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Unformatted text preview: MATH 31B SECTION 2 FIRST 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. #1 #2 #3 #4 #5 Total SID: Name: TA: Section(circle): Tuesday Thursday 1 MATH 31B SECTION 2 FIRST MIDTERM 2 Problem 1. Let S be the sphere of radius 1. Set up and then evaluate the integral expressing the volume of S using (a) the method of crosssections and (b) the method of cylindrical shells. Solution. (a) Consider the plane parallel to the y z coordinate plane and intersecting the xaxis at the point x. The intersection of this plane with the interior of the sphere is a disk of radius 1
£ (b) The shpere can be obtained by rotating the halfcircle y 1 x2 about the yaxis. If we ﬁx a value of x and rotate the resulting line, we get a cylinder of radius x and height 2 1 x2 . If we let x vary between 0 and 1, the resulting cylinders combine to ﬁll up the whole sphere. Thus the volume is
1 0 0 1 0 0 © © ¢© ¢ ¢ £ ¢ V 2π udu 2π ¤ § ¤ £ ¢ § £ ¢ Substituting u 1 x2 , du 2xdx gives us
1 udu £ 2πx 2 1
x2 dx 43 πu 3 2 1 4 π 3 £ ¤ £ ¨© © ¤ ¢ § ¨ 1 1 1 ¢ © ¦ £ ¥ ¢ ¦ £ ¥ ¢ 1 A x dx
¦¥ 1 π1 x2 dx ¦ £ x2 . Hence its area A x is π 1
¥ § ¦¥ ¨ § x2 and the volume of the sphere is πx 13 x 3
1 £ ¢ ¡ y2
¤ z2 1 x2 4 π 3 MATH 31B SECTION 2 FIRST MIDTERM ex dx. 3 Solution. See practice midterm. Problem 2. Evaluate the indeﬁnite integral ex
MATH 31B SECTION 2 FIRST MIDTERM 4 Problem 3. Consider the deﬁnite integral
1 For which values of n can you guarantee that the Midpoint Rule would approximate the integral to 1 within 3 10 4 ? Justify your answer. Solution. Denote the error by E . We know that Kb a3 K3 13 K E 2 2 24n 24n 3n2 where K is the maximum value of the absolute value of the second derivative of f x x on 1 12 1 32 ,f x . We see that f x is monotone increasing 1 3 . We compute f x x x 2 4 on 1 3 and therefore its maximum is attained at x 3 and its minimum is attained at x 1.Since 1 1 and f 1 f3 4 , we see that
4 27 1 1 4 3n2 12n2 We thus want to see for which n the inequality 1 1 10 4 2 12n 3 is satisﬁed. Rewriting, we have 1 1 12 10 4 n2 4 10 4 n2 3 104 n2 4 104 100 n 50 4 2 Thus n 25.
¨ ¨ ¢ ¢ ¢ E ¨ ¢ ! 4 3 3 ¢ Hence we can take K 1 4 2# 1 ¦ ¥ '$ $ fx " 1 x 4 and our estimate becomes
0 13 ¤ ¢¦¥ ¢ ¦ %'$ ¥$ ¢ ¦ £ ¢ ¨ ¥ ¢ £ ¦ ¢ I
¨ ¤ 3 § xdx ¥$ ¢ ¦ %&$ £ ¥ ! ¨ $ ¢ ¦ ¥ )$ ¥ ¢ ¦ %$ ( $ ¢ ¦ ¥ &$ # 3 #" " MATH 31B SECTION 2 FIRST MIDTERM 5 Problem 4. Let f x x 1 cos x. (a) Show that the function f is monotone and conclude that it 2 is 11. (b) Find the value of the derivative f 1 π 1 . 2
¢¦ £ ¥
1 π
1 2 £ 1 1 2 sin π ¢ ¢ ¦ ¥$ ¢ ¦ 6¦ £ ¥ ¨ ¥$ ¢¦ £ ¥$¦ f
¥ 1 π 1 2 1 f f 1π 5 3 ¨ £ £ ¢¦ ¥ ¢ ¦ ¥$ Solution. (a) Note that f x 11. (b) We note that f π π
¨ 1 1 sin x 2 1 , so that f 2 1 2 1 2 0. Thus f is strictly increasing and hence is π. Hence 1 fπ 1 1 ¦ £ ¥$¦ ¨ ¥ ¡ ¢¦¥ MATH 31B SECTION 2
2 FIRST MIDTERM 6 Problem 5. Sketch the graph of the function f x e x . Indicate the limits at inﬁnity, maxima, minima and inﬂection points. Please use the coordinate axes drawn below.
1.4 1.2 1 0.8 0.6 0.4 0.2 0 0.2 0.4 2 1 0 Solution. The graph is the “bell curve”. 2 We have that f x 2xe x , which is zero at x 0. So there is only one extreme point. Also, the function is increasing on ∞ 0 and is decreasing 0 ∞ , since the ﬁrst derivative is postitive on ∞ 0 and negative on 0 ∞ . 2 2 2 Since f x 2e x 2x 2xe x 2 4x2 e x we see that f x 0 if 4x2 2, i.e., x 2. Thus the inﬂection points are at x 2 and x 2. Since f 0 0, it follows that the only extreme point is a maximum. On the interval ∞ 2,f x 0, so the function is concave up. On the interval 2 2, fx 0 and so the function is concave down. Finally, on the interval 2 ∞ the second derivative is again positive, and the function is concave up. 2 Lastly, as x ∞, x2 ∞ and so e x 0. Thus
¨C DC
x ∞ x ∞ ¢¦¥ ¢¦¥ lim f x lim f x 0 ¦ ¤ ¢ ¤ £ 8¥ ¦ $ ¢ ¦ ¥ &$ ¡ ¤ ¥ ¦ ¡ ¥ ¤ £ ¢ ¨ ¢ ¨ ¦ ¢¦¥ ¡ exp(x*x) 1 2 ¤ B ¢ £ 7¥ ¢ ¦ ¨ ¦ ¡ ¥ ¦ £ 7¥ ¨ 5 ¦ %)$ ¥$ £ 8¥ ¨ £ ¦ B £ £ ¤ £ ¢ ¦ ¥$ £ ¨ $ £ ¢ ¦ ¥ )$ ¦ £ 7¥ £ 8¥ B $ 9 @¦ ¥ )$ $ 9 A¦ ¥ '$ ¤ ¢ ...
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 Fall '09
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