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 | 1. Explain why each approximation of length in
Example 8.2 is less than the actual length. |
| | 2. To estimate the
slope of f (x) = x2 + 1 at x = 1 , you would compute the slopes of various secant lines. Note that y = x2 + 1 curves up. Explain why the
secant line connecting (1,2) and (1.1,2.21) will have
slope greater than the slope of the tangent line. Discuss how the slope of the
secant line between (1,2) and (0.9,1.81) compares to the
slope of the tangent line. |
| In exercises 3-14, estimate the
slope (as in
Example 8.1) of y = f (x) at x = a.
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| 3. f (x) = x2 + 1, a = 1 | 4. f (x) = x2 + 1, a = 2 |
| 5. f (x) = cos x, a = 0 | 6. f (x) = cos x, a =  /2 |
| 7. f (x) = x3 + 2, a = 0 | 8. f (x) = x3 + 2, a = 1 |
| 9.  | 10. 
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| 11. f (x) = ex, a = 0 | 12. f (x) = ex, a = 1 |
| 13. f (x) = ln x, a = 1 | 14. f (x) = ln x, a = 2
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| In exercises 15-22, estimate the length of the curve y = f (x) on the given interval using (a) n = 4 and (b) n = 8 line segments. (c) If you can program a calculator or computer, use larger
n's and conjecture the actual length of the curve.
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| 15. f (x) = x2 + 1, 0 x 2 | 16. f (x) = x3 + 2, 0 x 1 |
| 17. f (x) = cos x, 0 x /2 | 18. f (x) = sin x, 0 x /2 |
| 19.  | 20. f (x) = 1/x, 1 x 2 |
| 21. f (x) = x2 + 1, - 2 x 2 | 22. f (x) = x3 + 2, - 1 x 1
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| In exercises 23-42, write out the first four terms of the sequence and estimate  .
|
| 23.  | 24.  |
| 25.  | 26.  |
| 27.  | 28.  |
| 29.  | 30.  |
| 31.  | 32.  |
| 33.  | 34.  |
| 35. an = e - n, n 1 | 36. an = sin (1/n), n 1 |
| 37. an = cos (1/n), n 1 | 38. an = e1/n, n 0 |
| 39.  | 40.  |
| 41.  | 42. 
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| 43. If a1 = 1-0.9 , a2 = 1-0.99 , a3 = 1-0.999 etc., determine  . Based on this limit, argue that  . |
| 44. As in exercise 43, argue that  . |
 | 45. The central concepts of calculus have been introduced in this section. An important aspect of our future development of calculus is to derive simple techniques for computing quantities such as
slope and length. In this exercise, you will learn how to directly compute the
slope of a curve at a point. Suppose you want the slope of y = x2 at x = 1. You could start by computing slopes of secant lines connecting the point (1,1) with nearby points. Suppose the nearby point has x - coordinate 1 + h , where h is a small (positive or negative) number. Explain why the corresponding y - coordinate is (1 + h)2. Show that the
slope of the secant line is  2 + h. As h gets closer and closer to 0, this
slope better approximates the slope of the tangent line. By setting h equal to 0, show that the
slope of the tangent line equals 2. In a similar way, show that the slope of y = x2 at x = 2 is 4 and find the
slope of y = x2 at x = 3. Based on your answers, conjecture a formula for the
slope of y = x2 at x = a , for any unspecified value of a. |
| | 46. The Sierpinski triangle is an example of a fractal, a geometric shape with an infinite amount of detail. To construct the Sierpinski triangle, start with an equilateral triangle (Figure A). Mark the midpoints of each of the sides and connect the midpoints with line segments (Figure B). Remove the middle “upside-down” triangle. There are now three equilateral triangles. For each one, mark the midpoints, connect with line segments and remove the middle triangle (Figure C). Continue on forever in this fashion (a third stage is shown in Figure D). The eventual result is the Sierpinski triangle. For this exercise, you will track two different sequences. First, let a0 be the area of the original triangle. If each side of the original triangle has length 1, show that a0 =  /4. Let a1 be the area of the first stage of construction, as shown in Figure B. Show that a1 = 3/16. Let a2 be the area of the second stage of construction, as shown in Figure C. Find a2. Continue to find the areas a3 and a4 and conjecture  . Next, let p0 be the perimeter of the original triangle. Show that p0 = 3. Let p1 be the perimeter of the first stage of construction, p2 the perimeter of the second stage of construction, and so on. Determine  . Fractals are often intricate and beautiful. They are also challenging mathematically because of apparent contradictions such as you've found: how can something have zero area and infinite length?
Figure A
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Figure B
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Figure C |
Figure D
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| | 47. We close this section with one of the threads that runs throughout this book: there are surprising connections to be made among the different areas of mathematics. One such “coincidence” involves the Sierpinski triangle of exercise 46 and a game we'll call the chaos game. Start with the vertices of an equilateral triangle labeled A, B and C. Pick one vertex at random. Again choose a vertex at random, move halfway from your starting vertex to the new vertex, and mark your location. Then repeat this sequence of steps: pick a vertex at random, move halfway from your current position to the vertex, and mark your new location. A small number of steps are shown in the figure, but yours will look different since you will choose different random vertices. What will the picture look like if you mark 1000 points? Program a graphing calculator or computer and find out! Seeing the screen fill up with dots is a great surprise to most people.
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