APPM 1340
Exam 3
Fall 2015
INSTRUCTIONS: Books, notes, and electronic devices are not permitted. This exam is worth 100 points.
Box your final answers. Write neatly, top to bottom, left to right, one problem per page. A correct answer
with incorrect or no supporting work may receive no credit. SHOW ALL WORK
1. ( 24 points;6,6,6,6) Derivate the following functions with respect to x:
p
√
1 − cos x
(a) y = x x
(b) y =
(c) y = (x + x−1 )2
(d) y = 3x2 − 5x − 8x
x
Solution:
√
√
3
3 1
3 x
(a) y = x x = x 2 =⇒ y 0 = x 2 =
.
2
2
(b) y 0 =
0
x sin x + cos x − 1
(x sin x) − (1 − cos x)
=
2
x
x2
(c) y = 2(x + x
−1
)(1 − x
−2
1
1
)= 2 x+
1− 2
x
x
1
1
(d) y 0 = (3x2 − 5x)− 2 (6x − 5) − 8
2
2. (18 points;6,6,6) Consider the following functions:
f (x) =
|x+1|
x
and
p
g(x) = (x2 − 1)
 2
x ≤ −1
 x ,
−x2 + 1, −1 < x < 1
h(x) =

−1 + x2 , x > 1
and
(a) Find f 0 (−2).
(b) Find g 0 (1).
(c) Find h0 (−1).
Solution:
(a) For x = −2 we have f (x) =
−(x+1)
x
= −1 −
1
x
so f 0 (x) =
and f 0 (−2) =
1
x2
1
4
(b) Since g(x) is undefined for x < 1, we know g 0 (1) does not exist .
1
g 0 (x) = 12 (x2 − 1)− 2 (2x) = √xx2 −1 =⇒ g 0 (1) = ∅
(c) h(x) 
is two different functions to the left and the right of x.
x < −1
 2x,
−2x, −1 < x < 1
h0 (x) =

2x,
x>1
Note that
lim h0 (x) = lim (2x) = −2 while
x→−1−
x→−1−
lim h0 (x) = lim (−2x) = 2, therefore h0 (x) = ∅
x→−1+
x→−1+
3. (18 points;6,6,6) Consider the following graph of f (x) = 2x3 − 3x2 + 3 with secant line P Q:
f (x)
Q
Secant
f (a + h) − f (a)
P
f (a + h)
h
f (a)
a
h
x
a+h
(a) Find the average rate of change of f (x) between points P and Q when a = h = 1.
(b) Find the instantaneous rate of change of f (x) at point P when a = 1 and h = 0.
(c) What is the rate of change of f (x) at x = a?
Solution:
(a) P (1, 2) and Q(2, 7) means
∆y
∆x
=
7−2
2−1
= 5
(b) f 0 (x) = 6x2 − 6x =⇒ f 0 (1) = 6 − 6 = 0
(c) f 0 (a) = 6a2 − 6a = 6a(a − 1)
4. (7 points) Find constants A, B, and C such that the function y = Ax2 + Bx + C satisfies the differential
equation y 00 + y 0 − 2y = (x + 1)(x − 8).
Solution: y = Ax2 + Bx + C =⇒ y 0 = 2Ax + b =⇒ y 00 = 2A
y 00 + y 0 − 2y = (x + 1)(x − 8) =⇒ 2A + 2Ax + B − 2Ax2 − 2Bx − 2C = x2 − 7x − 8
(−2A)x2 + (2A − 2B)x + (2A + B − 2C) = (1)x2 + (−7)x + (−8)
− 2A = 1 and 2A − 2B = −7 and 2A + B − 2C = −8
A=−
1
and B = 3 and C = 5
2
5. (8 points) Balloon Advertising Company sells a giant spherical balloon that can be inflated to vary from
t3
a 2 foot diameter to a 6 foot diameter. Gas is pumped into the balloon at 3 fsec
. How fast is the radius
32π
3
growing at the moment the volume is 3 f t ?
Solution: V = 34 πr3 =⇒
dV
dt
V = 43 πr3 =
32π3
3π4
dr
dt
=
3
4π(2)2
32π
3
=
=⇒ r3 =
3
16π
= 4πr2 dr
dt = 3 therefore,
=⇒ r3 =
32
4
dr
dt
=
3
4πr2
= 8 =⇒ r = 2
MORE ON THE BACK
6. (7 points) Consider the following solid. All corners are 90◦ angles. Suppose x is continuously increasing.
How fast is the volume of the object changing with respect to x when the volume is 21?
x
2x − 1
x+1
x
x+3
Solution: V = 3x(2x−1)+x2 (x+1) = 6x2 −3x+x3 +x2 = x3 +7x2 −3x =⇒
dV
= V 0 (x) = 3x2 +14x−3
dx
When Volume equals 21 we have x3 + 7x2 − 3x = 21 or
√
x3 + 7x2 − 3x − 21 = 0 =⇒ x2 (x + 7) − 3(x + 7) = 0 =⇒ (x2 − 3)(x + 7) = 0 =⇒ x = 3
√
√
We only consider x = 3 since x = −7 and x = − 3 would create sides of negative length.
√
√
√
√
3
Now, V ( 3) = 3( 3)2 + 14( 3) − 3 = 6 + 14 3 , units
time
7. (18 points;6,6,6) Answer the following:
(a) Consider the relationship 3x2 + 3y 2 + 6x − 12y = 0. What is
(b) Find
dy
dθ
dy
dx ?
if y = (tan θ sin θ + csc θ).
A−P
, suppose T and A are constants, while r and P are variable. Describe how
(c) Given that T =
Pr
fast P is changing with respect to r.
Solution:
(a) Implicit differentiation produces: 6x + 6yy 0 + 6 − 12y 0 = 0
y 0 (6y − 12) = −6x − 6
y0 =
−6(x + 1)
(x + 1)
x+1
=−
=
6(y − 2)
y−2
2−y
(b) y 0 = tan θ cos θ + sin θ sec2 θ − csc θ cot θ or perhaps, sin θ + tan θ sec θ − csc θ cot θ
(c) T =
A−P
Pr
TPr = A − P
TPr + P = A
P (T r + 1) = A
P (r) =
P 0 (r) =
A
T r+1
(T r + 1) ∗ 0 − At
−At
=
2
(T r + 1)
(T r + 1)2
Alternatively, P (r) = A(T r + 1)−1 =⇒ P 0 (r) = −A(T r + 1)−2 =
END of Exam
−At
.
(T r + 1)2