chapter2ProblemsAndSolutions

chapter2ProblemsAndSolutions - Math 103 Problems Chapter 1...

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Unformatted text preview: Math 103 Problems Chapter 1 1.1 Areas in the plane 1 (a) Compute the area of the staircase shown in Figure 1.1. (b) What would be the area of that region if, instead of the ten steps shown, it consisted of 100 steps, each of width 0.1 and with heights 0.1, 0.2, ..10 ? (c) If there are a very large number of steps of very small width, and very small height increments, what would be the approximate area of the region shaded? 0.0 10.0 0.0 10.0 Figure 1.1: Detailed Solution: (a) The steps in Figure 1.1 have width 1 and heights 1, 2, 3, up to 10. Thus, the area of the staircase is given by S = 1(1 + 2 + 3 + ··· + 10) = 10 summationdisplay k =1 k = 10(11) 2 = 55 . (b) If there were 100 steps of width 0.1 and heights incremented by 0.1 per step, then the total area would be S = 0 . 1(0 . 1 + 0 . 2 + 0 . 3 + ··· + 10 . 0) = 0 . 01(1 + 2 + 3 + . . . 100) S = 0 . 01 100 summationdisplay k =1 k = 0 . 01 100(101) 2 = 0 . 01(5050) = 50 . 5 v.2005.1 - January 3, 2006 1 Math 103 Problems Chapter 1 (c) As the number of (small) steps increases, we expect to get approximately the area of a tri- angular region, which is exactly half of the rectangle, i.e 100 / 2 = 50. We see that the two estimates made here indeed approach this value as the number of steps increases. 1.2 Find the area bounded by the x-axis, the y-axis, and the graph of the function y = f ( x ) = 1- x . (See Figure 1.2.) x y y=f(x)=1-x 1 1 Figure 1.2: Figure for problem 1.2 (a) By using your knowledge about the area of a triangular region. (b) By setting up the problem as a sum of the areas of N rectangular strips, using the appropriate summation formula, and letting the number of strips ( N ) get larger and larger to arrive at the result. Show that your answer in (b) is then identical to the answer in (a). Detailed Solution: (a) The function is a straight line of slope -1 and intersects the x and y axes at the points (1 , 0) , (0 , 1). The triangular region thus has height 1 and base 1 so its area is 1 / 2. (b) Split the interval [0 , 1] in to N equal pieces (each has width 1 /N ). This generates points along the x axis whose coordinates are , 1 N , 2 N , .. ( N- 1) N (The k th coordinate is x k = k/N .) Here we use the left corner of each interval so created to define our rectangles. The height of a given rectangle is then of the form f ( x k ) = 1- x k = 1- k N v.2005.1 - January 3, 2006 2 Math 103 Problems Chapter 1 The area of this rectangle is A k = f ( x k )Δ x = parenleftbigg 1- k N parenrightbigg 1 N . We sum these to get A total = N- 1 summationdisplay k =0 (1- k N ) 1 N = 1 N N- 1 summationdisplay k =0 1- 1 N 2 N- 1 summationdisplay k =0 k. We compute this with summation formulae, using some care: The first sum consists of a total of N “ones” (starting at k = 0 and ending at k = N- 1). The second consists of the sum of the first N- 1 (not N ) integers - we have to modify the formula to take this into account....
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This note was uploaded on 09/25/2008 for the course MATH 103 taught by Professor Israel during the Spring '07 term at UBC.

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chapter2ProblemsAndSolutions - Math 103 Problems Chapter 1...

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