B U Department of Mathematics
Math 102 Calculus II
Fall 2006 Midterm 1
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θ
1. [7] (AntonBivensDavis, p.729, q.51) Sketch the curve given by r = cos in polar coordinates.
2
State explicitly the symmetries of the graph and the slopes while approaching the origin. In
your sketch, one must observe these easily.
Solution:
The first thing to do is to give values for θ to collect data for r. The reasonable thing
to do is to go up to 4π since that is the period for f (θ) = cos θ2 . Then it is easy to
see that the graph looks like
√1
2
−1
1
− √12
Then make sure about the symmetries. Since cos is an even function, we get the
symmetry with respect to x-axis.
It is not easy to detect the other symmetries. Think as follows: as θ varies from 0
to 2π, the curve traverses the upper half of the figure above. We can then get the
symmetry with respect to the origin by observing that
f (2π + θ) = −f (θ).
Since there is symmetry with respect to the x-axis and the origin, there is symmetry
with respect to the y-axis.
Finally, the slopes while approaching 0 are always 0:
dy
dy/dθ
r ′ sin θ
(π) =
(π) = ′
(π) = tan(π) = 0.
dx
dx/dθ
r cos θ
Above computation is valid since at π (or 3π) the denominators are never zero.
2. In the figure below, C1 and C2 are circles with radii r1 and r2 and with centers at the origin O
and at (r2 , 0) respectively. Suppose that 2r2 > r1 so that C1 and C2 intersect each other at two
distinct points. Let P be the point of intersection in the first quadrant and H be the shaded
crescent.
(a) [1+1] Write down the two equations in polar coordinates that describe C1 and C2 .
(b) [2] Find the angle P OB in terms of rr12 . Easiest way is to use the answer to part (a).
(c) [4] Write down an integral in polar coordinates for the area of H. Then compute the area
in terms of rr12 .
P
r1
B
O
r2
C2
C1
Solution:
(a) C1 : r = r1 , C2 : (x − r2 )2 + y 2 = r22 ⇒ r 2 = 2r2 x ⇒ r = 2r2 cos θ
(b) P is a point of intersection of C1 and C2 . Hence,
θ0 = \
P OB = cos−1 (
1 r1
) > 0.
2 r2
(c)
area(H) =
=
=
=
=
=
=
θ0
1
(2r2 cos θ)2 − (r1 )2 ) dθ
2
Z−θθ00
1 + cos 2θ
− r12 )dθ
(4r22
2
0
sin 2θ
(2r22 (θ +
) − r12 θ)θ00
2
2r22 θ0 + r22 sin 2θ0 − r12 θ0
(2r22 − r12 )θ0 + 2r22 cos p
θ0 sin θ0
r1 4r22 − r12
(2r22 − r12 )θ0 + 2r22
r2
q2r2
2
2
2
(2r2 − r1 )θ0 + r1 4r2 − r12 .
Z
3. [6] (AntonBivensDavis, p.823, q.44) Let a, b, c and d be four vectors that are parallel to a fixed
plane. Show that (a × b) × (c × d) = 0. Be as simple as possible but be precise!
Solution:
Let n be a normal to the plane. Given that a, b, c, d ⊥ n, we have a × b = kn and
c × d = ln for some k, l ∈ R. So
(a × b) × (c × d) = (kn) × (ln) = (kl)n × n = 0
4. Let D1 and D2 be two distinct, non-parallel planes in R3 and let α be the angle between them.
Consider two vectors v1 and v2 parallel to D1 and D2 respectively. Let θ be the angle between
v1 and v2 . Assume 0 ≤ α, θ ≤ π. Determine whether the following statements are true or false.
For each, if answer is yes, prove the claim. If no, give an example where the claim is not true.
(a) [2] It is always true that θ ≥ α.
(b) [2] It is always true that θ ≤ α.
Solution:
Answer is no for both cases. Let L be the line of intersection and v be a vector
parallel to L. Then v is parallel to both D1 and D2 .
(a) Set v1 = v2 = v. Then θ = 0 < α.
(b) Set v1 = v, v2 = −v. Then θ = π > α.
5. [10] Using vector calculus, find the point on the line x + y − z = 2, 3x + y = 1 closest to the
point (1, 1, 1). You are not allowed to answer this question by finding the global minimum of an appropriate
function!
Solution:
The case in question is exactly the same as the previous question: two planes intersecting along a line (a line in R3 is given by two independent linear equations).
There are several ways to solve this question. Here, we first find the plane D passing
through the point P = (1, 1, 1) and normal to the line. A direction n along the line
is:
n = h1, 1, −1i × h3, 1, 0i = h1, −3, −2i
The equation for the plane D is
D : h1, −3, −2i · hx − 1, y − 1, z − 1i = x − 3y − 2z + 4 = 0
The closest point Q = (x, y, z) on L to P is the point of intersection of L and D;
hence Q satisfies:
x + y − z = 2, 3x + y = 1, −x + 3y + 2z = 4
One can then compute that Q = (x, y, z) = (−
8
3 23
, , − ).
14 14 14