Math 125
Carter
Test 2 Fall 2012
General Instructions: Do all your work and write your answers inside the blue book. Do
not write on the test. Write your name on only the outside of the blue book. Please put the
test sheet inside the blue book as you leave. There are problems on both sides of this
page.
To hard boil eggs place several in a pot of cold water, turn on the heat, when the water
comes to a boil cook for no more than 8 minutes. Take the eggs from the hot water and
immediately immerse in iced water for 1 minute. If you want to peel them now, place each
egg in the hot water for 10 seconds and peel.
1. Compute the derivatives for each of the following expressions (5 points each). Do not
simplify your results!
(a)
y = x3 + 2x2 − 10x + 4.
Solution.
y 0 = 3x2 + 4x − 10.
(b)
y = (x2 + 2x − 5)(x − 4).
Solution.
y 0 = (x2 + 2x − 5) + (2x + 2)(x − 4).
(c)
ex
y=
.
sin (x)
Solution.
y0 =
sin (x)ex − cos (x)ex
.
(sin (x))2
(d)
y = tan (x2 + 1).
Solution.
y 0 = 2x sec2 (x2 + 1).
(e)
y = cos (ex ).
Solution.
y 0 = −ex sin (ex ).
1
(f)
f (t) = ln (sin (t)).
Solution.
f 0 (t) =
cos (t)
.
sin (t)
(g)
f (θ) =
cos (θ)
1 + sin (θ)
!3
.
Solution.
cos (θ)
f (θ) = 3
1 + sin (θ)
0
!2
−(1 + sin (θ)) sin (θ) − cos2 (θ)
·
(1 + sin (θ))2
− cos2 (θ)
=3
(1 + sin (θ))3
!
!!
(h)
y = ln (arcsin (x)).
Solution.
!
y0 =
!
1
1
· √
.
arcsin (x)
1 − x2
(i)

x(x − 2)
y = ln 
x2 + 1
!3 
.
Solution. Since
h
i
y = 3 ln (x) + ln (x − 2) − ln (x2 + 1) ,
one obtains,
y0 =
1
2x
1
+
− 2
.
x x−2 x +1
2. (10 points) Compute the equation of line tangent to the curve x2 + y 2 = 25 at the
point (−3, 4). Solution.
2xx0 + 2yy 0 = 0.
Thus
2x + 2yy 0 = 0,
or
y 0 = −x/y.
At the point in question, y 0 (−3, 4) = 3/4. The equation is, therefore,
4(y − 4) = 3(x + 3),
or
4y − 3x = 25.
2
3. (10 points) There are two points at which the tangent to the curve f (x) = x3 − x is
horizontal. Determine the coordinates of each.
Solution.
f 0 (x) = 3x2 − 1.
The tangent is horizontal when the derivative is 0. Solving
0 = 3x2 − 1,
one obtains
1
1
1
x = ±√ = ±
3
3
The value of the function, f at these points is

1
1
f ±
3



1 3
2
1
±


2
=
3
1
+ ∓
3
1
2
=∓
2
.
3
1
=±
3
1
1
3
2
2
1
+ ∓
3
1
2
1 1
=±
3
2
1
−1
3
2
.
3
4. (10 points) Determine the equation of the line tangent to the curve y = sin (x) at
the point x = π6 . Solution. Since y 0 = cos (x), at the point in question (x = π/6),
√
y 0 = 3/2, and y = 1/2. The equation of the tangent line is
√
y − 1/2 = 3/2(x − π/6).
5. (10 points) A large coffee cone 10 centimeters in radius and 15 centimeters tall drips
coffee at a rate of 2 cubic centimeters per second. How fast is the height of the coffee
dropping when the height is 3 centimeters? The volume of a cone is V = π3 r2 h where
r denotes the radius and h denotes the height.
Solution.
• Let r denote the radius of the coffee.
• Let h denote the height of the coffee.
• Let V denote the volume of the coffee.
• Given V = π3 r2 h.
dV
= −2
dt
dh
| .
dt h=3
• Given
• Find
cubic centimeters per second.
Since the coffee in the cone is proportional to the cone itself,
r
10
2
=
= .
h
15
3
3
One obtains r = 2h
. Substitute this into the expression for volume to get volume as a
3
function of height alone.
π
π
V = r2 h =
3
3
2h
3
!2
h=
22 π 3
h.
33
Differentiate with respect to time:
22 π
22 π dh
dV
dh
= 3 3h2
= 2 h2 .
dt
3
dt
3
dt
Solve for
dh
dt
:
dh
dV
32
=
· 2 2.
dt
dt 2 πh
At the time in question,
32
−1
dh
= −2 · 2 2 =
.
dt
2 π3
2π
6. (10 points) The radius of a spherical ball of ice is decreasing at a rate of 2 centimeters
per minute. How fast is the volume decreasing when the radius is 10 centimeters? The
volume of a sphere is given by V = 4π
r3 where r denotes the radius. Solution.
3
• Let r denote the radius of the sphere.
• Let V denote the volume enclosed by the sphere.
• Given V =
dr
= −2
dt
dV
|
.
dt r=10
• Given
• Find
4π 3
r .
3
cm/min.
Differentiate
V =
4π 3
r
3
to get
dV
dr
= 4πr2 .
dt
dt
Substitute,
dV
= 4π102 (−2) = −800π.
dt
7. (10 points) Determine the maximal and minimal values for the function f (x) = 25 − x2
over the interval −6 ≤ x ≤ 2. Solution. Compute
f (−6) = 25 − 36 = −11.
f (2) = 25 − 4 = 21.
f 0 (x) = −2x.
The critical point(s) occur at x = 0. At this point, f (0) = 25.
The Maximum value of 25 is achieve when x = 0.
The minimum value of −11 is achieved when x = 25.
4