Ghai Aditi M23bQ22014AditiGhai

8/12/2019 Ghai Aditi M23bQ22014AditiGhai

1/10

Name: Aditi Ghai

Section instructor: Tyler Morrison Section time: Friday, 1:00 PM

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

March 14, 2014

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

Total 20

1


2/10

To speed up grading, please transcribe your answers to questions 1 and 2 intothe 1st Try column on the front page.

1. Consider the change of variables

uv = uv

u uv. Then R 2 f

xy |dxdy | is equal to

(a) R 2 f uv |dudv |

(b) R 2 (f )uv |dudv |

(c)

R

2 f uv

u uv |dudv

|(d) R 2 uf

uvu uv |

dudv |(e) R 2 uf

uvu uv |

dudv |Solution: (e)

Note, we are interested in the quantity det D uv and not det D uv .

2. The area enclosed by the curve specied in polar coordinates byr 2 = sin , 0 is(a) 12(b) 4(c) 1

(d) 2(e) 2

Solution: (c)The iterated integral to be evaluated is

0

sin 0 rdrd .

2


3/10

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

v1 =1111

, v2 =1111

, and v3 =1100

.

Solution:

The volume of the parallelepiped in R4 is given by detT T T , where T isthe matrix formed by the column vectors v1, v2, v3.

T T T =4 0 20 4 02 0 2

.

The operation of adding a multiple of one column to another column doesnot change the determinant. To make the calculation of the determinanteasier, add 1 times column 1 to column 3.

T T T =4 0 20 4 02 0 0

.Developing on the third column, we nd

detT T T = 2det0 42 0 = 2(0 8) = 16.

So the volume of the parallelepiped is detT T T = 16 = 4.

3


4/10

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 of the half-ball is 23 .

Solution:

Note, the expected value of the of the z -coordinate is equivalent to the of z -coordinate for the center of mass. Let us rst calculate the numeratorfor this z -coordinate in spherical coordinates. Then, we will divide by thevolume of the half-ball, 23 .

Numerator:

2

0

2

0 1

0r sin r 2 cos drdd

= 2

2

0sin cos d

1

0r 3dr

We can evaluate

2

0 sin cos d using u-substitution.u = sin ,du = cos d

= 2 r 4

4

1

0

1

0udu

= 2 14

u2

2

1

0

du

= 2

12

= 4

Now, let divide by the volume of the half-ball, 23 .4

23

= 38

The expected value of the z -coordinate is 38 .

4


5/10

5. (3 points) Invent a parametrization for the curved surface of the cone de-ned by z 2 = 4( x2 + y2), 0

z

1, using parameters r and . Set up

and evaluate a double integral over r and to determine the area of thissurface. (Since this surface will lie at if unrolled, it is possible to con-rm your answer by elementary geometry, but you must get the answer byintegration!)

: r r cos r sin

2r =

xyz

D =cos r sin sin r cos

2 0

det D T

D = det 5 00 r 2 = 5r2

det D T D = 5r 2 = 5r

2

0 1

2

0

5rdrd

= 2 5 r2

2

1

2

0

= 5

4

5


6/10

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

Now that you know that

0 eu 2 du = 2 ,you can evaluate other impossible-looking denite integrals by making asubstitution to convert them to this one.

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

0

eax

x dx =

a

.

(b) Use an alternating series argument to show that

0 sin x x dxexists as an improper integral but not as a Lebesgue integral.

Solution:a.

0

eax

x dx =

aUsing u-substitution, with u = ax and du = a2 x , we may write

2 a 0 eu 2 du = 2 2 a = a

.b.

0 sin x x dxThe graph of the function

sin x

x is shown below. Note, that the areasbounded by the curve and the x-axis, denoted I k , decrease as x grows.That is, I 1 > I 2 > I 3 > > I k and as k , I k 0. Furthermore,A = k =1 = I 1 = I 2 + I 3 I 4 + ...As such, we have satised the three conditions of the alternating seriestest and have demonstrated that the area under the curve converges to

6


7/10

a nite A < . That the sequence of partial sums converges justiesthe existence of

0sin x x dx as an improper integral:

lima

a

0

sin x x dx

Graph of sin x x

However, 0sin x x dx does not exist as a Lebesgue integral. To see why,

consider

f k (x) = sin x x for (k 1) < x < k

I k = k

(k1)sin x

x dx 1

k k

(k1) |sin x|dx 2

k .I k is at least as large as the constructed lower bound 2 k , term byterm. Because the innite sum k =1

2 k is divergent, the innite

sum k =1 I k k =12

k must also be divergent.

Therefore, sin x x dx fails the Lebesgue-integrability test and the Lebesgueintegral does not exist.

7


8/10

7. (4 points) Suppose that f k (x) is an innite sequence of functions from theinterval [0,1] to R . Suppose further that the sequence is uniformly bounded,

in the sense that |f k (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 |f k (y) f k (x)| < for all k.Prove the Ascoli-Arzela theorem, which states that under these conditionsthere is a subsequence of f k (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:

f k (x) is an innite sequence of functions from the interval [0 , 1] to R . x,k|f k (x)| < R . > 0, > 0 s.t. |y x| < |f k (y) f k (x)| < ,k.

PROVE:

a subsequence of f k (x) that converges to a function f (x). The function f (x) is continuous.

PROOFFirst, we will prove that we can extract a subsequence of f k (x) that con-verges to f (x),x Q [0,1]. Next, we will prove that we can simialrly wecan extract a subsequence of f k (x) that converges to f (x),x Q [0,1]. Wewill conclude by showing that the f (x) to which the f k (x) converge mustbe continuous if the sequence f k (x) is equicontinuous.

1. Extract a convergent subsequence of f k (x),x Q [0 ,1 ]The Q [0, 1] are countably innite (Proof 2.1). We may enumerate themas follows: x1 = 0 , x 2 = 1 , x 3 = 12 , x 4 = 13 , x 5 = 23 , x 6 = 14 , x 7 = 34 ,...

Now, consider f 1(x1), f 2(x1), f 3(x1)... .

Since f k (x) is uniformly bounded , all f n (x1) are in the compact interval[R, R ]. So by the Bolzano-Weierstrass Theorem, we can extract a conver-gent subsequence of f i (x1) which converges to some limit f (x1) [R, R ].

We will denote this as f 1,1(x1), f 1,2(x1), f 1,3(x1)... f (x1).

8


9/10


10/10

|f n,n (x) f m,m (x)| < 3

+ 3

+ 3

= .

Therefore the sequence {f n,n (x)} is a Cauchy sequence and converges tosome value f (x) for any irrational x.

The sequence {f n,n (x)} coverges to f (x) for all rational values in the domain[0, 1] and all irrational values in the domain [0 , 1]. The sequence {f n,n (x)}coverges coverges to f (x) for all values in the domain [0, 1].

3. The function f (x) is continuous.

If the sequence f k (x) is equicontinuous, then the f (x) to which the f k (x)converge must be continuous. By equicontinuity of f k (x) , > 0, suchthat if |y x| < , then |f n,n (y) f n,n (x)| < 3 ,n .

Furthermore, since our sequence converges both at x and at y, we can chooseN so large that for all n > N , |f n,n (y) f (y)| < 3 and |f n,n (x) f (x)| < 3.By the triangle inequality,

|f (y) f (x)| |f (y) f n,n (y)| + |f n,n (y) f n,n (x)| + |f n,n (x) f (x)|.

|f (y) f (x)| < 3

+ 3

+ 3

It follows that for our chosen , |f (y) f (x)| < . Thus the denition of continuity is always satised with respect to f (x) and f (x) is necessarilycontinuous.

10

Documents

Ghai Aditi M23bQ22014AditiGhai