Math 116 Practice for Exam 2

Size: px

Start display at page:

Download "Math 116 Practice for Exam 2"

Dylan Cook
5 years ago
Views:

1 Math 116 Practice for Exam 2 Generated October 12, 215 Name: SOLUTIONS Instructor: Section Number: 1. This exam has 8 questions. 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. 2. Do not separate the pages of the exam. If any pages do become separated, write your name on them and point them out to your instructor when you hand in the exam. 3. Please read the instructions for each individual exercise carefully. One of the skills being tested on this exam is your ability to interpret questions, so instructors will not answer questions about exam problems during the exam. 4. Show an appropriate amount of work (including appropriate explanation) for each exercise so that the graders can see not only the answer but also how you obtained it. Include units in your answers where appropriate. 5. You may use any calculator except a TI-92 (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 3 5 note card. 6. If you use graphs or tables to obtain an answer, be certain to include an explanation and sketch of the graph, and to write out the entries of the table that you use. 7. You must use the methods learned in this course to solve all problems. Semester Exam Problem Name Points Score Fall Winter Fall Winter heart 12 Fall birds 14 Winter alpacas 11 Fall soup 11 Winter spinning 16 Total 99 Recommended time (based on points): 89 minutes

2 Math 116 / Exam 2 (November 16, 211) page 3 2. [12 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 2, y(t) = t 2 2t, for 3 t 3. Show all your work to receive full credit. a. [3 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 =,± 2 and dx = 4t3 8t = 4t(t 2 2) = = 2t 2 = at t = 1. Since there are no times at which dx never comes to a stop. and are simultaneously zero, the particle b. [2 points] For what values of t does the path of the particle have a vertical tangent line? Vertical tangent lines occur when dx this is true at t =,± 2. = and c. [3 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 3 t 3. The only critical point for y(t) was found above at t = 1. Since = 2 < and t= = 2 >, t=2 the First Derivative Test tells us that t = 1 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(1),y(1)) = ( 3, 1). d. [2 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 = 2,,2, while x(t) = t 4 4t 2 = t 2 (t 2 4) = t 2 (t 2)(t+2) = y(t) = t 2 2t = t(t 2) = gives t =,2. Thus, the particle passes through the origin at times t = and t = 2. Fall, 211 Math 116 Exam 2 Problem 2 Solution

3 Math 116 / Exam 2 (November 16, 211) page 4 e. [2 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. y x 2 From the given graph and above calculation, we know that the loop is traced out over the time interval t 2. The arclength of the loop is given by 2 (dx ) 2 + ( ) 2 = 2 (4t 3 8t) 2 +(2t 2) 2. Fall, 211 Math 116 Exam 2 Problem 2 Solution

4 Math 116 / Exam 2 (March 19, 212) page 4 3. [1 points] The motion of a particle is given by the following parametric equations y x x(t) = a(t2 1) t 2 +1 y(t) = t3 t t for < t < and a positive constant a. Show all your work to receive full credit. a. [3 points] Find the values of t at which the particle passes through the origin. We need to solve simultaneously x(t) = and y(t) =. x(t) =, then a(t2 1) t 2 +1 =. This is only possible if t 2 1 =. Hence t = ±1. y(t) =, then t3 t t 2 +1 =. This is only possible if t3 t =. Hence t =,±1. Times at the origin t = ±1. b. [5 points] Find the value of t at which the curve defined by the parametric equations has a vertical tangent line. Also, give the (x, y) coordinates of this point. [ (t x 2 +1)(2t) (t 2 ] 1)(2t) (t) = a (t 2 +1) 2 = and (x(),y()) = ( a,). 4at (t 2 +1) 2 then x (t) = at t =. c. [2 points] The curve has a vertical asymptote. Find the equation of this asymptote. a(t 2 1) lim x(t) = lim t t t 2 +1 asymptote is x = a. = lim t at 2 = a. Then the equation of the vertical t2 Winter, 212 Math 116 Exam 2 Problem 3 Solution

5 Math 116 / Exam 2 (November 14, 212) page 8 6. [13 points] A particle moves along the path given by the parametric equations x(t) = acost y(t) = sin2t for t 2π. where a is a positive constant. The graph of the particle s path in the x-y plane is shown below. In the questions below, show all your work to receive full credit. a. [2 points] At which values of t 2π, does the particle pass through the origin? = x(t) = acost: t = π 2, 3π 2. = y(t) = sin2t: 2t =,π,2π,3π,4π t =, π 2,π, 3π 2,2π. Particle passes through origin: t = π 2, 3π 2. b. [5 points] For what values of a are the two tangent lines to the curve at the origin perpendicular? Hint: Two lines are perpendicular if the product of their slopes is equal to 1. x (t) = asint, y (t) = 2cos2t. 1 = 2 a t = π 2 t = 3π 2 dx dx = a = a = 2 = 2 dx = 2 a ( 2 a dx = 2 a ) = 4 a 2 a = 2. c. [4 points] At what values of t 2π, does the curve have horizontal tangents? = y (t) = 2cos2t 2t = π 2, 3π 2, 5π 2, 7π 2,..., t = π 4, 3π 4, 5π 4, 7π 4. d. [2 points] Find an expression that computes the length of the curve. 2π a 2 sin 2 t+4cos 2 2t. Fall, 212 Math 116 Exam 2 Problem 6 Solution

6 Math 116 / Exam 2 (March 2, 213) page 8 5. [12 points] A particle moves according to the following parametric equations x = x(t) and y = y(t) for 2 t 2, where the graphs of x(t) and y(t) are shown below. a. [2 points] Is there a value of t at which the particle is at the point (,2)? If so, find the value of t where this happens. t = 1. b. [3 points] At which value(s) of t is the particle on the x-axis? t = 2,,2. c. [4 points] At what points (x,y) does the curve traveled by the particle have a horizontal tangent line? Include the times for each point. y (t) = when t = 1, (x,y) = (,2) and t = 1, (x,y) = (, 2). d. [3 points] For which of values of t is the slope of the tangent line to the curve positive? Slope= y (t) x (t) > if x and y have the same sign. This occurs at (,1), ( 1.5, 1) and (1.5,2). Winter, 213 Math 116 Exam 2 Problem 5 (heart) Solution

7 Math 116 / Exam 2 (November 13, 213) page 4 3. [14 points] The x and y positions of two birds in flight, Bird I and Bird II, are graphed below as functions of time t (see figures labeled Bird I and Bird II on the left). To the right, there are four parametric curves, A,B,C,D, showing flight paths of several birds in the x-y plane. 1 x 1 y 1 y 1 y Bird I.5 1 t 1 x 1 y.5 1 t A..5 1 x 1 y B..5 1 x 1 y Bird II.5 1 t.5 1 t C..5 1 x D..5 1 x a. [2 points] Is the horizontal velocity of bird I zero at any time < t < 1? If so, give an approximate t value. t =.5 b. [2 points] Based on the plots shown for bird II, consider a parametric curve for the flight path for bird II in the x-y plane. Would the slope of the tangent line to the flight path curve at time t =.9 be positive, negative, or zero? Justify. Positive slope since dx = y (t) x (t) and x (.9) and y (.9) both are negative. c. [4 points] One of the parametric curves A,B,C,D corresponds to bird I and another corresponds to bird II. Indicate which ones by circling the correct answers: Bird I corresponds to: A b c d Bird II corresponds to: a b c D Fall, 213 Math 116 Exam 2 Problem 3 (birds) Solution

8 Math 116 / Exam 2 (November 13, 213) page 5 d. [6 points] A third bird flies according to the following parametric equations x(t) = 1 t 3 y(t) = t 2 t. 1. Find the time(s) at which the bird is flying straight horizontally right or left. Show all your work. The bird is flying straight horizontally right or left if y (t) = 2t 1 =. Hence t = Find the speed of the bird at t = 1. Show all your work. Speed= (d(1 t 3 ) At t = 1, speed= 9+1 = 1. ) 2 ( ) + d(t 2 2 t) = ( 3t 2 ) 2 +(2t 1) 2. Fall, 213 Math 116 Exam 2 Problem 3 (birds) Solution

9 Math 116 / Exam 2 (March 24th, 214) page 3 2. [11 points] Abby and Brenda are alpacas running around in the xy-plane. Abby s position t minutes after she starts running is (cos(πt), 1) and Brenda s position t minutes after she starts running is ( t 2,e1 (t/2)2 ). Both alpacas begin running at the same time. a. [3 points] Do Brenda and Abby ever collide? If so at what time(s) does this occur? ForAbbyandBrendatocollidewemustsolvetheequations1 = e 1 (t/2)2 and t 2 = cos(πt). For the first equation we must have 1 (t/2) 2 = therefore t = ±2. Negative time doesn t make sense in this probelm so we only take t = 2. Plugging 2 into the second equation both sides are equal. So Abby and Brenda collide when t = 2. b. [5 points] Does Brenda or Abby ever stop moving at any time in the interval [2.5,4.5]? If so, which alpaca stops and at what time(s) does this occur? A (t) = ( πsin(πt),) so Abby stops moving whenever sin(πt) = so whenever t is an integer. Thus Abby stops when t = 3 or 4. B (t) = ( 1 2,te1 (t/2)2 ) the first coordinate can never be zero so Brenda is always moving. c. [3 points] Write an integral which gives the distance traveled by Brenda in the first 5 minutes she is running. Please circle your answer. Using the parametric arc length formula we get 5 ( 1 4 +t2 e 2 2(t/2)2 ) 1/2. Winter, 214 Math 116 Exam 2 Problem 2 (alpacas) Solution

10 Math 116 / Exam 1 (October 8, 214) page 7 6. [11 points] Franklin, your robot, is zipping around the kitchen making his famous Definitely Not Poison! soup. His coordinates in the xy-plane are given by the parametric equations x = t 2 t y = sin(πt) t seconds after he starts making soup. Assume that both x and y are measured in meters. a. [2 points] Calculate dx and. dx = 2t 1 = πcos(πt) b. [2 points] Find all times t when Franklin s velocity is zero. Franklin comes to a stop at all times t when both dx = 2t 1 = when t = 1/2. dx = πcos(πt) = when t = 1/2,3/2,5/2, etc. So Franklin comes to a stop when t = 1/2. = and =. t = 1/2 c. [3 points] Find Franklin s speed when t = 2 seconds. Include units. When t = 2: dx = 2(2) 1 = 3 Franklin s speed = = πcos(2π) = π ( dx )2 +( )2 Franklin s speed when t = 2 is 3 2 +π meters per second. Franklin s speed= 3 2 +π m/s d. [4 points] Write an integral which gives the distance traveled by Franklin during his first five seconds of zipping around. Do not evaluate this integral. 5 ( dx )2 +( )2 = 5 (2t 1) 2 +( πcos(πt)) 2 Fall, 214 Math 116 Exam 1 Problem 6 (soup) Solution

11 Math 116 / Exam 2 (March 23, 215) page 2 1. [16 points] Carla and Bobby run a race after spinning in circles for a good amount of time to make themselves dizzy. They start at the origin in the xy-plane and they race to the line y = 5. Assume the units of x and y are meters. Bobby s position in the xy-plane t seconds after the races starts is ( 3tcost, 1 3 tsint) and Carla s position in the xy-plane t seconds after the race starts is (tsint, tcost). a. [4 points] Write an integral that gives the distance that Carla travels during the first two seconds of the race. Do not evaluate your integral. We have dx = sin(t)+tcos(t), = cos(t)+tsin(t). The distance traveled by Carla in the first two seconds of the race is then given by 2 (sin(t)+tcos(t)) 2 +(tsin(t) cos(t)) 2. b. [3 points] Find Carla s speed at t = π. We have that Carla s speed is given by the function (sin(t)+tcos(t)) 2 +(tsin(t) cos(t)) 2, and so we need only plug in t = π which gives us the value below. Carla s speed at t = π is π 2 +1 m/sec c. [4 points] Carla and Bobby are so dizzy that they run into each other at least once during the race. Find the first time t > that they run into each other, and give the point (x,y) where the collision occurs. Setting the x and y coordinate functions equal gives us that collisions will occur when tan(t) = 3. The first time for t > when this occurs is 2π/3. Plugging this t value into either the equations for Bobby s or Carla s position will give the (x,y) coordinates given below for where the collision occurs. They first run into each other at t = The collision occurs at (x,y) = ( 2π π, π 3 ) d. [5 points] Bobby s phone flies out of his pocket at t = π/2. It travels in a straight line in the same direction as he was moving at t = π/2. Find the equation of this line in Cartesian coordinates. Winter, 215 Math 116 Exam 2 Problem 1 (spinning) Solution

12 Math 116 / Exam 2 (March 23, 215) page 3 Plug in t = π to the parametric equations for Bobby s position to get that 2 ( ) π Bobby is at the point P =, 2 at t = π. We can find the slope of the curve at 3 2 that point dx = 1 t=π dx P = 3 sin( π 2 )+ π 2 3 cos(π 2 ) t=π 3cos( π 2 )+ 3π 2 sin( π 2 ) = 2 3π. Since the line that gives the path of the phone is the same as the tangent line to the curve giving Bobby s motion at the point P, the equation of the line we want is that given below. The equation for the line is y = 2 3π x+ π 2 3 Winter, 215 Math 116 Exam 2 Problem 1 (spinning) Solution

Math 116 Practice for Exam 2

Math 116 Practice for Exam 2 Generated October 12, 2015 Name: Instructor: Section Number: 1. This exam has 8 questions. Note that the problems are not of equal difficulty, so you may want to skip over