Calculus III
Math 143 Spring 2008
Professor Ben Richert
Exam 2
Solutions
−
Problem 1. (20pts) Consider the points A = (1, 2, 3) and B = (2, 3, 5), and the line ⇀
r (t) = h2, 2, 3i + th1, 2, 5i.
−−
⇀
(a–5pts) Give parametric equations for the line containing A and parallel to AB.
Solution. The line through (a, b, c) in direction hd, e, f i is ha, b, ci + thd, e, f i, giving parametric equations
x(t)
=
a + dt
y(t) =
z(t) =
b + et
c + ft
−−
⇀
We use the point (1, 2, 3) and the direction AB = h1, 1, 2i which we get by subtracting the point B from A. So our
line is:
x(t)
=
1+t
y(t) =
2+t
z(t) =
3 + 2t
−
(b–5pts) Give the equation of a plane which is perpendicular to ⇀
r (t) and contains B.
Solution. The equation of a plane containing (x0 , y0 , z0 ) with normal ha, b, ci is
a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0.
−−
⇀
−
−
⇀
Since the plane we are looking for is perpendicular to AB, its normal must be parallel to AB = h1, 1, 2i, that is, we
can use h1, 1, 2i as its normal. With the point B = (2, 3, 5), we get the plane
(x − 2) + (y − 3) + 2(z − 5) = 0.
−
(c–10pts) Give the equation of the plane containing ⇀
r (t) and the point B.
⇀
Solution. Let C = (2, 2, 3), and note that C is a point on −
r (t). Then the normal to our plane will be perpendicular
−
−
⇀
⇀
−
to both h1, 2, 5i, the direction of r (t), and CB = h0, 1, 2i. We know that the
−
⇀
⇀ −
−
⇀
j
k
i
h1, 2, 5i × h0, 1, 2i = 1 2 5 = h−1, −2, 1i
0 1 2
is normal to both those directions. Thus the plane in question is −(x − 2) − 2(y − 3) + (z − 5) = 0.
Problem 2. (10pts) Decide if the slope of the tangent to the curve x(t), y(t) is positive, negative, or zero at t = 1 given the
graphs of the parametric equations x = x(t) and y = y(t) below.
yHtL
yHtL
1.0
1.00
0.95
0.5
0.90
0.85
t
1
2
3
0.80
0.75
-0.5
t
1
2
3
4
5
-1.0
1
4
5
dy
y ′ (t)
= ′ . From the graph, we see that x′ (1) > 0 and y ′ (1) < 0 (since the slope of the tangent to
dx
x (t)
dy
negative
x(t) at t = 1 is positive while the slope of the tangent to y(t) at t = 1 is negative), so that
=
must be negative
dx
positive
at t = 1.
Solution. We know that
Problem 3. (10pts) A workhorse applies a constant force to an apple cart. Unfortunately, the cart is on a ramp which only
allows it to move in the direction h1, 2, 3i, while the horse pulls in the direction given by the vector h2, 1, 1i.
(a–10pts) What is the cosine of the angle between the direction the cart is moving and the direction the horse is pulling?
⇀
−
⇀
u ·−
v
−
−
where ⇀
u = h1, 2, 3i and ⇀
v = h2, 1, 1i. So
Solution. We use the formula cos θ = ⇀
−
⇀
−
| u || v |
cos θ = √
2+2+3
h1, 2, 3i · h2, 1, 1i
7
√
= √ √ = √
.
12 + 22 + 32 22 + 12 + 12
14 6
14 · 6
(BONUS–2pts) If it requires 100 slugs of work to move the cart 10 feet, with how much force was the horse pulling (in the direction
h2, 1, 1i)?
⇀
−
⇀
−
⇀
−
Solution. Let F be the force vector applied by the horse. Then F is parallel to h2, 1, 1i, so F = ch2, 1, 1i for some
constant c. Now the force applied in the direction of the ramp is
⇀
−
F · h1, 2, 3i
⇀
−
ch2, 1, 1i · h1, 2, 3i
c(2 + 2 + 3)
7
√
comph1,2,3i F =
= √
=
= c√ .
2
2
2
|h1, 2, 3i|
1 +2 +3
14
14
Since the work done was 100 slugs, and work is force (in the right direction) times distance, we have that
7
100 = c √ 10
14
√
10 14
so that c =
. Thus the force applied in the direction h2, 1, 1i is
7
√
√
√
√
⇀
−
10 14
10 14 √
10 14
10 14 p 2
2
2
| F | = |ch2, 1, 1i| = |
2 +1 +1 =
6
h2, 1, 1i| =
|h2, 1, 1i| =
7
7
7
7
pounds.
Another way to do this (which makes less sense to me as it requires memorizing yet another formula), is recalling
that
work = displacement · force = |displacement||force| cos θ,
⇀
−
⇀
−
so writing F to be the force in the direction h2, 1, 1i and d to be the displacement of 10 ft in the direction h1, 2, 3i,
we get
⇀ −
−
⇀ −
−
⇀
⇀
⇀
−
7
100 = d · F = | d || F | cos θ = 10| F | √
14 · 6
and hence
√
10 14 · 6
pounds
|F | =
7
as before.
Problem 4. (15pts) What is the area inside r = 2 cos θ and outside r = 1?
Solution. The picture here looks like
1.0
0.5
-1.0
0.5
-0.5
1.0
1.5
2.0
-0.5
-1.0
and we see that r = 2 cos θ and r = 1 meet when 2 cos θ = 1, that is when cos θ =
1
π
, so when θ = ± .
2
3
Thus we are interested in the area inside 2 cos θ for −π/3 ≤ θ ≤ π/3 and outside r = 1 for −π/3 ≤ θ ≤ π/3. Using the
Z β
[f (θ)]2
dθ, we get
equation
2
α
Z π/3
Z π/3 2
(2 cos θ)2
1
area =
dθ −
dθ
2
−π/3
−π/3 2
Z π/3
Z π/3
π/3 −π/3
1 + cos(2θ)
θ π/3
dθ −
+
=
2
=
2 cos2 θ dθ − 2 −π/3
2
2
2
−π/3
−π/3
π/3
sin(2θ) sin(2π/3)
sin(−2π/3)
=θ+
− π/3 = (π/3) +
− (−π/3) −
− π/3
2 2
2
−π/3
sin(2π/3) sin(2π/3)
+
2
2√
3
.
= π/3 + sin(2π/3) = π/3 +
2
= (π/3) +
Problem 5. (15pts) How far does Billy the Bee fly if his path can be described by the parametric equations x(t) = 3t2 ,
y(t) = 2t3 for 0 ≤ t ≤ 2 where x and y are in feet and t is in seconds?
Solution. We want to compute the arclength of the path describing Billy’s flight. In general, the formula for arclength is
Z bp
AL =
(x′ (t))2 + (y ′ (t))2 dt
a
′
′
assuming that x (t) and y (t) are both continuous and the curve is traversed exactly once. That x′ (t) = 6t and y ′ (t) = 6t2
are continuous is obvious. The curve is traversed exactly once since y ′ (t) > 0 for all t 6= 0 which means that the y component
is always increasing on [0, 2], and thus the curve can never repeat a given y value.
So arclength is:
Z 2p
Z 2 p
Z 2p
(6t)2 + (6t2 )2 dt =
36t2 (1 + t2 ) dt =
6t 1 + t2 dt.
0
0
0
du
Let u = 1 + t2 whence the lie is that
= t dt. Note that u = 1 when t = 0 while u = 5 when t = 2. Now
2
Z 5
Z 2 p
5
√
3 u du = 2u3/2 = 2(5)3/2 − 2
6t 1 + t2 dt =
0
feet.
1
1
Problem 6. (8pts) Match the following graphs with their equations. You do not need to show any work for this problem.
Equation
Graph
(A) r = sin(2θ)
III
(B) r = cos(2θ)
I
(C) r = 1/2 + sin(θ)
II
(D) r = 1 + sin(θ)
IV
1.0
1.5
0.5
1.0
-1.0
0.5
-0.5
1.0
0.5
-0.5
(I)
(II)
-1.0
0.5
-0.5
2.0
0.5
1.5
1.0
0.5
-0.5
0.5
-0.5
-1.0
(III)
(IV)
-0.5
0.5
1.0