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
|
= |
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.
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(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:
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
|
|
= (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:
|
|
= 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
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
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
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:
|
|
|
|
= |
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.