Worksheet 8, Tuesday, November 5, 2013, Answer Key

Transcription

1 Math 105, Fall 2013 Worksheet 8, Tuesay, November 5, 2013, Answer Key Reminer: This worksheet is a chance for you not to just o the problems, but rather unerstan the problems. Please iscuss ieas with your partners. Your solutions shoul be focuse moreso on presentation than on numerical values. For your Relate Rates problems, you shoul ouble-check if your answer makes sense. If your quantity is ecreasing or shrinking, then you shoul get a negative rate of change. (Why?) If your quantity is increasing or growing, then you shoul get a positive rate of change. (Why?) Position-Velocity Problems 1. A stone is roppe from a brige that is 576 feet above a river. The stone s position above the water is given in feet at time t by s(t) = 16t How long oes it take for the stone to impact the water (fixe at position 0 here)? The stone impacts the water when s(t) = 0 = 16t That is, when t 2 = = 36. Thus, t = ±6, an we will only consier the positive time t = 6 secons here. The stone impacts the water at t = 6 secons. What is the stone s velocity when it impacts the water? The velocity is given by v(t) = s (t) = 32t. So the velocity of the stone when it impacts the water, at t = 6 secons, is given by v(6) = 32(6) = 192 ft sec. 2. Suppose a falling balls position is given by s(t) = t 2 feet at t secons. What is the balls initial position above the groun? Initial position is s(0) = 256 feet. Fin the average velocity of the ball uring the initial two s econs of its rop. Average velocity= V ave = change in istance change in time = s(2) s(0) = 32 ft 2 sec Fin the velocity at 2 secons an 3 secons respectively. v(t) = s (t) = 32t so v(2) = 64 ft sec an v(3) = 96 ft sec = (256 16(2)2 ) Unerstan why the velocity is negative here. The ball is falling in the negatively oriente irection. So the position s(t) is ecreasing ց which means its erivative s (t) = v(t) is negative. How much time passe before the ball hit the groun? First fin when the ball hits the groun. Set s(t) = 0 an solve for time t. s(t) = t 2 = 0 = t 2 = = t = ±4 = t = 4 secons, since we are consiering positive time here. So the ball hits the groun in 4 secons. =

2 What was the balls velocity when it hit the groun? The ball s velocity, when it hits the groun, is v(4) = 32(4) = 128 ft sec Finally, fin the balls acceleration at 3 secon a(t) = v (t) = s (t) = 32 (note, it s constant here, acceleration ue to gravity) So, a(3) = 32 ft sec 2 Relate Rates: For the following problems I expect you to follow the guielines set in class. You must show your work outline as follows, with full labelling: Given Information Equation relating variables Differentiate Substitute (given information) Solve (for esire quantity or rate) Answer the original questions (in wors!!) 3. Suppose a snowball remains spherical while it melts with the raius shrinking at one inch per hour. How fast is the volume of the snowball ecreasing when the raius is 2 inches? (cross-sectione in 2 imensions here) r Let r = raius of the sphere at time t Let V = volume of the sphere at time t Fin V =? when r = 2 feet an r = 1in hr Volume V = 4 3 πr3 (V) = ( ) 4 3 πr3 = V = 4 r π 3r2 3 = V = 4πr2r 2 (Relate Rates!)

3 V = 4π(2)2 ( 1) V = 16πin hr Answer the question that was aske: The volume of the snowball is ecreasing at a rate of 16π inches every hour at that moment. 4. An oil spill occurs at sea. The oil gushes out from an offshore errick an forms a circle whose area increases at a rate of 100 ft 2 /min. How fast is the raius of the spill increasing when the spill is 20 feet across? Let r = raius of the spill at time t Let A = area of the spill at time t Fin r r =? when iameter = 20 r = 10 ft an A ft2 = 100 min Area A = πr 2. (A) = (πr2 ) = A = 2πrr (Relate Rates!) A = 2πrr 100 = 2π(10) r = 20πr r = π = 5 ft πmin Answer the question that was aske: The raius of the oil spill is increasing at a rate of 5 π feet every minute at that moment. 3

4 5. The sies of a rectangle change with respect to time. The wih is increasing at a rate of 2 in/sec. while the length is ecreasing at a rate of 3 in/sec. How fast is the area of the rectangle changing when the wih is 6 inches an the length is 8 inches? y x Let x = length of the rectangle at time t Let y = wih of the rectangle at time t Let A = the area of the rectangle at time t Fin A x =? when x = 8 inches, y = 6 inches, The Area of the rectangle is given by length times wih A = xy. = 3 in/sec an y = 2 in/sec (A) = A (xy) = = xy +yx (Relate Rates!) A = xy +yx = 8(2)+6( 3) A = = 2 in2 /sec. Answer the question that was aske: The area is ecreasing at a rate of 2 inches per secon at that moment. 6. A box has a square base, an its height is always 10 inches. If the ege of the base is increasing at a rate of 2 in/min, how fast is the volume of the box increasing when the ege is 8 inches? 10 x x 4

5 Let x = length of base of box at time t Let V = volume of the box at time t Fin V x =? when x = 8 in, an = 2 in/min We use the volume formula for a box. Volume equals length times wih times height. Here we have the length equal to the wih, since it s a square base. V = 10x 2. (V) = (10x2 ) = V = 20xx (Relate Rates!) V = 20xx = 20(8)(2) V = 320 in3 /min Answer the question that was aske: The volume of the box is increasing at a rate of 320 cubic inches every minute at that moment. 7. A conical tank, 14 feet acorss the entire top an 12 feet eep, is leaking water. The raius of the water level is ecreasing at the rate of 2 feet per minute. How fast is the water leaking out of the tank when the raius of the water level is 2 feet? *** Recall the volume of the cone is given by V = 1 3 πr2 h. The cross section (with water level rawn in) looks like: 7 w r at 12 e h r Let r = raius of the water level at time t Let h = height of the water level at time t Let V = volume of the water in the tank at time t Fin V =? when r = 2 feet an r ft = 2 min 5

6 Volume= V = 1 3 πr2 h Extra solvable information: Note that h is not mentione in the problem s info. But there is a relationship, via similar triangles, between r an h. We must have r 7 = h 12 = h = 12r 7 After substituting into our previous equation, we get: V = 1 ( ) 12r 3 πr2 = πr3 (V) = ( ) 4 7 πr3 = V = 12 7 πr2 r (Relate Rates!) V = 12 7 πr2 r = 12 7 π(2)2 ( 2) V = 96π ft 3 7 min Answer the question that was aske: The water is leaking out at a rate of 96π 7 cubic feet every minute at that moment. 8. The top of a ten foot laer is sliing own a vertical wall at the rate of one foot every secon. How fast is istance between the bottom of the laer an the wall changing when the top of the laer is three feet above the groun? 10 y x Let x = istance between bottom of laer an wall at time t Let y = istance between top of laer an groun at the eep en at time t =? when y = 3 ft an y ft = 1 sec Fin x We have x 2 +y 2 = (10) 2. 6

7 Or x 2 +y 2 = 100 (x2 +y 2 ) = (100) = 2xx +2yy = 0 (Relate Rates!) Extra solvable information: Here we are not given the length of the base at the key moment, but we can use the Pythagorean Theorem to compute that value. x 2 +y 2 = 100 so x = 100 y 2 = = x +2(3)( 1) = x 6 = 0 x = = 3 91 Answer the question that was aske: The istance between the bottom of the laer 3 an the wall is increasing at a rate of feet every secon at that moment A kite starts flying 300 feet irectly above the groun. The kite is being blown horizontally at 10 feet per secon. When the kite has blown horizontally for 40 secons, how fast is the string running out? (a) 300 The picture at arbitrary time t is: groun Let x = istance kite has travelle horizontally at time t Let y = length of the string let out at time t Fin y =? when t = 40 sec. an x ft = 10 sec We have x 2 +(300) 2 = y 2 by the Pythagorean Theorem. 7 x kite string y

8 (x2 +(300) 2 ) = (y2 ) = 2x x = 2yy Extra solvable information: = x x = yy (Relate Rates!) At the key instant when t = 40, using the original rate of the horizontal travel of the kite, we have Distance =Rate times Time. So at that moment, 40 secons later x = x (t) = (10)(40) = 400 feet. We also nee to fin the length of the string at that key moment: We can use the Pythagorean Theorem again y = (300) 2 +(400 2 = = = 500 or you coul use the triangle rule. (400)(10) = (500) y y = = 8 ft sec Answer the question that was aske: The string is running out at a rate of 8 feet per secon at that moment. 8