bc mock exam 2020 solutions - afhs calculus · workers at a local office building are expected to...

Written by Bryan Passwater Special Thanks to T. Gott, T. Record, and M. Kiraly Scoring Guide by Ted Gott

𝑡 0 25 30 45 65 80

𝑊(𝑡) 2 10 13 30 60 81

𝐁𝐂𝟏:Workersatalocalofficebuildingareexpectedtobeatworkby9AM(𝑡 = 60)eachday.

Letthetwicedifferentiablefunction𝑦 = 𝑊(𝑡)betheamountofworkersinthebuildingat

time𝑡, inminutes, satisfyingthelogisticdifferentialequation𝑑𝑦𝑑𝑡

=115𝑦 S1 −

𝑦100

U

where𝑊(0) = 2.Selectedvaluesof𝑊(𝑡)aregiveninthetablewheret = 0is8AM.

(a)Find𝑊!(65)and𝑊!!(65).Usingcorrectunits, interpretthemeaningof𝑊!!(65)incontextof

thisproblem.

′W 65( ) = 115W 65( ) 1−

W 65( )100

⎛

⎝⎜

⎞

⎠⎟ =

115

60( ) 1−60( )

100

⎛

⎝⎜

⎞

⎠⎟ =

85

′′W t( ) = d2 ydt2

= 115

dydt

1− y100

⎛⎝⎜

⎞⎠⎟+ y − 1

100dydt

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

′′W 65( ) = 115

85

1− 60100

⎛⎝⎜

⎞⎠⎟+ 60( ) − 1

10085

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥= − 8

375

′′W 65( ) means the rate of the amount of workers in the

building is changing at a rate of − 8375

workers per minute

per minute at t = 65 or at 9:05 AM.

5:

1: ′W 65( )2 : ′′W t( )

0 / 2 if no product or chain rule

1: ′′W 65( )1: meaning in context

⎧

⎨

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪


𝑡 0 25 30 45 65 80

𝑊(𝑡) 2 10 13 30 60 81




=115𝑦 S1 −

𝑦100

U


(b)Evaluate[ 𝑊!(3𝑡 + 25)𝑑𝑡"

#.Showtheworkthatleadstoyouranswer.

′W 3t + 25( )dt0

∞

⌠⌡ = lim

b→∞′W 3t + 25( )dt

0

b

⌠⌡

= limb→∞

13W 3t + 25( )⎡

⎣⎢

⎤

⎦⎥

0

b

= 13

limb→∞

W 3b+ 25( )−W 25( )⎡⎣ ⎤⎦

= 13

limb→∞W 3b+ 25( )−10⎡

⎣⎤⎦ =

13

100−10⎡⎣ ⎤⎦ = 30

W t( ) is a logistic curve with horizonal asymptote y = 100

so limb→∞W 3b+ 25( ) = 100.

3:1: improper integral1: use horizontal asymptote1: value

⎧

⎨⎪

⎩⎪


𝑡 0 25 30 45 65 80

𝑊(𝑡) 2 10 13 30 60 81




=115𝑦 S1 −

𝑦100

U


(c)UsealeftRiemannsumwiththreesubintervalsindicatedbythedatainthetabletoapproximatethe

valueof145[ 𝑊(𝑡)𝑑𝑡.$%

#Usingcorrectunits, explainthemeaningof

145[ 𝑊(𝑡)𝑑𝑡$%

#inthecontextofthe

problem.

(d)UseEuler!smethodwithtwostepsofequalsizestartingat𝑡 = 45minutestoapproximate

thenumberofworkersinsidetheofficebuildingby9: 25AM(𝑡 = 85minutes).

145

W t( )dt0

45

⌠⌡ ≈ 1

452( ) 25( )+ 10( ) 5( )+ 13( ) 15( )⎡⎣ ⎤⎦ =

29545

145

W t( )dt0

45

⌠⌡ means the average number of workers

in the building from t = 0 (8 AM) to t = 45 (8:45 AM)

is about 659

workers.

3:1: left Riemann sum1: approximation1: meaning in context

⎧

⎨⎪

⎩⎪

t W t( ) dy = 115y 1− y

100⎛⎝⎜

⎞⎠⎟dx

45 30 11530( ) 1− 30100

⎛⎝⎜

⎞⎠⎟20( )= 2( ) 7

10⎛⎝⎜

⎞⎠⎟20( )=28

65 58 11558( ) 1− 58100

⎛⎝⎜

⎞⎠⎟20( )= 58

15⎛⎝⎜

⎞⎠⎟42100

⎛⎝⎜

⎞⎠⎟20( )=32.48

85 90.48

2 :1: Euler's method1: approximation⎧⎨⎩


𝑡 0 25 30 45 65 80

𝑊(𝑡) 2 10 13 30 60 81




=115𝑦 S1 −

𝑦100

U


(e)Isthereatime𝑐, suchthat25 < 𝑐 < 45,when𝑊!(𝑐) = 1.Justifyyouranswer.

(f)Let𝑃&(𝑡)betheseconddegreeTaylorpolynomialfor𝑊(𝑡)centeredat𝑡 = 65.Find𝑃&(𝑡)and

useittoapproximatethenumberofworkersinsidetheofficebuildingat9: 25AM(𝑡 = 85).

′W c( ) =W 45( )−W 25( )45− 25

= 30−1020

= 1

W t( ) is twice differentiable which mean W t( ) is continuous

so by the Mean Value Theorem there is a time c between

25 and 45 where ′W c( ) = 1, the average rate of change

W 45( )−W 25( )45− 25

.

2 : 1:W 45( )−W 25( )

45− 251: justification using MVT

⎧⎨⎪

⎩⎪

P2 t( ) =W 65( )+ ′W 65( ) t − 65( )+ ′′W 65( )2!

t − 65( )2

P2 t( ) = 60+ 85 t − 65( )− 82! 375( ) t − 65( )2

W 85( ) ≈ P2 85( ) = 60+ 85 20( )− 82! 375( ) 20( )2

= 60+ 32− 6415

= 131615

= 87.7333…

3:2 : second degree

Taylor polynomial1: approximation

⎧

⎨⎪

⎩⎪


𝐁𝐂𝟐:Thefunction𝑓istwicedifferentiablefor𝑥 ≥ −6.Aportionofthegraphof𝑓isgiveninthe

figureabove.Theareasformedbytheboundedregionsbetweenthegraphof𝑓andthe𝑥axisare

labeledinthefigure.Letthetwicedifferentiablefunction𝑔bedefinedby𝑔(𝑥) = [ 𝑓(𝑡)𝑑𝑡'

&.

(a)Find lim'→)&

𝑔(2𝑥) − 4𝑥𝑒'!)$ + 𝑥 + 1

.

limx→−2

g 2x( )− 4x⎡⎣ ⎤⎦ = g −4( )− −8( ) = f t( )dt2

−4

⌠⌡ +8

= 8− f t( )dt−4

2

⌠⌡ = 8− 10+ −2( )( ) = 0

limx→−2

ex2−4 + x +1⎡

⎣⎤⎦ = e

0 + −2( )+1= 0

limx→−2

g 2x( )− 4x

ex2−4 + x +1

= limx→−2

2 ′g 2x( )− 4

ex2−4 2x( )+1

l'Hospital's Rule! "## $##

=2 ′g −4( )− 4e0 −4( )+1

=2 f −4( )− 4

−3( ) =2 0( )− 4

−3( ) = 43

3:

1: limx→−2

g 2x( )− 4x⎡⎣ ⎤⎦1: applies L'Hospital's rule1: answer

⎧

⎨⎪⎪

⎩⎪⎪





&.

(b)Findtheabsolutemaximumof𝑔ontheinterval[−6,2].Justifyyouranswer.

′g x( ) = f x( ) = 0⇒ x = −4,0

x g x( )−6 f t( )dt

2

−6

∫ = − f t( )dt−6

2

∫ = − −5+10− 2( ) = −3

−4 f t( )dt2

−4

∫ = − f t( )dt−4

2

∫ = − 10− 2( ) = −8

0 f t( )dt2

0

∫ = − f t( )dt0

2

∫ = − −2( ) = 2

2 f t( )dt2

2

∫ = 0

The absolute maximum value of g is 2 when x = 0

3:

1: ′g x( ) = 0

1: candidates x = −4,01: answer with justification

⎧

⎨⎪⎪

⎩⎪⎪





&.

(c)Evaluate[ 𝑥𝑓!(2𝑥)𝑑𝑥*

)+.

u = x⇒ du = dx dv = ′f 2x( )dx⇒ v = 12f 2x( )

x ′f 2x( )dx−3

1

⌠⌡ = 1

2x f 2x( )⎡

⎣⎢

⎤

⎦⎥−3

1

− 12

f 2x( )dx−3

1

⌠⌡

= 12f 2( )⎛

⎝⎜⎞⎠⎟− −3

2f −6( )⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ −14

f 2x( )u!

2dx( )du!

−3

1

⌠

⌡⎮⎮

= 120( )⎛

⎝⎜⎞⎠⎟− −3

2−5( )⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ −14

f u( )du−6

2

⌠⌡

= −152

⎡

⎣⎢

⎤

⎦⎥ −14

−5( )+ 10( )+ −2( )⎡⎣ ⎤⎦ = −152− 34= − 33

4

3:2 : integration by parts1: answer⎧⎨⎩





&.

(d)TheregionRisthebaseofasolidwhosecrosssectionsperpediculartothe𝑥axisaresquares.

Write, butdonotevaluate, anintegralexpressionthatgivesthevolumeofthesolid.

(e)For𝑥 ≥ 5, thefunction𝑓isnotexplicitlygiven, but𝑓!(𝑥) = −4 m12n

')%.If𝑎, = 𝑓!(𝑛),

findq 𝑎,

"

,-%

orshowthattheseriesdiverges.

volume = f x( )⎡⎣ ⎤⎦2dx

2

5

⌠⌡⎮ 1: integral expression

−4 12

⎛⎝⎜

⎞⎠⎟

n−5

n=5

∞

∑ = −4 12

⎛⎝⎜

⎞⎠⎟

n−5

n=5

∞

∑ = −4 1

1− 12

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= −4 22−1

⎡

⎣⎢

⎤

⎦⎥ = −8 2 :

1: recognizes infinite geometric series

1: sum

⎧

⎨⎪

⎩⎪

bc mock exam 2020 solutions - afhs calculus · workers at a local office building are expected to...

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