bc calculus: true or false? - mastermathmentor.com...bc calculus: true or false? 29. time (60...
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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
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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
BC Calculus: True or False?
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27. Time (60 seconds)
6x
0
1
∫ dx = 6
x31
∞
∫ dx
Solution
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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
BC Calculus: True or False?
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28. Time (60 seconds)
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
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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
BC Calculus: True or False?
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29. Time (60 seconds)
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
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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
BC Calculus: True or False?
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30. Time (40 seconds)
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
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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
BC Calculus: True or False?
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31. Time (30 seconds)
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
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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
BC Calculus: True or False?
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32. Time (60 seconds)
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
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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
BC Calculus: True or False?
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33. Time (60 seconds)
13n
− 13
⎛⎝⎜
⎞⎠⎟
n
n=1
∞
∑n=1
∞
∑ is convergent
Solution
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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
BC Calculus: True or False?
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34. Time (30 seconds)
cos−1 x dx = xcos−1 x∫ − x
1− x2dx∫
Solution
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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
BC Calculus: True or False?
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35. Time (45 seconds)
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
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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
BC Calculus: True or False?
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36. Time (60 seconds)
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
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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
BC Calculus: True or False?
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37. Time (40 seconds)
The slope of a function is given by dydx
= y 100− y( ).The maximum slope of the function is 50.
Solution
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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.
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
BC Calculus: True or False?
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38. Time (40 seconds)
The arc length of y = ln cos x( ) on −π3
,π3
⎡
⎣⎢
⎤
⎦⎥
is given by L = 2 1+ tan2 x dx0
π 3
∫
Solution
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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
BC Calculus: True or False?
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39. Time (60 seconds)
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
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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( ). ⎛
⎝⎜
⎞
⎠⎟
BC Calculus: True or False?
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40. Time (20 seconds)
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
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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
BC Calculus: True or False?
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41. Time (45 seconds)
If F x( ) = 2x2 − 6x +8
dx∫ , the value of F 3( ) that
would eliminate the constant of integration is 0.
Solution
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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
BC Calculus: True or False?
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42. Time (60 seconds)
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
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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
BC Calculus: True or False?
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43. Time (45 seconds)
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
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43.
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
∫ .
I. The only way a series can converge is if the nth term converges to zero.
II. If f x( )dx converges, then 0
∞
∫ accumulated area under the curve must be
a finite number. Then the partial accumulated area f x( )dx must be finite too0
k
∫ .
True
BC Calculus: True or False?
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44. Time (30 seconds)
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
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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
BC Calculus: True or False?
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45. Time (45 seconds)
Both the p-series test and ratio test can determine
whether 2x( )nn
converges at x =n=1
∞
∑ 14
Solution
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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
BC Calculus: True or False?
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46. Time (60 seconds)
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
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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
BC Calculus: True or False?
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47. Time (40 seconds)
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
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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
BC Calculus: True or False?
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48. Time (15 seconds)
ddx
et
et +1dt =
−∞
x2
∫ ex2
ex2
+1
Solution
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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
BC Calculus: True or False?
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49. Time (40 seconds)
The rate at which the arc length of y = 23t3 2
is increasing on 0,x⎡⎣ ⎤⎦ is given by 1+ x dxdt
Solution
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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
BC Calculus: True or False?
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50. Time (30 seconds)
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
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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