midterm1-practice-solutions

midterm1-practice-solutions - MATH 31B SECTION 2 FIRST...

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Unformatted text preview: MATH 31B SECTION 2 FIRST PRACTICE MIDTERM Please note: Show your work. Correct answers not accompanied by sufficent explanations will receive little or no credit. Please call one of the proctors if you have any questions about a problem. No calculators, computers, PDAs, cell phones, or other devices will be permitted. If you have a question about the grading or believe that a problem has been graded incorrectly, you must bring it to the attention of the professor within 2 weeks of the exam. Problem 1. Let S be the cone obtained by rotating the line triangle bounded by the lines )2 = x, x = 1, y = 0 about the x—axis. Set up and then evaluate the integral expressing the volume of S using (a) the method of cross-sections and (b) the method of cylindrical shells. Solution. (a) Let us fix the value of x. The cross—section of the cone by the plane parallel to the _V., Z plane and intersecting the x-axis at x is a disk of radius x (see Figure 121). Therefore, the area of the cross—section We then find that the volume is 1 l ] V =/ A(x)dx :/ tuxde = —n. 0 o 3 (b) The cone can be seen to be filled with cylinders of radii r varying from 0 to 1; the height of the cylinder of radius r is 1 — r (see Figure 1b). We then get 1 '1 2 1 V 2/ 27tr(l — r)dr:/ 27tr-27tr2drz (1 — —)tt= —7t. 0 0 3 3 Problem 2. Let S be the region consisting of those points (x,y, z) that lie inside of the sphere of radius 1, and which satisfy 0 g x g 0.3. Find the volume of S. \-‘f‘ (/1) FIGURE 1. Cros—setions 0f the cone. 1 Ix) MATH 31B SECTION 2 FIRST PRACTICE MIDTERM 2 FIGURE . The sphere. amp bout 1. uni Mk1 0 OQS 0.5 0.16‘ I FIGURE 3. The function. Solution. We’ll use the method of cross-sections. Consider the cross—section of S by a plane parallel to the y,z coordinate place and intersecting the x axis at x. The cross—section is a circle of radius \/1 — x2 (see Figure 2). The area A(x) of this cross—section is 1t(\/l —x2)2 : 7t(l — x2), Since 0 g x g 0.3, the volume is given by 0.3 1 v =/ 1t(l—x2)dx = 0.37: — g(03)va: = 7c(0.3 + 0.009) : 30091:. 0 Problem 3. Give an example of a function f (x) for which the trapezoidal rule approximation with n = 2 to the integral f0] f (x)dx is exact (i.e., the error is zero), while the trapezoidal rule approximation to the same integral with n : 4 gives a nonzero error. (No formula is necessary; a graph of the function, with explanations, will suffice). Solution. The trapezoid rule with n = 2 involves dividing the interval into 2 subintervals and approximating the function by straight lines on these two subintervals. Thus this approximation only cares about f(0), [(0.5) and f(l) (i.e., the values off at the endpoints ofthese two intervals). The trapezoid rule with n : 4 involves doing the same, but over 4 intervals. So the approximation now also looks at f(0.25) and f(0.75). Now the idea is to make a function twhich is zero, except for two “bumps”. The bumps are chosen so that (1) two areas of the bumps cancel each other, so that the total intergral of f is zero; (2) f is zero at the ends and the midpoint of the interval; (3) f is also zero at 0.75, but f is strictly positive at 0.25. See Figure 3. Because of (I), the total intergral off is zero. MATH 31B SECTION 2 FIRST PRACTICE MIDTERM 3 Because of (2), the values of f at the endpoints of the intervals involved in the midpoint rule for n : 2 are zero, and so the approximation gives zero for the value of the interval. Because of (3), the trapezoids involved in the approximation of the graph of f have area zero on the intervals [05,075] and [075,1], but have a strictly positive area on both [0.0.25] (where the top of the trapezoid connects O : f(0) and f(0.25) > 0) and [025,05] (where the top connects f(0.25) > 0 and f(0.5) : 0]. So the trapezoid rule would give a strictly positive answer instead of zero. Problem 4. Compute the integral f e"+"‘dx. Solution. Using the formula 2“” = eaebwe get I . X \' /e"+€ dx = /exee dx. Substituting Lt : 6"“, du : e‘dx gives us /exeeidx = / e”st = e” +C I e"r +C. Problem 5. Let f(x) = \/ x2 +41, x 2 0. (a) Show that f is one-to-one on [0, +00); (b) compute the inverse of f(x); (c) find the derivative of the inverse of f (x) at the point Solution. (a) The function x2 is strictly increasing on [0, +00); hence so are the functions x2 + l and \/.x2 + 1. It follows that f is 1—]. (b) We have y = x/x2+l y2 : x2 +1 x2 = y2 — 1 x : i y2 e 1. Now, since x 2 0, we know that we must take the plus sign, and so x : y2 — 1. Thus f#1(x) : x2— 1. (c) Note thatf(l) : fl, so f’1(\/§) = 1. Since f’(x) = x , we have that \/x2+1 ‘ f"’(f"‘(\/i)) 1 (You can also compute the derivative of the inverse directly in this case). fl(\/§)__ 1 Problem 6. Graph the function f (x) = e" + 6“. Solution. The graph is presented below (Figure 4). We have that f ’ (x) : ex — e“, and so f(x) 2 0 ifx Z 0 and f(x) 3 0 ifx S 0. Hence f is decreasing on (—oo,0[ and increasing on [0, +00). Since f”(x) : e" + e" > 0 (since 9" > 0 for all x) we find that there are no inflection points and that f is concave u . Finall , lim,H 009‘ : 00, limx_,_we‘ = 0, we find that limxaim x = +00. P y + MATH 313 SECTION 2 FIRST PRACTICE MlDTERM —X FIGURE 4. The graph of ex + e ...
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This note was uploaded on 02/22/2010 for the course MATH Math 31B taught by Professor Houdayer during the Fall '09 term at UCLA.

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midterm1-practice-solutions - MATH 31B SECTION 2 FIRST...

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