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Related Rates

In this topic we will be dealing with two or more quantities that are either increasing or decreasing with respect to time. Related Rates. y 2 = a 2 + b 3. ( ) 2 = ( ) 2 + ( ) 3. 2. ( ). dy/dt. =. 2. ( ). da/dt. + 3( ) 2 db/dt.

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Related Rates

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  1. In this topic we will be dealing with two or more quantities that are either increasing or decreasing with respect to time. Related Rates

  2. y2 = a2 + b3 ( )2 = ( )2 + ( )3 2 ( ) dy/dt = 2 ( ) da/dt + 3( )2 db/dt

  3. All variables below are functions of t. 2x3 + y = 5z3 6 x2 dx/dt + dy/dt = 15 z2 dz/dt

  4. y and  are functions of t. y = sin  cos  d/dt dy/dt =

  5. Quantities that changes over time are functions of t. Example: The population of a town increase by 500 people every year. This year there are 3000 P = 3000 + 500t dP/dt = 500

  6. Things to Remember: 1. A rate is a derivative with respect to time. Thus, units such as cm/min, miles per hour and the like implies that a derivative is involve. 2. If a quantity is increasing, then its derivative is positive. If it is decreasing, its derivative is negative.

  7. 1. A 25-foot ladder is leaning against a wall. The base of the ladder is sliding away from the wall at the rate of 2 ft per minute. How fast is the top of the ladder sliding down when the base of the ladder is 7 feet away from the wall ?

  8. 2 ft/min dx/dt = dy/dt = ? 25 ft x = 7 ft y x x2 + y2 = 252 (2x) (dx/dt) (2y) (dy/dt) = 0 + 7 2 ?

  9. y = ? x = 7 ft 72 + y2 = 252 49 + y2 = 625 25 ft y y2 = 576 y = 24 x x2 + y2 = 252 (2x) (dx/dt) (2y) (dy/dt) = 0 + 7 2 ?

  10. y = ? x = 7 ft 72 + y2 = 252 49 + y2 = 625 25 ft y y2 = 576 y = 24 x x2 + y2 = 252 (2x) (dx/dt) (2y) (dy/dt) = 0 + 7 2 24

  11. (2)(7)(2) + (2)(24)(dy/dt) = 0 28 + 48 (dy/dt) = 0 48 (dy/dt) = - 28 48 48 dy/dt = - 7/12 ft/min

  12. 2. 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 ? A B

  13. dx/dt = 25 mi/h dy/dt = 60 mi/hi x  B dz/dt = ? y z x2 + y2 = z2 A At what rate is the distance between the cars increasing after 2 hours?

  14. At what rate is distance between the two cars increasing after 2 hours? x A y z B x2 + y2 = z2 dz/dt = ? (2x) (dx/dt) + (2y) (dt/dt) = (2z) (dz/dt) ? 50 ? ? 60

  15. After 2 hours: ? B  ? A

  16. After 2 hours: 502 + 1202 = z2 x 50 miles 16900 = z2 B  z = 130 120 miles z y 130 A

  17. 2x dx/dt + 2y dy/dt = 2z dz/dt 50 25 120 60 130 ? 2 (50) ( 25 ) + 2 (120) (60) = 2 (130) dz/dt 2500 + 14400 = 260 dz/dt 16900 = 260 dz/dt 260 260 dz/dt = 65 mi/h

  18. 3. The height of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm2/min. At what rate is the base of the triangle changing when the height is 10 cm and the area of 100 cm2 ? dh/dt = 1 cm/min db/dt = ? dA/dt = 2 cm/min h = 10 A = 100

  19. dh/dt = 1 cm/min h = 10 cm dA/dt = 2 cm2/min A = 100 cm2 db/dt = ? A = b h A = ½ bh 2 (½) (db/dt) (h) + (dh/dt) (½ b) dA/dt = ? 1 10 2

  20. dh/dt = 1 cm/min h = 10 cm dA/dt = 2 cm2/min A = 100 cm2 db/dt = ? A = b h A = ½ bh 2 100 = ( b ) (10) 2 b = 20 100 = 5b

  21. dh/dt = 1 cm/min h = 10 cm dA/dt = 2 cm2/min A = 100 cm2 db/dt = ? A = b h A = ½ bh 2 (½) (db/dt) (h) + (dh/dt) (½ b) dA/dt = ? 1 10 2

  22. dh/dt = 1 cm/min h = 10 cm dA/dt = 2 cm2/min A = 100 cm2 db/dt = ? A = b h A = ½ bh 2 (½) (db/dt) (h) + (dh/dt) (½ b) dA/dt = 20 1 10 2

  23. (½) (db/dt) (h) + (dh/dt) (½ b) dA/dt = 20 1 10 2 2 = (½) (db/dt)(10) + (1)(20) 2 = 5 (db/dt) + 20 2 – 20 = 5 (db/dt) - 18 = 5 (db/dt) 5 5 db/dt = - 3.6 cm/min

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