KEY TO CALCULUS II PRACTICE EXAM


(1)
The differential equation is separable; consequently, the solution is found by integrating. Note that we are taking t to be the independent variable (this is purely arbitrary since the independent variable is not specified).
dy
dt
=   ____
Ö1-y2
 
Þ ó
õ
dy
  ____
Ö1-y2
 
= ó
õ
 dt Þ Arcsin(y) = t+C

Thus, we see that y(t) = sin(t+C) is the general solution. Since we know that y(p) = -1, we can find the constant C:

-1 = sin(p+C) Þ Arcsin(-1) = p+C Þ -p/2 - p = C

Thus, C = -3p/2.

(2)
Observe that
ó
õ
1

0 
x-1 dx =
lim
h®0+ 
ó
õ
1

h 
x-1 dx =
lim
h®0+ 
(ln(1) - ln(h)) = +¥

Thus, this integral diverges.

(3)
This integral is best evaluated using the double angle formula for the cosine:
ó
õ
cos2(x) dx = 1
2
ó
õ
(1+cos(2x)) dx =   x  
2
+ sin(2x)
4
+ C

It can also be done by parts, but we will leave that to the reader.

(4)
This definite integral must be done by parts. Letting u = x and dv = exdx, we have
ó
õ
1

0 
xex dx = xex|01 - ó
õ
1

0 
ex dx = xex|01 - ex|01 = e - (e-1) = 1

(5)
This differential equation is solved by integrating twice. Note that, by the Fundamental Theorem of Calculus,
y¢ = ó
õ
y¢¢ dy = ó
õ
cos(t) dt = sin(t) + B

Integrating a second time gives us

y = ó
õ
y¢ dy = ó
õ
(sin(t) + B) dt = -cos(t) + Bt + A

(6)
The Taylor series representation for this function can be easily obtained if we observe that
f(t) = 2
1+(3t)
= 2
1 - (-3t)
= 2
1-u
= +¥
å
j = 0 
2u^j

Where, of course, u = -3t. This representation is based on the geometric series and is therefore valid for -1 < u < 1. Remembering now that u = 3t, we see that, for -1/3 < t < 1/3, we have

f(t) = +¥
å
j = 0 
2·(-3t)^j  = +¥
å
j = 0 
2 (-1)^j (3t)^j

(7)
The method of washers can be used to find this volume. According to this method, the volume of the solid is
V = p ó
õ
2

1 
(x)2- (Öx)2 dx = = p(x3/3 - x2/2 )|12 = p(8/3-4/2) - p(1/3-1/2) = 5p/6

(8)
On the interval [0,1], the function f(x) = 1-x2 is decreasing. Consequently, on any subinterval [xj,xj+1], we have f(xj) > f(xj+1). Therefore, every left-hand sum estimate of this integral will overestimate the actual value.
(9)
The fact that the rate of change of Q is directly proportional to the difference between Q and Q0 is all we need to construct the differential equation:
Q¢ = k(Q-Q0)

Note that Q-Q0 = -(Q0-Q). Since the negative sign can be absorbed into the constant k, it is actually immaterial which order we use in the difference.

(10)
This series can be evaluated by breaking it up into two convergent geometric series. Observe
+¥
å
j = 0 
3 - 2j
42j
= (1/16) +¥
å
j = 0 
3-2j
4j
= (1/16) +¥
å
j = 0 
æ
ç
è
3
4j
- æ
ç
è
2
4
ö
÷
ø
j

 
ö
÷
ø
= (1/16) +¥
å
j = 0 
3
4j
- (1/16) +¥
å
j = 0 
æ
ç
è
1
2
ö
÷
ø
j

 
= (1/16) æ
ç
è
3
1-(1/4)
- 1
1-(1/2)
ö
÷
ø
= 1/8

(11)
The n degree Taylor polynomial for a function f at a point x = a is given by
Tn(x) = f(a) + n
å
j = 1 
f^(j) (a)· (x-a)^j
j!

where f^(n)(a) denotes the nth derivative of f evaluated at x = a. With this in mind, it is easy to construct the desired polynomial:

T2(x) = f(1) + f¢(1) (x-1)1
1!
+ f¢¢(1) (x-1)2
2!
= 2 + (1/2) (x-1)
1
- (1/3) (x-1)2
2

(12)
This indefinite integral is best solved using partial fraction decomposition. Observe that
5x-7
(x-1)(x-2)
= A
x-1
+ B
x-2

implies that

5x-7 = A(x-2) + B(x-1) = Ax - 2A + Bx - B = (A+B)x - (2A + B)

equating coefficients gives us 5 = A+B and 7 = 2A + B. This system of equations tells us that A = 2 and B = 3. Consequently,

ó
õ
5x-7
(x-1)(x-2)
 dx = 2 ó
õ
dx
x-1
+ 3 ó
õ
dx
x-2
= 2ln| x-1 |+ 3ln| x-2 |+ C

(13)
This indefinite integral is best solved using two integrations by parts:
ó
õ
x2cos(x) dx
= x2sin(x) - 2 ó
õ
xsin(x) dx
= x2sin(x) -2 æ
è
-xcos(x) + ó
õ
cos(x) dx ö
ø
= x2sin(x) +2xcos(x) -2sin(x) + C

(14)
The trick to constructing the desired integral is to imagine adding up the volumes of thin cross sections cut through the solid perpendicular to the x-axis. Partition the interval [0,p] into n subintervals, using 0 = x0, x1, x2, ..., xn-1, xn = p as the endpoints of the partition. If n is large (so that the width Dxj of the jth subinterval is small for all 0 £ j < n), then a vertical cross section of the solid taken perpendicular to the x-axis on each subinterval is approximately a half cylinder having volume
Vj » p Rj2
2
Dxj

where Rj is the radius of the semicircle at xj. This radius is just half the distance between the curve y = sin(x) and the x-axis; hence, Rj = sin(xj)/2. Thus, the volume of the solid may be approximated by the Riemann sum

V » n-1
å
j = 0 
Vj = n-1
å
j = 0 
 p   
 2
· æ
ç
è
sin(xj)
2
ö
÷
ø
2

 
·Dxj = (p/8) n-1
å
j = 0 
sin2(xj)Dxj

Taking the limit as Dxj goes to zero gives the desired integral, namely

V = (1/8) ó
õ
p

0 
sin2(x) dx

(15)
The desired integral is obtained simply by using the formula for arc-length. If f is a differentiable function on an interval [a,b], then the arc-length of f on this interval is
ó
õ
b

a 
  
Ö

1 + [f¢(x)]2
 
 dx

in this case, since f(x) = x2, we know f¢(x) = 2x. The desired result follows.

(16)
Since density is constant, the mass of the liquid is determined by the volume of the cone. To find this volume, we will imagine thin slices taken through the cone perpendicular to the vertical axis. To this end, set a vertical axis in the center of the cone with x = 0 corresponding to the top of the cone and x = 2 corresponding to the bottom. Divide this axis into n subintervals using x0 = 0, x1, x2, ..., xn-1, xn = 2 as the endpoints of these subintervals. If n is large (so that the width Dxj of the jth subinterval is small for all 0 £ j < n), then a cross section of the solid taken perpendicular to the axis on each subinterval is approximately a cylinder having volume
Vj » pRj2 Dxj

where Rj is the radius of the circular cross section at xj. Using similar triangles, we find that Rj = (2-xj)/2. Consequently, the volume of the cone is approximated by

V » n-1
å
j = 0 
p æ
ç
è
2-xj
2
ö
÷
ø
2

 
Dxj = (p/4) n-1
å
j = 0 
(2-xj)2Dxj

Taking the limit as Dxj goes to zero gives us the desired volume integral:

V = (p/4) ó
õ
2

0 
(2-x)2 dx = -(p/4) æ
ç
è
(2-x)3
3
ö
÷
ø
|02 = 2p/3

The mass of the liquid is therefore 2pr/3 kilograms.

(17)
We use the ratio test to find the radius of convergence for this series. Indeed, we know that if
+infty
å
j = a 
Aj(x-b)^j

is a power series, and if we let

r =
lim
j®+¥ 
| Aj+1 |
| Aj |

then the radius of convergence for the series is R = 1/r; that is, the series will converge for all | x-b | < R. The series will diverge for all | x-b | > R, and may converge or diverge when | x-b | = R. In our case, note that

·
Aj = 3^j / j!
·
Aj+1 = 3^j+1 / (j+1)!

Observe that

Aj+1
Aj
= (3j+1)/(j+1)!
(3j/j!)
= æ
ç
è
3j+1
(j+1)!
ö
÷
ø
æ
ç
è
j!
3j
ö
÷
ø
= 3
 j+1  

using the fact that (j+1)! = j!(j+1). Therefore, it is easy to see that r = 0 in this case. Consequently, R = 1/r = +¥, and the radius of convergence is infinite.

(18)
This problem is worked the same way as Problem 17.
(19)
We are dealing with an improper integral in this problem. Observe that
ó
õ
+¥

1 
dx
1+x2
=
lim
h®+¥ 
ó
õ
h

1 
dx
1+x2
=
lim
h®+¥ 
(Arctan(h) - Arctan(1)) = p/2 - p/4 = p/4

(20)
Euler's method for estimating the value y begins with an initial point; in this case (0,1). This point uniquely determines a solution y to the differential equation. Using a uniform step size (in this case Dxj = .25) we move from this point to the point (1,y(1)) by increments (xj+Dxj, yj + Dyj), where Dyj = y¢Dxj. We find
·
STEP 0 : (x0,y0) = (0,1) and y¢ = Ö2
·
STEP 1 : (x1,y1) = (.25, 1+.25Ö2) = (.25,1.3536) and y¢ = 1.307
·
STEP 2 : (x2,y2) = (.50, 1.3536+.25(1.307)) = (.50,1.68035) and y¢ = 1
·
STEP 3 : (x3,y3) = (.75, 1.68035+.25(1.000)) = (.75,1.93035) and y¢ = .5412
·
STEP 4 : (x4,y4) = (1.00, 1.93035+.25(.5412)) = (1.00,2.06565) and y¢ = 0

Consequently, we find y(1) » 2.06565.

(21)
The desired series is obtained by integrating the series representation for f(x) = 1/x:
ln(x)
= ó
õ
x

1 
t ^-1 dt
= ó
õ
x

1 
+¥
å
j = 0 
(-1)^j (t-1)^j dt
= +¥
å
j = 0 
ó
õ
x

1 
(-1)^j (t-1)^j dt
= +¥
å
j = 0 
(-1)^j (x-1)^j+1
j+1

The interval of convergence is the same as that for the original series.


File translated from TEX by TTH, version 2.00.
On 27 Nov 2000, 15:44.