Related Rates. Introduction

Transcription

1 Relate Rates Introuction We are familiar with a variet of mathematical or quantitative relationships, especiall geometric ones For eample, for the sies of a right triangle we have a 2 + b 2 = c 2 or the area of a circle is given b A = πr 2 In the problems we will now iscuss, one or more of these quantities will var with time, that is, the are implicitl functions of time EXAMPLE 251 When ou 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 usuall we will not write r(t) eplicitl 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 = π2r r = 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 eamples of familiar geometric relations Assume that the quantities are varing (implicitl) with time Let s practice fining relationships among how the these quantities change EXAMPLE 252 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 253 The area of a rectangle is given b: A = lw so using the prouct rule (A) = A (lw) = w l + l w EXAMPLE 254 The volume of a cliner is given b: 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 b: P = 2l + 2w Fin the relation among the rates YOU TRY IT 252 The volume of a bo with a square base is given b V = 2 h Fin the relation among the rates YOU TRY IT 253 The volume of a sphere is given b V = 4 3 πr3 Fin the relation among the rates EXAMPLE 255 Here s a slightl ifferent one: In the triangle below: tan θ = using the chain rule 10 so θ 10 (tan θ) = ( ) sec 2 θ θ 10 = 1 10 Another wa to think about this problem is that if tan θ = 10, then θ = arctan( 10 ) so (θ) = [ ( )] arctan θ 1 10 = ( 10 )2 = = Relate Rates Problems In this section we will put the relationships that we have practice ifferentiating into contet an solve so-calle relate rates problems EXAMPLE 256 (Classwork Eample 2) In warm weather, the raius of a snowman s abomen ecreases at 2cm/hr How fast is the volume changing when the raius of the abomen is 80cm? SOLUTION Let s write in mathematical notation the rate(s) we are given an the rates we want to fin Given Rate: r = 2 cm/hr Notice that the rate is negative because the raius is ecreasing V r=80 To solve the problem we nee to fin a relationship between the volume an the raius of a sphere an then ifferentiate it implicitl with respect to time, t

3 math 130, calculus i 3 Relation: V = 4 3 πr3 Rate-if: Substitute: To solve the problem we ifferentiate the relation implicitl with respect to t an then substitute in the given particular information at the last step V r = 4πr2 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 257 (Classwork Eample 1) Two stuents finish a conversation an walk awa 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 Given Rates: = 4 ft/s an = 3 ft/s z t=10 Relation: To solve the problem we nee to fin a relationship between,, an z Use the Pthagorean theorem z 2 = Rate-if: To solve ifferentiate implicitl with respect to t an then substitute in the given particular information 2z z = z z = + Substitute: This is the general relation between the unknown an known rates Now we are intereste in what is happening at time t = 10 s Notice that we are NOT given the values of,, or z eplicitl at this time But the are eas to figure out Remember, for constant rates, we have rate time = istance So when t = 10, we get = 4 10 = 40 m an = 3 10 = 30 m, so z = = 50 m Now we substitute in this information z z = t= z = 40(4) + 30(3) = 250 t=10 z = 250 t=10 50 = 5 m/s EXAMPLE 258 (Classwork Eample 3) A kite 100 feet above the groun moves horizontall at a rate of 8ft/s At what rate is the angle between the string an the vertical

4 4 irection changing when 200 ft of string have been let out? Kite 100 θ Person SOLUTION Let enote the (horizontal) sie opposite θ an let z be the hpotenuse (Mark both now) Given Rates: = 8 ft/s θ z=200 Relation: Use trig to relate θ an : tan θ = 100 Rate-if: Differentiate implicitl with respect to time to obtain This means sec 2 θ θ = θ = 1 1 sec 2 θ 100 = cos2 θ Substitute: Now substitute in the known values From the triangle, when z = 200, cos θ = = 1 2, so θ = z=200 ( ) (8) = = 1 50 ra/s EXAMPLE 259 (Classwork Eample 4) As ou walk awa from a 20 foot high lampost, the length of our shaow changes If ou are 6 feet tall an walking at 3ft/sec, at what rate is the length of our shaow changing? Light 20 Person 6 SOLUTION Let s enote the length of the shaow an let 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 s 6 = +s 20 It makes sense to solve for s in terms of Cross-multipl an then simplif to get: s = 6 14 Rate-if an Substitute: Differentiate implicitl 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 2510 (Classwork Eample 5) A surface ship is moving in a straight line at 10 km/hr An enem sub maintains a position irectl below the ship while iving at an angle of 20 to the surface How fast is the sub moving? s

5 math 130, calculus i 5 Ship 20 Sub SOLUTION Let enote the position of the surface ship (horizontal sie) let z be the hpotenuse (Mark both now) Given Rates: = 10 km/hr z Relation: Use trig to relate z an cos 20 = z or z = cos 1 20 = (sec 20 ) Note: sec 20 is constant Rate-if an Substitute: Differentiate implicitl with respect to time to obtain z = sec 20 = sec 20 (10) 1064 km/hr YOU TRY IT 254 How fast is the epth of the sub changing in the previous eample?