Lesson 13: Related Rates of Change
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Lesson 13: Related Rates of Change

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In this section we look at problems where changing quantities are related. For instance, a growing oil slick is changing in diameter and volume at the same time. How are the rates of change of......

In this section we look at problems where changing quantities are related. For instance, a growing oil slick is changing in diameter and volume at the same time. How are the rates of change of these quantities related? The chain rule for derivatives is the key.

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  • 1. Section 2.7 Related Rates V63.0121.027, Calculus I October 20, 2009 Announcements Midterm average 57.69/75 (77%), median 59/75 (79%), standard deviation 11% Solutions soon. . . . . . .
  • 2. “Is there a curve?” Midterm Mean was 77% and standard deviation was 11% So scores average are good Scores above 66/75 (88%) are great For final letter grades, refer to syllabus . . . . . .
  • 3. 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. . . . . . .
  • 4. 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? . . . . . .
  • 5. A solution The volume of the disk is V = π r2 h . . r . dV We are given , a certain h . dt value of r, and the object is dr to find at that instant. dt . . . . . .
  • 6. Solution Solution Differentiating V = π r2 h with respect to time we have 0 dV dr dh¡ ! = 2π rh + π r2 ¡ dt dt ¡dt . . . . . .
  • 7. Solution Solution Differentiating V = π r2 h with respect to time we have 0 dV dr dh¡ ! dr 1 dV = 2π rh + π r2 ¡ =⇒ = · . dt dt ¡dt dt 2π rh dt . . . . . .
  • 8. Solution Solution Differentiating V = π r2 h with respect to time we have 0 dV dr dh¡ ! dr 1 dV = 2π rh + π r2 ¡ =⇒ = · . dt dt ¡dt dt 2π rh dt Now we evaluate: dr 1 10, 000 L = · dt r=1 km 2π(1 km)(20 cm) s . . . . . .
  • 9. Solution Solution Differentiating V = π r2 h with respect to time we have 0 dV dr dh¡ ! dr 1 dV = 2π rh + π r2 ¡ =⇒ = · . dt dt ¡dt dt 2π rh 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 . . . . . .
  • 10. Outline Strategy Examples . . . . . .
  • 11. 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) . . . . . .
  • 12. Strategies for Related Rates Problems . . . . . .
  • 13. Strategies for Related Rates Problems 1. Read the problem. . . . . . .
  • 14. Strategies for Related Rates Problems 1. Read the problem. 2. Draw a diagram. . . . . . .
  • 15. 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) . . . . . .
  • 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) 4. Express the given information and the required rate in terms of derivatives . . . . . .
  • 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 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. . . . . . .
  • 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. 6. Use the Chain Rule to differentiate both sides with respect to t. . . . . . .
  • 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. 7. Substitute the given information into the resulting equation and solve for the unknown rate. . . . . . .
  • 20. Outline Strategy Examples . . . . . .
  • 21. 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? . . . . . .
  • 22. Diagram 4 . ft/sec m . . . . . . . .
  • 23. Diagram 4 . ft/sec m . . 5 . 00 w . 4 . ft/sec . . . . . .
  • 24. Diagram 4 . ft/sec . s m . . 5 . 00 w . 4 . ft/sec . . . . . .
  • 25. Diagram 4 . ft/sec . s m . . 5 . 00 w . w . 5 . 00 4 . ft/sec . . . . . .
  • 26. Diagram 4 . ft/sec √ s .= (m + w)2 + 5002 . s m . . 5 . 00 w . w . 5 . 00 4 . ft/sec . . . . . .
  • 27. 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 We want to know when m = 4800, w = 3600, = 4, and dt dt dw = 4. dt . . . . . .
  • 28. Differentiation We have ( ) ds 1( )−1/2 dm dw = (m + w)2 + 5002 (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) . . . . . .