Engineering Math I – Fall 2014
Problem 1 (4 pts). Let f (x) =
Quiz #6 Solutions
√
x2 − 3x. Find f 00 (4).
Solution. The correct answer is (a). First differentiate f (x) using the chain rule to get:
1
2x − 3
1
.
f 0 (x) = (x2 − 3x)− 2 (2x − 3) = √
2
2 x2 − 3x
Next, differentiate f 0 (x) using both the quotient rule and the chain rule to get:
√
2
x2 − 3x · 2 − (2x − 3) · √2x−3
x2 −3x
00
√
f (x) =
(2 x2 − 3x)2
4(x2 − 3x) − (2x − 3)2
√
=
4(x2 − 3x) x2 − 3x
Hence
f 00 (4) =
16 − 25
4(42 − 3 · 4) − (2 · 4 − 3)2
9
√
√ =− .
=
2
2
32
4(4 − 3 · 4) 4 − 3 · 4
16 4
Problem 2 (4 pts). Suppose x2 + y 3 = 1. Find y 00 by implicit differentiation.
Solution. The correct answer is (b). First we differentiate the equation implicity to get
2x + 3y 2 y 0 = 0. Solving for y 0 yields
y0 = −
2x
.
3y 2
(1)
Next differentiate 2x + 3y 2 y 0 = 0 implicitly to get 2 + 6yy 0 + 3y 2 y 00 = 0. Solving for y 00
we get
−6yy 0 − 2
y 00 =
.
3y 2
Substituting equation (1) for y 0 yields
2x
−6y − 3y2 − 2
y 00 =
=
3y 2
4x
y
−2
3y 2
=
4x − 2y
.
3y 3
Remark. The answer (d) “None of the above.” is also acceptable for Problem 2 if you
first solved for y 0 and then differentiated equation (1) above to get
y 00 =
−6y 2 − 8x2
.
9y 5
MATH 151:549-551 – Fall 2014 Quiz #6 Solutions
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Problem 3 (6 pts). Suppose the position of a particle is given by the vector function
r(t) = ht3 , t2 i.
(a) Sketch the curve traced by r(t) for −2 ≤ t ≤ 2. Indicate with arrows the direction
in which t is increasing.
(b) Find the velocity and acceleration of the particle when t = 2.
Solution. Velocity is given by r(t) and acceleration is given by r00 (t). Using the
rules of differentiation for vector functions, we have
r0 (t) = h3t2 , 2ti
and
r00 (t) = h6t, 2i.
Therefore, when t = 2 we have
velocity = r0 (2) = h3 · 22 , 2 · 2i = h12, 4i
acceleration = r00 (2) = h6 · 2, 2i = h12, 2i.
(c) Is the particle speeding up or slowing when t = −1? Explain your answer.
Solution. When t = −1, the velocity is r0 (−1) = h3 · (−1)2 , 2 · (−1)i = h3, −2i and
acceleration is r00 (−1) = h6 · (−1), 2i = h−6, 2i. The acceleration vector is pointing
away from the velocity vector, so the particle is slowing down.
MATH 151:549-551 – Fall 2014 Quiz #6 Solutions
3
Problem 4 (4 pts). A satellite completes one orbit of Earth along the equator at an
altitude of 2000 km every 2 hours. Find the velocity of the satellite for any given time.
[Hint: Assume the satellite has a circular orbit and parametrize the orbit of the satellite.
Use the value 6600 km for the Earth’s radius; use hours as the unit for time.]
Solution. We consider the satellite to be orbiting the earth in a circle whose center is
at the origin (which is also the center of the earth) and has radius 6600 km + 2000
km = 8600 km. Parametrizing with respect to the angle θ = θ(t) made between the
position vector of any point on the circle and the x-axis, we get x = 8600 cos(θ(t)) and
y = 8600 sin(θ(t)).
We are told that θ(0) = 0 and θ(2) = 2π (the satellite completes an orbit every 2
hours). Thus θ(t) = πt. We may now say that the position of the satellite at a given time
t is described by the vector function r(t) = h8600 cos(πt), 8600 sin(πt)i. Hence velocity is
r0 (t) = h−8600π sin(πt), 8600π cos(πt)i.
Bonus Problem (2 pts). At what point does the curve s(t) = ht(t2 − 3), 3(t2 − 3)i cross
itself? Find two unit vectors in the directions of both tangents at that point.
Solution. The curve crosses itself if there are two different values for t, say t1 and
√ t2 ,
such that s(t√
1 ) = s(t2 ). Upon inspection of the vector function s(t), we see that s( 3) =
h0, 0i = s(− 3). So
√ the curve√crosses itself at (0, 0). The unit tangent vectors are given
s0 (t)
by |s0 (t)| for t = − 3 and t = 3. So we have:
s0 (t) = h(t2 − 3) + t(2t), 3(2t)i = h3t2 − 3, 6ti
and
|s0 (t)| =
p
(3t2 − 3)2 + (6t)2 .
√
Thus when t = − 3, we have for the unit vector
√
√
√
√
√
√
h6, −6 3i
6h1, − 3i
h3(− 3)2 − 3, 6(− 3)i
1
s0 (− 3)
√
q
√
=q
=
=
=
h1,
−
3i
√
√
√
2
|s0 (− 3)|
6
4
2
2
2
2
2
(3(− 3) − 3) + (6(− 3))
6 + (−6 3)
√
and when t = 3 we have (using the above
√ computation and noting the only thing that
changes is the negative sign in from the 3)
√
s0 ( 3)
1 √
√
= h1, 3i.
2
|s0 ( 3)|