This week
1 Review Session Thursday 12/9 at 4:00 p.m. in SMI 1022 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 examand 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 yourcalculator is in radian mode.
6 Unless otherwise instructed, ALWAYS GIVE YOURANSWERS IN EXACT FORM. For example, 3π,
√2 and
ln(2) are in exact form; the corresponding approximations9.424778, 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 = 0seconds. It bounces up and down for the next 2π secondsaccording to the following function:
s(t) = 10 e−2t sin2(
t +π
2
)0 ≤ t ≤ 2π
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 anddetermine the value of s(t) at those points.
3 Determine the global maximum of s(t).
Professor Christopher Hoffman Math 124
s(t) = 10 e−2t sin2(
t +π
2
)0 ≤ t ≤ 2π.
v(t) = 10(−2e−2t sin2(t +
π
2) + e−2t2 sin(t +
π
2) cos(t +
π
2))
Setting v(t) = 0 we get
0 = 10(−2e−2t sin2(t +
π
2) + e−2t2 sin(t +
π
2) cos(t +
π
2))
0 = −2e−2t sin2(t +π
2) + e−2t2 sin(t +
π
2) cos(t +
π
2)
0 = 2e−2t sin(t +π
2)(− sin(t +
π
2) + cos(t +
π
2))
So either
sin(t +π
2) = 0 or − sin(t +
π
2) + cos(t +
π
2) = 0
Professor Christopher Hoffman Math 124
sin(t +π
2) = 0
gives ust +
π
2= nπ
In the range of 0 ≤ t ≤ 2π we get π/2 and 3π/2.
Professor Christopher Hoffman Math 124
− sin(t +π
2) + cos(t +
π
2) = 0
sin(t +π
2) = cos(t +
π
2)
tan(t +π
2) = 1
t +π
2= tan−1(1) + nπ
t = −π2
+π
4+ nπ
t = −π4
+ nπ
In the range of 0 ≤ t ≤ 2π we find 3π/4 and 7π/4. Thus ourfour critical numbers are π/2, 3π/4,3π/2 and 7π/4.
Professor Christopher Hoffman Math 124
To see which critical points are local maxima we use either thefirst or second derivative tests.
s′′(t) = e(−2 t) sin(
12π + t
)2
+ 20 e(−2 t) cos(
12π + t
)2
−80 e(−2 t) sin(
12π + t
)cos
(12π + t
)
s′′(12π) = 20 e(−π), s′′(
34π) = −20 e(− 3
2 π)
s′′(32π) = 20 e(−3π) s′′(
74π) = −20 e(− 7
2 π)
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 orat one of the endpoints. Checking the values it is clear thats(0) = 10 is the absolute maximum.
Professor Christopher Hoffman Math 124
Differentiate the following functions. You need not simplify youranswers.
1 y = x3+ex
ln(x2+e) quotient rule
2 y = exπx logarithmic differentiation
3 y = arctan(√
x) = tan−1(√
x) chain rule
Professor Christopher Hoffman Math 124
Differentiate the following functions. You need not simplify youranswers.
1 y = x3+ex
ln(x2+e)
y ′ =(3x2 + ex ) ln(x2 + e)− (x3 + ex )(2x/(x2 + e))
(ln(x2 + e))2
2 y = exπx
ln(y) = ln(exπx ) = ln(e) + ln(xπx ) = 1 + πx ln(x)
y ′
y= 0 + π(x
1x
+ ln(x))
y ′ = yπ(1 + ln(x)) = exπxπ(1 + ln(x))
Professor Christopher Hoffman Math 124
Certain empirical studies model the number of different animalspecies, N, that exist in a tropical forest with area S, by theequation
N = aS0.25
where a is a constant. If observations indicate that S isdecreasing by 1.8% per year, estimate the yearly percentagedecrease in the number of species.
Let ∆N be the yearly change in N and ∆S be the yearlychange in S. We want to find
100∆NN
.
We are given that
100∆SS
= −1.8 or ∆S = −.018S
Professor Christopher Hoffman Math 124
Use linear approximation.
N = aS0.25
N ′(S) = .25aS−.75
Linear approximation says
∆N ≈ N ′(S)(∆S) = .25aS−.75(−.018S).
100∆NN≈ 100
.25aS−.75(−.018S)
aS.25 = 100(.25)(.018) = −.45
The number of species is dropping by about .45% per year.
Professor Christopher Hoffman Math 124
Two people start walking from the same point. The first walkswest at a rate of 5 feet per second. Ten seconds later thesecond walks northeast at a rate of 4 feet per second. How fastis the distance between them changing twenty seconds afterthe first person starts walking?
Professor Christopher Hoffman Math 124
Let θ be the angle between the two paths. Let a be the distancefrom the starting point to the first person. Let b be the distancefrom the starting point to the second person. Let c be thedistance between the two people.
Law of Cosines
a2 + b2 = c2 − 2ab cos(θ).
We want to find dcdt twenty seconds after the first walker began.
2adadt
+ 2bdbdt
= 2cdcdt− 2 cos(θ)
(a
dbdt
+ bdadt
)because θ is not changing. So we need to find
a,dadt, b,
dbdt, c
20 seconds after the first walker began.Professor Christopher Hoffman Math 124
After 20 seconds θ = 3π/4 and is not changing.
dadt
= 5 and a = 100
dbdt
= 4 and b = 40.
To find c we plug in
a2 + b2 = c2 − 2ab cos(θ).
(100)2 + (40)2 = c2 − 2(100)(40) cos(3π/4).
11600− 4000√
2 = c2.
c =
√11600− 4000
√2.
Professor Christopher Hoffman Math 124
2adadt
+ 2bdbdt
= 2cdcdt− 2 cos(θ)
(a
dbdt
+ bdadt
)
2(100)(5) + 2(40)4 = 2√
11600− 4000√
2dcdt
− 2(−√
2/2) (100(4) + 40(5)) .
dcdt
=1320− 600
√2
2√
11600− 4000√
2≈ 7.33.
Professor Christopher Hoffman Math 124
Answer the questions based on the following graph of y = g(x).
Circle the largest.
g′′(−2)12
g(3) g′(1.6) g′(1.9) g′(−1)
g′(1.6)
Professor Christopher Hoffman Math 124
Answer the questions based on the following graph of y = g(x).
Find a number b such that g′(b) is the absolute minimumfor g′(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′(c) is the absolute maximumfor g′(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′(d) = 0.
d = −2, 0, 2, 3.5, or 4.5
Professor Christopher Hoffman Math 124
Answer the questions based on the following graph of theDERIVATIVE y = f ′(x).
Find all critical number(s) of f in −3 ≤ x ≤ 5 and classifythem 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→0x sin x
1−cos x L’Hospital twice
2 limx→0x sin x
1+cos x = 02 Plug in x = 0. Don’t use L’Hospital!!
3 limx→0(1− 7x)3/2x take logs to use L’Hospital
4 Always check your work with a calculator!!
Professor Christopher Hoffman Math 124
c = limx→0
(1− 7x)3/2x
ln(c) = limx→0
ln(
(1− 7x)3/2x)
ln(c) = limx→0
(3/2x) ln (1− 7x)
ln(c) = limx→0
3 ln (1− 7x)
2xApply L’Hospital
ln(c) = limx→0
3 −71−7x
2.
ln(c) = limx→0
−212(1− 7x)
.
ln(c) =−21
2or lim
x→0(1− 7x)3/2x = c = e−21/2.
Professor Christopher Hoffman Math 124
Answer the questions based on the following table of values.x f (x) f ′(x) f ′′(x)
0 8 −1 21 6 −3 52 0 −8 103 −2 0 34 1 4 −1
1 limx→1f (3−x)2
f (3−2x)
2 limx→2x2f (x)
1−f (2x)
3 Let h(x) = (2 + 1x )f (x). Find h′(1).
4 Let r(x) = f (f (x + 3)). Find r ′(−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 ′(x) f ′′(x)
0 8 −1 21 6 −3 52 0 −8 103 −2 0 34 1 4 −1
1 limx→1f (3−x)2
f (3−2x)
Plugging in x = 1 we get
limx→1
f (3− x)2
f (3− 2x)=
f (2)2
f (1)=
02
6.
Professor Christopher Hoffman Math 124
Answer the questions based on the following table of values.x f (x) f ′(x) f ′′(x)
0 8 −1 21 6 −3 52 0 −8 103 −2 0 34 1 4 −1
2 2 limx→2x2f (x)
1−f (2x)
Here we use L’Hospital’s rule.
limx→2
x2f (x)
1− f (2x)=
2xf (x) + x2f ′(x)
−2f ′(2x)
=2(2)f (2) + 22f ′(2)
−2f ′(4)=
2(2)(0) + 4(−8)
−2(4)= 4
Professor Christopher Hoffman Math 124
Answer the questions based on the following table of values.x f (x) f ′(x) f ′′(x)
0 8 −1 21 6 −3 52 0 −8 103 −2 0 34 1 4 −1
3 (2 pts) Let h(x) = (2 + 1x )f (x). Find h′(1).
We use the product rule
h′(x) = (2 +1x
)f ′(x) + (−1/x2)f (x)
h′(1) = (2 +11
)f ′(1) + (−1/12)f (1)
h′(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 ′(x) f ′′(x)
0 8 −1 21 6 −3 52 0 −8 103 −2 0 34 1 4 −1
4 Let r(x) = f (f (x + 3)). Find r ′(−1).
By the chain rule we have
r ′(x) = f ′(f (x + 3))f ′(x + 3).
r ′(1) = f ′(f (4))f ′(4) = f ′(1)(4) = (−3)(4) = −12.
Professor Christopher Hoffman Math 124
Answer the questions based on the following table of values.x f (x) f ′(x) f ′′(x)
0 8 −1 21 6 −3 52 0 −8 103 −2 0 34 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 linearapproximation about (a, f (a)).f (1) = 6 and f ′(1) = −3 thus the equation of the tangent line is
y = f (1) + f ′(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) or x = 29/30.
Professor Christopher Hoffman Math 124
A cylinder is to be made of an elastic material. To support thestructure, a wire of fixed length L wraps around the cylinderonce as shown below. (A section of vacuum cleaner hose is agood example of this.) Notice, the cylinder can be constructedfrom a rectangular piece of material as pictured. Varying thedimensions of the rectangle changes the volume of the cylinder.Find the largest volume. Your final answer should be a functionof L. (A right circular cylinder has volume V = πr2h.)
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 = πr2h.
The variables r and h are related by
r2 + h2 = L2 or r2 = L2 − h2.
Substituting in for r2 we get
V = πr2h = π(L2 − h2)h = π(L2h − h3).
dVdh
= π(L2 − 3h2).
To find the maximum we set V ′ = 0 and get
dVdh
= 0 = π(L2 − 3h2).
L2 = 3h2 or h =L√3.
Professor Christopher Hoffman Math 124
This is a maximum because
d2Vdh2 = −3πh < 0.
The maximum volume is
V (L√3
) = π
(L2 L√
3−(
L√3
)3)
= L3(
23√
3
).
Professor Christopher Hoffman Math 124
One point on the curve
x4 + 3xy + y4 = 5
is (x , y) = (1,1). Use linear approximation to estimate the yvalue of a nearby point on the curve with x = 1.1.
x4 + 3xy + y4 = 5
4x3 + 3(xy ′ + y) + 4y3y ′ = 0
y ′(3x + 4y3) = −3y − 4x3
y ′ =−3y − 4x3
3x + 4y3
At the point (1,1) we get y ′ = −7/7 = −1.
Professor Christopher Hoffman Math 124
y ′ =−3y − 4x3
3x + 4y3
To estimate y when x = 1.1 we use tangent line approximationnear 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) = tb cos(2πt)y(t) = tb sin(2πt)
For what value of b will the tangent line to the curve through thepoint (1,0) be y = 5x − 5.
First we find t that makes x(t) = 1 and y(t) = 0. Equation ofthe tangent line at (1,0) is
y = 0 +y ′(t)x ′(t)
(x − 1)
This isy = 5x − 5
when y ′(t)x ′(t) = 5. Finally we solve for b that makes y ′(t)
x ′(t) = 5.
Professor Christopher Hoffman Math 124
To find t guess and check is easiest. t = 0? no t = 1? yes.
1 = x(t) = tb cos(2πt).
0 = y(t) = tb 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 ′(t) = tb(− sin(2πt))(2π) + btb−1 cos(2πt).
x ′(1) = 1b(− sin(2π(1)))(2π)+b1b−1 cos(2π(1)) = 0+b(1) = b.
y ′(t) = tb cos(2πt)(2π) + btb−1 sin(2πt).
y ′(1) = 1b cos(2π(1))(2π)+1¯
b−1 sin(2π(1)) = 1(1)(2π)+0 = 2π
y ′(t)x ′(t)
=2πb
Setting this equal to 5 we get
5 =y ′(t)x ′(t)
=2πb
or b =2π5
Professor Christopher Hoffman Math 124