Fall 2004 Math 151 3 Derivatives 3 10 Related Rates c 2004 Art Belmonte Mon 11 Oct Solution Draw a diagram Let x be the distance of the man from the building along the ground Let y be the height of his shadow above the ground Stewart 219 8 4 wall Summary 3 shadow 2 A given quantity Q t may be related to one or more other quantities perhaps geometrically or conceptually Typically t represents time The equation expressing the relationship may be explicit or implicit Differentiating the equation with respect to t and solving for d Q dt reveals how the rate of change of Q is related to the rates of change of the other quantities 1 y spotlight man 0 x 12 x 1 2 3 In the examples that follow we ll often find that drawing a diagram helps to formulate a relationship between the quantities involved We ll examine a variety of examples some from Stewart others from Gilat 0 3 6 By similar triangles dy dt Hand Examples 219 6 If a snowball melts so that its surface area decreases at a rate of 1 cm2 min find the rate at which the diameter decreases when the diameter is 10 cm 9 12 15 2 y or y 24 12 x 1 So 12 12 x dx 2 24 12 x dt 24 dx 2 12 x dt 24 1 6 12 4 2 0 6 m s So the man s shadow is decreasing at a rate of 0 6 m s 220 21 Solution Gravel is being dumped from a conveyor belt at a rate of 30 ft3 min and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal How fast is the pile s height increasing when the pile is 10 ft high Let r be the radius of the snowball and L its diameter Then r L 2 The surface area is S 4 r 2 4 L 2 2 L 2 that is dL dS 2 L whence S L 2 Thus dt dt Solution Let r be the base radius of the cone and h its height Since the height and diameter of the cone are always equal and the radius is half the diameter we have r 12 h 1 dS 1 1 dL cm min 1 dt 2 L dt 2 10 20 The diameter decreases at a rate of 1 0 0159 cm min 20 The volume of a cone is 2 1 h 3 that is V V 13 r 2 h 13 12 h h 12 1 dV dh h 2 whence Thus dt 4 dt 219 8 1 3 12 h 4 dV 4 6 dh ft min 30 2 2 dt 5 h dt 10 A spotlight on the ground shines on the wall of a building 12 m away If a man 2 m tall walks from the spotlight toward the building at a speed of 1 6 m s how fast is his shadow on the building decreasing when the man is 4 m from the building So the height increases at a rate of 1 6 0 38 ft min 5 b At t 12 s what is the rate at which sr is changing This is a supplemental question 220 22 A kite 100 ft above the ground moves horizontally at a speed of 8 ft s At what rate is the angle between the string attached to the kite and the horizontal decreasing when 200 ft of string have been let out Solution Solution We ll use a symbolic vector approach then float things at the end 1 Note that 54 mi h 79 15 ft s and 28 mi h 41 15 ft s Here is a picture depicting the car and train after they ve gone through the crossing Draw a diagram Let the origin be where the kid flying the kite is standing on the ground Let x be the distance along the ground from the kid to a perpendicular from the kite to the ground Let be the angle between the string and the ground Gilat 72 Sample Problem 3 5 100 train 75 Stewart 220 20 50 kite y 100 car 25 100 50 0 kid crossing 25 25 0 25 50 75 100 125 150 x 0 x 0 50 100 150 200 a Since this is a sample problem in your lab manual please read Gilat s explication therein We reproduce a version of the table he gives It is unformatted since this is a sample problem from Chapter 3 Formatting is covered in Chapter 4 100 or x 100 cot Via trigonometry we have tan x dx d Thus 100 csc2 When 200 ft have been let out dt dt 1 Solving for d dt and using d x dt 8 sin 100 200 2 yields b At t 12 s the rate at which sr is changing is 2 99 ft s2 Herewith a full MATLAB diary file d x dt sin2 100 2 8 1 100 2 1 0 02 rad s 50 So the angle decreases at a rate of 0 02 rad s or 1 146 s d dt Gilat 72 Sample Problem 3 5 augmented Rate of change of speed of train relative to car at t 12 seconds syms t Position of car rc and train rt rc 2 t 2 41 1 15 t 200 0 pretty rc MATLAB Examples rt 0 Gilat 72 Sample Problem 3 5 augmented 2 616 2 t t 200 15 79 1 5 t 400 pretty rt 0 396 5 t 400 Velocity of car vc train vt and of train relative to car vr vc diff rc t pretty vc A train and a car are approaching a road crossing At t 0 the train is 400 ft south of the crossing traveling north at a constant speed of 54 mi h At the same time the car is 200 ft west of the crossing traveling east at a speed of 28 mi h and accelerating at 4 ft s2 616 4 t 15 vt diff rt t pretty vt a Determine the positions of the train and the car the distance between them and the speed sr of the train relative to the car for the next 10 seconds This is what Gilat requested 0 vr vt vc pretty vr 2 396 5 0 0 616 4 t 15 396 5 Speed of train relative to car sr sr len vr pretty sr 2 4 15 225 t 1 2 4620 t 111925 Rate of change of speed of train relative to car and its value in ft s 2 when t 12 seconds dsr dt diff sr t pretty dsr dt 450 t 4620 2 15 2 1 2 225 t 4620 t 111925 dsr dt 12 subs dsr dt t 12 dsr dt 12 2 9891 Gilat s entities t time s x x coordinate of car position vector ft y y coordinate of train position …
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