d dx x = d dx (10,000 - p2 ) 1/2 dx [10,000 - p2 ] p' = dv = 0 dl dv V + n Things to remember: dt dt ; dy dt = 3

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1 45. x 0,000 " p 2 (0,000 - p 2 ) /2 d x d (0,000 - p2 ) /2 2 (0,000 - p2 ) -/2 d [0,000 - p2 ] 2(0,000 " p 2 ) 2 (!2pp') p' "p p # 0,000 " p 2 " 0,000 " p2 p 47. (L + m)(v + n) k (L + m)() + (V + n) dl 0 dl EXERCISE 4-6 -(L + m) V + n Things to remember:. SUGGESTIONS FOR SOLVING RELATED RATE PROBLEMS Step. Sketch a figure. Step 2. Identify all relevant variables, including those whose rates are given those whose rates are to be found. Step 3. Express all given rates rates to be found as derivatives. Step 4. Find an equation connecting the variables in step 2. Step 5. Implicitly differentiate the equation found in step 4, using the chain rule where appropriate, substitute in all given values. Step 6. Solve for the derivative that will give the unknown rate.. y x Differentiating with respect to t: 2x ; 2(5)(3) when x 5, 3 EXERCISE 4-6 9

2 3. x 2 + y 2 Differentiating with respect to t: 2x + 2y 0 2x -2y - y x when x -0.6, y 0.8, -4 ; (!0.6) (-4) -6 3, 5. x 2 + 3xy + y 2 Differentiating with respect to t: 2x + 2x + 3y + 2y 0 (3x + 2y) -(2x + 3y) + 3y) -(2x 3x + 2y ; when x, y 2, 2 -(2! + 3! 2) (3! + 2! 2) xy 36 x Given: d(xy) d(36) + y 0 4 when x 4 y 9. Therefore, 4 + 9(4) The y coordinate is decreasing at 9 units per second. 9. z rope x y 4 2x dz 2z or x z dz From the triangle, x 2 + y 2 z 2 or x z 2, since y CHAPTER 4 ADDITIONAL DERIVATIVE TOPICS

3 Given: dz -3. Also, when x, z2 or z 96. Therefore, 96(-3) "3 96 " 96 0 " feet/second. [Note: The negative sign indicates that the distance between the boat the dock is decreasing.]. Area: A πr 2 da d!r2 π 2R dr Given: dr 2 ft/sec da 2πR 2 4πR 3. V 4 3 πr3 4 3 π3r2 dr 4πR2 dr Given: dr 3 cm/min 4πR2 3 2πR 2 da R 0 ft 4π(0) 40π ft 2 /sec 26 ft 2 /sec R 0 cm 2π(0) 2 200π 3768 cm 3 /min 5. P T k () P kt k dt Given: dt 3 degrees per hour, T 250, P 500 pounds per square inch. From (), for T 250 P 500, k Thus, we have 2 dt 2(3) 6 Pressure increases at 6 pounds per square inch per hour. 7. By the Pythagorean theorem, x 2 + y or x 2 + y () y 2x + 2y 0 x EXERCISE

4 Therefore, - x. Given: y From (), y x 2, when x 6, y Thus, y 8 when x 6, (6,8)!3(6) 8!8 8!9 4 ft/sec. 3. Thus,!3x y. 9. y length of shadow x distance of man from light z distance of tip of shadow from light We want to compute dz. Triangles ABE CDE are similar triangles; thus, the ratios of corresponding sides are equal. Therefore, z 20 y 5 z! x [Note: y z - x.] 5 z or 20 z! x 5 z 4(z - x) z 4z - 4x 4x 3z 4 3 dz dz 4 3 Given: dz 5. Thus, 4 20 (5) 3 3 ft/sec. A 20 ft B 5 ft x z C D y E 2. V 4 3 πr3 () Since 4πr2 4 cu ft/sec, 4!r 2 4!r 2 (4)!r 2 ft/sec (2) At t minute 60 seconds, V 4(60) 240 cu ft, from (), r 3 3V 4! 3(240) 80 4!! ; r " 80 $ /3 #! % CHAPTER 4 ADDITIONAL DERIVATIVE TOPICS

5 From (2)!(3.855) ft/sec At t 2 minutes 20 seconds, V 4(20) 480 cu ft r 3 3V 4! 3(480) 4! From (2), 360! ; r " 360 $ /3 #! % 4.857!(4.857) ft/sec To find the time at which!r 2 00 r 2 00! r 00" 0 " Now, when r 0 ", V 4 3 π # & % $ 0 " ' ( " 750 " 00 ft/sec, solve Since the volume at time t is 4t, we have 4t t 750 " 00 " secs. EXERCISE

6 23. y e x + x + ; 3. ex + ex (3) + 3 3(e x + ) To find where the point crosses the x axis, use a graphing utility to solve e x + x + 0 The result is x Now, at x -.278, 3(e ) units/sec. 25. C 90,000 + x () R 0x - x2 (2) P R - C (3) (A) Differentiating () with respect to t: dc d(90,000) + d(x) dc Thus, dc (500) " $ # 500 % ' & $5,000 per week. Costs are increasing at the rate of $5,000 per week at this production level. (B) Differentiating (2) with respect to t: dr d(0x) - 0 # 0 " x & % $ 5' Thus, dr d x2-2x ( # 6000& " % 0 " ((500) $ x 6000, $ 5 ' # 500 % ' & (-00)500 -$50,000 per week. Revenue is decreasing at the rate of $50,000 per week at this production level. 96 CHAPTER 4 ADDITIONAL DERIVATIVE TOPICS

7 (C) Differentiating (3) with respect to t: dr - dc Thus, from parts (A) (B), we have: -50,000-5,000 -$65,000 Profits are decreasing at the rate of $65,000 per week at this production level. 27. S 60,000-40,000e x Differentiating implicitly with respect to t, we have ds 20e x ds -40,000( )e x Now, for x , we have ds 20(0)e (2000) 6000e - 2,207 Thus, sales are increasing at the rate of $2,207 per week. 29. Price p dem x are related by the equation 2x 2 + 5xp + 50p 2 80,000 () Differentiating implicitly with respect to t, we have 4x dp + 5x + 5p dp + 00p (A) From (2),!(5x + 00p) 4x + 5p Setting p in (), we get 2x x + 45,000 80,000 or x x - 7,500 0 Thus, x "75 ± (75)2 + 70,000 2 Since x 0, x 00 Now, for x 00, p dp![5(00) + 00()] " 2 4(00) + 5() 0 (2) dp!75 ± , we have , The dem is decreasing at the rate of units/month. EXERCISE