Unformatted text preview: Midterm IIIb Solutions MAT 21B Name: Student Id: Section: No notes. No calculators. All incidents of cheating or the appearance thereof will be brought to the attention of Student Judicial Affairs. 1) 2) 3) 4) 5) 6) Total (out of 50): 1 1) (True/False) Label the following statements as either true or false. (No explanation required. Each correct label is worth two points.) a: ln6 = ln2 + ln3 True. ln 2 + ln 3 = ln(2 3) = ln 6. b: e6 = (e3 )(e2) False. (e3)(e2) = (e e e)(e e) = e5 = e6 . c: lnxdx = False. We check that d: sin(a2 ) + cos(a2 ) = 1 False. Consider a = cos = 2 = 1. 4 e: f (x)g(x)dx = ( f (x)dx)( g(x)dx) False. Consider f (x) = g(x) = 1. Then, f (x)g(x) dx = dx = x+C, whereas, ( f (x) dx)( g(x) dx) = ( dx)( dx) = (x + C1)(x + C2) = x2 + (C1 + C2)x + C1 C2.
2 2 . d dx 1 x 1 +C x 1 + C =  x2 = ln x. Then sin(a2 ) + cos(a2 ) = sin + 4 2) (8 points) Find 3 dx, x x x2 1 >1 We use Trigonometric Substitution. Let x = sec , where 0 < . Then dx = sec tan d. So, 2 3 dx = x x2  1 3 (sec tan d) sec sec2  1 We use the identity sec2  1 = tan2 , and simplify, to get 3 dx = x x2  1 3 sec tan d = 3 sec tan2 tan d  tan  Now, if x = sec , we imagine a reference right triangle with hypotenuse length x, one angle , the leg adjacent to length 1. Then the third leg has length x2  1. We can then determine, tan = x2  1. Notice that for x > 1, this tells us that tan > 0. So we have, 3 tan d = 3  tan  d = 3 + C. Substituting x back in, we have, 3 dx = 3 sec1 x + C 21 x x 3 3) (8 points) Find x3ex dx We use integration by parts three times, because the power of x is 3. Let u = x3, dv = ex dx. Then du = 3x2 dx, and v = ex . So, x3ex dx = x3ex  3x2ex dx. Now, let u = 3x2, d~ = ex dx. Then d~ = 6x dx, v = ex . ~ v u ~ And, 3x2ex dx = 3x2ex  6xex dx. So, x3 ex dx = x3ex  3x2ex  6xex dx . Finally, let u = 6x, and d^ = ex dx. Then d^ = 6, and ^ v u v = ex dx. Thus, ^ 6xex dx = 6xex  And we find, x3ex dx = x3ex  3x2ex  (6xex  6ex + C) = x3 ex3x2ex +6xex6ex+C = ex x3  3x2 + 6x + 6 + C. 6ex dx = 6xex  6ex + C. 4 4) (8 Points) Find dx x2 +2x dx We use the method of Partial Fractions. Suppose there are A and B such that, A B 1 = + . x2 + 2x x x+2 Then finding a common denominator for the right hand side, we get A(x + 2) + Bx 1 = . x2 + 2x x2 + 2x So, 1 = (A + B)x + 2A. Equating the coefficients of powers of x gives us the two equations A + B = 0 and 2A = 1. In other words, A = 1 , 2 and B =  1 . So, 2 dx dx = x2 + 2x =
1 2 1 + dx = x x+2 2 1 2 1 1 dx  x 2 1 dx x+2 1 1 ln x  ln x + 2 + C 2 2 5 5) (8 points) Estimate the minimum number of subintervals needed to approximate the integral
2 0 sinxdx with an error of magnitude less than 104 when using the Trapezoidal Rule and when using Simpson's Rule. The error when using the Trapezoidal Rule is bounded by M(b  a)3 ET  , 12n2 where M is an upper bound for the second derivative. So, d2 d (sin x) = (cos x) =   sin x 1. 2 dx dx So we can take M = 1. And our interval is [a, b] = [0, 2], so the interval length is b  a = 2. Then, we want to find how large we must take n so that 23 < 104 2 12n 12 104 3 104 1 < = n2 8 2 2 n2 > 3 104 2 n> 3 104
2 So we need approximately a minimum of 3104 subintervals in order to approximate the integral to within 104 using the Trapezoidal Rule. 6 The error when using Simpson's Rule is bounded by M(b  a)5 ES  , 180n4 where M is an upper bound for the fourth derivative. So, d4 d3 d2 d (sin x) = (cos x) = ( sin x) = ( cos x) = sin x 1 dx4 dx3 dx2 dx So again we can take M = 1. Our interval is still length 2. And now want to find how large we must take n so that 25 < 104 4 180n 1 180 104 45 104 < = n4 32 8 8 n4 > 45 104 n>
4 8 45 104 8 So we need approximately a minimum of 4 45104 subintervals in order to approximate the integral to within 104 using Simpson's Rule. 7 6) (8 points) A truncated conical container is 7 inch deep, 2.5 inches across at the base, and 3 inches across at the top. It is full of chocolate milkshake weighing 4/9 oz/in3. A straw sticks up one inch above the top. Set up an integral that calculates how much work is done drinking the entire milkshake through the straw (neglecting friction). We want to consider horizontal crosssections of the container of infinitesimal width. To do so, we must figure out the radius of such a crosssection in terms of height. Referring to the figure, we set up a coordinate system with y = 0 at the tip of the cone. To find the distance from the tip of the cone to the bottom of our container, we solve for h by using Proportional Triangles. 7+h h = 1.5 1.25 7+h
3 2 = h
5 4 2(7 + h) 4h = 3 5 10(7 + h) = 12h h = 35. Then, we have the radius, r at height y satisfies r 1.25 = y 35
5 r = 4 y 35 8 r 5 1 = = y 140 28 1 r= y 28 So the crosssectonal slab at height y has volume Vy = 1 ( 28 y)2 dy = 784 y 2 dy in3.
4 Now, the weight of this slab is Wy = ( 784 y 2 dy in3)( 9 oz/in3) = 2 1764 y dy oz. And the distance the slab is moved is (h + 7 + 1  y) = (43  y), so the work done lifting this slab to the top of the glass, and then through the straw is 2 y (43  y) dy 1764 And finally, the total work done is the integral over all such crosssectional slabs as y varies from the bottom of the truncated cone (at y = h = 35) to the top (at y = h + 7 = 42): W =
42 35 2 y (43  y) dy 1764 9 ...
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This note was uploaded on 04/16/2008 for the course MATH 21B taught by Professor Vershynin during the Spring '08 term at UC Davis.
 Spring '08
 Vershynin
 Math

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