S. F. Ellermeyer
MATH 2203 — Exam 2 Solutions
February 16, 2004
Name
Instructions. This exam contains seven problems, but only six of them will be graded. You
may choose any six to do. Please write DON’T GRADE on the one that you don’t want me
to grade. In writing your solution to each problem, include sufficient detail and use correct
notation. (For instance, don’t forget to write “=” when you mean to say that two things are
equal.) Your method of solving the problem should be clear to the reader (me). If I have to
struggle to understand what you have written, then you might not get full credit for your
solution even if you get a correct answer.
1. Make sure to show all details of your computations here. You may express angles in
terms of radians or degrees (whichever you choose) and you may approximate angles
by rounding to two decimal places.
(a) Find cylindrical coordinates for the point whose rectangular coordinates are (x, y, z) =
(−4, 8, −2).
Solution:
q
p
√
√
2
2
r = x + y = (−4)2 + (8)2 = 80 = 4 5.
Also, tan (θ) = y/x = −2 and θ is in the second quadrant of the xy plane, so
θ = arctan (−2) + π ≈ 2.0344 ≈ 116.56◦ . Cylindrical coordinates of this point are
thus
´
³ √
(r, θ, z) = 4 5, arctan (−2) + π ≈ 116.56◦ , −2 .
Let us check our answer:
x = r cos (θ)
√
= 4 5 cos (arctan (−2) + π)
√
= −4 5 cos (arctan (−2))
¶
µ
√
1
= −4 5 √
5
= −4
and
y = r sin (θ)
√
= 4 5 sin (arctan (−2) + π)
√
= −4 5 sin (arctan (−2))
¶
µ
√
−2
= −4 5 √
5
= 8.
1
(b) Find the rectangular
coordinates
for the point that has cylindrical coordinates
¡ π
¢
are (r, θ, z) = 5, 4 , −5 .
Solution: Note that this point is in the room below us.
³ π ´ 5√2
=
x = r cos (θ) = 5 cos
4
2
³ π ´ 5√2
=
y = r sin (θ) = 5 sin
4
2
so the rectangular coordinates are
à √
!
√
5 2 5 2
(x, y, z) =
,
, −5 .
2
2
(c) Find spherical coordinates for the point whose rectangular coordinates are (x, y, z) =
(−4, 8, −2).
Solution:
p
ρ = x2 + y 2 + z 2
q
= (−4)2 + (8)2 + (−2)2
√
= 84
√
= 2 21.
Also,
so
√
21
z
cos (φ) = = −
ρ
21
à √ !
21
φ = arccos −
= 1.7908 ≈ 102.61◦ .
21
In addition,
cos (θ) =
x
ρ sin (φ)
−4
³
³ √ ´´
= √
2 21 sin arccos − 2121
√
−2 21
³
³ √ ´´
=
21 sin arccos − 2121
√
−2 21
³ √ ´
=
21 2 21105
√
5
=−
5
2
and θ is in the second quadrant of the xy plane, so
à √ !
5
≈ 2.0344 ≈ 116.56◦ .
θ = arccos −
5
Spherical coordinates for this point are thus
à √ !!
Ã
à √ !
√
5
21
, arccos −
(ρ, θ, φ) = 2 21, arccos −
5
21
´
³ √
≈ 2 21, 116.56◦ , 102.61◦ .
Let us check our work:
x = ρ sin (φ) cos (θ)
Ã
à √ !!
Ã
à √ !!
√
21
5
= 2 21 sin arccos −
cos arccos −
21
5
!Ã √ !
à √
√
5
2 105
−
= 2 21
21
5
= −4.
and
y = ρ sin (φ) sin (θ)
Ã
à √ !!
Ã
à √ !!
√
21
5
sin arccos −
= 2 21 sin arccos −
21
5
à √
!Ã √ !
√
2 105
2 5
= 2 21
21
5
=8
and
z = ρ cos (φ)
Ã
à √ !!
√
21
= 2 21 cos arccos −
21
à √ !
√
21
= 2 21 −
21
= −2.
(d) Find the rectangular
¡
¢ coordinates for the point that has spherical coordinates are
(ρ, θ, φ) = 5, π4 , 3π
.
4
3
Solution: Note that this point is in the room below us.
x = ρ sin (φ) cos (θ)
µ ¶
³π ´
3π
= 5 sin
cos
4
4
Ã√ ! Ã√ !
2
2
=5
2
2
=
5
2
and
y = ρ sin (φ) sin (θ)
µ ¶
³π ´
3π
= 5 sin
sin
4
4
Ã√ ! Ã√ !
2
2
=5
2
2
=
5
2
and
µ
3π
z = ρ cos (φ) = 5 cos
4
¶
à √ !
√
5 2
2
=5 −
=−
2
2
so the rectangular coordinates of this point are
Ã
√ !
5 5 5 2
, ,−
.
(x, y, z) =
2 2
2
2. Here are six parametric descriptions of curves labelled 1—6 and six pictures of curves
labelled A—F. Match each equation with the curve that it describes.
4
1) x = cos (t) , y = sin (t) , z = t
2) x = t cos (t) , y = t sin (t) , z = t
3) x = t cos (t) , y = t sin (t) , z = t2
4) x = cos (t) , y = sin (t) , z = 1
5) x = t cos (t) , y = t sin (t) , z = 1
6) x = sin (t) , y = t, z = t2
Equations 1 match curve F.
5
Equations 2 match curve E.
Equations 3 match curve D.
Equations 4 match curve A.
Equations 5 match curve B.
Equations 6 match curve C.
3. Sketch the plane curve defined by the vector function
r (t) = t3 i + t2 j.
Make your sketch on the grid provided below. It might be helpful to make a table of
values so that you can get a good idea of what the curve looks like.
Find the general equation for the tangent vector, r0 (t), and use this to find the tangent
vector to the curve r (t) at the point corresponding to the parameter value t = 1. (In
other words, find r0 (1).) Illustrate this tangent vector in your picture.
Solution: r0 (t) = 3t2 i + 2tj so r0 (1) = 3i + 2j.
6
4. Find the curvature function, κ (x), for the curve f (x) = ex . Then determine the point
on the graph of f at which this graph has maximum curvature. (Show all details.)
Solution: For the curve f (x) = ex , we have
f 0 (x) = ex
f 00 (x) = ex
so the curvature is
ex
|f 00 (x)|
κ (x) = ¡
=
¢
3/2
(1 + e2x )3/2 .
1 + (f 0 (x))2
We would like to find at what value of x the curvature is maximum: Since
³
´
2x 3/2 x
2x 1/2
2x
x 3
(1 + e ) e − e 2 (1 + e ) (2e )
κ0 (x) =
(1 + e2x )3
³
´
3/2
1/2
(1 + e2x ) ex − ex 3 (1 + e2x ) (e2x )
=
(1 + e2x )3
1/2
ex (1 + e2x ) (1 − 2e2x )
=
(1 + e2x )3
ex (1 − 2e2x )
=
(1 + e2x )5/2
1 − 2e2x = 0. Solving this equation for x gives us e2x = 1/2
we see that κ0 (x) = 0 when
√
2
or (ex ) = 1/2 or ex = 2/2 or
Ã√ !
2
x = ln
.
2
7
Since this x gives us the only critical point of κ and since it is clear that the curvature
must be maximum at some point for the graph of f (x) = ex , we conclude that the
curvature is maximum at the point
à Ã√ ! √ !
2
2
ln
,
≈ (−0.34637, 0.70711) .
2
2
5. The position function of a moving particle is
¡
¢
r (t) = t2 i + 5tj+ t2 − 16t k.
Find the time, t, at which the speed of the particle is at its minimum. (Show all
details.)
Solution:
®
­
position=r (t) = t2 , 5t, t2 − 16t
velocity=r0 (t) = h2t, 5, 2t − 16i
speed= |r0 (t)|
q
= 4t2 + 25 + (2t − 16)2
√
= 8t2 − 64t + 281
To find when the speed is minimum, we compute
´
d ³√ 2
8t − 32
8t − 64t + 281 = √ 2
dt
8t − 64t + 281
This derivative is equal to zero when t = 4. Also, this derivative is negative when t < 4
and positive when t > 4. Thus t = 4 is where the minimum speed occurs.
8
6. Here are six parametric descriptions of surfaces labelled 1—6 and six pictures of surfaces
labelled A—F. Match each equation with the curve that it describes.
1) r (u, v) = hcos (u) , sin (u) , vi
2) r (u, v) = hu, v, v2 i
3) r (u, v) = hv cos (u) , v sin (u) , v2 i
4) r (u, v) = h1 + u, 1 + v, u + vi
5) r (u, v) = hv cos (u) , v sin (u) , vi
9
6) r (u, v) = hv cos (u) , v sin (u) , 1i
Equation 1 match surface F.
Equation 2 match surface A.
Equation 3 match surface C.
Equation 4 match surface B.
Equation 5 match surface D.
Equation 6 match surface E.
10