Mosaphir Drisana M23bQ2-2014DrisanaMosaphir

Name: Drisana Mosaphir

Section instructor: Tyler Section time: Friday 1-2

MATHEMATICS 23b/E-23b, SPRING 2014Practice Quiz # 2

March 14, 2014

This quiz includes questions from the 2012 quiz.On multiple-choice items 1 and 2, a correct answer on the first try was worth

2 points. If your first answer was wrong, a correct answer on the second trywas worth 1 point. During the real quiz, please transcribe your multiple-choiceanswers into the 1st Try column, and bring your paper to the front of the roomfor grading, preferably before 7:15PM but surely before 8:30 PM.

If you write solutions, upload your pdf file to the dropbox on the PracticeQuiz 2 Web page before 10 PM on Monday, March 24. If, after looking at othersolutions the next morning, you realize that you have made an error, uploadcorrected solutions, under a different file name, by 7 PM on Tuesday, March 25.

Suggested file name formats:

M23bQ2-2014PaulBamberg.pdf

M23bQ2-2014PaulBambergRevised.pdf

Use your own name, of course! Do not use unusual characters like quotes inyour file name. The Harvard server is fussier than Windows. Underscores andhyphens are probably OK.

You can include a .jpg or .png file in your .pdf file. See outline 12 for anexample.

Problem 1st Try 2nd Try Points Score1 E D 22 C E 23 xxx xxx 34 xxx xxx 35 xxx xxx 36 xxx xxx 37 xxx xxx 4

Total 20

1

3. (3 points) Find the volume of the parallelepiped in R4 that is spanned bythe vectors

~v1 =

1111

, ~v2 =

1111

, and ~v3 =

1100

.Answer: In order to find the volume in R4, we take the parallelepipedspanned by our given three vectors and turn it into a parallelepiped spannedby the three aforementioned vectors and an orthonormal unit vector ~v4(found using the Gram-Schmidt process).

Creating an orthonormal basis ~1, ~2, ~3:

~1 can be created simply by turning ~v1 into a unit vector; since the original

has length 2, ~1 =

12121212

.To figure out ~2, we will first figure out an orthogonal vector ~y2, then scaleit to be the unit vector ~2.

~y2 = ~v2 (( ~v2 ~1) ~1) =

1111

(since the dot product = 0). Scaled to a

unit vector, ~2 =

12121212

.

Similarly, ~y3 = ~v3 (( ~v3 ~1) ~1 + ( ~v3 ~2) ~2)) =

12121212

= ~3 (since itsalready a unit vector).

Now, we need a unit vector ~v4 such that its dot product with any of ourgiven vectors is 0. Letting ~v4 be made up of terms w, x, y, and z, we endup solving the systems of equations

w + x+ y + z = 0

,w x y + z = 0

,w + x y z = 0.

3

Doing so, we find that y = w and x = z = w. Setting w = 12

(as doing so

will produce a unit vector), we get ~v4 =

12121212

.

To find the volume, we take | det(T4)| = | det

1 1 1 1

2

1 1 1 12

1 1 0 12

1 1 0 12

| = |4| = 4.

4

4. (3 points)

A point is chosen at random in the northern hemisphere of the unit ball,without privileging any part of the ball. Using spherical coordinates, calcu-late the expected value of its z-coordinate. If time permits, use cylindricalcoordinates to check your answer. You may use the fact that the volume ofthe half-ball is 2

3pi.

Expected point = 10 dr

2pi0 d

pi20 rsin r

2cos d2pi3

=pi42pi3

= 38

Check (cylindrical coordinates): 10 r dr

r0 z dz

2pi0 d

2pi3

= 38

5

5. (3 points) Invent a parametrization for the curved surface of the cone de-fined by z2 = 4(x2 + y2), 0 z 1, using parameters r and . Set upand evaluate a double integral over r and to determine the area of thissurface. (Since this surface will lie flat if unrolled, it is possible to con-firm your answer by elementary geometry, but you must get the answer byintegration!)xyz

= (r

)=

r cosr sin2 r

T =

cos rsinsin rcos2 0

det(T TT ) = 2r 12

02r dr

2pi0

d = pi2

6

6. (3 points - shortened version of section problem 19.4)

Now that you know that 0

eu2

du =

pi

2,

you can evaluate other impossible-looking definite integrals by making asubstitution to convert them to this one.

(a) Use this approach to show that for a > 0, 0

eaxxdx =

pi

a.

(b) Use an alternating series argument to show that 0

sinxxdx

exists as an improper integral but not as a Lebesgue integral.

Answer:

(a) Set u2 = axu =ax =

ax

12

du =a12x12 dx =

a

2xdx

0eu

2

x

2xadu = 2

a

0eu

2= 2

api2

=

pi9

(b) Alternating series test:

i. sin(x) is bounded butx is unbounded so the function 0 as

xii. Between zeroes, the function alternates in sign

iii. The magnitudes of the areas of the regions decrease monotonically(with the regions being the areas in between points where thefunction has zeroes)

so it exists as an improper integral.

However, to pass the Lebesgue criteria, there must be convergence for

the sum of the unsigned (absolute) areas. Ik = (k+1)pikpi

|sin(x)|x

dx 1kpi

(k+1)pikpi

|sin(x)| dx = 2kpi

, a power series ib k with p < 1, whichdiverges.

7

7. (4 points) Suppose that fk(x) is an infinite sequence of functions from theinterval [0,1] to R. Suppose further that the sequence is uniformly bounded,in the sense that |fk(x)| < R for all x and all k, and the sequence isequicontinuous at every x, meaning that for any > 0, there exists > 0such that if |y x| < , then |fk(y) fk(x)| < for all k.Prove the Ascoli-Arzela theorem, which states that under these conditionsthere is a subsequence of fk(x) that converges to a continuous function f(x).You may use the fact that any convergent sequence is Cauchy and that anyCauchy sequence of real numbers is convergent. You need not prove thatthe convergence of the subsequence is uniform.

Given: fk(x) is an infinite sequence of functions all defined on [0, 1], all ofwhich are real-valued. The sequence is uniformly bounded (kx, |fk(x)|

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