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..9 y)::: MATH 1013 Applied Calculus Section D Fall 2012 Test 2/ November 5 2012 Student Name: IP-No.: You have 50 minutes to solve the following problems: Show your complete work The maximal mark on this test, including any bonus points, is 100. Permitted aid: help sheet o f standard letter size with notes and formulas NO CALCULATOR graphical or nongraphical) is allowed on this test. Problem 1) (24 marks 2 .s.ec C...r I ~ ( x).::: r L,( ):: . S ~ f - 6 4 e) L,cs.).::. s 2 2 3 2 2 ~ 5 .f- 0) . s·Cs - 2)-1·(> -b) s J- "2 c 2 ) Z g) fc .x > :: s V ?, ex) f ~ X = L O ) ~ . . t ) c ~ d) f C X) .. --{ '; X - ~ I f) fcx) .e >\. s; V l x--) f ~ x : : > 5 ; VI K) { ...e.. · L > ; ? X ) -t X r ~ : H c . k ~ h) L .) ) = X 2. c ..e.) ) L/( X):::. 2 X. 1 -l- )( ) ~ -e ..J-K ·-e
8

MATH1013_TT2_2012F.pdf

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Page 1: MATH1013_TT2_2012F.pdf

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..9 y) : : :

MATH 1013 Applied Calculus Section D Fall 2012

Test 2/ November 5 2012

Student Name: IP-No.:

You have 50 minutes to solve the following problems: Show your completework The maximal mark on this test, including any bonus points, is 100.

Permitted aid: help sheet of standard letter size with notes and

formulas

NO CALCULATOR graphical or nongraphical) is allowed on this test.

Problem 1)Differentiate each function with respect to the independent variable:

(24 marks

2

.s.ec C...r

I~ ( x).:::

r

L, ( ) : :

. S ~ f - 6 4e) L,cs.).::. s 2

2 3 2 2~ 5 .f- 0).

s·Cs - 2)-1·(> -b)

s J- "2 c t 2 ) Z

g) fc .x > :: s V ?, ex)

f ~ X = L O ) ~ . . t )

d) f CX ) .. --{ '; X - ~ I

f) fcx) .e >\. s ; V l x--)

f ~ x : :> •5 ; VI K) {

...e.. · L > ; ? X ) -t X r ~ : H c . k ~

h) L .) ) = X 2. c ..e.) )

L/(X):::. 2 X . 1 -l- )( )-e ..J-K ·-e

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Problem 2) (20 marks)

You are given a formula for the derivative f (x):

I

f c x = 3x+- t

Use the formula to

a) locate any/all critical points off (i.e. points x with f (x) = 0 .

b determine the intervals over which f is decreasing, and

c) sketch a graph off based on your answers from a) and b .

a.)rex) :. 0 = > X; : : - k

3

S) ~ ~ cf v -C. ~ D e . · 5 v a. creo : ::) w ~ er.e..vu-

 f(_.,t. r p ~ o 1w lo:::_jPv;{ {o ~ ~ e _

3ropL, wf._ ,Q ,· .C _ y : v ~ ~ 67 r ~ r t ) t 's

c . . ~ a-t.; ve.

I 3 x o ~ ~ 0 ~k ) < for KC::5

=. f ;s rA ere o. =..J ~ C-eo - ~r 3 .

fclc )c) v

-3X

~

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Problem : Multiple Choice: Underline the right answer 20 marks)

a) Suppose y (2) > 0. Then the object whose position is y(t) is ...

i) speeding up

ii) slowing down

iii) moving toward the origin

iv) moving away from the origin

v) none of the above

b) Suppose y = sin-1(x). Then ...

r -2i) f = -S • '? X)

I

ii) 7 - cscc.x) ·co f cx

Iiv) L-1 -   ======/ -  .c; {-)( _l

I

iii) 7 =

v) none of the above

c) Suppose (xo, yo) is on the graph of the invertible function f. Then (f1) (yo) =

2 I

i) - f C0

) · Ko)

(

iii) iv)

v) none of the above

d) Suppose, 7 .:: VI 3x ¥- I . Then ...

ii)f j_ l

7 I3 .Xi)

iii)

I

I =iv)

7'.::: 3

IV (]x-+1)

v) none of the above.

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Problem 4) 20 marks).

Use implicit differentiation to show that the x-intercept andy-intercept of any tangent line

to the curve 1-

 J7-fC' sum up to c, where c is any positive

constant.

.0:_ ( Ck -C7{k) 7 d= dx - 2 )o(..x

/ ~_ } _ I c{ 1 02-fX' 2 -iyrx)

7::::

o<x

_a 1 I

::::._ -   ~ _-0;~ --..[ X.,- ~

ey l / IU 1--;oc- of fw {o :::;Rr. { ( ;c..-e_

p 0 ; V I { X I 1 ) ) ...., C { b (/ ( 0 ~ 5 ..( 0

I

fc - a ,.A

11- / .A; ( cJ..

¥ _,e. cc,...- rv l l . :

(

( = to + 1 CK0 ) • X - ~ } r W ; {c__, 1 o )

,0 - Yo f-  ( (Yo ) · x - k

0} ->

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Problem 5: (16 marks

Use linearization at the specified a to approximate the following numbers:

a) a .:: b4 - b)0.1

= 0e aI I

cr) f_cy):::

I

Lr x > = f a > t f )) , - o )fca + ro) · { o

fc 0 ) :: :1.[6<;wL,ve. {c x ::

--\

-   .e. I

-

[I Kc K :: efcx) for f x :: [7 ,.

fc 0/ :::

0

=f K) =

..=-) .e.

I I

.3 ,.-t2- 3 f {6 = .e :: I

f a - 1 I= > L o.t) =£

0 0-

36 4-.2/3 • 12. • Co.t-- o

- l 1-- I· 0.1- 1

3 ( r; ct t-- 1) 2 - f.t

= l - _I

1-

l o c.,.g

-   I I6 f 3 . 6 v . 6s -6 r)

Lr f I . I4 6

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Bonus question 10 points)

A tank in the shape of a lower hemisphere is 10 feet in diameter. When the fluid in the

tank is b feet deep, its volume is

V== ' i 'rb

6

. ::> 2. 2. JQ +b

where a is the radius ofthe circular surface ofthe fluid. The tank is drained at a rate of5

cubic feet per minute from an outlet at the center of its lower base. At what rate is the

height of fluid changing when there are 12 cubic feet of water in the tank?

D = t f f.

//:. ~ fi

I.. r ~ ..QI d.. ; c...-<;. Q ~ r {C ..t£~ f p / : : : : _ . x p r ~ ~ >

s . u ~ face ; -. { IV ....1...., 1) o 0. r:-, d._ b

r

2 2a_ +( r - ) =

fr 'Jd (

2.

o

~ > a = r2

- (o-& ) ~ 2 .-   2 ~ b<. ~

r f<; ~ V ? O W c S f l ) = >

2 v.d.. ~ . p p : l A p r ~ ~ S V i ~ 1 IV I .A - '> o r- o e. .ci.. b Ot J {7

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-==)> X :::- Y. - lto0

I

'1 (Xo)

<=c-rc) - c

(

V = · : · 3 2 r - t/ ) -1- b2

) - ~ b ·C3 b - 2 S )

- s > ~ . & - i7- s]3

; 2 ~ ]3rJ.. slep: Vc,t.) = l;'))--'( 'v)) - ~ Scf

1 {v 3 ~ { t, c · ) ev-.J v d v vt.--1 e V c ·) Cue { ....c i 2 e c 0 Ih. '- LA l { . 6 -ff . t.. t ~ o t e o --7 ~ ~ ~ . ;d_Rr- :

o{V

O(..f

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J c1f t -- 1-.?;f 1

wL,ve ; )

I

c 1 > a . tfG.t_.e vc.. e

o.e-.c;{ / ;) V(v) =: 12. {f 3.

I 2

=--::> - 5 = c .f · C t0 rr · be.; - · c.,;.)) ~ )

23~ ·

3 0

IL,;<;. :s. j > 0 > ~ 7 b / e _ b7 vJ\{ue_ o{ VL.,e_ J c - . . . f < V v v ? ~ d . r : o { eVaL v.e_ ~ . R Ore Vt,., , o s. V, .ll r-}3(,{ l-,oc- d s: ld.e_ f ~

/r . r rCOt..-.hL-, v v{ t'--1 ;t7

ar-.cf..._ J r r?Q.. f V

IC,Gi l -

/ 2 LO r e ~ v o . o S;,.,..

- j~ 0 r b ::: t . ~ h e { ~ ,.. bo I IV s . . e { I ~ ~I

o,,_,c( ~ e i o l v e f-or- bCJ.): ~ , r c / f . v C-,tZ 1 -cAie..

or / WC...,lM -1-G- t. f tv;d l ~ v c ( {c;,/(s; IN(__,pc-t . . ( ~v o{ v ~ t. /:;. f 2 ~ f 3.