Fall 2004 Math 1513 Derivatives3.10 Related RatesMon, 11/Octc2004, Art BelmonteSummaryA given quantity Q(t) may be related to one or more otherquantities, perhaps geometrically or conceptually. Typically trepresents time. The equation expressing the relationship may beexplicit or implicit. Differentiating the equation with respect to tand solving for dQ/dt reveals how the rate of change of Q isrelated to the rates of change of the other quantities.In the examples that follow, we’ll often find that drawing adiagram helps to formulate a relationship between the quantitiesinvolved. We’ll examine a variety of examples, some fromStewart, others from Gilat.Hand Examples219/6If a snowball melts so that its surface area decreases at a rate of1cm2/min, find the rate at which the diameter decreases when thediameter is 10 cm.Solution• Let r be the radius of the snowball and L its diameter. Thenr = L/2.• The surface area is S = 4πr2= 4π(L/2)2= π L2;thatis,S=πL2. ThusdSdt= 2πLdLdt, whencedLdt=12πLdSdt=12π(10)(−1)=−120πcm/min.The diameter decreases at a rate of120π≈ 0.0159 cm/min.219/8A spotlight on the ground shines on the wall of a building 12 maway. If a man 2 m tall walks from the spotlight toward thebuilding at a speed of 1.6m/s, how fast is his shadow on thebuilding decreasing when the man is 4 m from the building?Solution• Draw a diagram. Let x be the distance of the man from thebuilding along the ground. Let y be the height of his shadowabove the ground.−3 0 3 6 912 15−2−101234Stewart 219/8spotlightyx12 − xman wallshadow• By similar triangles,y12=212 − xor y = 24(12 − x)−1.Sodydt=−24(12 − x)−2−dxdt=24(12 − x)2dxdt=24(12 − 4)2(−1.6)=−0.6m/s.So the man’s shadow is decreasing at a rate of 0.6m/s.220/21Gravel is being dumped from a conveyor belt at a rate of30 ft3/min and its coarseness is such that it forms a pile in theshape 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?Solution• Let r be the base radius of the cone and h its height. Sincethe height and diameter of the cone are always equal and theradius is half the diameter, we have r =12h.• The volume of a cone isV =13πr2h =13π12h2h =112πh3;thatis,V =112πh3.ThusdVdt=14πh2dhdt, whencedhdt=4πh2dVdt=4π(10)2(30)=65πft/min.So the height increases at a rate of65π≈ 0.38 ft/min.1220/22A kite 100 ft above the ground moves horizontally at a speed of8ft/s. At what rate is the angle between the string attached to thekite and the horizontal decreasing when 200 ft of string have beenlet out?Solution• Draw a diagram. Let the origin be where the kid flying thekite is standing on the ground. Let x be the distance along theground from the kid to a perpendicular from the kite to theground. Let θ be the angle between the string and the ground.0 50100150200050100Stewart 220/20kite kidθx100• Via trigonometry, we have tanθ =100xor x = 100 cotθ.Thusdxdt=−100 csc2θdθdt. When 200 ft have been let out,sin θ =100200=12. Solving for dθ/dt and using dx/dt = 8yieldsdθdt=−dx/dt100sin2θ=−8100122=−150=−0.02 rad/s.So the angle θ decreases at a rate of 0.02 rad/sor1.146◦/s.MATLAB ExamplesGilat: 72 / Sample Problem 3-5, augmentedA train and a car are approaching a road crossing. At t = 0thetrain is 400 ft south of the crossing traveling north at a constantspeed of 54 mi/h. At the same time the car is 200 ft west of thecrossing traveling east at a speed of 28 mi/h and accelerating at4ft/s2.(a) Determine the positions of the train and the car, the distancebetween them, and the speed srof the train relative to the carfor the next 10 seconds. [This is what Gilat requested.](b) At t = 12 s, what is the rate at which sris changing? [This isa supplemental question.]SolutionWe’ll use a symbolic vector approach, then float things at the end.Note that 54 mi/h = 7915ft/s and 28 mi/h = 41115ft/s. Here is apicture depicting the car and train after they’ve gone through thecrossing.−25 0 25 50 75 100 125 150−250255075100xyGilat 72 / Sample Problem 3−5 traincar crossing(a) Since this is a sample problem in your lab manual, pleaseread Gilat’s explication therein. We reproduce a version ofthe table he gives. (It is unformatted, since this is a sampleproblem from Chapter 3. Formatting is covered in Chapter 4.)(b) At t = 12 s, the rate at which sris changing is 2.99 ft/s2.Herewith a full MATLAB diary file.%--------------------------------------------------% 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*tˆ2 + (41 + 1/15)*t - 200, 0]; pretty(rc)[ 2 616 ][2 t + --- t - 200 0][15 ]rt = [0, (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)[ 616 ][4 t + --- 0][15]vt = diff(rt, t); pretty(vt)[0 396/5]vr = vt - vc; pretty(vr)2[ 616 ][-4 t - --- 396/5][15 ]%% Speed of train relative to car (sr)%sr = len(vr); pretty(sr)2 1/24/15 (225 t + 4620 t + 111925)%% Rate of change of speed of train relative to car% and its value (in ft/sˆ2) when t = 12 seconds%dsrdt = diff(sr, t); pretty(dsr dt)450 t + 46202/15 -----------------------------2 1/2(225 t + 4620 t + 111925)dsrdt 12 = subs(dsr dt, t, 12)dsrdt 12 =2.9891%% Gilat’s entities% t time (s)% x x-coordinate of car position vector (ft)% y y-coordinate of train position vector (ft)% d distance from car to train, || rt - rc || (ft)% sc speed of car = || vc || = vc(1),% since vc(2) = 0 and t is positive (ft/s)% sr speed of train relative to car (from above)% [Gilat calls this speedtrainRcar] (ft/s)x = rc(1); pretty(x)2 6162 t + --- t - 20015y = rt(2); pretty(y)396/5 t - 400d = len(rt - rc); pretty(d)4 3 2 1/22/15 (225 t + 9240 t + 402700 t - 4488000 t + 11250000)sc = vc(1); pretty(sc);6164t+---15pretty(sr)2 1/24/15 (225 t + 4620 t + 111925)% Replicate [a version of] Gilat’s table: FLOAT ‘EM!t=0:10;x = eval(vectorize(x));y = eval(vectorize(y));d = eval(vectorize(d));sc = eval(vectorize(sc));sr = eval(vectorize(sr));%table = [t’ x’ y’ d’ sc’ sr’]table =0 -200.0000 -400.0000 447.2136 41.0667 89.21391.0000 -156.9333 -320.8000 357.1284 45.0667 91.12432.0000 -109.8667 -241.6000 265.4077 49.0667 93.16753.0000 -58.8000 -162.4000 172.7171 53.0667 95.33474.0000 -3.7333 -83.2000 83.2837 57.0667 97.61785.0000 55.3333
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