Section 9.2 Calculus with Parametric Curves
Calculus with Parametric Curves
Suppose f and g are differentiable functions and we
parametric curve x = f (t), y = g(t) where y is also a
gives
dy
dy
=
dt
dx
If dx/dt 6= 0, we can solve for dy/dx:
dy
dy
= dt
dx
dx
dt
if
want to find the tangent line at a point on the
differentiable function of x. Then the Chain Rule
·
dx
dt
dx
6= 0
dt
(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
Section 9.2 Calculus with Parametric Curves
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:
√
3. Therefore, the point (3, 0) on C arises from two values of the
(a) Notice that √
x = t2 = 3 when
t
=
±
√
parameter, t = 3 and t = − 3. This indicates that the curve C crosses itself at (3, 0). Since
dy
1
dy/dt
3t2 − 3
3
t−
=
=
=
dx
dx/dt
2t
2
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.) The
corresponding point on C is (0, 0).
(c) To determine concavity we calculate the second derivative:
d dy
3
1
1+ 2
3(t2 + 1)
d2 y
dt dx
2
t
=
=
=
dx
dx2
2t
4t3
dt
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
Section 9.2 Calculus with Parametric Curves
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
dy/dθ
r sin θ
sin θ
=
=
=
dx
dx/dθ
r(1 − cos θ)
1 − cos θ
When θ = π/3, we have
π
=r
− sin
x=r
3
3
π
√ !
π
3
,
−
3
2
π r
y = r 1 − cos
=
3
2
and
√
dy
sin(π/3)
3/2 √
=
=
= 3
dx
1 − cos(π/3)
1 − 12
√
Therefore the slope of the tangent is 3 and its equation is
√ !
√
rπ r 3
π
r √
y− = 3 x−
or
+
3x − y = r √ − 2
2
3
2
3
(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 tangents
at those points. We can verify this by using l’Hospital’s Rule as follows:
lim +
θ→2nπ
dy
sin θ
cos θ
= lim +
= lim +
=∞
dx θ→2nπ 1 − cos θ θ→2nπ sin θ
A similar computation shows that dy/dx → −∞ as θ → 2nπ − , so indeed there are vertical tangents when
θ = 2nπ, that is, when x = 2nπr.
3
Section 9.2 Calculus with Parametric Curves
Areas
We know that the area under a curve y = F (x) from a to b is A =
Z
a
b
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 an
area formula by using the Substitution Rule for Definite Integrals as follows:
A=
or
Z
Z
b
ydx =
a
Z
β
g(t)f ′(t)dt
α
α
′
g(t)f (t)dt if (f (β), g(β)) is the leftmost endpoint
β
EXAMPLE 3: Find the area under one arch of the cycloid
x = r(θ − sin θ),
y = r(1 − cos θ)
4
Section 9.2 Calculus with Parametric Curves
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
Z 2πr
Z 2π
A=
ydx =
r(1 − cos θ)r(1 − cos θ)dθ
0
0
=r
=r
2
2
Z
Z
2π
(1 − cos θ)2 dθ
0
2π
(1 − 2 cos θ + cos2 θ)dθ
0
2π
1
1 − 2 cos θ + (1 + cos 2θ) dθ
=r
2
0
2π
1
2 3
=r
θ − 2 sin θ + sin 2θ
2
4
0
3
· 2π = 3πr 2
= r2
2
2
Z
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, if
F ′ is continuous, then
s
2
Z b
dy
L=
1+
(2)
dx
dx
a
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 to
β and f (α) = a, f (β) = b. Putting Formula 1 into Formula 2 and using the Substitution Rule, we obtain
s
s
2
2
Z b
Z β
dy
dy/dt dx
L=
1+
1+
dt
dx =
dx
dx/dt dt
a
α
Since dx/dt > 0, we have
L=
Z
β
α
s
dx
dt
2
5
+
dy
dt
2
dt
(3)
Section 9.2 Calculus with Parametric Curves
Even if C can’t be expressed in the form y = F (x), Formula 3 is still valid but we obtain it by polygonal approximations. We divide the parameter interval [α, β] into n subintervals of equal width ∆t. If
t0 , t1 , t2 , . . . , tn are the endpoints of these subintervals, then xi = f (ti ) and yi = g(ti) are the coordinates
of 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 polygons
as n → ∞:
n
X
L = lim
|Pi−1 Pi |
n→∞
i=1
The Mean Value Theorem, when applied to f on the interval [ti−1 , ti ], gives a number t∗i in (ti−1 , ti ) such
that
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−1 Pi | =
and so
p
(∆xi )2 + (∆yi )2 =
p
p
2 =
2
[f ′ (t∗i )∆t]2 + [g ′ (t∗∗
)∆t]
[f ′ (t∗i )]2 + [g ′(t∗∗
i
i )] ∆t
L = lim
n→∞
n
X
p
2
[f ′ (t∗i )]2 + [g ′ (t∗∗
i )] ∆t
(4)
i=1
p
The sum in (4) resembles a Riemann sum for the function [f ′ (t)]2 + [g ′(t)]2 but it is not exactly a
′
′
Riemann sum because t∗i 6= t∗∗
i in general. Nevertheless, if f and g are continuous, it can be shown that
∗
∗∗
the limit in (4) is the same as if ti and ti were equal, namely,
L=
Z
β
α
p
[f ′ (t)]2 + [g ′ (t)]2 dt
Thus, using Leibniz notation, we have the following result, which has the same form as (3).
6
Section 9.2 Calculus with Parametric Curves
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 length
of C is
s
Z β 2 2
dy
dx
L=
+
dt
dt
dt
α
Notice that the formula in this Theorem is consistent with the general formulas
Z
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
s
Z 2π
Z 2π 2 2
Z 2π p
dy
dx
2
2
sin t + cos tdt =
dt = 2π
L=
+
dt =
dt
dt
0
0
0
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
s
Z 2π 2 2
Z 2π p
Z 2π
dx
dy
2
+
dt =
4 cos2 2t + 4 sin 2tdt =
2dt = 4π
dt
dt
0
0
0
REMARK: Notice that the integral gives twice the arc length of the circle because as t increases from 0
to 2π, the point (sin 2t, cos 2t) traverses the circle twice. In general, when finding the length of a curve
C from a parametric representation, we have to be careful to ensure that C is traversed only once as t
increases 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
Section 9.2 Calculus with Parametric Curves
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
= r(1 − cos θ) and
dθ
we have
L =
=
=
Z
Z
Z
=r
2π
0
2π
0
2π
0
Z
0
s
dx
dθ
2
+
dy
dθ
dy
= r sin θ
dθ
2
dθ
q
r 2 (1 − cos θ)2 + r 2 sin2 θdθ
q
r 2 (1 − 2 cos θ + cos2 θ + sin2 θ)dθ
2π
p
2(1 − cos θ)dθ
1
To evaluate this integral we use the identity sin2 x = (1 − cos 2x) with θ = 2x, which gives
2
1 − cos θ = 2 sin2 (θ/2)
Since 0 ≤ θ ≤ 2π, we have 0 ≤ θ/2 ≤ π and so sin(θ/2) ≥ 0. Therefore
q
p
2(1 − cos θ) = 4 sin2 (θ/2) = 2| sin(θ/2)| = 2 sin(θ/2)
and so
L = 2r
Z
2π
sin(θ/2)dθ = 2r[−2 cos(θ/2)]2π
0 = 2r[2 + 2] = 8r
0
EXAMPLE 6: Find the length of the astroid x = cos3 t, y = sin3 t, 0 ≤ t ≤ 2π.
8
Section 9.2 Calculus with Parametric Curves
EXAMPLE 6: Find the length of the astroid x = cos3 t, y = sin3 t, 0 ≤ t ≤ 2π.
Solution: Since
dy
= 3 sin2 t cos t
dt
dx
= 3 cos2 t(− sin t) and
dt
we have
Z
π/2
0
s
dx
dt
2
+
dy
dt
2
dt =
=
=
=
=
It follows that
Z
Z
Z
Z
Z
π/2
0
π/2
0
π/2
0
p
9 cos4 t sin2 t + 9 sin4 t cos2 tdt
q
9 sin2 t cos2 t(cos2 t + sin2 t)dt
p
9 sin2 t cos2 tdt
π/2
|3 sin t cos t|dt
0
π/2
0


1
Z 1
sin t = u
3u2
3


3 sin t cos tdt = d(sin t) = du =
3udu =
=
2 0 2
0
cos tdt = du
Length of first-quadrant portion =
Z
0
Therefore the length of the astroid is four times this: 4 ·
9
π/2
s
3
= 6.
2
dx
dt
2
+
dy
dt
2
dt =
3
2