Solutions to Homework 1

Transcription

1 Solutions to Homework 1 1. Let f(x) = x 2, a = 1, b = 2, and let x = a = 1, x 1 = 1.1, x 2 = 1.2, x 3 = 1.4, x 4 = b = 2. Let P = (x,..., x 4 ), so that P is a partition of the interval [1, 2]. List the following: each of the subintervals, each of the values of x i, where x i = x i x i 1. Let t i be the midpoint of the subinterval [x i 1, x i ]. Compute the value of the following Riemann sum: f(t 1 ) x f(t 4 ) x 4 = 4 f(t i ) x i. i=1 Finally, compare the value of the Riemann sum to the value of f(x) dx. Explain why one value is larger than the other. 2 1 Solution: The subintervals are [1, 1.1], [1.1, 1.2], [1.2, 1.4], and [1.4, 2]. The lengths of these intervals are, respectively, x 1 =.1, x 2 =.1, x 3 =.2, and x 4 =.6. The midpoints of the subintervals are, respectively, t 1 = 1.5, t 2 = 1.15, t 3 = 1.3, and t 4 = 1.7. The value of the Riemann sum is therefore equal to the following: (1.5) 2 (.1) + (1.15) 2 (.1) + (1.3) 2 (.2) + (1.7) 2 (.6) = The value of 2 1 f(x) dx may be computed using the fundamental theorem of calculus. The function F (x) = 1 3 x3 is an anti-derivative of f(x) = x 2. So, the value of the integral is equal to 1 3 (2)3 1 3 (1)3 = 7 3, which, in decimal form, is equal to 2.3. The value of the integral can be seen to be larger for geometric reasons. Consider the figure on the next page. The thick purple curve is the curve y = x 2, i.e. the graph of f(x) = x 2. The thin purple line is the line tangent to y = x 2 at the point (1.7, (1.7) 2 ). The blue lines and the x-axis enclose an area equal to the value of f(t 4 ) x 4 = (1.7) 2 (.6). The area enclosed by this rectangle is also equal to the area of the light purple shaded region. (To see this, use the fact that the two triangles formed by the tangent line, and the lines which contain the blue line segments are congruent triangles.) And the difference between this area and the area under the curve y = x 2 is equal equal to the area of dark purple sectors. That there is an excess, namely, the sum of the areas of the two dark purple 1

2 sectors, shows that the value of the integral is larger than the value of the Riemann sum (since the same argument applies not just to the term f(t 4 ) x 4 but to each of the terms that appears in the Riemann sum). A cool fact that I learned this past week is that a midpoint Riemann sum of a non-negative concave up function is always less than or equal to the value of the integral. So, for example, the same would be true if we had used the function f(x) = x 4. This cool fact is indeed correct because of the reasons explained above: if f(x) is concave up then the curve lies above any tangent line to the curve. 2. Suppose that f(x) is a continuous function on [, ) and that f(x). Suppose that the area of the region which lies below the graph of y = f(x), above the x-axis, and between the lines x = and x = a, where a, is always equal to a 2. Can you determine the values of f(x) for each x [, )? Explain. 2

3 Solution: The hard part is to translate the above into an equation. The area of the region which lies below the graph of y = f(x), above the x-axis, and between the lines x = and x = a, where a is equal to the integral of f(x) over the interval [, a]. (This would not be true if the graph of f(x) dipped below the x-axis or if a < so these two hypotheses are needed.) We are given that this are is equal to a 2. Therefore, a f(x) dx = a 2 for every a. The question is asking if the function f(x) can be determined from this given information. The answer is yes. To see why and to determine f(x), we take the derivative of both sides of the above equation with respect to the variable a: d da a f(x) dx = d da a2. Since f(x) is continuous on [, a], the Fundamental Theorem of Calculus applies. There are two parts in the Fundamental Theorem of Calculus. The part which is needed here is the following (see Theorem 5.11 in the textbook on p. 325 in Section 5.4): Theorem: If f is continuous on an open interval I containng the value a, then d x f(t) dt = f(x). dx Applying the theorem, we deduce that d da a a f(x) dx = f(a) ( NOT f(x) ). On the other hand, d da a2 = 2a. So, f(a) = 2a. If we use a different letter to denote the independent variable, such as the letter x, then we can say that f(x) = 2x. Thus, f(x) is uniquely determined by the given conditions: it must be the function f(x) = 2x. 3. (You are permitted to use a calculator to perform the arithmetic required in this problem. Use exact values; do not round off.) 3

4 The velocity v(t) as a funciton of time t of a particle moving along the real line is partially described by the data in the table below. Using this data, construct a table which gives a partial description (estimate) of the displacement s(t) of the particle. Construct a second table which gives a similar discription of the acceleration a(t) of the particle. Finally display the data of all three tables in a graphical format of your choosing (e.g. bar chart). t v(t) Solution: There are several reasonable answers to this question. First, let s discuss one reasonable solution, and then we ll discuss some other choices that are reasonable. By definition, the displacement is the integral of the velocity. More precisely, if the displacement is written as a function of time t as s(t), then s(t) = t v(x) dx. Since we are not given complete information about v(t), we will have to estimate the value of this integral. Since we are given the precise values of v(t) when t {.,.1,.3,.6, 1., 1.5, 1.9, 2.2, 2.4, 2.5}, we use this data to construct a Riemann sum which estimates the value of the integral above. Let t =., t 1 =.1, t 2 =.3, t 3 =.6, and so forth so that t 9 = 2.5. Using the left endpoint to determine the heights (in quotes because this is not a height, but rather a function value the function can be negative) of the rectangles in the Riemann sum, we have that s(.1) v(t ) t 1 = (1.)(.1.) =.1 s(.3) v(t ) t 1 +v(t 1 ) (t 2 ) = (1.)(.1.)+(2.)(.3.1) =.5 Continuing in this fashion, we can construct the table of values below. We need to make one choice: what is s()? This is analagous to determining the constant of integration. I ll choose s() =, which means that I m assuming that the particle is initially located at x = on the real line. t s(t)

5 Another reasonable answer is to use the right endpoints. For example, s(.1) v(t 1 ) t 1 = (2.)(.1.) =.2 This yields the following table of values: t s(t) Note that we cannot determine s(2.5) using the right endpoints since we do not know the velocity at any times greater than 2.5. Yet another reasonable answer is to average the velocities at the left and right endpoints. For example, s(.1) [(v(t 1 ) + v(t ))/2] t 1 = [( )/2](.1.) =.15 This method yields the following values: t s(t) To estimate the acceleration, consider the averate rates of change in the velocity: a(1.) v(t 1) v(t ) = t 1.1. = 1. The following table of values is obtained: t a(t) Finally, the above data (the midpoint data is omitted for brevity) can be displayed graphically (see Figures 1 and 2 at the end of this document). 4. A surface having the shape of a doughnut is called a torus. More precisely, suppose that C is a circle lying in a plane P which itself lies in three dimensional space. Let L be a line in P which does not intersect C. Let S be the surface of revolution generated by revolving C about the line L. Then S is called a torus. Suppose that a circle C and a line L are given as above. Suppose that the circle C has radius r and the distance from the line L to the center of the circle C is equal to d. Determine the volume enclosed by the torus S generated by revolving C about L. 5

6 Solution: First choose suitable coordinates. Let L be the y-axis and let (d, ) be the center of C. It is clear that the distance from the center of C to L is d. The points of C are the locus of the equation (x d) 2 + y 2 = r. The surface S is the surface of revolution obtained by rotating C about the y-axis. We can use the washer (or shell) method to compute the volume. Since the volume enclosed by S is equal to twice the volume enclosed by the surface obtained rotating the semi-circle determined by the x-axis and the arc of C which lies above the x-axis, we ll compute this volume since the integral will be somewhat less complicated to set up. To use the washer method, consider a cross section which is perpendicular to the axis. Suppose the points of the cross section have y-coordinate equal to y. Then the larger radius R + of the washer is determined by solving for x in the equation for C and using the positive square root: R + = d + r 2 y 2. The smaller radius R of the washer is determined in the same way except one uses the negative square root: R = d r 2 y 2. The area of the cross section A(y) is equal to π(r 2 + R 2 ). The cross sections in this half of the solid enclosed by the torus vary between y = and y = r. So, the volume of this solid of revolution (one half of the filled in torus) is equal to r π[(d + r 2 y 2 ) 2 (d r 2 y 2 ) 2 dy. Expanding the integrand and simplifying, one obtains 4πd r r 2 y 2 dy = 4πd( 1 4 πr2 ) = π 2 dr 2. Doubling this value, we conclude that the volume of the region enclosed by the torus is equal to 2π 2 dr 2. If we used the shell method, then the volume of half of the solid enclosed by the torus is equal to d+r d r 2πx r 2 (x d) 2 dx 6

7 since the height of the shell containing points with x-coordinate equal to x is equal to r 2 (x d) 2. But this integral is much more difficult to compute. For instance, the substitution u = r 2 (x d) 2 seems reasonable, but then du = 2(x d) dx. I ll not go through the details here, but if you would like to compute this integral here is a strategy: first using the substitution u = x d, next separate the result into two integrals, and finally compute one integral as above (geometrically) and observe that the second integral is zero (since the integrand is odd and the limits of integration are from r to r). There is a third way to compute the volume enclosed by the torus. A theorem of Pappus states that the volume enclosed by a solid of revolution is equal to 2πdA, where A is the area of the region being revolved about an axis and d is the distance from the axis to the centroid of the region being revolved about the axis. In this problem, the centroid of the region enclosed by C is clearly the center of the circle and so the distance from the center at (d, ) to the axis (the y-axis) is equal to d. The area of the region enclosed by C is πr 2. Pappus s theorem implies that the volume of the solid of revolution is equal to 2πd(πr 2 ) = 2π 2 dr Estimate the volume of a large (as defined by the U.S. Department of Agriculture) chicken egg. Your solution must include a description of your methods and your rationale for making simplifying assumptions. Solution: There are many ways to approach this problem. The simplest would be to obtain an egg and to use a pencil to sketch an outline of the egg. If this sketch is transfered to graph paper then it is easy to determine the areas of the circular cross sections obtained by slicing the egg perpendicular to the longest axis of symmetry of the egg. Another method is to take a hard boiled egg and to slice the egg into disks. The cross sectional areas and the thickness of each slice can be measured and then the volumes can be summed. Still another approach might be to crack an egg into a graduated cylinder and read off the answer. A cleaner approach would be to fill a graduated cylinder with water and measure the displacement when an egg is placed in the water. 7

8 Figure 1: The blue curve displays the velocity as a function of time (assuming piecewise linear behavior beteween data points). The red and green curves show the displacement as a function of time using the left and right endpoints, respectively. Figure 2: The blue curve displays the velocity as a function of time (assuming piecewise linear behavior beteween data points). The black curve shows the acceleration as a function of time. 8