review session thursday 12/9 at 4:00 p.m. in smi 102 final

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.


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).


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


sin(t +π

2) = 0

gives ust +

π

2= nπ

In the range of 0 ≤ t ≤ 2π we get π/2 and 3π/2.


− 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.


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.


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.


Here is a portion of the graph of s(t)


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


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))


1 y = arctan(√

x) = tan−1(√

x)

dydx

=1

1 + (√

x)2

(1

2√

x

)


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


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.


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?


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.


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.


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)



Find a number b such that g′(b) is the absolute minimumfor g′(x) in the interval −3 ≤ x ≤ 0.

b = −1



Find a number c such that g′(c) is the absolute maximumfor g′(x) in the interval −1 ≤ x ≤ 4.

c = 4



Find a number d such that g′(d) = 0.

d = −2, 0, 2, 3.5, or 4.5


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.


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!!


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.


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).



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.



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



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



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.



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.


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.)


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.


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

).


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.


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.


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.


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.


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


review session thursday 12/9 at 4:00 p.m. in smi 102 final

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