Section 10.2 Calculus with Parametric Curves
Ruipeng Shen
March 6, 10
1
Derivatives of y respect to x
Tangents
Let us consider a curve given by a parametric equation
x = x(t)
y = y(t)
Here both x and y are differentiable functions of t. Let us assume that y is a differentiable
function of x. Chain rule tells us
dy dx
dy
=
·
.
dt
dx dt
dx
Thus if
6= 0, we have
dt
dy
dy
= dt .
(1)
dx
dx
dt
dx
dy
The tangent is vertical if
= 0 but
6= 0.
dt
dt
Second derivative d2 y/dx2
We can calculate
d
d2 y
=
2
dx
dx
dy
dx
=
d
dt
dy
dx
.
dx
dt
dy
dy/dt
=
. Using prime notation for the derivatives with respect to t and the equation
dx
dx/dt
(1), we can write
d2 y
y 00 (t)x0 (t) − x00 (t)y 0 (t)
[y 0 (t)/x0 (t)]0
=
=
.
2
0
dx
x (t)
[x0 (t)]3
Here
Example 1. A curve C is defined by the parametric equation
x = t2
y = t3 − 3t
(a) Show that C has two tangents at the point (3, 0) and find their equations.
1
(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.
3
t=-1
2
1
⎯⎯
t=±√3
-2
-1
t=0
0
1
2
3
4
5
6
7
-1
-2
t=1
-3
Figure 1: Curve Sketching for Example 1
Solution
(a) By solving the equation
x = t2 = 3
y = t3 − 3t = 0
√
we obtain two solutions t = ± 3. The slope of tangent is given as follows in term of t:
(t3 − 3t)0
3t2 − 3
dy
=
=
.
2
0
dx
(t )
2t
√
Plugging in t = ± 3, we have
√
dy = 3;
√
dx t= 3
√
dy = − 3.
√
dx t=− 3
√
Thus the equation of tangent lines are given by y = ± 3(x − 3).
(b) It has a horizontal tangent line if dy/dx = 0. This is equivalent to 3t2 − 3 = 0 =⇒ t = ±1.
Thus we have two points (1, 2) and (1, −2). In order to determine where the tangent line is
dx
vertical, we need to solve the equation
= 2t = 0. Thus we have only one point (0, 0).
dt
(c) The second derivative is given by
d dy
d 3t2 − 3
3
1
1+ 2
d2 y
dt dx
dt
2t
2
t
=
=
=
2
dx
d 2
dx
2t
(t )
dt
dt
2
This is positive if t > 0 and negative if t < 0. Therefore the curve is concave upward when t > 0
and downward when t < 0.
Example 2. Find the tangent to the cycloid
x = r(t − sin t)
y = r(1 − cos t)
at the point where t = t0 .
1
-4
-3
-2
-1
0
1
2
3
4
5
-1
Figure 2: The Cycloid
Solution
By our basic formula, we have
dy/dt
r sin t
sin t
dy
=
=
=
dx
dx/dt
r(1 − cos t)
1 − cos t
Thus when t = t0 , the tangent is
y − r(1 − cos t0 ) =
2
sin t0
[x − r(t0 − sin t0 )]
1 − cos t0
⇐⇒
y=
sin t0
rt0 sin t0
· x + 2r −
1 − cos t0
1 − cos t0
Area
Let us consider the parametric curve
x = x(t)
,
y = y(t)
α ≤ t ≤ β.
Assume x(t) increases and y(t) ≥ 0. Then we have that the area of the region bounded below
the curve is given by
Z x(β)
Z β
A=
y dx =
y(t)x0 (t) dt.
x(α)
α
Example 3. Find the area under one arch of the cycloid
x = r(t − sin t)
y = r(1 − cos t)
3
1.2
0.8
0.4
-0.8
-0.4
0
0.4
0.8
1.2
1.6
2
2.4
2.8
3.2
3.6
4
Figure 3: The Cycloid
Solution
Let us consider the arch starting at t = 0 and ending at t = 2π. We can integrate
Z 2π
Z 2π
0
A=
y(t)x (t) dt =
r(1 − cos t)[r(t − sin t)]0 dt
0
0
2π
Z
r(1 − cos t) · r(1 − cos t) dt = r
=
2
Z
0
=r2
2π
(1 − 2 cos t + cos2 t) dt
0
2π
Z
0
2
1
3
− 2 cos t + cos 2t
2
2
dt = r2
3
1
t − 2 sin t + sin 2t
2
4
2π
0
=3πr .
3
Arc Length
In a small piece of curve we have (see Figure 4)
p
p
|P\
(∆x)2 + (∆y)2 = [x(ti ) − x(ti−1 )]2 + [y(ti ) − y(ti−1 )]2
i−1 Pi | ≈
p
p
≈ [x0 (ti−1 )∆t]2 + [y 0 (ti−1 )∆t]2 = [x0 (ti−1 )]2 + [y 0 (ti−1 )]2 · ∆t
Taking a sum, we have
L≈
n
X
|P\
i−1 Pi | ≈
i=1
n p
X
[x0 (ti−1 )]2 + [y 0 (ti−1 )]2 · ∆t
i=1
Finally we can take a limit and obtain
Z
β
L=
α
s
dx
dt
2
+
dy
dt
2
dt.
Example 4. Use the following parametric equation of the unit circle to calculate its arc length.
x = r cos t
, t ∈ [0, 2π].
y = r sin t
4
Pi
ti=ti-1+Δt
Δy
Pi-1
ti-1
Δx
ti
ti-1
Figure 4: Arc Length
Solution
By the basic formula we have
s
Z 2π 2 2
Z 2π p
dx
dy
L=
(−r sin t)2 + (r cos t)2 dt
+
dt =
dt
dt
0
0
Z 2π
=
r dt = 2πr.
0
Example 5. Find the area under one arch of the cycloid
x = r(t − sin t)
y = r(1 − cos t)
Solution
we have
Let us consider the arch starting at t = 0 and ending at t = 2π. By the basic formula
2π
Z
s
L=
0
Z
=r
0
2π
√
dx
dt
2
+
dy
dt
2
Z
dt =
2π
p
[r(1 − cos t)]2 + [r sin t]2 dt
0
Z
2 − 2 cos t dt = r
0
2π
2π
t
t
= 8r.
2 sin dt = r −4 cos
2
2 0
5
4
Surface Area
Let C be a curve defined by a parametric equation
x = x(t)
, t ∈ [α, β].
y = y(t)
We can obtain a surface of revolution if we revolves the curve C about the x-axis. Its area can
be calculated as
s 2
Z
Z β
2
dx
dy
2πy(t)
S = 2πy ds =
+
dt.
dt
dt
α
Example 6. Show that the surface area of a sphere of radius r is 4πr2 . Note that the sphere
can be obtained by rotating the upper half circle
x = r cos t
, t ∈ [0, π].
y = r sin t
about the x-axis.
Solution
We can calculate
s 2
Z π
Z π
q
2
dy
dx
2
2
+
dt =
2πr sin t (−r sin t) + (r cos t) dt
S=
2πy(t)
dt
dt
0
0
Z π
2πr2 sin t dt = 4πr2 .
=
0
6