Related Rates - the Basics

Related Rates - the Basics In this section we exploe the way we can use deivatives to find the velocity at which things ae changing ove time. Up to now we have been finding the deivative to compae the change of the two vaiables in the function. Function Deivative Meaning y = x + x dy/dx = x + Compaes the change in y (ise) to the change in x (un) - gives us slope. s = t + t ds/ = t + Compaes the change in s (position) to the change in t (time) - gives us velocity. v = t + t dv/ = t + Compaes the change in v (velocity) to the change in t (time) - gives us acceleation. c = x + x dc/dx = x + Compaes the change in c (cost) to the change in x (units sold) - gives us maginal cost. Now we ae going to find a diffeent kind of change. The change in two vaiables with espect to the thid vaiable, time. In othe wods we will be taking the deivative with espect to TIME. Always befoe we have been taking the deivative with espect to x, y, o some othe vaiable in the equation. Now we will take the deivative with espect to a vaiable which is not in the equation - time. In doing this we will then find how fast x, o y, o volume, o aea, o distance is changing ove time. In calculus tems, this lets us find dx/, dy/, dv/, o dd/. When we do this, we call the pocess a Related Rates poblem. If I stat with an equation like x y 15 and I want to know how fast x and y ae changing ove time, I can take the deivative with espect to time and get this: dx dy x y 0 You might not be impessed yet, but this is going to open up a whole new spectum of possibilities fo us! Rates ae all aound us and now we know how to find equations to show us how they ae changing in elationship with each othe. Fun Stuff 1

Related Rates - the method So, now that we know what we ae doing - taking the deivative of all vaiables in the poblem with espect to time - let's look at methodology and some hints on how to do it coectly. Function Deivative Meaning y = x + x dy x dx dx Compaes both the change in y (vetical position) and the change in x (hoizontal position) to the change in time. Thee ae a numbe of key points to emembe as you do these poblems: 1. Evey time the poblem asks you to find "how fast is... changing..." you will take the deivative with espect to time. Evey time you ead "the... is deceasing/inceasing..." you ae talking about its deivative with espect to time. These ae the vebal clues to get you stated. Remembe this - evey AP exam has at least one of these in the fee esponse section. It is an impotant use of deivatives that is used in business, physics, physical science, eal life poblems. Memoize these vebal clues! The biggest poblem that students have with these is that they do not ecognize the poblems when they see them.. Be sue to teat constants as constants and vaiables as vaiables and take thei deivatives accodingly. Follow the quotient and poduct ules whee appopiate. Find how fast volume is changing fo the function V = AB whee B is a constant. dv da B Compaes both the change in volume and the change in A ove time. B is teated as a constant. 1

. Daw a pictue (if applicable) and wite all given infomation down, using pope notation, befoe you do anything else. A ight tiangle with legs a and b, whee b is always twice as long as a, has an aea that is inceasing at a ate of 4 in pe minute. How fast is leg a inceasing when a = 6 inches? da 4 ate at which the aea is inceasing b = a infomation given about b 4. Wite down what you ae tying to find, using pope mathematical notation. A ight tiangle with legs a and b, whee b is always twice as long as a, has an aea that is inceasing at a ate of 4 in pe minute. How fast is leg a inceasing when a = 6 inches? da? when a = 6

5. Find an equation which illustates the elationship between the given vaiables. You want you equation to involve only those vaiables whose ates of change ae given o ae to be detemined. If you have "extas" you will have to use given elationships and substitution to eliminate them. It will usually be easie if you can find an equation with only vaiables. A ight tiangle with legs a and b, whee b is always twice as long as a, has an aea that is inceasing at a ate of 4 in pe minute. How fast is leg a inceasing when a = 6 inches? A 1 ab b = a exta infomation we ae given about b 1 A aa a substitute the value of b and simplify aea fomula fo the tiangle 6. Take the deivative with espect to time. This means that evey time you take the deivative of a vaiable you will have a d*/. Remembe how we had to have a dy/dx evey time we took the deivative of the y in implicit diffeentiation? Now, since the vaiable we ae taking the deivative with espect to is time (t) and thee is no "t" in the equation, we will always have this notation fo all deivatives. da da a Anothe point on this subject. Because we ae taking the deivative with espect to time, the pime notation y ' has no meaning in elated ates poblems and you cannot use it. You can only use that when you ae taking the deivative of the dependent vaiable (usually y) with espect to the othe vaiable, the independent vaiable, that is in the poblem. So, if y 4 x 1x, you can use dy 4 x 1x dx o 4 y x 1x. 4 But if you have y x 1x and you take the deivative with espect to time, you MUST use the notation that shows that you ae taking the deivative with espect to time: dy dx 1x dx 1 1x 1 dx 7. Finally, sub in the values you wee given fo this instant in time and solve fo whateve you ae looking fo. NEVER put these values in until afte you have taken the deivative (the only exception is when one of the values is a constant). These poblems ae snapshots in time, giving instantaneous values, and theefoe only elate to deivatives, not to the oiginal equation. Putting instantaneous values into the equation befoe you have taken the deivative is a common eo, so be caeful.

da da a da When 4 da 1 in/min and a = 6, 4 6 da 8. Check to be sue you have the pope units on you answe. It will be simple to tell what they should be. Look at the deivative you solved fo. If you found da/, then you ae compaing the units of aea (squae somethings) to units of time (seconds, minutes, whateve you wee given). If you found dx/ and x was the length of the side of a squae, then you ae compaing the unit of length (feet, inches,...) to time. da 1 in/min These poblems take some pactice - give youself some time to do bunches of poblems. The moe you do, the easie it will become. 4

Related Rates - Examples Hee ae fou examples: A: Suppose that the adius and suface aea S 4 of a sphee ae diffeentiable functions of t. Wite an equation that elates ds/ to d/. Thee is no pictue hee that we need to daw. We ae given the equation fo the suface aea of a sphee, S 4, and we need to find an equation to elate ds/ to d/. In othe wods, we ae finding an equation that will show us how fast both suface aea and adius ae changing with espect to time. Any time we talk about change, we have to take the deivative. Change compaed to changing time means we take the deivative with espect to time. S ds 4 The 4 and ae both constants, so ou only vaiables ae S and. d 8 We took the deivative with espect to time. This is ou answe. In othe examples we continue and put values in at this point and solve fo eithe ds/ o d/. B: When a cicula plate of metal is heated in an oven, its adius inceases at the ate of 0.01 cm/min. At what ate is the plate's aea inceasing when the adius is 50 cm? I highlighted the key wods. They ae the ones that tell you what you know, and what you need to find. I will not daw a pictue hee it is not had to visualize ou cicle and adius! d 0.01 cm/min This is the infomation we ae given. The adius inceases at the ate of 0.01 cm/min. This is what we ae asked to find. da/ =? when = 50 How fast is the aea inceasing when the adius is 50? This is the stuff you do not use until afte you have taken the deivative. A Hee is the equation which elates the aea of a cicle to its adius. da d Take the deivative with espect to time. da 500.01 da cm / min Finally, sub in values and solve fo da/. Be sue to put the pope units in you answe. Since I found da/, I am compaing the instantaneous change in aea (which is measued in squae centimetes) to the change in time (which is measued in minutes). 1

C: The mechanics at Lincoln Automotive ae e-boing a 6-in deep cylinde to fit a new piston. The machine they ae using inceases the cylinde's adius one-thousanh of an inch evey min. How apidly is the cylinde volume inceasing when the boe (diamete) is.800 in? Again, look fo the clues that tell us that we will be taking the deivative with espect to time. The machine they ae using inceases the cylinde's adius one-thousanh of an inch evey min. How apidly is the cylinde volume inceasing when the boe (diamete) is.800 in? Fist we daw and label the pictue: 6 in.8 in d 0.01 in 1 min 000 h = 6 in in/min Wite down the infomation you ae given. The adius is changing at the ate of 0.001 in pe evey min The height is not changing, it is a constant. Wite down what you need to find. dv/ =? when d =.8 in dv/ =? when = 1.9 in V h h V 6 dv d 1 dv 1 1 1.9 000 dv 0.09 in / min How fast is volume changing when the diamete is.6? Find since we need adius fo the fomula. This is the fomula fo volume. Because h is a constant (the height of the can neve changes) we can put that value in ight now. Take the deivative (using the powe ule on the ight side of the equation). Sub in known values and solve fo dv/. Label the answe with the coect units.

D: A spheical balloon is inflated with helium at the ate of 100 ft /min. (a) How fast is the balloon's adius inceasing at the instant the adius is 5 ft? (b) How fast is the suface aea inceasing at that instant? I don t eally need the pictue it is just a cicle with a adius. I am sue you can imagine it. a) How fast is the balloon's adius inceasing at the instant the adius is 5 ft? dv 100 ft / min Wite down the infomation you ae given. The sphee is inceasing in size, hence the volume is inceasing. Wite down what you need to find. d/ =? when = 5 ft 4 V dv d 4 100 45 d d 1 ft/min How fast is the adius inceasing when the adius is 5 ft? This is the equation fo volume of a sphee. Take the deivative with espect to time. Sub in values and solve. b) How fast is the suface aea inceasing? ds/ =? when = 5 ft S 4 ds d 8 ds 8 51 40 ft /min Wite down what you need to find. How fast is the suface aea inceasing when the adius is 5 ft? This is the fomula fo the suface aea of the sphee. Take the deivative with espect to time. Sub in values (the value of d/ came fom pat a). Solve and label the answe with the coect units. Follow the steps and you will be fine. I have found that the biggest poblem fo most students is not ealizing when they ae going to be taking the deivative with espect to time. You must dill into you head the fact that when we talk about how fast something is changing that we ae compaing its change in time to the passing of time (just like velocity is the change in distance compaed to the change in time). Pactice and you will be fine.

Related Rates - Avoid these Eos! Follow the steps in ou examples and you will be fine. I have found that the biggest poblem fo most students is not doing the poblem it is in not even ealizing that they have to take the deivative with espect to time. You must dill into you head the fact that when we talk about how fast something is changing that we ae compaing its change in time to the passing of time (just like velocity is the change in distance compaed to the change in time). Pactice and you will be fine. Avoid these fou most common eos. Common Eo #1: Using the notation y ' = o A ' =. The pime notation is not valid when we ae taking the deivative with espect to time. You must use egula notation because it is the only way to show that you ae compaing the change in you vaiable with espect to the change in time. dy/ =.dx/ o da/ =. d/ Common Eo #: putting given instantaneous values into the equation befoe you have taken the deivative If something is always constant, you can put it in ight away. Fo instance, the length of a ladde that is leaning against the house (its length is not changing) o the adius of a can (the can is not shinking). Howeve, if you ae given instantaneous values, they cannot be used until afte you have taken the deivative (emembe that deivatives give you instantaneous change). You can ecognize these values in the poblem - they will be given in sentences like "find how fast is changing when h = 5" The "how fast is changing" tanslates to d/; it is what you ae going to find. The "when h = 5" means h is changing too (so thee is pobably going to be a dh/ in the deivative) and we ae going to stop time at the instant when h = 5 and look to see how fast is changing. 1

Common Eo #: fogetting you units o putting the wong units on you answes. When you solve fo whateve ate you ae looking fo, that notation will give you the units. dv/ compaes volume to time, so units will be cm /sec o ft /min o (units used to find volume) units of time used d/ compaes the units that you adius is measued in to the units of time you used: cm/sec o ft/min o miles/h dc/ compaes cost to change in time: $/day o pennies/sec Common Eo #4: fogetting to use the pope deivative ules. If you equation is V = x y then when you take the deivative you will have to use the poduct ule on x * y. When you take the deivative of the x it will have a dx/; when you take the deivative of the y, you will have a dy/. When you take the deivative of V, you get dv/: dv dy dx x y x If you equation is x + y = 15, you should use implicit diffeentiation. Do not solve fo y fist. dx dy The deivative of the constant is zeo. Don't foget you dx/ and dy/. x y 0

Homewok Examples #1,, 5 #1: The adius and aea A of a cicle ae elated by the equation expession that compaes da/ to d/. A. Wite an A Hee is the equation which elates the aea of a cicle to its da d adius. Take the deivative with espect to time. This is all we have to do fo this one! #: The adius, height h, and volume V of a ight cicula cylinde ae elated by the equation V h. a) How is dv/ elated to dh/ if is constant? is constant V dv h dh Wite down what you know. This tells that when we take the deivative we will teat just like the othe constant,. Hee is the equation which elates the volume of a cylinde to its adius and height. Take the deivative with espect to time. b) How is dv/ elated to d/ if h is constant? h is constant Wite down what you know. This tells that when we take the deivative we will teat h just like the othe constant,. V h Hee is the equation which elates the volume of a cylinde to its adius and height. It helps to goup all of the constants togethe out in font. Take the deivative with espect to time. dv d h 1

c) How is dv/ elated to d/ and dh/ if neithe h no is constant? V h Hee is the equation which elates the volume dv dh h d of a cylinde to its adius and height. Take the deivative with espect to time. This time we have to use the poduct ule because both and h ae vaiables. #5: If x, y, and z ae lengths of the edges of a ectangula box, the common length of the box's diagonals is s x y z. How is ds/ elated to dx/, dy/, and dz/? s x y z 1/ s x y z ds 1 1/ x y z x dx y dy z dz You can stop thee o simplify a bit: ds 1/ dx dy dz x y z x y z Hee is the fomula that elates s to x, y, and z. We need to ewite it with exponents so that we will be able to take its deivative. Take the deivative with espect to time. We will be using the geneal powe ule, so don't foget that you have to multiply by the deivative of the inside of the ( ).

Homewok Examples #1 #1: A 1-foot ladde is leaning against a house when it base stats to slide away. By the time the base is 1 ft fom the house, the base is moving at the ate of 5 ft/sec. Find the following: a) How fast is the top of the ladde sliding down the wall at that moment? Hee is a pictue: y 1 Ladde = 1 ft Wite down what you know. When x = 1 ft, dx/ = 5 ft/sec dy/ =? when x = 1 ft Wite down what you want to find. y x 1 169 Hee is the equation which elates the sides of the tiangle fomed by the ladde and side of the house. Remembe dy dx y x 0 y x 1 169 So when x = 1, y = 5 dy 5 1 5 0 dy 1 ft/sec x that the ladde has a constant length; it will not change. Take the deivative with espect to time. Since the ladde is a constant length, its deivative is zeo. We wee not given the length of y when x = 1, so we will have to find it fist. Then, put in the values you know and solve fo dy/. Use pope units when you give the answe. The negative means that the height up the wall is deceasing, which makes sense if the base of the ladde is going away fom the wall. 1

b) How fast is the aea of the tiangle fomed by the ladde, wall, and gound changing at that moment? Hee is a pictue: y 1 x Ladde = 1 ft When x = 1 ft, dx/ = 5 ft/sec da/ =? when x = 1 ft 1 A xy da 1 1 x y dx da 1 1 x dy y dx da 1 1 1 1 5 5 da 59.5 ft / sec Wite down what you know. Wite down what you want to find. Hee is the equation which elates the aea of the tiangle to its sides. Take the deivative with espect to time. We have to use the poduct ule on the ight side. Put in the values you know and solve fo da/. The values fo y and dy/ wee found in pat (a). Use pope units when you give the answe. The negative means that the aea is deceasing.

c) At what ate is the angle between the ladde and the gound changing at that moment? Hee is a pictue: y 1 x Ladde = 1 ft When x = 1 ft, dx/ = 5 ft/sec d/ =? when x = 1 ft tan sec y x dy dx x y d x When x = 1, y = 5, ladde = 1, so sec = 1/1 1 d (1)( 1) (5)(5) 1 (1) d 1 adians/sec Wite down what you know. Wite down what you want to find. Hee is the equation which elates the tangent of the angle to its sides. Take the deivative with espect to time. We have to use the quotient ule on the ight side, although you could ewite the equation as tan = y x -1 and use the poduct ule instead. Put in the values you know and solve fo d/. The values fo y and dy/ wee found in pat (a). You have to find the value of the secant of the angle fom you tiangle. Use pope units when you give the answe. The negative means that the angle is deceasing.

Homewok Examples #17, 7 #17: Wate is flowing at the ate of 50 m /min fom a concete conical esevoi (vetex down) of base adius 45 m and height 6 m. Find the following and give the answes in cm/min. a) How fast is the wate level falling when the wate is 5 m high? Hee is a pictue: 45 Wate level 6 h dv/ = -50 m /min cone base = 45 m, cone height = 6 m 45 h so 45 h 6 6 dh/ =? when h = 5 m 1 V h 45 h so 6 1 45 05 V h h h 6 108 dv 05 dh h 6 05 dh 50 (5) 6 dh 8 m/min 5 dh 8m 100cm cm/min 5 min 1m 9 Wite down what you know. We can find a elationship between the adius and height at wate level by using simila tiangles (they have popotional sides). Wite down what you want to find. Hee is the equation which elates the volume of wate to the adius and height. Use the elationship to eliminate the fom the equation. We don't know what d/ is and we aen't looking fo it, so it is bette to eliminate that vaiable. Take the deivative with espect to time. Put in the values you know and solve fo dh/. We wee told to give the answe in cm/min, so you have to convet units. The negative means the depth of the wate is falling. Thus the wate is falling at a ate of /9 cm/min 1

b) How fast is the adius of the wate's suface changing when the wate is 5 m high? Hee is a pictue: 45 Wate level 6 h Wite down what you know. We use the same value fo that we found peviously using popotions. dv/ = -50 m /min 45 h 6 d/ =? when h = 5 m Wite down what you want to find. 45 h 6 d 45 dh 6 d 45 6 9 d 80 cm/min We ae going to take a cleve shot cut hee, instead of stating with the volume fomula again. 45 We know that h and we found dh/ in pat 6 (a). That means we can use this simple equation to find d/. Take the deivative with espect to time. Put in the values you know and solve fo d/. Ou units ae aleady coect.

#7: A paticle moves fom ight to left along the paabolic cuve y x in such a way that its x-coodinate (in metes) deceases at the ate of 8 m/sec. How fast is the angle of inclination of the line joining the paticle to the oigin changing when x = -4? y (x, y) x Hee is a pictue: The angle of inclination,, is measued fom the positive x-axis. But I am going to look at the tiangle with as the efeence tiangle to get the tig atios that I need. They will change at the same ate except that while gets lage, will decease.

dx/ = -8 m/sec d/ =? when x = -4 m tan tan y x dy dx x y d sec x When x = -4, y ( 4), hypotenuse = 0 5, so 5 sec sec y x 1/ dy so 1 x 1/ ( 1) At x = -4, dx/ = -8, and dy 1 4 1/ (8) dx 5 d ( 4)() ()( 8) ( 4) d 8 4 adians/sec 16 5 5 y x tan x x 1/ 1/ 1/ x x x tan 1/ x x x x d 1 sec / dx x 5 d 1 / 4 8 d 11 4 8 ad/sec 8 5 5 1/ Wite down what you know. Wite down what you want to find. This equation shows the elationship between x, y, and the tangent of the angle. Take the deivative with espect to time. Use the quotient ule on the ight side. Put in the values you know and solve fo d/. This one is moe complicated because we have to go back to ou oiginal infomation to find values fo y, the secant of the angle, and dy/. Thee is a slightly diffeent way to do this poblem by simplifying the tangent atio befoe taking the deivative. Be caeful as you simplify this one! The bases ae not the same so we can't simply subtact exponents. Now we can diffeentiate and don't have to deal with the y. Sub in with the values we found befoe and solve fo d/. 4

Homewok Examples #9 #9: The length l of a ectangle is deceasing at a ate of cm/sec while the wih is inceasing at a ate of cm/sec. When l = 1 cm and w = 5 cm, find the following ates of change. a) How is the aea changing? Hee is a pictue: w As you wite down what we know, emembe that a ate of change is a deivative! dl cm/sec Wite down what you know. dw cm/sec da? Wite down what you want to find. when l = 1 cm, w = 5 cm A wl Hee is the equation which elates the aea of a ectangle to its wih and length. da dl dw Take the deivative with espect to time. We have to w l use the poduct ule on the ight side because both w and l changing. da 5 1 Put in the values you know and solve fo da/. Use pope units when you give the answe. da 14 cm /sec l 1

b) How is the peimete changing? Hee is a pictue: w l dl cm/sec dw cm/sec dp? Wite down what you know. Wite down what you want to find. when l = 1 cm, w = 5 cm Pwl Hee is the equation which elates the peimete of a ectangle to its wih and length. Take the deivative with espect to time. dp dw dl dp dp 0 cm/sec Put in the values you know and solve fo dp/. Use pope units when you give the answe.

c) How is the length of a diagonal changing? Hee is a pictue: d w l dl cm/sec Wite down what you know. dw cm/sec dd? when l = 1 cm, w = 5 cm Wite down what you want to find. d w l Hee is the equation which elates the Method one: d w l d dd w dw l dl when l = 1 cm, w = 5 cm, d w l so d 5 144 169 d = 1 dd 1 5 1 dd 14 cm/sec 1 Method two: 1/ d w l dd 1 1/ w l w dw l dl dd 1 1/ 5 144 (5)() (1)( ) dd 1 1/ 14 169 8 cm/sec 1 diagonal of a ectangle to its wih and length. This fist method has an easie deivative but moe wok in the next step. Take the deivative with espect to time. Put in the values you know and solve fo dd/. Use pope units when you give the answe. The negative means that the diagonal will be deceasing. This second method has a moe complicated deivative but does not entail solving fo the value of d in the evaluation step. I took the deivative with the geneal powe ule, put in the values we know and solved fo dd/, using pope units. The negative means that the diagonal will be deceasing.