MATHEMATICS 184 (Section 921) - TERM EXAM #3
SOLUTIONS
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INSTRUCTIONS: Notes, books and calculators are not permitted. Answers may
receive no credit if accompanying work is not provided. When indicated, simplify
answers as much as possible, and be sure to clearly indicate the final answer for
each question. Please use the back of the pages if you need extra space.
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1. A spherical scoop of ice cream is melting in such a way that its radius is decreasing at a con3
stant rate of 0.1 cm/minute. When the volume of the scoop is 32π
3 cm , at what rate is its
volume decreasing?
Hint: Volume of a sphere = 43 πr3 , where r = radius in cm, note that here radius is a function of
time.
V = 43 πr3
Here, the radius and consequently the volume are changing with time, so
V (t) = 43 π(r(t))3 where r(t) is the radius at time t.
Differentiate with respect to time
dV
dt
dV
dt
We are given that
4
dr
π(3)(r(t))2
3
dt
dr
= 4π(r(t))2
dt
=
(by chain rule)
dr
= −0.1 cm/min, so
dt
dV
= 4π(r(t))2 (−0.1)
dt
We are also given that V =
32π
3
at the time of interest, so at that time
1
3
V)3
4π
3 32π 1
= (
)3
4π 3
1
= (8) 3
= 2
r(t) = (
Therefore,
dV
dt
= 4π(2)2 (−0.1)
= −1.6π
so the volume of the ice cream is decreasing at a rate of −1.6π cm3 /min when the volume is
32π
3
3 cm .
Page 2
8
2. A lighthouse is on a small island 4 km. away from the nearest point P of a straight shoreline.
The light of this lighthouse makes 2 revolutions per minute, moving a beam of light along the
shoreline as it revolves. How fast is the beam of light moving along the shoreline when it is 2
km.from the point P?
Hint: One revolution of the light is 2π radians (do not use degrees)
Let x(t) be the distance between the point P and the beam of light on the shoreline at time t.
Then, from this picture
we can see that
tan(θ(t)) =
x(t)
4
where θ(t) is the angle between the beam of light and the line connecting P and the lighthouse.
There are two nice ways to do this:
Method 1:
x(t) = 4tan(θ(t))
Differentiating with respect to time we have
dx
=
dt
Since we are given that
4
cos2 (θ(t))
·
dθ
dt
dx
dθ
= 4π per min. and we are looking for
when x(t) = 2 we have
dt
dt
dx
=
dt
4
cos2 (θ(t))
· 4π
and when x =2, using pythagoras and
cos(θ) =
ad jacent
opposite
=
we finally have
dx
=
dt
16π
16
20
Page 3
= 20π
√4
20
Method 2:
θ(t) = arctan( x(t)
4 )
So
dθ
=
dt
Again,
1
x(t)
1+( 4 )2
· 14
dx
dt
dθ
= 4π and x(t) = 2, so
dt
1
dx
2 2 dt
1 + (4)
dx
5
= 16π · = 20π
dt
4
16π =
So either way, we have that the beam of light is moving at a speed of 20 π km per minute when
it is 2 km from the point P.
Page 4
11
3. Find the equations of the tangent lines to the curve
x2 − xy + y2 = 4
at the points where the curve crosses the x-axis.
First, this curve crosses the x-axis when y = 0, so when
x2 − x(0) + 02 = 4
x2 = 4
x = ±2
So it crosses at (2, 0) and (−2, 0).
Next, we need the slope of the tangents to these points. So implicitly differentiate:
2x − (1 · y + x · y) + 2y · y0 = 0
2y · y0 − x · y0 = y − 2x
y0 (2y − x) = y − 2x
y − 2x
y0 =
2y − x
At (2, 0):
y = mx + b
m=
0−2(2)
2(0)−2
=
−4
−2
=2
So now,
y = 2x + b
plug in (2, 0)
0 = 2(2) + b
b = −4
Page 5
At (−2, 0):
m=
0−2(−2)
2(0)−(−2)
=
4
2
=2
So now,
y = 2x + b
plug in (−2, 0)
0 = 2(−2) + b
b = 4
So the equations of the tangent lines are y = 2x − 4 and y = 2x + 4
Page 6
4. Differentiate the following functions: (Do not simplify your answers)
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(a) f (x) = tan(x) + ln(x) + arctan(x) + ex
f 0 (x) =
3
2
1
1
+ 1x + 1+x
2
cos2 (x)
2
+ ex (2x)
(b) y = x · sin2 (x)cos(x)
d
(sin2 (x)cos(x))
dx
2
= sin (x)cos(x) + x[2sin(x)cos(x)cos(x) + sin2 (x)(−sin(x))]
y0 = sin2 (x)cos(x) + 2xsin(x)cos2 (x) − xsin3 (x)
y0 = 1 · (sin2 (x)cos(x)) + x ·
(last step not necessary)
2
(c) y = 5 · ln(7x3 ) + 3 · arcsin(6x − 2)
y0 = 5 · 7x13 · 21x2 + 3 · √
1
1−(6x−2)2
Page 7
·6
2
(d) f (x) = ee
x
x
f 0 (x) = ee · ex
x
f 0 (x) = ee +x
3
(e) g(t) =
(2t−5)4
(t 2 +1)3
2 Ways to do This:
Product Rule:
g(t) = (2t − 5)4 (t 2 + 1)−3
g0 (t) = 4(2t − 5)3 (2) · (t 2 + 1)−3 + (2t − 5)4 · (−3)(t 2 + 1)−4 (2t)
Quotient Rule:
g0 (t) =
4
4(2t−5)3 (2)·(t 2 +1)−3 −(2t−5)4 ·(−3)(t 2 +1)−4 (2t)
(t 2 +1)6
(f) ey = ln(xy) + 2x2
Implicit Differentiation
1
(1 · y + x · y0 ) + 4x
xy
x · y0
y
ey · y0 −
=
+ 4x
xy
xy
1
1
y0 (ey − ) =
+ 4x
y
x
1
+ 4x
y0 = x y 1
e −y
ey · y0 =
Page 8