0.1 Related Rate In many phyical ituation we have a relationhip between multiple quantitie, and we know the rate at which one of the quantitie i changing. Oftentime we can ue thi relationhip a a convenient mean of meauring the unknown rate of change of one of the other quantitie, which may be very difficult to meaure directly. Such a ituation i called a related rate problem. The key to olving related rate problem i uing the known relationhip between the quantitie to find a relationhip between their rate of change, by differentiating the relationhip between the quantitie themelve. Example 1. Suppoe that a pherical balloon i being inflated, with the radiu of the balloon increaing at.1 m/. Find the rate of change of the volume of the balloon with repect to time. Solution The firt tep to olving thi problem i aigning variable to the quantitie of interet - the radiu and volume of the balloon. Let u refer to the radiu a r and the volume a V. Now we need to find a relationhip between thee quantitie. For a phere, we know that V = 4 πr. Since the balloon i being pumped up, clearly both V and r are changing with time. We are given / = 0.1 m/, and our goal i to find the time rate of change of the volume, /. We do o by differentiating the relationhip between V and r with repect to time, applying the differentiation operator d/ to both ide of our relationhip between V and r. Doing o we find d V = d ( 4 πr) = 4 π d r = 4 π r2 = 4πr 2 The above expreion give u a relationhip between the time rate of change of volume and radiu of the balloon. In thi particular ituation we are given a pecific value for /, o we find that = 4πr2 0.1 m = 0.4πr 2 m/. Note that in the above expreion we would need to ubtitute with the radiu of the balloon in meter in order to have the unit of / be given in m /. Example 2. Conider a right-circular cylinical tank of water of contant radiu r with a tap at it bottom from which water i flowing at a contant rate of 000 L/min. At what rate i the height of the water in the tank decreaing? Solution The firt tep i to identify the variable of interet. If we let h denote the height of the water in the tank, then what we are intereted in finding i /. From the problem tatement we are given the rate of change of the volume of the water, in L/min, which we will denote a /. For a right-cylinder, volume i given by the product of the bae area and height of the cylinder. Since we have a right-circular cylinder, volume, radiu, and height are related by the expreion V = πr 2 h. 1
We differentiate both ide of the equation with repect to t to find = πr2. Here it i important to note that only V and h are changing with time, and that r i imply a contant, o we do not need to worry about uing the product rule. Solving for / we find = 1 πr 2. In thi ituation we need to be a little bit careful with unit. We mut note that 1L = (10cm), o we have that = 000(10cm) = 10 6 cm min min. Now in order to have the unit match on both ide of the expreion we need give r in cm, o h will be expreed in cm, and / will be expreed in cm/min. Finally, we arrive at the expreion = 106 πr 2 cm min. Note that maller the radiu i, the fater the height op. Thi i becaue with a mall radiu 000 L of water leaving per minute would correpond to a volume of water with a large height, wherea when r i larger the volume would have a relatively maller height. The above example illutrate the importance of keeping track of unit when olving related rate problem. In general we can aociate the unit a a part of the variable in the relationhip between the quantitie of interet, and differentiate to relate the rate of change without paying regard to unit. However, once we want to ubtitute numerical value for different quantitie in the relationhip we need to make ure our unit match. In more complicated ituation it can be very helpful to aw a picture of the ituation at hand. Keeping thi in mind the generally trategy for olving a related rate problem i to identify the quantitie of interet and aign appropriate variable name. Having done o one aw a picture of the phyical ituation, and identifie the relationhip between the previouly defined variable, and whatever additional contant are a part of the ituation. Thee relationhip generally tem from a geometric or phyical conideration. Finally one differentiate both ide of thi relationhip to relate the rate of change of the variable of interet, and olve for the unknown rate of change in term of known quantitie. Example. Conider a riing hot-air balloon with a completely vertical acent. The height of the balloon i being tracked by a rangefinder 500ft from the liftoff point. At the moment the rangefinder elevation angle i π/4, the angle i increaing at the rate of 0.14 rad/min. How fat i the balloon riing at that intant? Solution Firt we identify our variable of interet - the angle the range finder make with the ground θ, and the height of the balloon y. In order to relate thee quantitie we mut conider the phyical ituation. The rangefinder i on the ground a ditance 500ft from the liftoff point from the balloon, and ince the the acent of the balloon i completely vertical, the balloon i ome height y above the liftoff point. Thee two ditance form two ide of a right triangle, with a hypotenue completed by the laer of the rangefinder. Thi i a right triangle with a bae of 500 ft, height of y ft, and with angle θ adjacent to the bae. Given thi right triangle we can relate the angle on the rangefinder and the height of the balloon. We find that tan(θ) = y 500 ft. 2
Differentiating both ide of the equation with repect to t, we find ec 2 (θ) dθ = 1 dy 500 ft. Thu, dy = 500 ec2 (θ) dθ ft. We know that dθ/ θ=π/4 = 0.14 rad/min, o evaluating thi expreion at thi angle of interet we find dy = 500 ec 2 (π/4) 0.14 ft θ=π/4 min = 70( 2) 2 ft ft = 140 min min. Example 4. Suppoe a police cruier i chaing a peeding car. The police cruier i heading outh approaching an interection, and the peeding car ha already paed the interection and i heading eat. When the police cruier i 0.6 mi north of the interection, the car i 0.8 mi to the eat, and the police radar determine that the ditance between them and the car i increaing at 20mph. If the cruier i moving at 60mph when thi meaurement i made, what i the peed of the car? Solution Let y denote the poition of the police cruier, and x the poition of car. It i convenient to define our origin at the interection, o that the poition of the cruier will fall entirely on the vertical axi, and the poition of the car on the horizontal axi. Let be the ditance between the cruier and the peeding car. In thi cae we have three variable of interet, which happen to be related a the three ide of a right triangle. Our triangle ha bae x, height y, and hypotenue, o uing the Pythagorean theorem, it follow Differentiating with repect to time we find 2 = x 2 + y 2. 2 d = 2xdx dy + 2y. In the above expreion both and dx/ are unknown. theorem again we can rewrite in term of x and y, finding Fortunately, uing the Pythagorean = x 2 + y 2. Solving for dx/, the time rate of change of the peeding car we find dx = 1 ( d x y dy ) = 1 ( x x 2 + y 2 d y dy ). Thi expreion relate the rate of change of the police car, the ditance between the two car, and the peeding car. To find the peed of the chaed car we mut ue the numerical data given in the problem tatement. Here we mut be a careful in order to make ure that our interpretation of thee data are conitent with the coordinate ytem we have etup. Namely, ince the police cruier i moving downward, we find that dy/ = 60 mph, a negative rate of change. Since the ditance between the two car i increaing we find d/ = 20 mph. Now we can ubtitute the known quantitie into the above expreion finding dx = 1 ( 0.8 0.8 2 + 0.6 2 20 0.6 ( 60)) mph = 70 mph.
Example 5. Conider an oil well located near the middle of a calm, hallow lake leaking oil onto the urface of the lake at the rate of 16 cubic feet per minute. Since oil and water don t mix, a the oil leak onto the urface of the lake it pread out into a uniform layer, called an oil lick, that float on top of the water. The lick i circular, centered at the leak, and 0.02 ft thick. Find the rate at which the radiu of the lick i increaing when the radiu i both 500 ft and 700 ft. Why would you expect the rate of change of the radii to be different, even though the oil i being added at a contant rate? Solution The variable of interet are the volume V and the radiu r of the oil lick. Since the urface of the oil lick i circular, and it ha a height of 0.02 ft, we are dealing with a very thin, right-circular cylinder. The relationhip between the volume and radiu of the cylinder i given by V = πr 2 h = 0.02πr 2 ft. Differentiating both ide of the equation with repect to t we find = 0.04πr ft. We know that / = 16 ft /min, and want to find /. Solving for / we find that = 1 0.04πr 1 ft = 400 πr ft2. Subtituting the appropriate value for r we find = 400 0.255 ft and = 400 0.182 ft. r=500 500π r=700 700π One way of interpreting the increaing ize of the oil lick i that the exiting portion of the lick doe not change, but concentric hollow dic are being added around the initial circular urface. From thi perpective, the larger r i, the larger the volume of uch a dic i, whereby the radiu r need to increae by le in order for a contant volume of oil to be added to the lick. Example 6. Suppoe we have two right circular cone, cone A and cone B. The height of cone A and the diameter of cone B both change at a rate of 4 cm/, while the diameter of cone A and the height of cone B are both contant. At a particular intant, both cone have the ame hape: h = d = 10 cm where h i height and d i diameter. Find the rate of change of the volume of the two cone at thi time. Why would you expect the volume of the cone to be changing at different rate? Solution Before proceeding it i helpful to viualize the ituation. Since the height of cone A i increaing, cone A i becoming a tall and kinny cone. Since cone B become wider and wider, it tart to look hort and fat. Of interet here are the height h, diameter D (uing capital D a d would create a notation iue), and volume V for the two cone. For a right-circular cone, the volume i given by V = 1 πr2 h. Since our given the diameter rather than radiu, we rewrite thi equation V = 1 π(d/2)2 h = π 12 D2 h. Before proceeding we need to recognize that we in fact have two related rate problem here - one for cone A and another for cone B. We will ue the ubcript a and b to ditinguih between the 4
two cone. For cone A, D will be a contant and h will vary with time, while for cone B, h will be a contant and D will vary with time. Thu, we mut differentiate thi expreion eparately for each cone, becaue different quantitie will depend on time. At the particular intant of interet, we know h a = h b = D a = D b = 10 cm. We alo know that a / = 4 cm/ and dd b / = 4 cm/. For cone A, differentiating with repect to t, we find the relationhip a = π a 12 D2 a = π 12 102 4 cm = 100π cm Now for cone B, differentiating with repect to t, we find the relationhip b = π 12 h b 2D b dd b = π 10 10 4cm 6 = 200π 104.72 cm. cm 209.44 cm. We expect that the two cone will have different rate of change, becaue in one cae, the volume i related to the firt power of a changing quantity, and in the econd it i related to the quare of the changing quantity. 5