horizontal when dy/dx = 0, which occurs when sin θ = 0 and 1 − cos θ 6= 0, that is,θ = (2n − 1)π, n an integer. The corresponding point on the cycloid is ((2n − 1)πr, 2r).

When θ = 2nπ, both dx/dθ and dy/dθ are 0. It appears from the graph that there are vertical tangentsat those points. We can verify this by using l’Hospital’s Rule as follows:

limθ→2nπ+

dy

dx= lim

θ→2nπ+

sin θ

1 − cos θ= lim

θ→2nπ+

cos θ

sin θ= ∞

A similar computation shows that dy/dx → −∞ as θ → 2nπ−, so indeed there are vertical tangents whenθ = 2nπ, that is, when x = 2nπr.

3


Areas

We know that the area under a curve y = F (x) from a to b is A =

∫ b

a

F (x)dx, where F (x) ≥ 0. If the

curve is given by the parametric equations x = f(t) and y = g(t), α ≤ t ≤ β, then we can calculate anarea formula by using the Substitution Rule for Definite Integrals as follows:

A =

∫ b

a

ydx =

∫ β

α

g(t)f ′(t)dt

[

or

∫ α

β

g(t)f ′(t)dt if (f(β), g(β)) is the leftmost endpoint

]

EXAMPLE 3: Find the area under one arch of the cycloid

x = r(θ − sin θ), y = r(1 − cos θ)

4


EXAMPLE 3: Find the area under one arch of the cycloid

x = r(θ − sin θ), y = r(1 − cos θ)

Solution: One arch of the cycloid is given by 0 ≤ θ ≤ 2π. Using the Substitution Rule with y = r(1−cos θ)and dx = r(1 − cos θ)dθ, we have

A =

∫

2πr

0

ydx =

∫

2π

0

r(1 − cos θ)r(1 − cos θ)dθ

= r2

∫

2π

0

(1 − cos θ)2dθ

= r2

∫

2π

0

(1 − 2 cos θ + cos2 θ)dθ

= r2

∫

2π

0

[

1 − 2 cos θ +1

2(1 + cos 2θ)

]

dθ

= r2

[

3

2θ − 2 sin θ +

1

4sin 2θ

]2π

0

= r2

(

3

2· 2π

)

= 3πr2

Arc Length

We already know how to find the length L of a curve C given in the form y = F (x), a ≤ x ≤ b. In fact, ifF ′ is continuous, then

L =

∫ b

a

√

1 +

(

dy

dx

)2

dx (2)

Suppose that C can also be described by the parametric equations x = f(t) and y = g(t), α ≤ t ≤ β,where dx/dt = f ′(t) > 0. This means that C is traversed once, from left to right, as t increases from α toβ and f(α) = a, f(β) = b. Putting Formula 1 into Formula 2 and using the Substitution Rule, we obtain

L =

∫ b

a

√

1 +

(

dy

dx

)2

dx =

∫ β

α

√

1 +

(

dy/dt

dx/dt

)2dx

dtdt

Since dx/dt > 0, we have

L =

∫ β

α

√

(

dx

dt

)2

+

(

dy

dt

)2

dt (3)

5


Even if C can’t be expressed in the form y = F (x), Formula 3 is still valid but we obtain it by polyg-onal approximations. We divide the parameter interval [α, β] into n subintervals of equal width ∆t. Ift0, t1, t2, . . . , tn are the endpoints of these subintervals, then xi = f(ti) and yi = g(ti) are the coordinatesof points Pi(xi, yi) that lie on C and the polygon with vertices P0, P1, . . . , Pn approximates C.

As in Section 7.4, we define the length L of C to be the limit of the lengths of these approximating polygonsas n → ∞:

L = limn→∞

n∑

i=1

|Pi−1Pi|

The Mean Value Theorem, when applied to f on the interval [ti−1, ti], gives a number t∗i in (ti−1, ti) suchthat

f(ti) − f(ti−1) = f ′(t∗i )(ti − ti−1)

If we let ∆xi = xi − xi−1 and ∆yi = yi − yi−1, this equation becomes

∆xi = f ′(t∗i )∆t

Similarly, when applied to g, the Mean Value Theorem gives a number t∗∗i in (ti−1, ti) such that

∆yi = g′(t∗∗i )∆t

Therefore

|Pi−1Pi| =√

(∆xi)2 + (∆yi)2 =√

[f ′(t∗i )∆t]2 + [g′(t∗∗i )∆t]2 =√

[f ′(t∗i )]2 + [g′(t∗∗i )]2∆t

and so

L = limn→∞

n∑

i=1

√

[f ′(t∗i )]2 + [g′(t∗∗i )]2∆t (4)

The sum in (4) resembles a Riemann sum for the function√

[f ′(t)]2 + [g′(t)]2 but it is not exactly aRiemann sum because t∗i 6= t∗∗i in general. Nevertheless, if f ′ and g′ are continuous, it can be shown thatthe limit in (4) is the same as if t∗i and t∗∗i were equal, namely,

L =

∫ β

α

√

[f ′(t)]2 + [g′(t)]2dt

Thus, using Leibniz notation, we have the following result, which has the same form as (3).

6


THEOREM: If a curve C is described by the parametric equations x = f(t), y = g(t), α ≤ t ≤ β, where f ′

and g′ are continuous on [α, β] and C is traversed exactly once as t increases from α to β, then the lengthof C is

L =

∫ β

α

√

(

dx

dt

)2

+

(

dy

dt

)2

dt

Notice that the formula in this Theorem is consistent with the general formulas

L =

∫

ds and (ds)2 = (dx)2 + (dy)2

of Section 7.4.

EXAMPLE 4: If we use the representation of the unit circle

x = cos t, y = sin t 0 ≤ t ≤ 2π

then dx/dt = − sin t and dy/dt = cos t, so the above Theorem gives

L =

∫

2π

0

√

(

dx

dt

)2

+

(

dy

dt

)2

dt =

∫

2π

0

√

sin2 t + cos2 tdt =

∫

2π

0

dt = 2π

as expected. If, on the other hand, we use the representation

x = sin 2t, y = cos 2t 0 ≤ t ≤ 2π

then dx/dt = 2 cos 2t, dy/dt = −2 sin 2t, and the integral in the above Theorem gives

∫

2π

0

√

(

dx

dt

)2

+

(

dy

dt

)2

dt =

∫

2π

0

√

4 cos2 2t + 4 sin2 2tdt =

∫

2π

0

2dt = 4π

REMARK: Notice that the integral gives twice the arc length of the circle because as t increases from 0to 2π, the point (sin 2t, cos 2t) traverses the circle twice. In general, when finding the length of a curveC from a parametric representation, we have to be careful to ensure that C is traversed only once as tincreases from α to β.

EXAMPLE 5: Find the length of one arch of the cycloid x = r(θ − sin θ), y = r(1 − cos θ).

7


EXAMPLE 5: Find the length of one arch of the cycloid x = r(θ − sin θ), y = r(1 − cos θ).

Solution: From Example 3 we see that one arch is described by the parameter interval 0 ≤ θ ≤ 2π. Since

dx

dθ= r(1 − cos θ) and

dy

dθ= r sin θ

we have

L =

∫

2π

0

√

(

dx

dθ

)2

+

(

dy

dθ

)2

dθ

=

∫

2π

0

√

r2(1 − cos θ)2 + r2 sin2 θdθ

=

∫

2π

0

√

r2(1 − 2 cos θ + cos2 θ + sin2 θ)dθ

= r

∫

2π

0

√

2(1 − cos θ)dθ

To evaluate this integral we use the identity sin2 x =1

2(1 − cos 2x) with θ = 2x, which gives

1 − cos θ = 2 sin2(θ/2)

Since 0 ≤ θ ≤ 2π, we have 0 ≤ θ/2 ≤ π and so sin(θ/2) ≥ 0. Therefore

√

2(1 − cos θ) =√

4 sin2(θ/2) = 2| sin(θ/2)| = 2 sin(θ/2)

and so

L = 2r

∫

2π

0

sin(θ/2)dθ = 2r[−2 cos(θ/2)]2π0

= 2r[2 + 2] = 8r

EXAMPLE 6: Find the length of the astroid x = cos3 t, y = sin3 t, 0 ≤ t ≤ 2π.

8


EXAMPLE 6: Find the length of the astroid x = cos3 t, y = sin3 t, 0 ≤ t ≤ 2π.

Solution: Sincedx

dt= 3 cos2 t(− sin t) and

dy

dt= 3 sin2 t cos t

we have

∫ π/2

0

√

(

dx

dt

)2

+

(

dy

dt

)2

dt =

∫ π/2

0

√

9 cos4 t sin2 t + 9 sin4 t cos2 tdt

=

∫ π/2

0

√

9 sin2 t cos2 t(cos2 t + sin2 t)dt

=

∫ π/2

0

√

9 sin2 t cos2 tdt

=

∫ π/2

0

|3 sin t cos t|dt

=

∫ π/2

0

3 sin t cos tdt =

sin t = ud(sin t) = ducos tdt = du

=

∫

1

0

3udu =3u2

2

]1

0

=3

2

It follows that

Length of first-quadrant portion =

∫ π/2

0

√

(

dx

dt

)2

+

(

dy

dt

)2

dt =3

2

Therefore the length of the astroid is four times this: 4 · 3

2= 6.

9

calculus with parametric curves

Documents