Question: 1. People enter a line for an escalator at a rate modeled by the function r given by

r(t) = 44(t/100)³(1-(t/300)⁷ for 0≤t≤300

r(t) = 0 for t>300

where r(t) is measured in people per second and t is measured in seconds. As people get on the escalator, they exit the line at a constant rate of 0.7 person per second. There are 20 people in line at time t = 0.

(a) How many people enter the line for the escalator during the time interval 0≤t≤300 ?

(b) During the time interval 0≤t≤300, there are always people in line for the escalator. How many people are in line at time t = 300 ?

(c) For t > 300, what is the first time t that there are no people in line for the escalator?

(d) For 0≤t≤300, at what time t is the number of people in line a minimum? To the nearest whole

number, find the number of people in line at this time. Justify your answer.

Answer: (a) ∫{0,300}r(t)dt=270

According to the model, 270 people enter the line for the escalator during the time interval 0 ≤ t ≤ 300.

(b) 20 + ∫{0,300}(r(t) − 0.7) dt = 20 + ∫{0,300} r(t) dt − 0.7⋅300 = 80

Initial + (RateIn-RateOut)

According to the model, 80 people are in line at time t = 300.

(c) Based on part (b), the number of people in line at time t = 300 is 80. The first time t that there are no people in line is 300 + 80/0.7 = 414.286 (or 414.285) seconds.

80(people)/0.7(people/sec) = sec until no people after 300 sec

(d) The total number of people in line at time t, 0 ≤ t ≤ 300, is modeled by

20 + ∫{0,t} r(x) dx − 0.7t

r(t) − 0.7 = 0 ⇒ t₁ = 33.013298, t₂ = 166.574719

The number of people in line is a minimum at time t = 33.013 seconds, when there are 4 people in line. Find where f’=0 using CALC solver

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Question: 2. Researchers on a boat are investigating plankton cells in a sea. At a depth of h meters, the density of plankton cells, in millions of cells per cubic meter, is modeled by p(h) = 0.2h²e^(−0.0025h²) for 0≤h ≤30 and is modeled by f (h) for h ≥ 30. The continuous function f is not explicitly given.

(a) Find p’(25). Using correct units, interpret the meaning of p’(25) in the context of the problem.

(b) Consider a vertical column of water in this sea with horizontal cross sections of constant area 3 square meters. To the nearest million, how many plankton cells are in this column of water between h = 0 and h = 30 meters?

(c) There is a function u such that 0≤f(h)≤u(h) for all h ≥ 30 and ∫{30,∞} u(h) dh = 105. The column of in Part B is K meters deep, where K > 30. Write an expression involving one or more integrals that gives the number of plankton cells, in millions, in the entire column. Explain why the number of plankton cells in the column is less than or equal to 2000 million.

(d) The boat is moving on the surface of the sea. At time t ≥ 0, the position of the boat is (x(t), y(t)), where x’(t) = 662 sin(5t) and y’(t) = 880 cos(6t). Time t is measured in hours, and x(t) and y(t) are measured in meters. Find the total distance traveled by the boat over the time interval 0≤t≤1.

Answer: (a) p′(25) = −1.179

At a depth of 25 meters, the density of plankton cells is changing at a rate of −1.179 million cells per cubic meter per meter. Use CALC nDerive

(b) ∫{0,30}3p(h) dh = 1675.414936

There are 1675 million plankton cells in the column of water between h = 0 and h = 30 meters.

Use CALC fnINT

(c) ∫{30,K} 3f(h) dh represents the number of plankton cells, in millions, in 30

the column of water from a depth of 30 meters to a depth of K meters. The number of plankton cells, in millions, in the entire column of

water is given by ∫{0,30}3p(h)dh+∫{30,K}3f(h)dh.

Because 0≤ f(h)≤u(h) forall h≥30,

3∫{30,K}f(h)dh ≤ 3∫{30,K}u(h)dh ≤ 3∫{30,∞}u(h)dh = 3⋅105 = 315.

(d) ∫{0,1}√[(x′(t))² + (y′(t))²] dt = 757.455862

The total distance traveled by the boat over the time interval 0 ≤ t ≤ 1 is 757.456 (or 757.455) meters. Integral of rate of change of distance in distance formula is the distance

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Question: 3. (NO CALC) The graph of the continuous function g, the derivative of the function f, is shown above. The function g is 2 piecewise linear for −5≤x<3, and g(x) =2(x − 4)² for 3≤x≤6.

(a) If f(1)=3, what is the value of f(−5) ?

(b) Evaluate ∫{1,6} g(x) dx.

(c) For −5<x<6, on what open intervals, if any, is the graph of f both increasing and concave up? Give a reason for your answer.

(d) Find the x-coordinate of each point of inflection of the graph of f. Give a reason for your answer.

Answer: (a) f(-5) = f(1) + ∫{1,-5}g(x)dx=f(1)-∫{-5,1}g(x)dx=3-(-9-(3/2)+1)=3-(-19/2)=25/2

(b) ∫{1,6}g(x)dx =∫{1,3}g(x)dx+∫{3,6}g(x)dx= ∫{1,3}2dx+∫{3,6}2(x −4)²dx

= 4+(2/3)(x −4)³ = 4+(16/3)-(-2/3)=0

(c) The graph of f is increasing and concave up on 0 <x< 1 and 4<x<6because f′(x)=g(x)>0 and f′(x)=g(x) is increasing on those intervals. When g(x) is positive and increasing

(d) The graph of f has a point of inflection at x=4 because f ′(x) = g(x) changes from decreasing to increasing at x=4.

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Question: 4. (NO CALC) The height of a tree at time t is given by a twice-differentiable function H, where H(t) is measured in meters and t is measured in years. Selected values of H(t) are given in the table above.

(a) Use the data in the table to estimate H’(6). Using correct units, interpret the meaning of H’(6) in the context of the problem.

(b) Explain why there must be at least one time t, for 2 < t < 10, such that H’(t) = 2.

(c) Use a trapezoidal sum with the four subintervals indicated by the data in the table to approximate the average height of the tree over the time interval 2≤ t≤10.

(d) The height of the tree, in meters, can also be modeled by the function G, given by G(x) = 100x/(1+x), where x is the diameter of the base of the tree, in meters. When the tree is 50 meters tall, the diameter of the base of the tree is increasing at a rate of 0.03 meter per year. According to this model, what is the rate of change of the height of the tree with respect to time, in meters per year, at the time when the tree is 50 meters tall?

Answer: (a) H’(6) ≈ [H(7)-H(5)] / [7-5] = (11-6) / 2 = (5/2)

H’(6) is the rate at which the height of the tree is changing, in meters per year, at time t=6 years

(b) [H(5)-H(3)] / [5-3] = (6-2)/2=2

Because H is differentiable on 3≤t≤5, H is continuous on 3≤t≤5. By the Mean Value Theorem, there exists a value c, 3 < c < 5, such that H′(c) = 2.

f is continuous over [a,b]

f is differentiable over (a,b)

secant and tangent lines will be equal (parallel)

(c) The average height of the tree over the time interval 2≤t≤10 is given by

(1/(10-2))∫{2,10}H(t )dt

use average and trapezoid formulas

(1/8)∫{2,10} H(t)dt≈(1/8)( [(1.5+2)/2](1) + [(2+6)/2](2) + [(6+11)/2](2) + [(11+15)/2](3))

=(1/8)(65.75)=(263/32)

Average×Years is each term

The average height of the tree over the time interval 2 ≤ t ≤ 10 is 263 meters.

(d) G(x) = 50 so x=1

(d/dt)(G(x)) = (d/dx)(G(x)) (dx/dt)

given (dx/dt) so chain rule

=[(1+x)100-100x(1)] / [(1+x)²] × (dx/dt)

=100/(1+x)² × (dx/dt)

let x=1, given (dx/dt)=.03

(d/dt)(G(x)) = 100/(1+1)²×0.03=3/4

According to the model, the rate of change of the height of the tree with respect to time when the tree is 50 meters tall is 3/4 meter per year.

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Question: 5. (NO CALC) The graphs of the polar curves r = 4 and r = 3 + 2 cos θ are shown in the figure above. The curves intersect at θ = π/3 and θ = 5π/3.

(a) Let R be the shaded region that is inside the graph of r = 4 and also outside the graph of r = 3 + 2cos θ, as shown in the figure above. Write an expression involving an integral for the area of R.

(b) Find the slope of the line tangent to the graph of r=3+2cosθ at θ=π/2.

(c) A particle moves along the portion of the curve r=3+2cosθ for 0 < θ< π/2. The particle moves in such a way that the distance between the particle and the origin increases at a constant rate of 3 units per second. Find the rate at which the angle θ changes with respect to time at the instant when the position of the particle corresponds to θ = π/3. Indicate units of measure.

Answer: (a) Area = (1/2)r²θ=∫(1/2)f(θ)²dθ

Area =(1/2)∫{π/3,5π/3} (4²-(3+2cosθ)²)dθ

(b) (dr/dθ)=-2sinθ so at θ=π/2, (dr/dθ)=-2

r(π/2)=3+2cos(π/2)=3

y=rsinθ=dy/dθ=(dr/dθ)sinθ+rcosθ

x=rcosθ=dx/dθ=(dr/dθ)cosθ-rsinθ

dy/dx at θ=π/2 is 2/3

(c) dr/dt=(dr/dθ)(dθ/dt)=-2sinθ(dθ/dt)

dθ/dt=(dr/dt)(1/(-2sinθ))

given (dr/dt)=3

(dθ/dt)=3(1/(-2sin(π/3)))=3/-√3=-√3 radians per second

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Question: The Maclaurin series for ln(1 + x) is given by

x-(x²/2)+(x³/3)-(x⁴/4)+…+(-1)ⁿ⁺¹(xⁿ/n)+…

On its interval of convergence, this series converges to ln(1 + x). Let f be the function defined by f(x)=xln(1+(x/3))

(a) Write the first four nonzero terms and the general term of the Maclaurin series for f.

(b) Determine the interval of convergence of the Maclaurin series for f. Show the work that leads to your answer.

(c) Let P (x) be the fourth-degree Taylor polynomial for f about x = 0. Use the alternating series error

bound to find an upper bound for |P (2) − f (2)| .

Answer: (a) The first four nonzero terms are (x²/3)-(x³/(2(3²))+(x⁴/3(3³))-(x⁵/4(3⁴))

The general term is (-1)ⁿ⁺¹(xⁿ⁺¹/n3ⁿ)

divide all terms by three they multiply by x

(b) Ratio Test

lim{n→∞}|[(-1)ⁿ⁺²(xⁿ⁺²/(n+1)3ⁿ⁺¹)]/[(-1)ⁿ⁺¹(xⁿ⁺¹/n3ⁿ)]| =

lim{n→∞}|(-x/3)(n/(n+1))|=|(x/3)|

(c) By the alternating series error bound, an upper bound for |P (2) − f (2)| is the magnitude of the next term of the alternating 4 series.

|P₄(2) − f (2)|<|-(2⁵)/(4(3⁴))|=8/81

to find 4th degree error calculate 5th term (this will be the max error

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