2018 FREE RESPONSE QUESTION TENTATIVE SOLUTIONS

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1 8 FREE RESPONSE QUESTION TENTATIVE SOLUTIONS L. MARIZZA A BAILEY Problem. People enter a line for an escalator at a rate modeled by the function, r given by { 44( t r(t) = ) ( t )7 ) t t > where r(t) is measured in people per second and t is measured in seconds. As people get on the escalator, the exit the line at a constant rate of.7 person per second. There are people in line at time t =. (a) How many people enter the line for the escalator during the time interval t. Solution. To find the number of people entering the line for the escalator during the time interval [, ], we need to integrate the piecewise defined function, r. Since the function, r, is defined to be r(t) = 44( t ) ( t )7 ) while t [, ], then r(t)dt = 7 Therefore, 7 people enter the escalator in the first seconds. (b) During the time interval t, there are always people in line for the escalator. How many people are in line at time t =? Solution. To find the number of people in line at the escalator at t =, we need to find the number of people who entered the line and subtract the number of people who exit the line at t =. Also, we cannot forget the people who were in line intially. To get this we need to evaluate the expression + r(t)dt Since we already evaluated the first integral, and we know that.7dt =.7() =. Therefore, the amount of people in line at t = is + 7 = 8. There were 8 people in line at t =..7dt (c) For t >, what is the first time t that there are no people in line for the escalator? Solution. First we need to realize that after t =, there are no more people entering the line for the escalator. Since the number of people getting on the escalator is less than one person per second, the line can t be empty prior to t =. This means we need to find when the number of people exiting the line reaches 8 which is the number of people in line at t =. So.7t = 8 at t = 4.86 or t = 8 seconds 7 after t =. So the first time ghere are no people in line is at seconds or 9 seconds. 7 If this does not satisfy, the amount of people in line is given by the piece-wise defined function, A(t) = {.7t + t r(x)dx t 8.7(t ) t > Since there is no solution for A(t) = for t, then the first solution must be when t >, which occurs at t = 9 7. Date: May 7, 8.

2 (d) For t, 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. Solution 4. The amount of people in line for t [, ] is given by A(t) =.7t + t r(x)dx. To find the absolute minimum, we must first find critical points for A(t) by finding t where A (t) =. This occurs when ( t ) ( t )7 = There are two solutions to this equation in the interval [, ]; at t =. and t = To see where the absolute minimum occurs we must evaluate A(t) at both interior critical points and the end point. t A(t) A() =. A(.) = A(66.575) = 58.7 A()=8 Therefore, the minimum number of people in line occur at. seconds and there are approximately 4 people in line at that time.

3 Problem. Researchers on a boat are investigating plankton cells in a sea. A a depth of h meters, the density of plankton cells, in millions of cells per cubic meter, is modeled by p(h) =.h e.5h for h and is modeled by f(h) for h. The continuous function f is not explicitly given. (a) Find p (5). Using correct units, interpret the meaning of p (5) in the context of the problem. Solution 5. p (5) =.79 which means that the density of plankton was decreasing by.79 millions of cells per cubic meter per meter of depth at the depth of 5 meters. (b) Consider a vertical column of water in this sea with horizontal cross sections of constant area square meters. To the nearest million, how many plankton cells are in this column of water between h = and h = meters? Solution 6. The number of plankton cells is the integral of the density function of plankton cells times the volume element dh p(h)dh = (c) There is a function u such that f(h) u(h) for all h and u(h)dh = 5. The column of water in part(b) is K meters deep, where K. 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 million. Solution 7. If the column is K meters deep, then we need to split up the integral between the depth on which p(h) is the density function, and the depth on which f(h) is the density function. p(h)dh + K f(h)dh = p(h)dh + K = (5) = u(h)dh (d) The boat is moving on the surface of the sea. At time t, the position of the boat is (x(t), y(t)), where dx dy = 66 sin(5t) and = 88 cos(6t). Time t is measured in hours, and x(t) and y(t) are measured in dt dt meters. Find the total distance traveled by the boat over the time interval t. Solution 8. Since the total distance that the boat travels along a parametric path is given by the arclength of the path, we need only evaluate the integral (x (t)) + (y (t)) dt

4 Problem. The graph of the continuous function, g, the derivative of the function, f, is shown above. The function g is piecewise linear for 5 x < and g(x) = (x 4) for x 6. (a) If f() =, what is the value of f( 5)? Solution 9. The fundamental theorem of calculus yields the integral equation (b) Evaluate 6 g(x)dx. f() f( 5) = 5 g(x)dx f( 5) = ( 4 + ( )) + () f( 5) = + f( 5) = 9 9 = f( 5) = f( 5) Solution. Since the function, g, is piecewise defined, we need to break up the integral into pieces; g(x)dx + 6 g(x)dx = + + () + = 6 (x 4) = + ( ) ( ) = = + 8 = + 6 = (x 4) 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. Solution. The graph of f is increasing when f >. Since f = g, this means it occurs when g is above the x-axis. The graph of f is concave up when f > which occurs when g is increasing. Therefore, we are looking for the intervals on the graph when g is increasing and above the x-axis. These intervals are (, ) (4, 6) (d) Find the x-coordinate of each points of inflection of the graph of f. Give a reason for your answer.

5 Solution. A point of inflection of f occurs when f changes concavity. Since g = for all of the values, x where < x <, then even though we have a concavity change, it does not occur at a point Therefore, the only point of inflection occurs when x = 4. t 5 7 (years) H(t) (meters)

6 Problem 4. 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. Solution. We can estimate H (6) using the slope of the graph in the interval (, 7). H (6) H(7) H(5) 7 5 = 6 Therefore, H (6) is approximately.5 meters per year which means the tree is growing approximately.5 meters per year. (b) Explain why there must be at least one time t for < t < such that H (t) =. Solution 4. Since H is twice differentiable on its domain, then it is continuous on [, 5] and differentiable on (, 5). This means, that by Mean Value Theorem, there is a time c, < < c < 5 < such that H (c) = H(5) H() 5 = 6 (c) Use a trapezoidal sum with four subintervals indicated by the data in the table to approximate the average height of the tree over the time interval t. Solution 5. The average height of the tree is given by This can be approximated by the trapezoidal sum 8 [ +.5 () H(t)dt + 6 () + () + = 5 = 5 + ()] (d) The height of the tree, in meters, can also be modeled by the function G, given by G(x) = x, where x is + x the diameter of the base of the tree, in meters. When the tree is 5 meters tall, the diameter of the base of the tree is increasing at a rate of. 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 5 meters tall? Solution 6. Since dg dx dx dt = dg dt and dg dx = ()( + x) (x)() ( + x) and G(x) = x dg = 5 when x =, =, then + x dx dg dt = 4 (.) which means the tree is growing by meters per year when it is 5 meters tall. 4

7 Problem 5. The graphs of polar curves r = 4 and r = + cos(θ) are shown in the figure above. The curves interesect at θ = π and θ = 5 π. (a) Let R be the shaded region that is inside the graph of r = 4 and also outside the graph of r = + cos(θ) as shown in the figure above. Write an expression involving an integral for the area of R. Solution 7. First we need to find the points at which the curves intersect. We can do this by solving the equation; 4 = + cos(θ) = cos(θ) = cos(θ) π, 5π = cos(θ) Therefore, the integral expression can be given by 5π π (4 ) ( + cos(θ)) dθ (b) Find the slope of the line tangent to the graph of r = + cos(θ) at θ = π. Solution 8. Since Since and dy dy dx = dθ dx dθ y = r sin(θ) and y = r sin(θ) + r cos(θ) x = r cos(θ) and x = r cos(θ) r sin(θ) We can see that r( π ) = + cos( π ) =, and r ( π ) = sin( π ) =. Therefore, y = () + () and x = ( )() ()() which means dy dx = =. (c) A particle moves along the portion of the curve r = + cos(θ) for < θ < π. The particle moves in such a way that the distance between the particle and the origin increases at a constant rate of 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. Indicate units of measure. Solution 9. The particle moving in such a way that the distance between the particle and the origin increases at a constant rate of units per second means dr =. Since dt dr dθ dθ dt = dr dt and dr dθ π = sin( π ) = then dθ dθ = which means dt dt =

8 Problem 6. The MacLaurin series for ln( + x) is given by x x + x x ( )n+ x n n + On its interval of convergence, this series converges to ln( + x). Let f be a function defined by f(x) = x ln( + x ). (a) Write the first four nonzero terms and the general term of the MacLaurin Series for f. Solution. Since then ln( + x ) = x x () + x () x4 4 (4) + + ( )n+ x n + n (n) x ln( + x ) = x x () + x4 () x5 4 (4) + + ( )n+ x n+ + n (n) (b) Determine the interval of convergence of the MacLaurin series for f. answer. Show the work that leads to your Solution. We can use ratio test, if we put absolute values around the sequence of the series. lim an+ = lim x n+ n n n+ (n + ) n (n) x n+ a n = x lim = x n n n + By the ratio test, the series converges when x <, which means < x <. However, now we need to consider the endpoints of the interval by plugging those x-values into the power series and checking for convergence. At x =, we get ( ) n+ n+ ( ) n+ = n (n) n n= which is an alternating harmonic series, which converges because it is a truly alternating series whose terms decrease in absolute value to zero. On the other hand, when we try x =, we get, n= n= ( ) n+ ( ) n+ = n (n) which is a harmonic series, which diverges by p-series test. Therefore, the interval of convergence is < x. (c) Let P 4(x) be the fourth-degree Taylor polynomial for f about x =. Use the alternating series bound to find an upper bound for P 4() f(). Solution. Since the fourth degree Taylor polynomial n= n= n P 4(x) = x x () + x4 () ( ) n+ n+ = ( ) n+ ( n (n) )n n n= then the alternating series is decreasing in absolute value to zero, which means we can use the alternating series error bound. This means that P 4() f() < 5 4 (4) = 8 8. Department of Mathematics, BASIS Scottsdale address: marizza.bailey@basised.com