Related Rates. Introduction. We are familiar with a variety of mathematical or quantitative relationships, especially geometric ones.

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1 Relate Rates Introuction We are familiar with a variety of mathematical or quantitative relationships, especially geometric ones For example, for the sies of a right triangle we have a 2 + b 2 = c 2 or the area of a circle is given by A = πr 2 In the problems we will now iscuss, one or more of these quantities will vary with time, that is, they are implicitly functions of time EXAMPLE 2501 When you rop a pebble in a pon, a circular wave front moves outwar with time r But r is a function of time: r = r(t) though usually we will not write r(t) explicitly Suppose we measure an foun that r was increasing at a constant rate of 02 m/s Using proper calculus notation, we coul write this as r = 02 Since A = πr 2, we can now ask, How is the area changing with respect to time? In other wors, what is A To etermine the answer we use implicit ifferentiation (A) = (πr2 ) A r = π2r = 2πr(02) A = 04πr m2 /s Notice that the rate of change in the area is not constant even though the rate of change in the raius was The change in the area epens on the particular value of r that we are intereste in For instance, when r = 5 m, then while if r were 3 m, then A = 04π5 = 2 m 2 /s r=5 A = 04π3 = 12 m 2 /s r=3

2 2 Differentiating Relationships Here are some aitional examples of familiar geometric relations Assume that the quantities are varying (implicitly) with time Let s practice fining relationships among how the these quantities change EXAMPLE 2502 For the sies of a right triangle: a 2 + b 2 = c 2 so (a2 + b 2 ) = (c2 ) 2a a b c + 2b = 2c EXAMPLE 2503 The area of a rectangle is given by: A = lw so using the prouct rule (A) = A (lw) = w l + l w EXAMPLE 2504 The volume of a cyliner is given by: V = πr 2 h so again using the prouct rule (V) = (πr2 h) V r h = 2πrh + πr2 YOU TRY IT 251 The perimeter of a rectangle is given by: P = 2l + 2w Fin the relation among the rates P (2l + 2h) = 2 l + 2 h (P) = answer to you try it 251 YOU TRY IT 252 The volume of a box with a square base is given by V = x 2 h Fin the relation among the rates answer to you try it 252 (V) = (x2 h) V = 2xh h + x2 YOU TRY IT 253 The volume of a sphere is given by V = 4 3 πr3 Fin the relation among the rates r = 4πr2 V ) ( 4 3 πr3 (V) = answer to you try it 253 EXAMPLE 2505 Here s a slightly ifferent one: In the triangle below: tan θ = y 10 so using the chain rule θ 10 y (tan θ) = ( y ) sec 2 θ θ 10 = 1 y 10 Another way to think about this problem is that if tan θ = y 10, then θ = arctan( y 10 ) so (θ) = [ ( y )] arctan θ 1 10 = 10 y 1 + ( y 10 )2 = 1 y 10 + y2 = 10 y y 2 10 Relate Rates Problems In this section we will put the relationships that we have practice ifferentiating into context an solve so-calle relate rates problems EXAMPLE 2506 (Classwork Example 2) In warm weather, the raius of a snowman s abomen ecreases at 2 cm/hr How fast is the volume changing when the raius of the abomen is 80 cm?

3 3 Solution Let s write in mathematical notation the rate(s) we are given an the rates we want to fin Given Rate: V r=80 r = 2 cm/hr Notice that the rate is negative because the raius is ecreasing To solve the problem we nee to fin a relationship between the volume an the raius of a sphere an then ifferentiate it implicitly with respect to time, t Relation: V = 4 3 πr3 Rate-ify: To solve the problem we ifferentiate the relation implicitly with respect to t an then substitute in the given particular information at the last step Substitute: V = 4πr2 r This is the general relation between the unknown an known rate Now we substitute in the given information r = 2 cm/hr an the fact that we are intereste in the the changing volume when r = 80 cm V = 4πr 2 r r=80 = 4π(80)2 ( 2) = 51, 200π cm 3 /hr EXAMPLE 2507 (Classwork Example 1) Two stuents finish a conversation an walk away from each other in perpenicular irections If one person walks at 4ft/sec an the other at 3ft/sec, how fast is the istance between the two changing at time t = 10 sec? Solution This time a iagram helps: z x y Given Rates: = 4 ft/s an y = 3 ft/s z Relation: Rate-ify: To solve the problem we nee to fin a relationship between x, y, an z Use the Pythagorean theorem z 2 = x 2 + y 2 To solve ifferentiate implicitly with respect to t an then substitute in the given particular information 2z z y = 2x + 2y z z = x + y y This is the general relation between the unknown an known rates

4 4 Substitute: Now we are intereste in what is happening at time t = 10 s Notice that we are NOT given the values of x, y, or z explicitly at this time But they are easy to figure out Remember, for constant rates, we have rate time = istance So when t = 10, we get x = 4 10 = 40 m an y = 3 10 = 30 m, so z = = 50 m Now we substitute in this information z z = x + y y 50 z = 40(4) + 30(3) = 250 z = = 5 m/s EXAMPLE 2508 (Classwork Example 3) A kite 100 feet above the groun moves horizontally at a rate of 8ft/s At what rate is the angle between the string an the vertical irection changing when 200 ft of string have been let out? x Kite 100 θ Person Solution Let x enote the (horizontal) sie opposite θ an let z be the hypotenuse (Mark both now) Given Rates: = 8 ft/s θ z=200 Relation: Use trig to relate θ an x: tan θ = 100 x Rate-ify: Differentiate implicitly with respect to time to obtain Substitute: This means sec 2 θ θ = θ = 1 1 sec 2 θ 100 = cos2 θ Now substitute in the known values From the triangle, when z = 200, cos θ = = 2 1, so ( ) θ = z= (8) = = 1 50 ra/s EXAMPLE 2509 (Classwork Example 4) As you walk away from a 20 foot high lampost, the length of your shaow changes If you are 6 feet tall an walking at 3ft/sec, at what rate is the length of your shaow changing? Light 20 Person 6 s x

5 5 Solution Let s enote the length of the shaow an let x enote the istance from the lamp Given Rates: = 3 ft/s s Note in this problem we want to fin this rate in general Relation: Use similar triangles to relate s an x s 6 = x+s 20 It makes sense to solve for s in terms of x Crossmultiply an then simplify to get: s = 6 14 x Rate-ify an Substitute: Differentiate implicitly with respect to time to obtain s = 6 14 = 6 14 (3) = 9 7 ft/s Notice that the rate of change is constant an oes not epen on where the person is EXAMPLE (Classwork Example 5) A surface ship is moving in a straight line at 10 km/hr An enemy sub maintains a position irectly below the ship while iving at an angle of π 9 (20 ) to the surface How fast is the sub moving? Ship = x π/9 Sub = z Solution Let x enote the position of the surface ship (horizontal sie) let z be the hypotenuse Given Rates: = 10 km/hr z Relation: Use trig to relate z an x cos π 9 = x z or z = 1 cos π x = 9 (sec π 9 )x Note: sec π 9 is constant Rate-ify an Substitute: Differentiate implicitly with respect to time to obtain z = sec π 9 = sec π (10) 1064 km/hr 9 YOU TRY IT 254 How fast is the epth of the sub changing in the previous example? answer to you try it 254 Let x enote the position of the surface ship (horizontal sie) let y be the epth Ship = x π/9 Depth = y Given Rates: = 10 km/hr y Relation: Use trig to relate y an x tan π 9 = y x or y = x tan π 9 )x Note: tan π 9 is constant Rate-ify an Substitute: Differentiate implicitly with respect to time to obtain y = tan π 9 = tan π 9 (10) 364 km/hr

6 6 EXAMPLE (Classwork Example 6) An ice block (raw an label a iagram) with a square base is melting in the sun at a steay rate of 48 cm 3 /hr If the height is ecreasing at a rate of 05 cm/hr, how fast is the ege of the base changing when the block is 10 cm in height an has an ege length of 6 cm? How fast is the surface area of the block changing at the same moment? Solution Let x enote the ege length of the base an h the height V Given Rates: = 48 cm 3 /hr an h = 05 cm/hr Notice that the rates are negative because the quantities are ecreasing h=10, x=6 Also if S is the surface area, we want S h=10, x=6 To solve the problem we nee to fin a relationship between the volume, height an the ege of a block an then ifferentiate it implicitly with respect to time, t Relation: V = x 2 h, since the base is square Rate-ify: Now ifferentiate the relation implicitly with respect to t V h = 2xh + x2 This is the general relation between the unknown an known rates Substitute: Now we substitute in the given information when h = 10 an x = 6 cm V = 48 = 2(6)(10)4 h=10, x=6 + (6) 2 ( 05) 30 = 120 h=10, x=6 h=10, x=6 = 025 cm/hr Relation: Rate-ify: For the surface area problem, the four sies are rectangles with area xh an the top an bottom are squares with area x 2 So S = 2x 2 + 4xh Differentiate S h = 4x + 4h + 4x Substitute We know all of the rates on the right sie of the equation when h = 10 an x = 6 So substituting in we fin: S h=10, x=6 = 4(6)( 025) + 4(10)( 025) + 4(6)( 05) = 22 cm 2 /hr