Excerpt from "Calculus" 2013 AoPS Inc.

Transcription

1 Excerpt from "Calculus" 03 AoPS Inc. Te term related rates refers to two quantities tat are dependent on eac oter and tat are canging over time. We can use te dependent relationsip between te quantities to determine a relationsip between teir rates of cange. Here is a basic example: Problem 4.30: A balloon in te sape of a spere is being inflated at te rate of cm 3 /sec. At te time at wic te radius of te balloon is 3 cm, ow fast is its radius increasing? Solution for Problem 4.30: Let r denote te radius of te balloon (in cm) and V be its volume (in cm 3 ). Tis already is a little bit deceptive, since bot of te quantities are canging over time. So wat we sould really do is let r(t) denote te radius and V(t) denote te volume, were r and V are functions of t, and were t denotes te time (in seconds). Te relationsip between te quantities is V(t) = 4 3 (r(t))3. Wat we want is an expression tat relates te rates of cange of tese quantities. For tat, we take te derivative of te entire equation wit respect to t; tat is, we take d V(t) = d 4 3 (r(t))3. Te left side of tis equation is just dv. On te rigt side, we must use te Cain Rule: Tus dv d 4 3 (r(t))3 = 4 3 d ((r(t))3 ) = 4(r(t)) dr. = 4(r(t)) dr. Now we can plug in our given data. Te rate of cange of te volume is given as a constant cm 3 /sec, so dv =. Also, at te time t we are interested in, we are given r(t) = 3. So we plug tese in: = 4(3) dr. Tus, we solve to get dr =. So te radius is increasing by 8 8 cm/sec. Tis is te basic premise beind all related rates problems. We ave two quantities, bot of wic are a function of te same independent variable (usually time). If tere is an equation relating te two quantities, ten we can take te implicitly di erentiate tat equation wit respect to te independent variable to get an equation relating te variables and teir derivatives. In related rates problems, it is customary not to write explicitly te (t) part of te notation. For example, in Problem 4.30, we would typically write te equation relating te quantities as simply V = 4 3 r3, omitting te mention of te variable t. Ten, te derivative of tis equation is V 0 = 4r r 0. However, it s important to remember tat in tis expression, V and r (and r 0 ) are quantities tat depend on t. Problem 4.3: An observer is kilometer away from a rocket launc pad. At time t = 0, te rocket lifts o straigt upwards, and at time t seconds as acieved an altitude of 0t meters. At 5 seconds after takeo, ow fast is te distance between te rocket and te observer increasing? Solution for Problem 4.3: Even toug tere is only one moving object te rocket tere are two quantities tat we are interested in tat are canging wit respect to time: te altitude of te rocket (wic we ll call ) and te distance from te rocket to te observer (wic we ll call x). 8

2 Excerpt from "Calculus" 03 AoPS Inc RELATED RATES Te picture at rigt sows ow tese quantities are related. But don t make te following mistake: Rocket Bogus Solution: x = +. 0t m We ave to be more careful about te units! Te observer is kilometer from te launc pad, but te eigt of te rocket is given in meters. So, if we want to express everyting in meters, we ave te equation km Observer x = + (000). Taking te derivative of bot sides wit respect to t, we get x dx = d. We are given t = 5. But t doesn t directly appear as a term in te above expression. So we need to figure out wat te quantities in te above equation are in terms of t. d Wen t = 5, we ave = 0t = 50. Tis means tat x = (50) + (000) = 7(50), so x = 50 p 7. Also, = 0t = 00. Tus, we ave x dx ) x dx ) (50 p 7) dx ) dx = d = d = (50)(00) = p 00 = 00 p So te distance between te rocket and te observer is increasing at a rate of 00 p 7 7 m/sec. Te next problem is a classic related rates problem: Problem 4.3: A 0-foot-tall ladder rests against a wall, but te foot of te ladder is slipping away from te wall at a rate of in/sec. Wen te ladder forms a 60 degree angle to te ground, ow fast is te top of te ladder sliding down? 0 ft in/sec 9

3 Excerpt from "Calculus" 03 AoPS Inc. Solution for Problem 4.3: We sould set tis up to ave as our two variables te functions wose rates we care about. Terefore, we sould use w (te distance from te wall) and (te eigt of te top of te ladder). We can ten write an equation relating tese quantities: w + = 00. But tis uses feet as te units on te rigt side. Since te rate is in in/sec, maybe we sould write te wole ting in inces: w + = Altoug it is always good practice to keep our units consistent, it doesn t really matter ere, because te next step is to take te derivative wit respect to time: ww = 0. Now we need to plug in te values of tese quantities so tat we can solve for 0. We know tat w 0 = (in/sec). We need to figure out w and. But te oter bit of information tat we aven t used yet is te 60 angle. Tis gives us w = 5 (feet) and = 5 p 3 (feet). But te rate of cange is in inces per second! So we need to use w = 60 (inces) and = 60 p 3 (inces). We ten plug tese values in to our related rates expression: (60)() + (60 p 3) 0 = 0. Solving for 0 gives 0 = p3. So te top of te ladder is falling at a rate of p3 inces per second. By te way, in Problem 4.3, wat does te function 0 (t) look like in general? We can solve for 0 in our related rates equation: 0 = w 0 w = w, since w 0 =. But we know wat te quantity w/ is: it s te cotangent of te angle tat te ladder makes wit te ground. So we ave 0 = cot, were is te angle tat te ladder makes wit te ground. Tis makes real world sense. At = /, te ladder is vertical, and cot(/) = 0, meaning tat te top of te ladder is stationary. As decreases, cot increases, so te top of te ladder moves faster and faster downwards. Let s look at one more related rates problem tat s a bit arder. Problem 4.33: A cone-saped filter as a ole at te top of radius 4 cm and a ole at te bottom of radius cm, and is 6 cm in eigt. Water flows out of te bottom at a rate of cm 3 /sec. If te filter begins completely filled at time t = 0, ow fast is te water level decreasing after 30 seconds? 6 4 Solution for Problem 4.33: We start by identifying te relevant quantities tat are canging wit respect to time. Tese are te eigt of te water, wic we ll call, and te volume of water in te filter, wic we ll call V. We need to determine ow tey are related. 0

4 Excerpt from "Calculus" 03 AoPS Inc RELATED RATES We see tat te filter is a frustum, wic is a cone wit a smaller cone copped o. Te volume of a frustum is most easily computed as te di erence in volume between te two cones. To compute te volumes of te cones, we need teir eigts. So we will be muc better o if our eigt is measured from te vertex of te imaginary cones, not from te bottom of te filter. Next, te formula for te volume of a cone is 3 r, were r is te radius and is te eigt. So we need to determine ow te eigt is related te radius. Tey are directly proportional, as we can see from our picture at rigt. In particular, decreasing te eigt by 6 causes a decrease of 3 in te radius, so we conclude tat /r =, and tus te radius at eigt is /. Note tat te top of our frustum is at = 8 (not = 6), and at te top we ave radius 8/ = 4, as expected. Te bottom of te frustum is at = =. So te volume of te water at eigt is: 4 V = 3 r 3 () () = 3! 3 () () = 3 3. We can di erentiate tis equation to get an equation relating te rates: V 0 = 4 0. We are told tat V 0 =, as te rate of decrease of volume is constant. But we also need to find te value of in order to solve for 0. So we need to find te volume at t = 30 and compute its eigt. We can start by finding te initial volume (at t = 0): Tis simplifies to 4. V(0) = (8)3 3. Wen t = 30, we will ave lost 30 = 60 cm 3 of water, so te volume will be V(30) = 4 tis in to our equation for volume to find at time t = 30: 60. We can plug Tis gives 3 = 4 60 = 3 3. r , and tus = Finally, we can go back to our related rates equation: and plug in te found value of. Tis gives: = 4 0, We solve for 0 to get our answer: = = 8 3p (5 70) 3 = 3p. (64 90) 3 Tis is approximately 0.059, and tus at time t = 30 te water level is decreasing at a rate of cm/sec.

5 Excerpt from "Calculus" 03 AoPS Inc. Exercises 4.6. A plane is flying overead at an altitude of 0 km and a speed of 00 m/sec. Te plane is 0 km away from a camera on te ground, and te plane is flying in a direction tat will take it directly over te camera. Te camera continuously rotates to keep te plane centered in its lens. Wen te plane is 5 km away from te camera, ow fast is te camera rotating? 4.6. Boyle s Law states (assuming tat te temperature is constant) tat te pressure and volume of a gas are inversely proportional. Suppose Gas X starts at atm (atmospere) of pressure and takes up 00 cm 3. If te pressure is increased by 0. atm/min, ten after 0 minutes, at wat rate is te volume decreasing? Hints: A sperical snowball is 0 cm in diameter, but is melting at te rate of 0.5 cm 3 per second. Wen te snowball is reduced to alf its original volume, at wat rate is its diameter decreasing? A meteorite as entered te eart s atmospere and is burning up at a rate tat is proportional to te meteorite s surface area. Wat can you determine about te rate tat te meteorite s radius is decreasing? Two cars start at an intersection at time t = 0 wit velocity 0. At t = 0, one car accelerates due nort at 8 m/sec, and te oter car accelerates due east at 5 m/sec. After 6 seconds, by wat rate is te distance between te cars increasing? 4.6.6? Te clock on te wall at my o ce as an our and tat s 6 cm long and a minute and tat s 0 cm long. At exactly :00, at wat rate is te distance between te tips of te two ands canging? Hints: 48, 67 Review Problems 4.34 Te following is te grap of te derivative f 0 for some function f. Sketc possible graps of f and f 00. Take into account were te function is increasing, decreasing, its concavity, and any inflection points. y x 4.35 Suppose tat f is a function wit domain R suc tat f 0 (x) = x 3x + 3 for all x R. Prove tat f as an inverse (a) (b) (c) Sow tat if f is a quadratic polynomial, ten te grap of f as no inflection points. Sow tat if f is a cubic polynomial, ten te grap of f as exactly one inflection point. Can you generalize?