Math 128 Final Exam Fall 2015 SPECIAL CODE: 121601

Name __________________________________

Signature: __________________________________

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Academic Honesty Statement: By signing my name above, I acknowledge that I understand each of the following behaviors:

using a calculator on a cell phone (or any other communication technology); referring to a piece of paper or object with helpful information on it (cheat sheet, crib sheet, bill of a baseball cap, etc...); looking at a test or answer sheet that is not my own; allowing another student to look at my test or answer sheet; communicating with other students (verbally or nonverbally); taking the test for another student; taking my bubble sheet of answers with me when I've finished; talking while waiting to hand in my test materials to the proctors

to be a form of academic dishonesty (cheating). I am also pledging not to engage in any of these behaviors. I understand that if I do engage in these behaviors, the consequences will be failure of the exam and a formal charge of academic dishonesty to the Ombuds Office. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Please shut off all cell phones, ear phones, computers, beepers, etc… Please put everything away except a #2 pencil and a calculator that is NOT your cell phone. You may write on the test. There are thirty multiple choice questions and each question is weighted equally. 1. On the bubble sheet, where it says “Name,” please print your last name, leave a space, and then print your first name in the rectangles. Then fill in the bubbles underneath. 2. On the bubble sheet, where it says “Identification Number,” please write your entire Student ID number in the rectangles and fill in the bubbles underneath. Please double check to make sure you bubbled in your ID # correctly. 4. On the bubble sheet, where it says “Special Codes,” please write the numbers: 121601 in the rectangles and fill in the bubbles underneath. Please double check to make sure you bubbled in the special code correctly. 5. Lastly, on the bubble sheet, in the margin above your name, please neatly print “Final Exam Fall 2015”, and sign your name. Please make sure you bubble in your answers carefully on the bubble sheet and circle your answers on your test booklet.

Key

1.) Evaluate the integral: ∫ 𝑥2 (1 + 5𝑥3 + 7𝑥4 + 4

𝑥3) 𝑑𝑥 (A) 1

3𝑥3(𝑥 + 5

4𝑥4 − 7

3𝑥3 − 2𝑥2) + 𝐶

(B) 1

3𝑥3 + 5

6𝑥6 − 7

𝑥+ 4 ln|𝑥| + 𝐶

(C) 2𝑥 + 25𝑥4 − 14

𝑥3 − 4𝑥2 + 𝐶

(D) 𝑥2 + 5𝑥5 + 7

𝑥2 + 4 ln|𝑥| + 𝐶 2.) Find 𝑓(2) if 𝑓′(𝑥) = 3𝑥4 + 𝑒6𝑥 with 𝑓(0) = 3. (A) 27,145 (B) 27,142 (C) 27,148 (D) 162,803 3.) Evaluate the integral. ∫ (𝑥3 + 3𝑥2) √𝑥4 + 4𝑥3𝑑𝑥

(A) 23

(14

𝑥4 + 𝑥3) (15

𝑥5 + 𝑥4)3/2

+ 𝐶 (B) 2(𝑥4 + 4𝑥3)3/2 + 𝐶 (C) 1

6(𝑥4 + 4𝑥3)3/2 + 𝐶

(D) 2

3(𝑥4 + 4𝑥3)3/2 + 𝐶

4.) Evaluate the integral. ∫ 𝑒4𝑥 cos(𝑒4𝑥) 𝑑𝑥 (A) 1

4sin(𝑒4𝑥) + 𝐶

(B) − 1

4sin(𝑒4𝑥) + 𝐶

(C) sin(𝑒4𝑥) + 𝐶 (D) 1

4𝑒4𝑥 sin(𝑒4𝑥) + 𝐶

=SX75x5tZat¥dx= 5×75×7752+45

'

.dx

O=L

,x' t Exo -75 't4lnlxltC

Approximate5

⇒ fix = Sf 'md*=ZXtfe×tC

gI 3=fH=0tItC ⇒ c=3-f=

⇒ ftp.Z.32tfekt = 27,147.839 .. .

4=1/4+4×3⇒ du -14×3+125)dx

⇒ tydu -4×735)dx=fu" t.tydu-ty.EU "

toO=z(x4t4x)" '

to

use"

⇒ du --4e"XdxO ⇒

tydu-exdx-fco.cat#du=tysinlultC=tqsinCe4MtC

5.) Evaluate the integral. ∫ 𝑥2𝑒6𝑥𝑑𝑥 (A) 1

18𝑒3𝑒6𝑥 + 𝐶

(B) 6𝑥2𝑒6𝑥 − 72𝑥𝑒6𝑥 − 864𝑒6𝑥 + 𝐶 (C) 1

6𝑥2𝑒6𝑥 − 1

18𝑥𝑒6𝑥 − 1

108𝑒6𝑥 + 𝐶

(D) 1

6𝑥2𝑒6𝑥 − 1

18𝑥𝑒6𝑥 + 1

108𝑒6𝑥 + 𝐶

6.) Evaluate the integral. ∫ 4𝑥 sin(5𝑥)𝑑𝑥 (A) − 4

5𝑥 cos(5𝑥) + 4

25sin(5𝑥) + 𝐶

(B) 4

5𝑥 cos(5𝑥) + 4

25sin(5𝑥) + 𝐶

(C) − 4

5𝑥 cos(5𝑥) − 4

25sin(5𝑥) + 𝐶

(D) − 2

5𝑥2 cos(5𝑥) + 𝐶

7.) A graph of the relative growth rate of a population is given in the following figure: By what percentage does the population change over the 30 year period?

30 𝑡 (years)

Relative Growth Rate

−0.03

(A) Increases by 64% (B) Decreases by 36% (C) Increases by 45% (D) Decreases by 45%

Next page ⇒

O

u=4X V= - tzcoscsx )

Odu --4dx du=siul5x)d×

=- 4zxcosC5x) - S -tgcoslsxlitdx

=- 4zXcosC5x)t¥ . 'Fsinl5xltC

= - IgXcosC5x)t4zsinC5x)tC

next ⇒

page

O

5. Sketch = ur - Suda

u=x'

v -

- te "

= tape '- SEE ?2xdx

du .. zxdx do -

- e "dX=

zjeox-zyxexdxJ.si/eTh=ExEItfEdxu=xv=Ie4=zxe'

- store'

to

du -- dx dv=e*d×

=Ex2e'

- Exeat # E' to

7. Inception ) -

- So

"

Pjftfdt = I. 30 . I -003 )

⇒ eh't:?) -

o.gs=

-0.45

= e

⇒Pcso ) = e-

" " ? Pco)

=0.637628 . . .

* Pco )

orthe population decreased by 236.390

"

,

8.) Given the supply curve 𝑝 = 64 + 𝑞2 and the demand curve 𝑝 = 192 − 𝑞2, find the producer surplus when the market is in equilibrium. (A) $1365 (B) $1024 (C) $683 (D) $341 9.) You take out a car loan for $15,000. Assuming a 5% annual interest rate and a monthly payment of $340, how long will the loan run? (A) Between 3 and 4 years (B) Between 4 and 5 years (C) Between 5 and 6 years (D) Between 6 and 7 years 10.) A person will need $100,000 in 18 years. Assuming a 6% annual interest rate, how much should you deposit monthly? (A) $757 (B) $157 (C) $463 (D) $258 11.) Exam scores are normally distributed with a mean of 84 and a standard deviation of 3. What is the probability that a randomly selected person will score between a 70 and 80? (A) 0.6827 (B) 0.1000 (C) 0.0912 (D) 0.1731 12.) The mean time at a stop light is exponentially distributed with a mean of 30 seconds. Find the probability that a person will wait more than 30 seconds. (A) 0.50 (B) 0.6321 (C) 0.0333 (D) 0.3679

⇒

•

o ⇒

⇒O

o

' II;"

193¥,e-"a÷"Id×= 0.091209 .

. .

¥=µ= 30

⇒k=43#OPr (

xZ3o)=§g÷oe¥oxo#

Improper

OR = I - Pr ( Xt 30 )

= I - S !°IoE%×d× re O . 367879

8. Find intersection P T

64+92=192 - q2

a-u⇒a=:÷¥¥,in

must be the

positiveroot

P=64+82=128 II:p,dnYer=

18128 -64+9449

2341.33

-

-- - -

. - -- - - -

--

. --

-- -

- - -

--

i -. -

--

.

9. Find M,

# yearsof loan :

15000--172.3405%7=126,3407 ( e'" " II )

W

⇒ m= Its In t IT = 4.0624 . .

.

±

. -

-

- - --

- - --

-

-

-

- --

-

-- - - -

10.

$100,000 in 18 yrsr= 0.06 S= amount depositedfear

too ,ooo=A 's

se-oo.tt/---sE" " '

&E- " It

'

100,000⇒ S=

,

= 3085.34

⇒ 5/12=257.11

13.) Assume X is a uniformly distributed random variable on 3 ≤ 𝑥 ≤ 15. Find 𝑃(3 ≤ 𝑋 ≤ 5). (A) 0.0833 (B) 0.1667 (C) 0.6667 (D) 0.50 14.) X is an exponentially distributed random variable with a mean of 7. Find the median. (A) 14.29 (B) 7 (C) 4.85 (D) 1.95 15.) Given the following PDF, find the median. 𝑦 1

5 PDF

10 𝑥 (A) 5; 10; 4.56; 9.02; 10.11; 6.47 (B) 5.55; 8.28; 7.07; 9.48; 2.34; 8.49 (C) 2.93; 10.09; 12.09; 11.15; 6.16; 3.34 (D) 9.047; 7.03; 5.66; 10.75; 9.78; 6.99

°are

: ⇒ aware .

= 116

OP " " { FewI ⇒

M= 'T⇒

74k or k= 't

F- HI T=Y,fI± 4.852030 . . . .

K

y -Fox

→

illIT

0 I = red area

= IT . IoT = I of

⇒ I TIE

⇒ 1-2=50 ⇒ T = IF = I 7. 0710678 . .

.

and we want the positive root.

16.) Given the PDF {0 𝑥 < 2

118

𝑥 − 19

2 ≤ 𝑥 ≤ 80 𝑥 > 8

, find the mean.

(A) 6 (B) 4 (C) 5.93 (D) 4.93 17.) A piece of electronic surveying equipment is designed to operate is specific temperatures and humidity levels. The higher the performance index, the better the performance. If the performance index of the function 𝑃 = 𝑓(ℎ, 𝑡) is given in the table below, estimate 𝑓(10,20).

(A) 0.81 (B) 0.88 (C) 0.91 (D) 0.18 18.) Given 𝑓(𝑥, 𝑦) = 3𝑥2𝑦 + 6𝑦2 − 5𝑥𝑒𝑦2 + 𝑦 cos(3𝑥), find 𝑓𝑥(𝑥, 𝑦). (A) 6𝑥𝑦 + 12𝑦 − 10𝑦𝑒𝑦2 − 3𝑦 sin(3𝑥) (B) 3𝑥2 + 12𝑦 − 10𝑥𝑦𝑒𝑦2 + cos(3𝑥) (C) 6𝑥 − 10𝑦𝑒𝑦2 − 3 sin(3𝑥) (D) 6𝑥𝑦 − 5𝑒𝑦2 − 3𝑦 sin(3𝑥)

Oµ

-

- sasxhtsxafdx-

- ÷ x'

- txt != # ( (83587-92%-24)=6

Ah -

- 10

O f ( 10,20) = f (0+10,20+0)At = O

I Flo,

20) t fhco,207 . Ah t ft ( 0,20 ) . At

f×=6xy to - 5eE3ysinl3x)

v.io?ne:.i.-esIinEeiom

.

⇒ fl 10,282 0.81 toff . 10 25

= 0.882

19.) Given 𝑓(𝑥, 𝑦) = sin(8𝑥𝑒7𝑥𝑦). Find 𝑓𝑦(𝑥, 𝑦). (A) −56𝑥2𝑒7𝑥𝑦 cos(8𝑥𝑒7𝑥𝑦) (B) cos(56𝑥2𝑒7𝑥𝑦) (C) (56𝑥2𝑒7𝑥𝑦 + 8𝑒7𝑥𝑦) cos(8𝑥𝑒7𝑥𝑦) (D) 56𝑥2𝑒7𝑥𝑦 cos (8𝑥𝑒7𝑥𝑦) 20.) Find 𝑓𝑥𝑦 if 𝑓(𝑥, 𝑦) = 8𝑦 sin(𝑥𝑦). (A) −8𝑦3 sin(𝑥𝑦) + 16𝑥𝑦 cos(𝑥𝑦) (B) 8𝑦 cos(𝑥𝑦) (C) −8𝑥𝑦2 sin(𝑥𝑦) + 16𝑦 cos(𝑥𝑦) (D) 8𝑦2 cos(𝑥𝑦) 21.) The function 𝑓(𝑥, 𝑦) = 𝑥2 + 𝑦3 − 3𝑥 + 5𝑦2 − 9 has a local minimum at what y value? (A) 𝑦 = 0 (B) 𝑦 = − 10

3

(C) 𝑦 = 0, 𝑦 = − 10

3

(D) 𝑦 = 3

2

22.) Find the 𝑥 coordinate of the critical value(s) of 𝑓(𝑥, 𝑦) = 𝑥𝑦 − 2𝑥3 + 𝑦2 (A) 𝑥 = 0 (B) 𝑥 = 0, 𝑥 = − 1

12

(C) 𝑥 = − 1

12, 𝑥 = 0, 𝑥 = 1

12

(D) 𝑥 = −8, 0, 8

Fy -_coslsixe " ) . ( 8×2×4

"

= cos (8×2×4) . 56×22×4

O

f×=8y . costly) . y=8y%sHy)

fxy -

- 8 fsinlxyl ) .Xt2ycosHyD

O = - 8xjs.su/XyltI6ycosHy)

Onext ⇒

page

oi

is;÷o⇒x=Iin

Xt 2.6×2=0 X=0

X ( 1+12×1=0⇒

.

or

Y wonx= - 1112

21 .

fix,

= X 't y'

- 3×+5×2-9

Fx = 2×-3=0⇒ X= 312

fy=3yIl0y=o⇒ y ( 3×+14=0 ⇒ 411013

crit .pt, :( 312,0) and C %,

- '%)-

2

Fxx = 2 fxy=O D= fxxfyy - fxy

Fyy'

- 6ytIO

2. (6.0+10) -02=20 >

°⇒( 312,0)local minDI ph ,d= Fxx > 0

D ( 312,

-" b) = 2. ( 6 't 'T )t10 ) - &

⇒ ( 3h,

- " Is )

= 2. (

'

-20 to ) =- 20 LO saddle

23.) Suppose 𝑓(𝑥, 𝑦) = 𝑥2 + 𝐴𝑥 + 𝑦2 + 𝐵𝑦 + 𝐶 has a local minimum value of 21 at the point (2,3). Find the value of A. (A) −6 (B) −4 (C) 2 (D) Not enough information to find A. 24.) A company sells two products, x and y. The profit the company makes is given by the function 𝑃(𝑥, 𝑦) = 30𝑥𝑦 − 4𝑥 − 3𝑦 + 2, where P is measured in hundreds of thousands of dollars. Find the maximum profit. (A) $160,000 (B) $170,000 (C) $320,000 (D) There is no maximum profit. 25.) Given the following information:

𝑓𝑥(3,2) = 0, 𝑓𝑦(3,2) = 0, 𝑓𝑥𝑥(3,2) = 9, 𝑓𝑦𝑦(3,2) = −4, 𝑓𝑥𝑦(3,2) = −7 𝑓𝑥(4,5) = 0, 𝑓𝑦(4,5) = 0, 𝑓𝑥𝑥(4,5) = −2, 𝑓𝑦𝑦(4,5) = −8, 𝑓𝑥𝑦(4,5) = 3

Which of the following is true? (A) There is a local minimum at (3,2) and a local maximum at (4,5). (B) There is a saddle point at (3,2) and a local maximum at (4,5). (C) There is a local minimum at (3,2) and a saddle point at (4,5). (D) There is a saddle point at (3,2) and a local minimum at (4,5).

0 next page ⇒

•over x

, yzo Pix, yl has ng

maximum since you can let x ,y→ to.

DO

cprftgical : {3/7}D= 9 . -4 - I -712 =

- 8520 SADDLE

D= - 2 . - 8 - 32 = 7 SO

Fxx 14,57 = - 2 coLocal Max

23. fix

, y ) = X 't Axt ya t By to has a local

min of 21 at (2,37 .

Fx = 2x t A = O ⇒ X =

- Ala = 2 ⇒ A

= -4

Fy = 2ytB = o ⇒ y =- 13/2=3 ⇒ B =

- 6

mini ffyx,22

F'"

⇒

⇒ D= fxx.fi/y-fxy= 2.2-0=4 > 0

⇒ Dso ,Fxx > 0

⇒ 12,37 is a local min.

26.) Suppose the quantity of items produced, 𝑞, depends on the number of two items, 𝑥 and 𝑦, needed to produce the item and can be modeled by the function:

𝑞 = 6𝑥3/4 𝑦1/4 If the cost is $18 per unit x, and $20 per unit y, with a total budget of $1728, find the optimal quantity produced. (A) 314 (B) 487 (C) 300 (D) 320 27.) Find the global minimum of the function. 𝑓(𝑥, 𝑦) = 𝑥2 − 12𝑥 + 𝑦2 − 16𝑦 with constraint 2𝑥 + 2𝑦 = 12. (A) −68 (B) −61 (C) −36 (D) −74 28.) Find the particular solution of the differential equation: 𝑥5 𝑑𝑦

𝑑𝑥= 𝑦 with 𝑦(1) = 3𝑒−1/4.

(A) 𝑦 = 𝑒− 1/4𝑥4−1/4 (B) 𝑦 = −3𝑒− 1/4𝑥4 (C) 𝑦 = 𝑒− 1/4𝑥4 +1/4 (D) 𝑦 = 3𝑒− 1/4𝑥4

next ⇒page

O

O next ⇒page

fTEDx ⇒ In ly I =- I x'

"t C

⇒ ly I = I ¥ " I CE"" "

Cc so )-4x4

O ⇒ y = Ce

I :3 e-' "

= ya , = c e

"=

CE" 4

⇒c =3

- ¥y Ix ) = 3 e

26. qcx , y ) = 6 X' "

y" 4

gcx ,y ) = 18×+204 = 1728

oh,= ¥ x'

" "

y" "=9zY = 118 gets

X' 14

9y=Ex""

y.

' ' '

Ez = a so Sy-

- 20

:÷I÷=÷o¥ ⇒ txt"÷=¥fn-

⇒ y= 3¥ . x= Fox4=311047

*

1728 = 18 x t zox= 24 x ⇒ X = 172241=72

*962,2161231971577y= F. 72=21.6

I" ¥::'

: :::: ⇒

".

constraint : 12=2×+2 ,= 2 Cy - 2) t2y=4y -4

⇒ .4y= 16 ⇒ f- 4 and X -

- y - 2--4-2=2

f (2,4 ) =22-12.2+4116-4

= 4-24+16 - 64

= - 68

29.) Find the particular solution to the differential equation: 𝑥𝑦′ = 4𝑥2𝑒2𝑥 with 𝑦(0) = 18. (A) 𝑦 = 2𝑥𝑒2𝑥 − 𝑒2𝑥 + 19 (B) 𝑦 = 4𝑥𝑒2𝑥 − 2𝑒2𝑥 + 20 (C) 𝑦 = 2𝑥𝑒2𝑥 − 𝑒2𝑥 + 18 (D) 𝑦 = 2𝑥𝑒2𝑥 − 2𝑒2𝑥 + 19 30.) A flower pot weighing 3 pounds falls from a balcony 150 feet above the ground. Using that acceleration due to gravity is −32𝑓𝑡/ s2 , how long does it take the flower pot to hit the ground? (A) 2.6 seconds (B) 2.8 seconds (C) 3.1 seconds (D) 3.5 seconds Second Derivative Test for Functions of Two Variables:

𝐷 = 𝐷(𝑎, 𝑏) = 𝑓𝑥𝑥(𝑎, 𝑏)𝑓𝑦𝑦(𝑎, 𝑏) − [𝑓𝑥𝑦(𝑎, 𝑏)]2

a) If 𝐷 > 0 and 𝑓𝑥𝑥(𝑎, 𝑏) > 0, then 𝑓(𝑎, 𝑏) is a local minimum. b) If 𝐷 > 0 and 𝑓𝑥𝑥(𝑎, 𝑏) < 0, then 𝑓(𝑎, 𝑏) is a local maximum. c) If 𝐷 < 0, then 𝑓(𝑎, 𝑏) is not a local maximum or minimum. d) If D =0, then the test is inconclusive.

O

next ⇒page

O next page⇒

This page is left blank for additional work.

29 . Xy'

= 4×2 EH

{ ya ,

⇒ {Y

'

= "¥=y×eax1/107=18

ylxi-fyixidx-4fxexdx-4-Exex-SEE.ch)

2X

u-

. x u-

- Ee =2×e" - e"

to

du -

- dx du -

- e' "

dx

I 18--4107=2.0 . I - Etc =- Itc ⇒/c=l9J

30 .

The weight of the object does nut matter ! ( weather.ws? )act ) =

- 32 ⇒ VH1 = Sandt = -32ft C

v 103=0 (falls ) O = No ) ⇒ c=O

sus-

- 150 ⇒ set )=SuAIdt=S→2tdtSH ) = -32ft 't C

I 150=5107 = Otc ⇒ SHI =- 161-2+150

when is 541=0 ? - 161-7150=0

F- bio ⇒ t ' IEE= I 3.06186 - -

-

and we want the positive root.