Math 6 Second Midterm November 6, Name: EXAM SOLUTIONS Instructor: Section:. Do not open this exam until you are told to do so.. This exam has pages including this cover. There are 9 problems. Note that the problems are not of equal difficulty, so you may want to skip over and return to a problem on which you are stuck.. Do not separate the pages of this exam. If they do become separated, write your name on every page and point this out to your instructor when you hand in the exam. 4. Please read the instructions for each individual problem carefully. One of the skills being tested on this exam is your ability to interpret mathematical questions, so instructors will not answer questions about exam problems during the exam. 5. Show an appropriate amount of work (including appropriate explanation) for each problem, so that graders can see not only your answer but how you obtained it. Include units in your answer where that is appropriate. 6. You may use any calculator except a TI-9 (or other calculator with a full alphanumeric keypad). However, you must show work for any calculation which we have learned how to do in this course. You are also allowed two sides of a 5 note card. 7. If you use graphs or tables to find an answer, be sure to include an explanation and sketch of the graph, and to write out the entries of the table that you use. 8. Turn off all cell phones and pagers, and remove all headphones. Problem Points Score 4 5 6 8 7 8 9 Total
Math 6 / Exam (November 6, ) page. [ points] Indicate if each of the following is true or false by circling the correct answer. No justification is required. a. [ points] The function y(t) = cos t + B sin t + 9 t is a solution of y + 9y = with y() =. True False y = 9 cos t 9B sin t 9y = 9 cos t + 9B sin t + t Hence y + 9y = t. b. [ points] The value of the integral used to compute the area enclosed by a curve r = f(θ) given in polar coordinates can be negative if f(θ). True False f(θ), then A = b a f(θ) dθ. c. [ points] If f(x) is a continuous function such that f(x)dx diverges, then f(x) dx must diverge. If f(x) = x, then x dx diverges but True x dx converges. False d. [ points] If P (x) is a cumulative distribution function for the probability density function p(x), then + P (x) is also a cumulative distribution function for p(x). True False A cumulative distribution function P (x) must satisfy lim x P (x) =, but lim x + P (x) =. Hence + P (x) can t be a cumulative distribution. e. [ points] All solutions to the differential equation y = + y 4 are increasing functions. True False Since + y 4 >, then y >. Then all the solution curves y must be increasing. f. [ points] Let P (t) be the cumulative distribution function of a probability density function p(t). If P () = then the median of p(t) is negative. True False If T is the median of p(t), then P (T ) =. Since P () = >, then T <.
Math 6 / Exam (November 6, ) page. [ points] Consider a particle whose trajectory in the xy-plane is given by the parametric curve defined by the equations x(t) = t 4 4t, y(t) = t t, for t. Show all your work to receive full credit. a. [ points] Is there any value of t at which the particle ever comes to a stop? Justify. No. For the particle to come a stop, its velocity in both the x- and y-direction must be zero. We have that at t =, ± and dx dt = 4t 8t = 4t(t ) = dy dt = t = at t =. Since there are no times at which dx dt never comes to a stop. dy and dt are simultaneously zero, the particle b. [ points] For what values of t does the path of the particle have a vertical tangent line? Vertical tangent lines occur when dx dt this is true at t =, ±. = and dy dt c. [ points] What is the lowest point (x, y) on the curve?. From the above calculation, We want to minimize the value of the y-coordinate over t. The only critical point for y(t) was found above at t =. Since dy dt = < and dy t= dt = >, t= the First Derivative Test tells us that t = is a local minimum, and thus a global minimum since it is the only critical point on the given interval. The lowest point on the curve is thus (x(), y()) = (, ). d. [ points] At what values of t does the particle pass through the origin? We set x(t) = and y(t) = and solve for t. gives that t =,,, while x(t) = t 4 4t = t (t 4) = t (t )(t + ) = y(t) = t t = t(t ) = gives t =,. Thus, the particle passes through the origin at times t = and t =.
Math 6 / Exam (November 6, ) page 4 e. [ points] The graph of the curve traced by these parametric equations is shown below. Find an expression for the length of the closed loop marked in the graph. 8 6 4 4 4 From the given graph and above calculation, we know that the loop is traced out over the time interval t. The arclength of the loop is given by ( dx ) ( ) dt + dy dt dt = (4t 8t) + (t ) dt.
Math 6 / Exam (November 6, ) page 5. [ points] In the late th century, Mom s Friendly Robot Company is the main global robot manufacturing company. The Bending Unit model is designed to contain a backup unit, effectively rendering it immortal. However, a small percentage of the robots suffer a manufacturing defect, in which the backup unit is malfunctional or not present. The function {, if t < p(t) =.4e t/c, if t gives the probability density for the lifetime of these defective Bending Units, where c is a positive constant and t is measured in years since the robots are activated. Show all your work to receive full credit. a. [ points] Interpret the quantity 4 p(t) dt gives the fraction of defective Bending Units that have a lifespan between and 4 years. OR 4 4 p(t) dt. p(t) dt gives the probability that a defective Bending Unit will have a lifetime between and 4 years. b. [4 points] Find the value of c. Since p(t) is a probability density function, we know that = b = lim b ( b = lim =.4c c = 5 p(t) dt = p(t) dt + p(t) dt = + [.4e t/c dt = lim c.4e t/c] b b.4ce b/c +.4ce ) = +.4c.4e t/c dt p(t) =. Thus, What is the mean (average) lifespan of a defective Bending Unit? c. [5 points]
Math 6 / Exam (November 6, ) page 6 Using c = 5 from above and that p(t) = for t <, t = = + tp(t) dt = tp(t) dt +.4te t/5 dt = lim b b tp(t) dt.4te t/5 dt. Integration by parts with u = t, dv =.4e t/5 (and du = dt, v = e t/5 ) gives [ t = lim te t/5 b + b = lim b [ = 5. b ] [ ] e t/5 dt = lim be b/5 + 5e t/5 b b be b/5 5e b/5 + 5e ] = + 5 Thus, the mean lifespan of a defective Bending Unit is 5 years.
Math 6 / Exam (November 6, ) page 7 4. [ points] A restaurant installs a kitchen ventilation system to control the amount of grease in the air due to cooking. The ventilation system reduces the amount of grease in the air by 9 percent every hour. Let Q(t) be the amount in grams of grease in the air t hours after the ventilation is activated. Then Q satisfies the differential equation dq dt = e 5t 9 Q, where e 5t is the rate at which the kitchen produces grease in grams per hour at time t. a. [ points] The slope field of the differential equation is shown below. Suppose that the air in the kitchen initially has.4 grams of grease. Sketch the solution curve in the slope field...8.6.4.....4.6.8. b. [7 points] Use Euler s method to estimate the values of the solution curve Q(t) through (,.4) for all values of t given in the table below. Show all your work. t Q(t).4.9466.7885.5757 Q n+ = Q n + (e 5t 9 Q n) t with t =. Q() =.4 = Q ( ) ( Q Q =.4 + 9 ) (.4).9466 ( ) ( ) Q Q =.9466 + e 5 9 (.9466).7885 ( ) Q () Q =.7885 + e 9 (.7885).5757. c. [ points] Does your approximation for Q() using Euler s method give an overestimate or an underestimate? Justify. The estimate is an overestimate since the solution curves are concave down.
Math 6 / Exam (November 6, ) page 8 5. [ points] Determine if the following integrals converge or diverge. If an integral diverges, explain why. If it converges, find the value to which it converges. Mathematical precision is important. a. [4 points] dx x The integrand is discontinuous as x =. Thus, substituting w = x, dw = dx we obtain b dx = lim ( x) / dx x so the integral converges. b w(b) = lim b w( ) w / dw = lim b w/ w(b) w( ) = lim ( = lim b + ( ) b =, x b b ) = + b. [4 points] 5 + sin(4θ) θ dθ Since sin(4θ), we know that 5 + sin(4θ) 7, and so that in particular θ 5 + sin(4θ) θ over the interval [, ). Since θ dθ diverges (p = ), we know that 5+ sin(4θ) θ dθ diverges by comparison. c. [ points] x + x dx x Since +x we know that the integral OR x x+x x +x as x (in particular, the integrand does not approach ), dx must diverge. x +x x +x for x. Hence dx dx diverges. x +x dx, by the comparison method
Math 6 / Exam (November 6, ) page 9 6. [8 points] Members of the recruitment committee for the Mars University (MU) chapter of the fraternity Epsilon Rho Rho are designing a pledge pin to distribute during Rush Week. The pin takes the shape of a cardioid with a circular hole in it. The cardioid is given by a polar equation of the form r = a + b cos(θ), while the circular hole has the polar equation r = cos(θ). The pin is pictured below, where the x- and y-axes are measured in inches. a. [5 points] The committee plans on coating one side of the pin in gold plating, which costs dollars per square inch. Give an expression representing the cost to plate one face of the pin in gold. Your answer may involve integrals and the constants a and b. We note that the cardioid r = a + b cos(θ) is traced out over θ π, while the circle r = cos(θ) is traced out over θ π. The area, in square inches, of the pin is then π (a + b cos(θ)) dθ π so that the total cost to plate one face of the pin is (cos(θ)) dθ, dollars. One could also note that the inner circle has radius π (, ) and the cost of the gold plating becomes π (a + b cos(θ)) dθ 4 π dollars. ) cos (θ) dθ, so that its area is just ( π (a + b cos(θ)) dθ π b. [ points] Find a and b.
Math 6 / Exam (November 6, ) page From the given graph, we see that when θ =, r = and when θ = π, r =. This yields the system of equations a + b cos() = a + b =, a + b cos( π ) = a =. Thus a =, b =. Or from the given graph, we see that when θ =, r = and when θ = π, r =. This yields the system of equations Thus a =, b =. a + b cos() = a + b =, a + b cos( π ) = a =.
Math 6 / Exam (November 6, ) page 7. [ points] A company designs an air filter for a ship s engine room that reduces the amount of fumes in the air by k percent every hour. The machinery in the engine room produces fumes at a rate of. kilograms per hour. Let Q(t) be the amount in kilograms of fumes in the room t hours after the engines are activated. Hence Q satisfies dq dt =. k Q. a. [9 points] Find a formula for Q(t). Suppose there are no fumes in the air when the engines are activated. k dq. k Q = dt. ln. k Q = t + C.. k Q = Ae k t Q(t) = k Q(t) = k (. Ae k t) ( e k t) using Q() =. b. [ points] What is the value of Q(t) in the long run? lim t Q(t) = k. c. [ points] Air safety regulations require that the concentration of fumes in the air not exceed 4 kilograms per liter at any time. If the volume of air in the engine room is liters, for what values of k are the safety regulations met at all times? Concentration= Q(t) Volume Hence k. k 4.
Math 6 / Exam (November 6, ) page 8. [ points] For α >, consider the family of spirals given by r = θ in polar coordinates. α a. [ points] Write down an integral that gives the length L of a spiral in this family for θ b. No credit will be given if you just write down the formula given in part (b). Using parametric equations: x = θ cos α and y = α θ sin α. α b ( ) ( L = + α or α cos θ θα+ θ α sin θ L = b sin θ + θα+ ( ) ( θ α + α ) dθ θ α+ ) θ α cos θ dθ b. [8 points] It can be shown that the length L of the spiral in part a) may also be written as b L = θ α + α θ dθ. Use this formula for L to find all values of α > for which the length of the spiral is infinite for θ. For which values of α is the length finite? Justify all your answers using the comparison test. θ α + α θ dθ behaves as Since θ α dθ θ α + α θ dθ then arc length is infinite of α. θ α + α θ dθ Hence the arc length is finite of α >. θ α dθ + α θ α dθ = + α θ α dθ
Math 6 / Exam (November 6, ) page 9. [ points] Match each of the following differential equations with its possible slope field. Circle your answers. No justification is required. (a) y = ay with a >. I II III IV (b) y = k(y a) with a > and k <. I II III IV I II III IV (a) y = x(y ax) with a >. V VI VII VIII (b) y = ay y with a >. V VI VII VIII V VI VII VIII