Math 21a
Homework 23 Solutions
Spring, 2014
1. Consider the curve parameterized by
r(t) =
√
2ti + et j + e−t k,
0 ≤ t ≤ 1.
(a) Sketch the projections of r(t) in the xy-, xz-, and yz-planes.
Solution:
Here are the projections of r(t) onto the three coordinate planes:
onto the xy-plane
onto the yz-plane
onto the xz-plane
(b) (Stewart 10.3 #3 ) Find the length of the curve.
2
Hint: What is et + e−t ?
√
√
Solution: Since r(t) = h 2t, et , e−t i, we find that r0 (t) = h 2, et , −e−t i and so
q√
p
p
|r0 (t)| = ( 2)2 + (et )2 + (−e−t )2 = 2 + e2t + e−2t = (et + e−t )2 = et + e−t .
(Notice that we can take this square root because et + e−t > 0.) Then the length of the curve is
Z 1
Z 1
i1
h
L=
|r0 (t)|dt =
(et + e−t )dt = et − e−t = e − e−1 .
0
0
0
2. (Based on Stewart 10.3 #8 ) Consider the curve parametrized by
r(t) = ht, ln t, t ln ti,
1 ≤ t ≤ 2.
(a) Plot r(t) using the command:
ParametricPlot3D[{t, Log[t], t Log[t]}, {t,1,2}]
Note: Be careful to leave a space between “t” and “Log[t]” in the third component of r(t). Mathematica allows
you to define variables and functions with names longer than a single character. Two letters together, such as
“xy”, would be interpreted as such a name (as would “tLog[t]”), rather than the product of “x” and “y”. For
the product, you need to type “x y” (or “x*y”).
Solution:
(b) The command you used in part (a) draws the curve inside a “bounding box”. Compute the length of the diagonal
of this box.
Solution: From the equation of the parametrized curve, we can see that the two endpoints of the diagonal are
(1, 0, 0) and (1, ln 2, 2 ln 2). We find the distance between the two using the distance formula:
p
p
p
distance = (2 − 1)2 + (ln 2 − 0)2 + (2 ln 2 − 0)2 = 12 + (ln 2)2 + (2 ln 2)2 = 1 + 5(ln 2)2 ≈ 1.845.
(c) Write down an integral for the length of the curve. You don’t need to compute the integral by hand...
Solution:
Thus
The curve is parameterized as r(t) = ht, ln t, t ln ti, so the velocity vector is r0 (t) = h1, 1t , 1 + ln ti.
s
r
2
1
1
0
2
2
|r (t)| = 1 +
+ (1 + ln t) = 2 + 2 + 2 ln t + (ln t)2 ,
t
t
and so the length of the curve is given by the integral
2
Z
L=
|r0 (t)| dt =
2
Z
r
2+
1
1
1
+ 2 ln t + (ln t)2 dt.
t2
(d) Find the length correct to five decimal places by using the command:
NIntegrate[t^2, {t,1,2}]
where “t^2” is replaced by the integrand you wrote in part (c). Compare your answer to the length you found in
part (b).
Solution: Using Mathematica, we get 1.8581. This is pretty similar to the answer we got in part (a)! That is,
the length of the curve is only slightly longer (under 1% longer) than the straight line connecting the endpoints.
3. (Based on Stewart 10.3 #12 ) Let C be the curve of intersection of the cylinder 4x2 + y 2 = 4 and the plane x + y + z = 2.
(a) Find a parameterization of C.
Hint: First parametrize x and y using the cylinder, then find z. What you’re doing in this approach is first
parametrizing the curve’s projection in the xy-plane.
Solution: Let C be the curve of intersection. The projection of C onto the xy-plane is the ellipse 4x2 + y 2 = 4 or
y2
x2 +
= 1. Then we can write x = cos t, y = 2 sin t, for 0 ≤ t < 2π. Since C also lies on the plane x + y + z = 2,
4
we have z = 2 − x − y = 2 − cos t − 2 sin t. Thus the parametric equations for C are x = cos t, y = 2 sin t,
z = 2−cos t−2 sin t, with 0 ≤ t ≤ 2π, and the corresponding vector equation is r(t) = hcos t, 2 sin t, 2−cos t−2 sin t i.
(b) Write down an integral for the length of C.
Solution: Since r(t) = hcos t, 2 sin t, 2 − cos t − 2 sin t i, we can compute r0 (t) = h− sin t, 2 cos t, sin t − 2 cos t i.
Thus the length of this velocity vector is
p
p
|r0 (t)| = (− sin t)2 + (2 cos t)2 + (sin t − 2 cos t)2 = 2 sin2 t + 8 cos2 t − 4 sin t cos t.
So the length of C is
2π
Z
L=
0
|r0 (t)| dt =
2π
Z
p
2 sin2 t + 8 cos2 t − 4 sin t cos t dt.
0
(c) Find the length correct to five decimal places by using the NIntegrate command given in the previous problem.
Solution:
Using Mathematica, we get the length to be about 13.5191.
x2
4. Find the line integral of the function f (x, y) = x around the curve C, which is the part of the ellipse given by
+y 2 = 1
4
that lies in the first quadrant. You should use Mathematica to find the numerical integral, again using the NIntegrate
command. (The integral you will almost certainly find is difficult to integrate by hand, but you might also try the
Integrate command.)
Solution: We can parameterize the curve (part of an ellipse) as r(t) = h2 cos t, sin t i, 0 ≤ t ≤
integral we want is
Z
Z b
f (x, y)ds =
f (r(t)) |r0 (t)| dt.
C
π
.
2
Then the line
a
So our line integral is
π/2
Z
q
2 cos(t) cos2 (t) + 4 sin2 (t) dt ≈ 2.76035.
0
5. (Stewart 13.2 #46 ) The base of a circular fence with radius 10m is given by x = 10 cos t, y = 10 sin t. The height of the
fence at position (x, y) is given by the function h(x, y) = 4 + 0.01(x2 − y 2 ), so the height varies from 3m to 5m. Suppose
that 1L of paint covers 100m2 . Sketch the fence and determine how much paint will be needed to paint both sides of
the fence.
Solution: Consider the base of the fence in the xy-plane, centered at the origin, with the height given by z = h(x, y).
The fence can be graphed using the Mathematica command
ParametricPlot3D[{10 Cos[u], 10 Sin[u], v (4 + Cos[2u])}, {u, 0, 2 Pi}, {v, 0, 1}]
which corresponds to the parametric equations x = 10 cos u, y = 10 sin u and
z = v[4 + 0.01((10 cos u)2 ) − (10 sin u)2 )] = v(4 + cos2 u − sin2 u) = v(4 + cos 2u).
Z
h(x, y)ds where C, the base of the fence, is given by r(t) = h10 cos t, 10 sin ti, (for 0 ≤ t < 2π).
The area of the fence is
C
Then
Z
2π
Z
h(x, y)ds =
C
h(r(t)) |r0 (t)| dt
0
2π
Z
=
4 + 0.01((10 cos u)2 ) − (10 sin u)2 )
p
(−10 sin t)2 + (10 cos t)2 dt
0
2π
Z
=
0
2π
√
1
(4 + cos 2t) 100dt = 10 4t + sin 2t
= 10(8π) = 80π m2
2
0
If we paint both sides of the fence, the total surface area to cover is 160π m2 , and since 1 L of paint covers 100 m2 , we
160π
= 1.6π ≈ 5.03 L of paint.
require
100
Reading: If you haven’t done so already, please read through Section 12.6 on surface area, and do the following problem,
because it’s easy and important!
6. (Optional: 1 extra credit point) We want you to make a few small corrections IN PEN in your textbook. On page
869, in the red box called “Definition 4”, the formula for surface area should say:
ZZ
A(S) =
|ru × rv | du dv
(that is, the dA should actually be a du dv).
D
(a) Make nearly the exact same change also on pageZ Z
870 right above the title “Surface Area of a Graph”. This time
the dA should be a dφdθ, so the integral is A =
|rr × rθ | dr dθ. Notice it is NOT “r dr dθ”.
D
(Continued on the next page)
(b) Make the same change again on page 950 in the red box numbered 2, and in two more equations on that page. (In
one of them dA should be du dv and in the other the dA should be dφ dθ.)
(c) Just a few more places. . . One at the bottom of page 952. Two at the bottom of page 955. One more on line 11 of
page 956. Once you’ve made these corrections, put a little note on your assignment letting us know that you did
that, and your CA will assign your point of extra credit!