This week
1
Review Session Thursday 12/9 at 4:00 p.m. in SMI 102
2
Final Exam Saturday December 11, 1:30-4:20 CMU 120
(10:30 section)
3
Final Exam Saturday December 11, 1:30-4:20 THO 101
(1:30 section)
Professor Christopher Hoffman
Math 124
Final Exam Instructions
1
Your exam contains 8 questions and 9 pages;
2
You have 2 hours and 50 minutes for this final exam.
3
Make sure to ALWAYS SHOW YOUR WORK; you will not
receive any partial credit unless all work is clearly shown.
If in doubt, ask for clarification.
4
If you need extra space, use the back pages of the exam
and clearly indicate this.
5
You are allowed one 8.5 × 11 sheet of handwritten notes
(both sides). Graphing calculators are NOT allowed;
scientific calculators are allowed. Make sure your
calculator is in radian mode.
6
Unless otherwise instructed, ALWAYS GIVE YOUR
√
ANSWERS IN EXACT FORM. For example, 3π, 2 and
ln(2) are in exact form; the corresponding approximations
9.424778, 1.4142 and 0.693147 are NOT in exact form.
Professor Christopher Hoffman
Math 124
A tennis ball is dropped from a height of 10 feet at time t = 0
seconds. It bounces up and down for the next 2π seconds
according to the following function:
π
s(t) = 10 e−2t sin2 t +
0 ≤ t ≤ 2π
2
where s(t) is the distance of the ball from the ground.
1
Find all the times when the velocity of the ball is zero.
2
Find all values of t for which s(t) is a local maximum and
determine the value of s(t) at those points.
3
Determine the global maximum of s(t).
Professor Christopher Hoffman
Math 124
π
s(t) = 10 e−2t sin2 t +
0 ≤ t ≤ 2π.
2
π
π
π v (t) = 10 −2e−2t sin2 (t + ) + e−2t 2 sin(t + ) cos(t + )
2
2
2
Setting v (t) = 0 we get
π
π
π 0 = 10 −2e−2t sin2 (t + ) + e−2t 2 sin(t + ) cos(t + )
2
2
2
π
π
π
) + e−2t 2 sin(t + ) cos(t + )
2
2
2
π π
π 0 = 2e−2t sin(t + ) − sin(t + ) + cos(t + )
2
2
2
So either
0 = −2e−2t sin2 (t +
sin(t +
π
)=0
2
or
− sin(t +
Professor Christopher Hoffman
π
π
) + cos(t + ) = 0
2
2
Math 124
sin(t +
π
)=0
2
gives us
π
= nπ
2
In the range of 0 ≤ t ≤ 2π we get π/2 and 3π/2.
t+
Professor Christopher Hoffman
Math 124
− sin(t +
π
π
) + cos(t + ) = 0
2
2
sin(t +
π
π
) = cos(t + )
2
2
tan(t +
t+
π
)=1
2
π
= tan−1 (1) + nπ
2
t =−
π π
+ + nπ
2 4
π
+ nπ
4
In the range of 0 ≤ t ≤ 2π we find 3π/4 and 7π/4. Thus our
four critical numbers are π/2, 3π/4,3π/2 and 7π/4.
t =−
Professor Christopher Hoffman
Math 124
To see which critical points are local maxima we use either the
first or second derivative tests.
2
2
1
1
(−2 t)
+ 20 e
cos
s (t) = e
sin
π+t
π+t
2
2
1
1
(−2 t)
−80 e
sin
π + t cos
π+t
2
2
00
(−2 t)
1
s00 ( π) = 20 e(−π) ,
2
3
s00 ( π) = 20 e(−3 π)
2
3
3
s00 ( π) = −20 e(− 2 π)
4
7
7
s00 ( π) = −20 e(− 2 π)
4
Thus 3π/4 and 7π/4 are the maxima.
Professor Christopher Hoffman
Math 124
The values of the functions at the local maxima are
s(3π/4) = 5e−3π/2 ≈ .0449
and
s(7π/4) = 5e−7π/2 ≈ .0000838.
The global maxima either occurs at one of the critical points or
at one of the endpoints. Checking the values it is clear that
s(0) = 10 is the absolute maximum.
Professor Christopher Hoffman
Math 124
Here is a portion of the graph of s(t)
Professor Christopher Hoffman
Math 124
Here is a portion of the graph of s(t)
Professor Christopher Hoffman
Math 124
Here is a portion of the graph of s(t)
Professor Christopher Hoffman
Math 124
Differentiate the following functions. You need not simplify your
answers.
x 3 +ex
ln(x 2 +e)
1
y=
2
y = ex πx
3
quotient rule
logarithmic differentiation
√
√
y = arctan( x) = tan−1 ( x)
Professor Christopher Hoffman
Math 124
chain rule
Differentiate the following functions. You need not simplify your
answers.
1
y=
x 3 +ex
ln(x 2 +e)
y0 =
2
(3x 2 + ex ) ln(x 2 + e) − (x 3 + ex )(2x/(x 2 + e))
(ln(x 2 + e))2
y = ex πx
ln(y ) = ln(ex πx ) = ln(e) + ln(x πx ) = 1 + πx ln(x)
y0
1
= 0 + π(x + ln(x))
y
x
y 0 = y π(1 + ln(x)) = ex πx π(1 + ln(x))
Professor Christopher Hoffman
Math 124
1
√
√
y = arctan( x) = tan−1 ( x)
dy
1
√
=
dx
1 + ( x)2
Professor Christopher Hoffman
Math 124
1
√
2 x
Certain empirical studies model the number of different animal
species, N, that exist in a tropical forest with area S, by the
equation
N = aS 0.25
where a is a constant. If observations indicate that S is
decreasing by 1.8% per year, estimate the yearly percentage
decrease in the number of species.
Let ∆N be the yearly change in N and ∆S be the yearly
change in S. We want to find
100
∆N
.
N
We are given that
100
∆S
= −1.8
S
Professor Christopher Hoffman
or
∆S = −.018S
Math 124
Use linear approximation.
N = aS 0.25
N 0 (S) = .25aS −.75
Linear approximation says
∆N ≈ N 0 (S)(∆S) = .25aS −.75 (−.018S).
.25aS −.75 (−.018S)
∆N
≈ 100
= 100(.25)(.018) = −.45
N
aS .25
The number of species is dropping by about .45% per year.
100
Professor Christopher Hoffman
Math 124
Two people start walking from the same point. The first walks
west at a rate of 5 feet per second. Ten seconds later the
second walks northeast at a rate of 4 feet per second. How fast
is the distance between them changing twenty seconds after
the first person starts walking?
Professor Christopher Hoffman
Math 124
Let θ be the angle between the two paths. Let a be the distance
from the starting point to the first person. Let b be the distance
from the starting point to the second person. Let c be the
distance between the two people.
Law of Cosines
a2 + b2 = c 2 − 2ab cos(θ).
We want to find
dc
dt
twenty seconds after the first walker began.
da
db
dc
db
da
2a
+ 2b
= 2c
− 2 cos(θ) a
+b
dt
dt
dt
dt
dt
because θ is not changing. So we need to find
a,
da
,
dt
b,
db
,
dt
20 seconds after the first walker began.
Professor Christopher Hoffman
Math 124
c
After 20 seconds θ = 3π/4 and is not changing.
da
=5
dt
db
=4
dt
To find c we plug in
and
a = 100
and
b = 40.
a2 + b2 = c 2 − 2ab cos(θ).
(100)2 + (40)2 = c 2 − 2(100)(40) cos(3π/4).
√
11600 − 4000 2 = c 2 .
c=
q
√
11600 − 4000 2.
Professor Christopher Hoffman
Math 124
2a
da
db
dc
db
da
+ 2b
= 2c
− 2 cos(θ) a
+b
dt
dt
dt
dt
dt
q
√ dc
2(100)(5) + 2(40)4 = 2 11600 − 4000 2
dt
√
− 2(− 2/2) (100(4) + 40(5)) .
√
dc
1320 − 600 2
= p
√ ≈ 7.33.
dt
2 11600 − 4000 2
Professor Christopher Hoffman
Math 124
Answer the questions based on the following graph of y = g(x).
Circle the largest.
g 00 (−2)
1
g(3)
2
g 0 (1.6)
g 0 (1.6)
Professor Christopher Hoffman
Math 124
g 0 (1.9)
g 0 (−1)
Answer the questions based on the following graph of y = g(x).
Find a number b such that g 0 (b) is the absolute minimum
for g 0 (x) in the interval −3 ≤ x ≤ 0.
b = −1
Professor Christopher Hoffman
Math 124
Answer the questions based on the following graph of y = g(x).
Find a number c such that g 0 (c) is the absolute maximum
for g 0 (x) in the interval −1 ≤ x ≤ 4.
c=4
Professor Christopher Hoffman
Math 124
Answer the questions based on the following graph of y = g(x).
Find a number d such that g 0 (d) = 0.
d = −2,
0,
2,
Professor Christopher Hoffman
3.5,
Math 124
or
4.5
Answer the questions based on the following graph of the
DERIVATIVE y = f 0 (x).
Find all critical number(s) of f in −3 ≤ x ≤ 5 and classify
them as local maxima, local minima or neither.
x = 3.5
neither a maxima nor a minima.
Professor Christopher Hoffman
Math 124
Compute these limits.
1
limx→0
x sin x
1−cos x
2
limx→0
x sin x
1+cos x
3
limx→0 (1 − 7x)3/2x
4
Always check your work with a calculator!!
L’Hospital twice
=
0
2
Plug in x = 0. Don’t use L’Hospital!!
take logs to use L’Hospital
Professor Christopher Hoffman
Math 124
c = lim (1 − 7x)3/2x
x→0
ln(c) = lim ln (1 − 7x)3/2x
x→0
ln(c) = lim (3/2x) ln (1 − 7x)
x→0
3 ln (1 − 7x)
2x
x→0
ln(c) = lim
Apply L’Hospital
−7
3 1−7x
.
2
x→0
ln(c) = lim
−21
.
x→0 2(1 − 7x)
ln(c) = lim
ln(c) =
−21
2
or
lim (1 − 7x)3/2x = c = e−21/2 .
x→0
Professor Christopher Hoffman
Math 124
Answer the questions based on the following table of values.
x f (x) f 0 (x) f 00 (x)
0
8
−1
2
1
6
−3
5
2
0
−8
10
3 −2
0
3
4
1
4
−1
1
limx→1
f (3−x)2
f (3−2x)
2
limx→2
x 2 f (x)
1−f (2x)
3
Let h(x) = (2 + x1 )f (x). Find h0 (1).
4
Let r (x) = f (f (x + 3)). Find r 0 (−1).
5
Use linear approximation to estimate f −1 (6.1).
Professor Christopher Hoffman
Math 124
Answer the questions based on the following table of values.
x f (x) f 0 (x) f 00 (x)
0
8
−1
2
1
6
−3
5
2
0
−8
10
3 −2
0
3
4
1
4
−1
1
limx→1
f (3−x)2
f (3−2x)
Plugging in x = 1 we get
f (3 − x)2
f (2)2
02
=
=
.
f (1)
6
x→1 f (3 − 2x)
lim
Professor Christopher Hoffman
Math 124
Answer the questions based on the following table of values.
x f (x) f 0 (x) f 00 (x)
0
8
−1
2
1
6
−3
5
2
0
−8
10
3 −2
0
3
4
1
4
−1
2
2 limx→2
x 2 f (x)
1−f (2x)
Here we use L’Hospital’s rule.
2xf (x) + x 2 f 0 (x)
x 2 f (x)
=
−2f 0 (2x)
x→2 1 − f (2x)
lim
=
2(2)(0) + 4(−8)
2(2)f (2) + 22 f 0 (2)
=
=4
−2(4)
−2f 0 (4)
Professor Christopher Hoffman
Math 124
Answer the questions based on the following table of values.
x f (x) f 0 (x) f 00 (x)
0
8
−1
2
1
6
−3
5
2
0
−8
10
3 −2
0
3
4
1
4
−1
3
(2 pts) Let h(x) = (2 + x1 )f (x). Find h0 (1).
We use the product rule
h0 (x) = (2 +
1 0
)f (x) + (−1/x 2 )f (x)
x
1
h0 (1) = (2 + )f 0 (1) + (−1/12 )f (1)
1
h0 (1) = 3(−3) + (−1)(6) = −15
Professor Christopher Hoffman
Math 124
Answer the questions based on the following table of values.
x f (x) f 0 (x) f 00 (x)
0
8
−1
2
1
6
−3
5
2
0
−8
10
3 −2
0
3
4
1
4
−1
4
Let r (x) = f (f (x + 3)). Find r 0 (−1).
By the chain rule we have
r 0 (x) = f 0 (f (x + 3))f 0 (x + 3).
r 0 (1) = f 0 (f (4))f 0 (4) = f 0 (1)(4) = (−3)(4) = −12.
Professor Christopher Hoffman
Math 124
Answer the questions based on the following table of values.
x f (x) f 0 (x) f 00 (x)
0
8
−1
2
1
6
−3
5
2
0
−8
10
3 −2
0
3
4
1
4
−1
5
Use linear approximation to estimate f −1 (6.1).
We want to find a that makes f (a) = 6 and then use linear
approximation about (a, f (a)).
f (1) = 6 and f 0 (1) = −3 thus the equation of the tangent line is
y = f (1) + f 0 (1)(x − 1) = 6 − 3(x − 1).
Now we plug in 6.1 for y and solve for x.
6.1 = 6 − 3(x − 1)
Professor Christopher Hoffman
or
Math 124
x = 29/30.
A cylinder is to be made of an elastic material. To support the
structure, a wire of fixed length L wraps around the cylinder
once as shown below. (A section of vacuum cleaner hose is a
good example of this.) Notice, the cylinder can be constructed
from a rectangular piece of material as pictured. Varying the
dimensions of the rectangle changes the volume of the cylinder.
Find the largest volume. Your final answer should be a function
of L. (A right circular cylinder has volume V = πr 2 h.)
Professor Christopher Hoffman
Math 124
Our variables our the radius r and the height h.
The equation we are finding the maximum of is
V = πr 2 h.
The variables r and h are related by
r 2 + h 2 = L2
or
r 2 = L2 − h2 .
Substituting in for r 2 we get
V = πr 2 h = π(L2 − h2 )h = π(L2 h − h3 ).
dV
= π(L2 − 3h2 ).
dh
To find the maximum we set V 0 = 0 and get
dV
= 0 = π(L2 − 3h2 ).
dh
L
L2 = 3h2 or h = √ .
3
Professor Christopher Hoffman
Math 124
This is a maximum because
d 2V
= −3πh < 0.
dh2
The maximum volume is
L
L
V ( √ ) = π L2 √ −
3
3
Professor Christopher Hoffman
L
√
3
3 !
Math 124
3
=L
2
√
3 3
.
One point on the curve
x 4 + 3xy + y 4 = 5
is (x, y ) = (1, 1). Use linear approximation to estimate the y
value of a nearby point on the curve with x = 1.1.
x 4 + 3xy + y 4 = 5
4x 3 + 3(xy 0 + y ) + 4y 3 y 0 = 0
y 0 (3x + 4y 3 ) = −3y − 4x 3
y0 =
−3y − 4x 3
3x + 4y 3
At the point (1, 1) we get y 0 = −7/7 = −1.
Professor Christopher Hoffman
Math 124
y0 =
−3y − 4x 3
3x + 4y 3
To estimate y when x = 1.1 we use tangent line approximation
near the point (1, 1). The equation of the tangent line is
y = 1 − 1(x − 1).
Plugging in x = 1.1 we get
y = 1 − 1(1.1 − 1) = .9.
Professor Christopher Hoffman
Math 124
Here are the parametric equations of a spiral curve:
x(t) = t b cos(2πt)
y (t) = t b sin(2πt)
For what value of b will the tangent line to the curve through the
point (1, 0) be y = 5x − 5.
First we find t that makes x(t) = 1 and y (t) = 0. Equation of
the tangent line at (1, 0) is
y =0+
y 0 (t)
(x − 1)
x 0 (t)
This is
y = 5x − 5
when
y 0 (t)
x 0 (t)
= 5. Finally we solve for b that makes
Professor Christopher Hoffman
Math 124
y 0 (t)
x 0 (t)
= 5.
To find t guess and check is easiest. t = 0? no t = 1? yes.
1 = x(t) = t b cos(2πt).
0 = y (t) = t b sin(2πt).
0 = sin(2πt).
2πt = nπ.
t = n/2.
1 = x(n/2) = (n/2)b cos(2πn/2).
(2/n)b = cos(nπ) = ±1.
So n = 2 and t = 1.
Professor Christopher Hoffman
Math 124
x 0 (t) = t b (− sin(2πt))(2π) + bt b−1 cos(2πt).
x 0 (1) = 1b (− sin(2π(1)))(2π)+b1b−1 cos(2π(1)) = 0+b(1) = b.
y 0 (t) = t b cos(2πt)(2π) + bt b−1 sin(2πt).
y 0 (1) = 1b cos(2π(1))(2π)+1b−1 sin(2π(1)) = 1(1)(2π)+0 = 2π
¯
y 0 (t)
2π
=
x 0 (t)
b
Setting this equal to 5 we get
5=
y 0 (t)
2π
=
0
x (t)
b
Professor Christopher Hoffman
or
Math 124
b=
2π
5