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Calculus Name: __________________________ Related Rates Packet (SOLUTIONS) – PART III Date: 12/6 Solutions to Related Rates Exercises (#11, 12, 13, 14) 11. A point, P, is moving along a curve whose equation is: y x 3 17 When P is at (2,5), y is increasing at a rate of 2 units/sec. How fast is x changing? dy 2 units s dt dx Find: ?? units when x = 2 and y = 5 s dt y 3 y x 17 1 dy 1 3 dx x 17 2 3x 2 dt 2 dt Plug in the “moment”: 1 1 3 dx 2 2 17 2 3 22 2 dt 5 dx 3 dt Conclusion: x is increasing at a rate of 5 3 units/sec at the moment when point P is at (2,5). x 12. A trough is 15 ft long and 4 ft across the top. Its ends are isosceles triangles with height 3 ft. Water runs into the trough at a rate of 2.5 ft3/min. How fast is the water level rising when it is 2 ft deep? 3 dV . 2.5 ft min dt 3 dh Find: when h = 2ft. ?? ft min dt Relationship(s) from Geo: 4 b Since 3 h V 1 h b bh length and 2 3 4 4h b and the length is 15ft, the volume will be 3 1 4h V h 15 2 3 Simplified, this is: V 10h2 . d dV dh V 10h2 20h dt dt dt dh dh Plug in your “moment”: 2.5 20 2 .0625 ft min dt dt Take the derivative: Conclusion: the water level is rising at a rate of 0.0625 ft/min when the water is 2 feet deep. Document1 Calculus Name: __________________________ Related Rates Packet (SOLUTIONS) – PART III Date: 12/6 13. If a spherical snowball melts so that its surface area decreases at a rate of 1 cm2/min, find the rate at which the diameter decreases when the diameter is 10 cm. [Surface area = 4r 2 ] 3 dSA 1 cm min dt Find: dd when d = 10cm ?? cm min dt d 2 2 d Using substitution, we combine these to get: SA 4 SA d 2 2 d dSA dd SA d 2 Take the derivative: 2d dt dt dt dd dd 1 cm Plug in the “moment: 1 2 10 0.016 cm min min dt dt 20 Important Relationships: SA 4r 2 and r Conclusion: the diameter is decreasing at a rate of 0.016 cm/min when the diameter is 10cm. 14. Two cars start moving from the same point. One travels south at 60 mi/h and the other travels west at 25 mi/h. At what rate is the distance between the cars increasing two hours later? dy dx and 60 mi 25 mi hr hr dt dt dn Find: when t = 2 hrs ?? mi hr dt 2 2 2 Relationships: d r t and x y n x n Take derivative: d dx dy dn x2 y 2 n2 2x 2y 2n dt dt dt dt Plug in the moment: When t = 2 hrs, x = -50 and y = -120, n 16900 dn 2 50 25 2 120 60 2 16900 dt dn 65 mi hr dt Conclusion: the distance between the cars is increasing at a rate of 65mi/hr two hours after they start moving from the same point. Document1 y