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$: BC Calculus: True or False? - MasterMathMentor.com...BC Calculus: True or False? 29. Time (60 seconds) x2 x2−16 ∫dx can be found by the technique of creating partial fractions:$
BC Calculus: True or False?

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26. Time (30 seconds)

A population is modeled by a function P that

satisfies the differential equationdPdt

= P500

5− P1000

⎛⎝⎜

⎞⎠⎟

.

If P 0( ) = 100,the population is growing the fastest when P = 1000.

Solution


26.

A population is modeled by a function P that

satisfies the differential equation dPdt

= P500

5− P1000

⎛⎝⎜

⎞⎠⎟

.

If P 0( ) = 100,the population is growing the fastest when P = 1000.

Logistic growth: limt→∞

dPdt

= 0⇒ 5− P1000

= 0⇒ P = 5000

The fastest growth occurs at half the carrying capacity or 0.5 5000( ) = 2500

ORdPdt

= P100

− P2

500000⇒ d 2Pdt2

= 1100

− P250000

⎛⎝⎜

⎞⎠⎟dPdt

= 0

P250000

= 1100

⇒100P = 250000⇒ p = 2500

False




6x

0

1

∫ dx = 6

x31

∞

∫ dx

Solution


27.

6x

0

1

∫ dx = 6

x31

∞

∫ dx

6x−1 2 dx = 12x1 2⎡⎣ ⎤⎦01= 12

0

1

∫6x−3 2

1

∞

∫ dx = −12x−1 2⎡⎣ ⎤⎦1∞= −12

x

⎡

⎣⎢

⎤

⎦⎥1

∞

= 0− −12( ) = 12

True




The rate of change of people waiting to download a popular book from the library is

given by w t( ) as shown in the figure below, where t is measured in weeks.

w t( )dt represents how many people are in line from week 0 to week 6.0

6

∫

Solution


28.

The rate of change of people waiting to download a popular book from the library is

given by w t( ) as shown in the figure below, where t is measured in weeks.

w t( )dt represents how many people are in line from week 0 to week 6.0

6

∫

w t( )dt0

6

∫ represents how many additional people are in line from week 0 to week 6.

The total number of people in line from weeks 0 to 6 is given by W0 + w t( )dt0

6

∫where W0 represents how many people were in line to begin with.

False




x2

x2 −16dx∫ can be found by the technique of creating partial fractions:

x2

x2 −16dx∫ = x2

x − 4( ) x + 4( ) dx∫ = 2x − 4

− 2x + 4

⎛⎝⎜

⎞⎠⎟dx∫ = 2ln x − 4 − 2ln x + 4 +C

Solution


29.

x2

x2 −16dx∫ can be found by the technique of creating partial fractions:

x2

x2 −16dx∫ = x2

x − 4( ) x + 4( ) dx∫ = 2x − 4

− 2x + 4

⎛⎝⎜

⎞⎠⎟dx∫ = 2ln x − 4 − 2ln x + 4 +C

Partial fractions will work but the denominator must have the highest power of x.

Doing long division: x2 −16 x2 = 1+ 16x2 −16

allows partial fractions to be used.

Answer is x + 2ln x − 4 − 2ln x + 4 +C

False




If the length of f x( ) on 1,8⎡⎣ ⎤⎦ is given by 1+ 1x2 dx,

1

8

∫the equation of f x( ) could be f x( ) = 8− ln x

Solution


30.

If the length of f x( ) on 1,8⎡⎣ ⎤⎦ is given by 1+ 1x2 dx,

1

8

∫the equation of f x( ) could be f x( ) = 8− ln x

′f x( )⎡⎣ ⎤⎦2= 1x2 ⇒ ′f x( ) = ± 1

x⇒ f x( ) = ± ln x +C

Since x is positive on 1,8⎡⎣ ⎤⎦ , f x( ) = 8− ln x is a possibility

True




In performing Euler's method, with f 0( ) = −2, Δx = 0.5, and for all points

dydx

= 2, then the approximation for f 3( ) = 10.

Solution


31.

In performing Euler's method, with f 0( ) = −2, Δx = 0.5, and for all points

dydx

= 2, then the approximation for f 3( ) = 10.

If dydx

= 2, then dy = 2dx = 2 0.5( ) = 1

f 0.5( ) = −2+1= −1, f 1( ) = −1+1= 0, f 1.5( ) = 0+1= 1

f 2( ) = 1+1= 2, f 2.5( ) = 2+1= 3, f 3( ) = 3+1= 4

False




The third degree Taylor-polynomial for f x( ) is given by −1( )k+1⋅ k +1( )!⋅ x −1( )k

k=0

3

∑The value of

′f 1( )f 1( ) −

′′′f 1( )′′f 1( ) = 10

Solution


32.

The third degree Taylor-polynomial for f x( ) is given by −1( )k+1⋅ k +1( )!⋅ x −1( )k

k=0

3

∑The value of

′f 1( )f 1( ) −

′′′f 1( )′′f 1( ) = 10

f x( ) ≈ −1+ 2 x −1( )− 6 x −1( )2+ 24 x −1( )3

f 1( ) = −1, ′f 1( ) = 2,′′f 1( )2!

= −6⇒ ′′f 1( ) = −12,′′′f 1( )3!

= 24⇒ ′′′f 1( ) = 144

′f 1( )f 1( ) −

′′′f 1( )′′f 1( ) =

2−1

− 144−12

= −2+12 = 10

True




13n

− 13

⎛⎝⎜

⎞⎠⎟

n

n=1

∞

∑n=1

∞

∑ is convergent

Solution


33.

13n

− 13

⎛⎝⎜

⎞⎠⎟

n

n=1

∞

∑n=1

∞

∑ is convergent

13n

= 13

n=1

∞

∑ 1n

which is a divergent p-series n=1

∞

∑ p ≤1( )

13

⎛⎝⎜

⎞⎠⎟

n

n=1

∞

∑ is a convergent geometric series r <1( )An infinite sum minus a finite sum is still infiniteFalse




cos−1 x dx = xcos−1 x∫ − x

1− x2dx∫

Solution


34.

cos−1 x dx = xcos−1 x∫ − x

1− x2dx∫

u = cos−1 x v = x

du = −1

1− x2dx dv = 1dx

cos−1∫ x dx = xcos−1 x + x

1− x2dx∫

False




Let R be the region bounded by the graph of y = x , y = 4,and the y-axis.That region is the base of a solid with cross-sections perpendicular to the x-axis being squares. The expression that represents the volume of that

solid is given by 4− x( )2

0

4

∫ dx.

Solution


35.

Let R be the region bounded by the graph of y = x , y = 4,and the y-axis.That region is the base of a solid with cross-sections perpendicular to the x-axis being squares. The expression that represents the volume of that

solid is given by 4− x( )2

0

4

∫ dx.

The graph is shown below. The integrand is correct as the side of the square

is given by 4− x . But x = 4⇒ x = 16 and the volume is 4− x( )2

0

16

∫ dx.

False




The graph of y = ′f x( ) is shown below with the area

of the shaded region as 9.5. x ⋅ ′′f x( )dx =−1

2

∫ − 0.5

Solution


36.

The graph of y = ′f x( ) is shown below with the

area of the shaded region as 9.5. x ⋅ ′′f x( )dx =−1

2

∫ − 0.5

Integration by parts: u = x v = ′f x( )du = dx dv = ′′f x( )dx

′′f x( )dx =−1

2

∫ x ⋅ ′f x( )⎡⎣ ⎤⎦−1

2− ′f x( )dx

−1

2

∫′′f x( )dx =

−1

2

∫ 2 ′f 2( )+1 ′f −1( )− 9.5= 2 4( )+1− 9.5= −0.5

True




The slope of a function is given by dydx

= y 100− y( ).The maximum slope of the function is 50.

Solution


37.

The slope of a function is given by dydx

= y 100− y( ).The maximum slope of the function is 50.

dydx

= 100y − y2 ⇒ d 2 ydx2 = 100− 2y( ) dy

dx= 100− 2y( ) y 100− y( ) = 0

Maximum slope occurs at either y = 0, y = 100, y = 50

At y = 0, y = 100,dydx

= 0,At y = 50,dydx

= 50 50( ) = 2500

Falsethis is logistic growth with carrying capacity 100 and maximum occurswhen y is one-half of the carrying capacity. But the problem asks for the slope and not the y-value.

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟




The arc length of y = ln cos x( ) on −π3

,π3

⎡

⎣⎢

⎤

⎦⎥

is given by L = 2 1+ tan2 x dx0

π 3

∫

Solution


38.

The arc length of y = ln cos x( ) on −π3

,π3

⎡

⎣⎢

⎤

⎦⎥

is given by L = 2 1+ tan2 x dx0

π 3

∫

′y = 1cos x

−sin x( ) = − tan x⇒ ′y( )2= tan2 x

Since the function is differentiable at x = 0,the arc length can be calculated from

x = 0 to x = π3

and doubled.

True




A 2nd-degree Maclaurin polynomial is used to approximatea function f . This Maclaurin polynomial is given by

P2 x( ) = 4+ 4x + 3x2. If f h( ) is approximated using this

Taylor polynomial and then with Euler's method with one step,

the difference in these approximations is 3h2.

Solution


39.

A 2nd-degree Maclaurin polynomial is used to approximatea function f . This Maclaurin polynomial is given by

P2 x( ) = 4+ 4x + 3x2. If f h( ) is approximated using this

Taylor polynomial and then with Euler's method with one step,

the difference in these approximations is 3h2.

f h( ) ≈ P2 h( ) = 4+ 4h+ 3h2

Euler: f 0( ) = 4,dydx

= 4⇒ dy = 4dx = 4h

f h( ) = 4+ 4h⇒ Difference = 4+ 4h+ 3h2 − 4+ 4h( ) = 3h2

TrueStudents must know that the constant term in the Maclaurin

polynomial is f 0( ). ⎛

⎝⎜

⎞

⎠⎟




A tree is growing with the thickness of its trunk obeying the relationship

rate thicknessis increasing

= k

thickness. The shape of the graph showing the

thickness of the tree trunk over time is exponential.

Solution


40.

A tree is growing with the thickness of its trunk obeying the relationship

rate thicknessis increasing

= k

thickness. The shape of the graph showing the

thickness of the tree trunk over time is exponential.

dTdt

= kT⇒T dt = k dt⇒ T dt = k dt⇒ T 2

2= kt +C∫∫

So the shape of the thickness curve is parabolic, not exponentialFalse




If F x( ) = 2x2 − 6x +8

dx∫ , the value of F 3( ) that

would eliminate the constant of integration is 0.

Solution


41.

If F x( ) = 2x2 − 6x +8

dx∫ , the value of F 3( ) that

would eliminate the constant of integration is 0.

2x2 − 6x +8

dx∫ = 2x − 4( ) x − 2( ) dx =

1x − 4

− 1x − 2

⎛⎝⎜

⎞⎠⎟dx∫∫

F x( ) = ln x − 4 − ln x − 2 +C

F 3( ) = ln1− ln1+C = 0⇒C = 0

True




The acceleration of a particle at rest moving on a straight line

is given by a t( ) = cos12t. The distance it travels from time

t = 0 to t = 2 is given by 2− 2cos1.

Solution


42.

The acceleration of a particle at rest moving on a straight line

is given by a t( ) = cos12t. The distance it travels from time

t = 0 to t = 2 is given by 2− 2cos1.

v t( ) = cos12t dt = 2sin∫ 1

2t +C

v 0( ) = 0+C = 0⇒C = 0⇒ v t( ) = 2sin12t

Since v t( ) > 0 on 0,2( ⎤⎦ , displacement = distance

Distance = 2sin12t dt = −4cos

12t⎡

⎣⎢

⎤

⎦⎥

0

2

∫0

2

= −4cos1+ 4cos0 = 4− 4cos1

False




Both statements are correct:

I. If a series ann=1

∞

∑ converges, the sequence an must converge to zero.

II. If f x( )dx converges, then 0

∞

∫ f x( )dx, where 0 < k < ∞, converges as well0

k

∫ .

Solution


43.

Both statements are correct:

I. If a series ann=1

∞

∑ converges, the sequence an must converge to zero.


∞

∫ f x( )dx, where 0 < k < ∞, converges as well0

k

∫ .

I. The only way a series can converge is if the nth term converges to zero.


∞

∫ accumulated area under the curve must be

a finite number. Then the partial accumulated area f x( )dx must be finite too0

k

∫ .

True




A company's net worth W changes over time. Based on the initial value of W ,in millions of dollars, the graph of W over time t, measured in years, can have several different shapes. Based on the graphs below, it is possible that the rate of change that describes this company's growth is given by dWdt

= kW 1000−W( ) where k is a constant.

Solution


44.

A company's net worth W changes over time. Based on the initial value of W ,in millions of dollars, the graph of W over time t, measured in years, can have several different shapes. Based on the graphs below, it is possible that the rate of change that describes this company's growth is given by dWdt

= kW 1000−W( ) where k is a constant.

The given differential equation signals logistic growth which would create anS-shaped curve in its growth curve. The curve would have to be concave down

and because limW→1000

kW 1000−W( )⎡⎣ ⎤⎦ = 0a logistic growth curve would have a

horizontal asymptote along W = 1000.False




Both the p-series test and ratio test can determine

whether 2x( )nn

converges at x =n=1

∞

∑ 14

Solution


45.

Both the ratio test and p-series test can determine

whether 2x( )nn

converges at x =n=1

∞

∑ 14

Ratio test: limn→∞

2x( )n+1

n+1⋅ n

2x( ) <1⇒ 2x <1

At x = 14

, the ratio test shows convergence

x = 14⇒

12

⎛⎝⎜

⎞⎠⎟

n

n= 1

2n nn=1

∞

∑ which is not a p-seriesn=1

∞

∑False




A Las-Vegas buffet has 200 people in line by the time the buffet opens.

People then join the line at the rate of J t( ) measured in people/hour as shown

by the red curve in the graph below. People leave the buffet line at the rate

of L t( ) as shown by the blue curve. The number of people in line at time t

when the line is getting longer is given by the expression:

200+ L t( )− J t( )⎡⎣ ⎤⎦0

1

∫ dt + J x( )− L x( )⎡⎣ ⎤⎦2

t

∫ dx, 2 < t < 4

Solution


46.

A Las-Vegas buffet has 200 people in line by the time the buffet opens.

People then join the line at the rate of J t( ) measured in people/hour as shown

by the red curve in the graph below. People leave the buffet line at the rate

of L t( ) as shown by the blue curve. The number of people in line at time t

when the line is getting longer is given by the expression:

200+ L t( )− J t( )⎡⎣ ⎤⎦0

1

∫ dt + J x( )− L x( )⎡⎣ ⎤⎦2

t

∫ dx, 2 < t < 4

It is correct to add the 200 original people and the last integral is correct

as between t = 2 and t = 4, J t( ) > L t( ). But from t = 0 to t = 1, the line

is getting shorter. So the first integral should be subtracted.

It can be written as 200− L t( )− J t( )⎡⎣ ⎤⎦0

1

∫ dt + J x( )− L x( )⎡⎣ ⎤⎦2

t

∫ dx

or 200+ J t( )− L t( )⎡⎣ ⎤⎦0

1

∫ dt + J x( )− L x( )⎡⎣ ⎤⎦2

t

∫ dx

False




The volume of the unbounded region lying between y = kx

with k a positive constant, x = 1, and the x − axis is rotated about the x − axis. The volume is a finite number.

Solution


47.

The volume of the unbounded region lying between y = kx

with k a positive constant, x = 1, and the x − axis is rotated about the x − axis. The volume is a finite number.

V = π k 2

x2

1

∞

∫ dx = −π k 2

x⎡

⎣⎢

⎤

⎦⎥

1

∞

= −π 0− k 2( ) = k 2π

True




ddx

et

et +1dt =

−∞

x2

∫ ex2

ex2

+1

Solution


48.

ddx

et

et +1dt =

−∞

x2

∫ ex2

ex2

+1

The integral could be found with u-substitution:

u = et +1,du = etdt,t = −∞

duu

= ln u ⇒∫ et

et +1dt =

−∞

x2

∫ ln et +1( )⎡⎣

⎤⎦−∞x2

= ln ex2

+1( )− ln1

ddx

ln ex2

+1( ) = ex2

⋅2xex

2

+1But using the 2nd FTC is faster. Just substitute x2 for t, remembering

to apply the chain rule by taking the derivatie of x2

False




The rate at which the arc length of y = 23t3 2

is increasing on 0,x⎡⎣ ⎤⎦ is given by 1+ x dxdt

Solution


49.

The rate at which the arc length of y = 23t3 2

is increasing on 0,x⎡⎣ ⎤⎦ is given by 1+ x dxdt

Using the arclength formula, ′y = t1 2 so L = 1+ t0

x

∫ dt

dLdt

= ddt

1+ t0

x

∫ dt = 1+ x dxdt

True




The first-quadrant region bounded by the x- and y-axes and the curvey = cos x is rotated about the y-axis. The resulting volume is given by

V = π cos−1 x( )2

0

1

∫ dx

Solution


50.

The first-quadrant region bounded by the x- and y-axes and the curvey = cos x is rotated about the y-axis. The resulting volume is given by

V = π cos−1 x( )2

0

1

∫ dx

As shown by the graph below, this is a disk problem with the disks stacked along

the y-axis. The radius = x = cos−1y. So V = π cos−1 y( )2

0

1

∫ dy. But the variable doesn't

matter as long as both are changed. So V = π cos−1 x( )2

0

1

∫ dx or π cos−1 z( )2

0

1

∫ dz.

True

bc calculus: true or false? - mastermathmentor.com...bc calculus: true or false? 29. time (60...

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