Hprec3 7

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Hprec3 7

  1. 1. 3.7: Rates of Change © 2007 Roy L. Gover (www.mrgover.com) Learning Goals: •Find the average rate of change of a function over an interval •Represent the avg. rate of change geometrically as the slope of a secant line •Use the difference quotient to find a formula for the average rate of change of a function
  2. 2. Average Rate of Change Time t 0 1 2 3 3.5 4 4.5 5 Distance d(t) 0 16 64 144 196 256 324 400 Average speed= Distance traveled time interval Find the average speed of the falling rock from t=3 to t=4.5 average speed= d(4.5)- d(3)= 4.5-3 324-144= 180=120 1.5 1.5
  3. 3. Try This Find the average speed of the Falling rock a.From t=1 to t=4 80 ft/se b.From t=2 to t=4.5 104 ft/sec Time t 0 1 2 3 3.5 4 4.5 5 Distance d(t) 0 16 64 144 196 256 324 400
  4. 4. Definition Let f be a function The average rate of change of f(x) with respect to x as x changes from a to b is the value Change in f(x) = f(b)-f(a) Change in x b-a
  5. 5. Example A balloon is being filled with water. Its approximate volume in gallons is V(x)= x³ 55 Where x is the radius of the balloon in inches. Find the average rate of change of the volume of the balloon as the radius increases from 5 to 10 inches
  6. 6. Solution Change in volume= V(10)-V(5) = Change in radius 10-5 18.18-2.27 = 15.91= 3.18 gallons per inch 10-5 5
  7. 7. Example A small manufacturing company makes specialty office desks. The cost (in thousands of dollars) of producing x desks is given by the function: C(x) = .0009x³-.06x²+1.6x+5 Find the average rate of change of cost a.From 0 to 10 desks b.From 10 to 30 desks c.From 30 to 50 desks
  8. 8. Solution a.15.9-5 = 10.9 = 1.09 10 10 b. 23.3-15.9= 7.4 = .37 30-10 20 c. 47.5 – 23.3 = 24.2 =1.21 50-30 20
  9. 9. Geometric Interpretation of Average Rate of Change Let f be a function. Average rage of change= f(b)-f(a) =Slope of secant line of f from x=a to x=b b-a joining (a,f(a)) and (b,f(b)) on the graph of f
  10. 10. The Difference Quotient Average rate of change is often computed over very small intervals, such as the rate from 4 to 4.01. Since 4.01= 4+.01 both cases are essentially the same: computing the rate of change over the interval from 4 to 4+h for some small quantity h. The average rate of change for any function f over a interval from x to x+h is:
  11. 11. The Difference Quotient Average Rate of Change= f(b) –f(a) = f(x+h)-f(x)= f(x+h)-f(x) b-a (x+h)-x h
  12. 12. Example Find the difference quotient of V(x)= x³ 55 And use it to find the average rate of change of V as x changes from 8 to 8.01
  13. 13. Solution V(x+h)-V(x)= (x+h) - x³ = 1 ((x+h)³ -x³ h 55 55 55 h h = 1 (x³ +3x²h+3xh²+h³-x³)= 3x² +3xh +h² 55 55 To find the average rate of change from 8 to 8.01 3x² +3xh+ h² = 3 ˣ 8² +3 ˣ 8(.01) + (.01)² = 55 55 3.495

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