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- 1. Section 2.7 Related Rates V63.0121.002.2010Su, Calculus I New York University May 27, 2010 Announcements No class Monday, May 31 Assignment 2 due Tuesday, June 1 . . . . . .
- 2. Announcements No class Monday, May 31 Assignment 2 due Tuesday, June 1 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 2 / 18
- 3. Objectives Use derivatives to understand rates of change. Model word problems . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 3 / 18
- 4. What are related rates problems? Today we’ll look at a direct application of the chain rule to real-world problems. Examples of these can be found whenever you have some system or object changing, and you want to measure the rate of change of something related to it. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 4 / 18
- 5. Problem Example An oil slick in the shape of a disk is growing. At a certain time, the radius is 1 km and the volume is growing at the rate of 10,000 liters per second. If the slick is always 20 cm deep, how fast is the radius of the disk growing at the same time? . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 5 / 18
- 6. A solution Solution The volume of the disk is V = πr2 h. . r . dV We are given , a certain h . dt value of r, and the object is to dr find at that instant. dt . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 6 / 18
- 7. Solution Differentiating V = πr2 h with respect to time we have 0 dV dr dh¡ ! = 2πrh + πr2 ¡ dt dt ¡dt . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 7 / 18
- 8. Solution Differentiating V = πr2 h with respect to time we have 0 dV dr dh¡ ! dr 1 dV = 2πrh + πr2 ¡ =⇒ = · . dt dt ¡dt dt 2πrh dt . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 7 / 18
- 9. Solution Differentiating V = πr2 h with respect to time we have 0 dV dr dh¡ ! dr 1 dV = 2πrh + πr2 ¡ =⇒ = · . dt dt ¡dt dt 2πrh dt Now we evaluate: dr 1 10, 000 L = · dt r=1 km 2π(1 km)(20 cm) s . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 7 / 18
- 10. Solution Differentiating V = πr2 h with respect to time we have 0 dV dr dh¡ ! dr 1 dV = 2πrh + πr2 ¡ =⇒ = · . dt dt ¡dt dt 2πrh dt Now we evaluate: dr 1 10, 000 L = · dt r=1 km 2π(1 km)(20 cm) s Converting every length to meters we have dr 1 10 m3 1 m = · = dt r=1 km 2π(1000 m)(0.2 m) s 40π s . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 7 / 18
- 11. Outline Strategy Examples . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 8 / 18
- 12. Strategies for Problem Solving 1. Understand the problem 2. Devise a plan 3. Carry out the plan 4. Review and extend György Pólya (Hungarian, 1887–1985) . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 9 / 18
- 13. Strategies for Related Rates Problems . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 10 / 18
- 14. Strategies for Related Rates Problems 1. Read the problem. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 10 / 18
- 15. Strategies for Related Rates Problems 1. Read the problem. 2. Draw a diagram. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 10 / 18
- 16. Strategies for Related Rates Problems 1. Read the problem. 2. Draw a diagram. 3. Introduce notation. Give symbols to all quantities that are functions of time (and maybe some constants) . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 10 / 18
- 17. Strategies for Related Rates Problems 1. Read the problem. 2. Draw a diagram. 3. Introduce notation. Give symbols to all quantities that are functions of time (and maybe some constants) 4. Express the given information and the required rate in terms of derivatives . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 10 / 18
- 18. Strategies for Related Rates Problems 1. Read the problem. 2. Draw a diagram. 3. Introduce notation. Give symbols to all quantities that are functions of time (and maybe some constants) 4. Express the given information and the required rate in terms of derivatives 5. Write an equation that relates the various quantities of the problem. If necessary, use the geometry of the situation to eliminate all but one of the variables. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 10 / 18
- 19. Strategies for Related Rates Problems 1. Read the problem. 2. Draw a diagram. 3. Introduce notation. Give symbols to all quantities that are functions of time (and maybe some constants) 4. Express the given information and the required rate in terms of derivatives 5. Write an equation that relates the various quantities of the problem. If necessary, use the geometry of the situation to eliminate all but one of the variables. 6. Use the Chain Rule to differentiate both sides with respect to t. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 10 / 18
- 20. Strategies for Related Rates Problems 1. Read the problem. 2. Draw a diagram. 3. Introduce notation. Give symbols to all quantities that are functions of time (and maybe some constants) 4. Express the given information and the required rate in terms of derivatives 5. Write an equation that relates the various quantities of the problem. If necessary, use the geometry of the situation to eliminate all but one of the variables. 6. Use the Chain Rule to differentiate both sides with respect to t. 7. Substitute the given information into the resulting equation and solve for the unknown rate. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 10 / 18
- 21. Outline Strategy Examples . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 11 / 18
- 22. Another one Example A man starts walking north at 4ft/sec from a point P. Five minutes later a woman starts walking south at 4ft/sec from a point 500 ft due east of P. At what rate are the people walking apart 15 min after the woman starts walking? . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 12 / 18
- 23. Diagram 4 . ft/sec . m . . P . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 13 / 18
- 24. Diagram 4 . ft/sec . m . . 5 . 00 P . w . 4 . ft/sec . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 13 / 18
- 25. Diagram 4 . ft/sec . . s m . . 5 . 00 P . w . 4 . ft/sec . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 13 / 18
- 26. Diagram 4 . ft/sec . . s m . . 5 . 00 P . w . w . 5 . 00 4 . ft/sec . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 13 / 18
- 27. Diagram 4 . ft/sec √ . . s m . s . = (m + w)2 + 5002 . 5 . 00 P . w . w . 5 . 00 4 . ft/sec . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 13 / 18
- 28. Expressing what is known and unknown 15 minutes after the woman starts walking, the woman has traveled ( )( ) 4ft 60sec (15min) = 3600ft sec min while the man has traveled ( )( ) 4ft 60sec (20min) = 4800ft sec min ds dm dw We want to know when m = 4800, w = 3600, = 4, and = 4. dt dt dt . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 14 / 18
- 29. Differentiation We have ( ) ds 1( 2 2 )−1/2 dm dw = (m + w) + 500 (2)(m + w) + dt 2 dt dt ( ) m + w dm dw = + s dt dt At our particular point in time ds 4800 + 3600 672 =√ (4 + 4) = √ ≈ 7.98587ft/s dt 2 + 5002 7081 (4800 + 3600) . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 15 / 18
- 30. An example from electricity Example If two resistors with resistances R1 and R2 are connected in parallel, as in the figure, then . . the total resistance R, measured in Ohms (Ω), is given R . 1 R . 2 by . . 1 1 1 = + R R1 R2 (a) Suppose R1 = 80 Ω and R2 = 100 Ω. What is R? (b) If at some point R′ = 0.3 Ω/s and R′ = 0.2 Ω/s, what is R′ at the 1 2 same time? . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 16 / 18
- 31. Solution Solution R1 R2 80 · 100 4 (a) R = = = 44 Ω. R1 + R2 80 + 100 9 (b) Differentiating the relation between R1 , R2 , and R we get 1 1 1 − 2 R′ = − R′ − 1 R′ 2 R R2 1 R2 2 So when R′ = 0.3 Ω/s and R′ = 0.2 Ω/s, 1 2 ( ) ( ) ′ 2 R′ 1 R′2 R2 R2 1 2 R′ 1 R′2 R =R + = + R2 R2 1 2 (R1 + R2 )2 R2 R2 1 2 ( )2 ( ) 400 3/10 2/10 107 = 2 + 2 = ≈ 0.132098 Ω/s 9 80 100 810 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 17 / 18
- 32. Summary Related Rates problems are an application of the chain rule to modeling Similar triangles, the Pythagorean Theorem, trigonometric functions are often clues to finding the right relation. Problem solving techniques: understand, strategize, solve, review. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 18 / 18

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