Math 190
Sample Test for Chapter 3
Answers
(not guaranteed correct)
1.
Find an equation for the line which is tangent to the curve
2x2 - 3x + 1
y=
at the point (1,0)
x2 + 3
2.
Find an equation for the line which is tangent to the curve y = (x + y)2/3
at the point (18,9).
3.
x - 4y = 1
-2x + 7y = 27
Find f’(x) for each of these functions
2
x - 2x + 2
f'(x) =
4x2 - 6x + 4
a)
f(x) = 2 x
b)
f(x) = 5x + sin3x + sinx3
f'(x) = 5 + 3sin2x cosx + 3x2 cosx3
c)
f(x) = secx tan2x
f'(x) = secx tanx (3sec2x - 1)
4.
Find
a)
 3x2 - 1
y =  2
 3x + 1
b)


x

y= 2
 3x + 2x + 1 
c)
sin2x
y=
tan6x
x2 - 2x + 2
dy
for each of the following
dx
-3



dy
-36x (3x2 + 1)2
=
dx
(3x2 - 1)4
1/3
dy
=
dx
3 x2/3
1 - 3x2
[3x2 + 2x + 1]4/3
dy
2 tan6x cos2x - 6 sin2x sec26x
=
dx
tan26x
5.
Find f’(x) for the following functions
a)
f(x) = x e-10x
f'(x) = e-10x (1 - 10x)
b)
f(x) = e-3x ln7x
1

f'(x) = e-3x  - 3 ln7x 
x

c)
f(x) = log8(3 – 5x)
f'(x) =
6.
Find
a)
y = x3
dy
= 3x2
dx
b)
y = 3x
dy
= 3x ln3
dx
c)
y = 33
dy
= 0
dx
d)
y = xx
dy
= xx + xx lnx
dx
7.
Find
a)
z = arcsin t 
b)
z = arctan(e
-5
(3 - 5x) ln8
dy
for each of the following
dx
dz
for each of the following
dt
dz
=
dt
t
1
)
2t
-1
t2 - 1
dz
2e2t
=
dt
1 + e4t
c)
z=
(cosh3t) (lnt)
2t2 + 1
dz
(2t2 + 1)cosh3t + 3t(2t2 + 1)sinh3t (lnt) - 4t2 cosh3t (lnt)
=
dt
t (2t2 + 1)2
8.
Using the definitions of the hyperbolic functions, show that
cosh2x = cosh2x + sinh2x
ex + e-x
Recall: coshx =
2
and sinhx =
ex - e-x
2
Thus:
2
2
cosh x + sinh x =
=
1 
  ( e2x + 2 + e-2x) + (e2x - 2 + e-2x)
4
=
1
  (2e2x + 2e-2x)
4
=
1 2x
  (e + e-2x)
2
=
=
9.
 ex + e-x   ex + e-x 
 ex - e-x   ex - e-x 


 +


2
2
2  2 




e2x + e-2x
2
cosh2x
A cup of hot chocolate has a temperature of 80 C in a room kept at 20 C.
After half an hour the hot chocolate has cooled to 60 C.
Let T = temperature of the chocolate and
t = # hours since the chocolate was 80 C.
Then (eventually, you will find) the equation to be used is
t
4
T = 20 + 609
a)
What is the temperature of the chocolate after another half hour?
After 1/2 hour the temperature of the chocolate is
b)
140
C or about 46.666… C.
3
When will the chocolate have cooled to 40 C?
The chocolate will have cooled to 40 C when t =
-ln3
ln(4/9)
or about 1.35 hours (1 hour and 21 minutes) after
its temperature was 80 C.
10.
A particle moves along a horizontal line so that its coordinate at any time t
is x =
a)
b2 + c2t2
, t ≥ 0, where b and c are positive constants.
Find the velocity
and acceleration
b)
v(t) =
a(t) =
c2t
b2 + c2t2
b 2c2
(b2 + c2t2)3/2
functions.
Show that the particle always moves in the positive direction.
The velocity function is always non-negative for non-negative
values of t. That means that the particle will always be moving
in the positive direction.
11.
An upper tank shaped like an inverted cone is draining into a lower tank
shaped like a cylinder. The conical tank has a height of 5 meters and an
upper radius of 4 meters. The cylindrical tank also has a height of 5
meters and a radius of 4 meters. If the water level in the conical tank
drops at a rate of 0.5 m/min, at what rate is the water level in the
cylindrical tank rising when the water level in the upper (conical) tank is 3
meters?
[For a cone: V =
1
3
 r2 h.
You should know the volume of a circular cylinder.]
At the instant that the water level in the upper, conical tank is 3 meters,
9
the water level in the lower, conical tank is increasing at a rate of
50
m/min.
12.
Use differentials to approximate the change in the atmospheric pressure
when the altitude increases from z = 2 km to z = 2.01 km. The formula
for pressure is: P(z) = 1000e-z/10.
As the altitude increases from 2.00 km to 2.01 km, the atmospheric
pressure drops about 0.82 (no units given).