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Section2_5 - Section 2.5 Derivatives as functions and...

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(3/19/08) Section 2.5 Derivatives as functions and estimating derivatives Overview: In this section we discuss more examples of derivatives as rates of change and show how approximate derivatives can be obtained from graphs and tables. We also look at the concept of marginal change from economics, and discuss acceleration and other higher-order derivatives. Topics: The derivative as a function Finding approximate derivatives from graphs Using a graph to find where a derivative has a particular value Estimating a derivative from a table The term “marginal” in economics Differentiablity and continuity Higher-order derivatives Acceleration The derivative as a function Imagine that a tank contains V ( t ) = 12 t - t 2 gallons of water at time t (minutes) for 0 t 12. As can be seen from the graph of V = V ( t ) in the tV -plane of Figure 1, the tank is empty at t = 0, water is added for six minutes and then drained out, and the tank is empty again at t = 12. t 2 4 6 8 10 12 V (gallons) 10 20 30 40 V = V ( t ) (minutes) t 2 4 6 12 10 r (gallons/minute) - 12 12 r = V prime ( t ) (minutes) FIGURE 1 FIGURE 2 Example 1 (a) Find a formula for the rate of change with respect to time of the volume of water in the tank and draw its graph. (b) What do the rates of change of the volume at t = 2 , t = 6, and t = 10 indicate about the water in the tank? Solution (a) The rate of change of the volume is the derivative V prime ( t ) = d dt (12 t - t 2 ) = 12 - 2 t . It is defined for 0 < t < 12 because V ( t ) is defined for 0 t 12. The graph of r = V prime ( t ) is the line segment in Figure 2. (b) Because V prime (2) = 12 - 2(2) = 8, water is flowing into the tank at the rate of 8 gallons per minute at t = 2. Because V prime (6) = 12 - 2(6) = 0, water is neither flowing into or out of the tank at t = 6. Since V prime (10) = 12 - 2(10) = - 8, water is flowing out of the tank at the rate of 8 gallons per minute at t = 10. square The values of r = V prime ( t ) at t = 2 , 6, and 10 are plotted in Figure 3, and the corresponding tangent lines to the graph of V are shown in Figure 4. The tangent line at t = 2 has slope 8 because the value of the derivative at t = 2 is 8. The tangent line at t = 6 is horizontal (has zero slope) because the value of the derivative at t = 6 is 0. The tangent line at t = 2 has slope - 8 because the value of the derivative at 118
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Section 2.5, Derivatives as functions and estimating derivatives p. 119 (3/19/08) t 2 6 12 r (gallons/minute) 8 - 8 r = V prime ( t ) 10 t 2 4 6 8 10 12 V (gallons) 10 20 30 40 V = V ( t ) FIGURE 3 FIGURE 4 t = 10 is - 8. Finding approximate derivatives from graphs If you are given a function by a drawing of its graph rather than by an exact formula, you cannot find exact values of its derivative. You can, however, find approximate values of the derivative, as in the next two examples, by drawing plausible tangent lines to the graph and determining their approximate slopes by estimating the coordinates of pairs of points on the lines.
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