Name: Solutions
Discussion Section Number: all
Homework 5 — Due 10/4/2010 in class
This cover sheet must be attached as the top page of your homework.
1. Let s(t) = t4 − 4t3 + 8t describe the position at time t of an object moving along a line.
(a) Find the velocity at time t.
(b) Find the acceleration at time t.
(c) Describe the motion of the object from time t = 0 to time t = 4. When is it advancing?
When is it retreating? Compute the total distance traveled by the object over the given
time.
(d) When is the object accelerating and when is it decelerating?
2. A bucket containing 5 gallons of water has a leak. After t seconds, there are
t 2
Q(t) = 5 1 −
25
gallons of water in the bucket.
(a) At what rate (to nearest hundredth gallon) is water leaking from the bucket after 2
seconds?
(b) How long does it take to for all of the water to leak out of the bucket?
(c) At what rate is the water leaking when the last drop leaks out?
3. Find the derivatives of the following functions:
(a) f (x) = (x3 − 3x)13
(b) g(x) = ln 4x4 + x2 − 2
(c) h(t) = ln (sec x + tan x)
(d) F (s) = sin(x2 ) cos(x2 )
(e) G(x) = tan−1 x1
4. Find an equation for the tangent line to the graph of y(x) at the point (1, 0), where y(x) is
given implicitly by the following equation:
sin−1 (xy) +
5. Find
dy
dx
π
= cos−1 (y).
2
for y(x) given implicitly by the following equation:
3x + log2 (xy) = 10
1
1. Error in question. Skipped. (See similar problems in WebWork and book examples.)
2. A bucket containing 5 gallons of water has a leak. After t seconds, there are
t
Q(t) = 5 1 −
25
2
gallons of water in the bucket.
(a) At what rate (to nearest hundredth gallon) is water leaking from the bucket
after 2 seconds?
dQ
1
t
−
= 10 1 −
dt
25
25
2
t
= −
1−
5
25
dQ 2
2
= −
1−
dt t=2
5
25
46
= −
gal/sec
125
(b) How long does it take to for all of the water to leak out of the bucket?
i.e., when is Q(t) = 0?
Q(t)
t 2
5 1−
25
t
1−
25
t
= 0
= 0
= 0
= 25 sec
(c) At what rate is the water leaking when the last drop leaks out?
dQ = 0 gal/sec
dt t=25
2
3. Find the derivatives of the following functions:
(a) f (x) = (x3 − 3x)13
(b) g(x) = ln 4x4 + x2 − 2
f 0 (x) = 13 x3 − 3x
12
= 39 x3 − 3x
12
3x2 − 3
x2 − 1
let g(u) = ln u and
u(x) = 4x4 + x2 − 2
then g(u(x)) = ln(4x4 + x2 − 2)
1 du
dg
=
dx
u dx
16x3 + 2x
=
4x4 + x2 − 2
(c) h(t) = ln (sec x + tan x)
(Note this should have been h(x) or t’s in the function. Otherwise, we’d get 0.)
let h(u) = ln u and u(x) = sec x + tan x
then h(u(x)) = ln (sec x + tan x)
dh
dh du
=
·
dx
du dx
sec2 x + tan x sec x
=
sec x + tan x
(d) F (s) = sin(x2 ) cos(x2 )
(This also should have been all s’s or all x’s.)
F 0 (x) = cos x2 (2x) cos x2 + sin x2 − sin x2 (2x)
= 2x cos2 x2 − 2x sin2 x2
(e) G(x) = tan−1
1
x
let G(u) = tan−1 u and let u(x) =
then
dG
dx
=
=
=
=
dG du
·
du dx 1
1
· − 2
x
1 + x12
2
−x
1
·
2
x +1
x2
−1
x2 + 1
3
1
x
4. Find an equation for the tangent line to the graph of y(x) at the point (1, 0), where
y(x) is given implicitly by the following equation:
sin−1 (xy) +
π
= cos−1 (y).
2
Using implicit differentiation, we get:
−y 0
y + xy 0
p
= p
1 − (xy)2
1 − y2
y
x
1
p
+ y0 p
= −y 0 p
2
2
1 − (xy)
1 − (xy)
1 − y2
!
x
1
−y
y0 p
+p
= p
2
2
1 − (xy)
1 − (xy)2
1−y

−y
p
·
√
1 − (xy)2
y0 =

1
+√1
x
1−(xy)2

1−y 2
At the point (x, y) = (1, 0) we have
0
=0
2
So using the point-slope formula with m = 0, we see that we have the horizontal tangent line
y0 =
y=0
5. Find
dy
dx
for y(x) given implicitly by the following equation:
3x + log2 (xy) = 10
Recall that 3x = eln 3
x
= ex ln 3 and log2 (xy) =
1 y + xy 0
3x ln 3 +
·
ln 2
xy
x
y
3x ln 3 + y 0
+
xy ln 2
xy ln 2
1
1
3x ln 3 + y 0
+
y ln 2
x ln 2
y
0
4
ln(xy)
ln 2 .
= 0
= 0
= 0
=
−3x ln 3 −
1
x ln 2
y
x
= −3 y ln 3 ln 2 −
x
· (y ln 2)