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8.1 Local Linearity and the Derivative 281 We’ll use the rate at which water is decreasing today to approximate the change in water level over the next two days. lim0 , so for small. If we let 2 and use 115, we have 0 115 2 230. This makes sense; the water is decreasing at a rate of 115 gallons/day so after two days we’d expect it to decrease by about 230 gallons. 2#0 230 Because is concave up, we expect the actual water level to be a bit higher than this. EXAMPLE 8.2 Approximate 16.8. Use a rst derivative to get a good approximation. SOLUTION Let’s begin by sketching and getting an off-the-cuff approximation of 16.8. This will help us see how a tangent line can be of use. We see that 16.8 is a bit larger than 4. y (16, 4) y = x 16 x Figure 8.3 Question: How do we know this? Answer: 16.8 is close to 16 and 16 4. Question: How do we know 16.8 is a tad more than 4? Answer: We know that is increasing between 16 and 16.8. Question: How can we estimate how much to add to 4 to get a good approximation of 16.8? 282 CHAPTER 8 Fruits of Our Labor: Derivatives and Local Linearity Revisited Answer: This depends on the rate at which is increasing near 16. This is where the derivative comes into the picture. The derivative of at 16 gives the rate of increase. 1 1 1 16 2 16 2 16 8 Here we can adopt one of two equivalent viewpoints. Viewpoint (i): The Derivative as a Tool for Adjustment. We begin with an off-the-cuff approximation of 16.8 grounded in a nearby value of that we know with certainty. Knowing the rate of increase of at 16 enables us to judge how to adjust our approximation (in this case, 4) to t a nearby value of . for small (because lim0 ). In this example 16.8 16 0.8. slope = dy y x y x 16 16.8 Figure 8.4 1 0.8 1 8 0.8 8 10 So 0.1 and we have the approximation 16.8 4 0.1 4.1. In other words, the “new” -value is approximately the “old” -value, 4, plus , where we approximate by . Viewpoint (ii): Tangent Line Approximation. Again we begin with an off-the-cuff ap-proximation grounded in a value of we know with certainty. We choose 16. The tangent line to at 16 is the best linear approximation of around 16, so we use that tangent line to approximate 16.8. slope = 1 tangent line at x = 16 y = x 16 16.8 Figure 8.5 8.1 Local Linearity and the Derivative 283 The -value corresponding to 16.8 on the tangent line is very close to the corre-sponding -value on the curve. The tangent line has slope 1 and passes through the point 16, 4. Its equation is of the form 1 1, from 1 where 8 and 1, 116, 4. 4 1 16 4 1 16 Notice that this just says that the “new” -value is approximately the “old” -value, 4, plus the adjustment term—as in the rst approach. The -value on the tangent line when 16.8 is 4 1 16.8 164.1. Is the approximation 16.8 4.1 an overestimate or an underestimate? We know that 2 1; thus, as increases the derivative decreases. Therefore is concave down, the tangent line lies above the graph of the function, and this approximation is a bit too big. Comparing our approximation, 16.8 4.1, with the calculator approximation, 16.8 4.09878, we see that the approximation is good. The error is about 0.00122; we are off by about 0.03%. REMARK If we wanted to further rene our approximation, we could use information supplied by the second derivative to make the required adjustment.1 Further successive renements involve higher-order derivatives and give us what is called a Taylor polynomial approximation to the function at 16. EXERCISE 8.1 Approximate 101 using a tangent line approximation. Compare your answer to that obtained using a calculator. EXERCISE 8.2 Suppose you use a tangent line approximation to estimate 4.8. Do you expect the dif-ference between your approximation and the actual answer to be greater or less than the analogous difference we found in Example 8.2? Explain, and then check your reasoning by computation and using a calculator. EXERCISE 8.3 We argued that is concave down by reasoning that is decreasing. Show that is concave down by showing that 0. Use the limit denition of derivative to compute . If you have trouble, look back at Section 5.1 to see how the derivative of was computed. Answers to Exercises 8.1, 8.2, and 8.3 are supplied at the end of this section. Generalizing Our Method To approximate $ using a tangent line approximation, choose an near $ for which the value of is known. Let’s call this -value . Find the tangent line to at . The slope 1 The larger the value of the second derivative at 16 the larger the rate of change of the slope near 16. 284 CHAPTER 8 Fruits of Our Labor: Derivatives and Local Linearity Revisited is and a point on the line is , . Therefore, the equation of the tangent line is or . Evaluating at $ gives the corresponding -value on the tangent line. Answers to Selected Exercises Answer to Exercise 8.1 101 10 2 100 101 100 10 20 110.5 Answer to Exercise 8.2 The slope of is changing more rapidly at 4 than at 16. In both cases we are using the tangent line to approximate the value of the function at a distance 0.8 from the known value. Therefore we expect the error to be greater for 4.8 than for 16.8. 4.8 2 214 4.8 4 2 1 0.82.2 The difference between the approximation, 2.2, and the actual value, 2.19089 """ , is approximately 0.009. Answer to Exercise 8.3 43 P R O B L E M S F O R S E C T I O N 8 . 1 1. Use the tangent line approximation (linear approximation) of at 25 to approximatethefollowing.Usethegraphof anditstangentlineat 25topredict the relative accuracy of your approximations. Check the accuracy using a computer or calculator. (a) 23 (d) 25.1 (b) 24 (c) 24.9 (e) 26 (f) 27 2. Suppose we want to use a tangent line approximation of at to approximate a particular square root numerically. Which values of should we choose to approximate each of the following? (a) 102 (b) 8 (c) 18 (d) 115.5 3. Use a tangent line approximation to at 2 to approximate 1.9. 8.1 Local Linearity and the Derivative 285 4. Approximate 98 using the appropriate rst derivative to help you. Explain your reasoning. 5. Use the fact that 13 3 23 to approximate 3 30. Do you expect your an- swer to be an over-approximation or an under-approximation? Explain. Compare your answer to the approximation supplied by your calculator. 6. A steamboat is traveling down the Mississippi River. It is traveling south, making its way from point St. Paul, Minnesota, to Dubuque, Iowa. At noon it departs St. Paul traveling at 10 mph and is accelerating. It continues to accelerate over the next 10 minutes. Between noon and 12:10 p.m. it has covered 5 miles. Let % be the distance the steamboat has traveled from point St. Paul, where is measured in minutes. We’ll use noon as benchmark time of 0. (a) Determine the sign of %0, %0, and %0. Which of these expressions are you given enough information to specify numerically? (b) Sketch % over the time interval [0, 10]. (c) Find good upper and lower bounds for the distance the boat has traveled between noon and 12:05 p.m. Use a sketch to illustrate your reasoning graphically. 7. Consider the solid right cylinder with a xed height of 10 inches and a variable radius. Let &’ be the volume of the cylinder as a function of ’, the radius, given in inches. Interpret &’ geometrically. Explain why your answer makes sense by looking at & geometrically. ... - tailieumienphi.vn