This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 crosssections 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 xaxis 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:/ 27tr27tr2drz (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 crosssections. 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 onetoone 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 ...
View
Full
Document
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.
 Fall '09
 HOUDAYER

Click to edit the document details