Integrated Calculus II Exam 2 3/24/3
Solutions to Practice Questions
Question 1
Calculate each of the following integrals.
If an integral does not exist, say why not.
R4 2
• 1 ( t33t−t−2t
2 +1 )dt.
We put u = t3 − t2 + 1, so du = (3t2 − 2t)dt.
The u-range is from u = 1−1+1 = 1, when t = 1, to u = 64−16+1 = 49,
when t = 4.
Z 49
Z 4
1
3t2 − 2t
)dt =
du = [ln(u)]49
( 3
1 = ln(49) = 2 ln(7) = 3.8918202981.
2
t
−
t
+
1
u
1
1
•
R 10
x2 ln(x)dx.
This integral has a bad point at x = 0.
We first do the indefinite integral, integrating by parts.
We put:
R
R
3
u = ln(x), du = x1 dx, dv = x2 dx and v = dv = x2 dx = x3 .
Z
Z
Z
Z 3
x3
x 1
2
x ln(x)dx = udv = uv − vdu =
ln(x) −
dx
3
3 x
Z 2
x
x3
ln(x) −
dx
=
3
3
0
=
x3
x3
x3
ln(x) −
+ C = (3 ln(x) − 1) + C.
3
9
9
Then we have:
Z
10
0
x2 ln(x)dx = lim+ [
= lim+ (
t→0
t→0
x3
(3 ln(x) − 1)]10
t
9
103
t3
(3 ln(10) − 1) − (3 ln(t) − 1))
9
9
103
(3 ln(10) − 1) = 656.4172532202.
9
Here we have used the fact that limt→0+ tk ln(t) = 0, whenever k > 0.
=
•
R∞
x(ln(x))2 dx.
This integral has a bad point at x = ∞.
We notice that in the given range [1, ∞), the integrand is positive or zero
and for x ≥ e, we have ln(x) ≥ 1, so we have:
Z ∞
Z ∞
Z ∞
Z t
2
2
x(ln(x)) dx ≥
x(ln(x)) dx ≥
xdx = lim
xdx
1
1
e
e
t→∞
e
x2 t
t2 e2
= lim [ ]e = lim ( − ) = ∞.
t→∞ 2
t→∞ 2
2
So the given integral diverges.
R1
4
• 0 x3 e−2x dx.
We put u = −2x4 , so du = −8x3 dx.
The u-range is from u = 0, when x = 0, to u = −2, when x = 1.
Z 1
Z −2
1
1
1
3 −2x4
−2
xe
dx =
(− )eu du = − [eu ]−2
0 = (1−e ) = 0.1080830896.
8
8
8
0
0
•
R1
2
2x3 e−x dx.
We put t = −x2 , so dt = −2xdx.
The t-range is from t = 0, when x = 0, to t = −1, when x = 1.
2
2
Also 2x3 e−x dx = (−x2 )e−x (−2xdx) = tet dt.
So we get:
Z
Z
0
1
−1
2
2x3 e−x dx =
0
tet dt.
0
Now we integrate by parts:
We put:
R
R
u = t, du = dt, dv = et dt, v = dv = et dt = et .
Then we have:
Z
Z
Z
Z
t
t
te dt = udv = uv − vdu = te − et dt = et (t − 1) + C,
Z
−1
tet dt = [et (t − 1)]−1
0
0
= e−1 (−2) − e0 (−1) = 1 −
2
2
= 0.2642411177.
e
•
Rπ
(cos5 (x) − cos3 (x))dx.
We have odd powers of cos(x), so we substitute u = sin(x), du = cos(x)dx.
When x = 0, u = 0 and when x = π2 , u = 1.
We have:
2
0
(cos5 (x) − cos3 (x))dx = (cos(x)4 − cos2 (x)) cos(x)dx
= cos2 (x)(cos2 (x) − 1) cos(x)dx
= (1 − sin2 (x))(− sin2 (x)) cos(x)dx
= (1 − u2 )(−u2 )du = (u4 − u2 )du.
So the given integral becomes:
Z 1
u5 u3
1 1
2
(u4 − u2 )du = [ − ]10 = − = − = −0.1333333333.
5
3
5 3
15
0
•
R3
1
dx.
2 x(x−2)
There is a bad point at x = 2.
We use partial fractions:
1
A
B
A(x − 2) + Bx
= +
=
.
x(x − 2)
x x−2
x(x − 2)
So 1 = A(x − 2) + Bx.
Putting x = 2, gives 2B = 1, so B = 12 .
Putting x = 0, gives −2A = 1, so A = − 12 .
So we have, for t > 2:
Z 3
Z
1
1 3 1
1
dx =
(
− )dx
2 t x−2 x
t x(x − 2)
1
= [ln(x − 2) − ln(x)]3t
2
1
= (− ln(3) + ln(t) − ln(t − 2)).
2
+
As t → 2 , the term − 21 ln(t − 2) → ∞, whereas the other terms remain
finite, so the given integral diverges to infinity.
3
Question 2
Solve the following differential equations and discuss the behavior of each solution as a function of the variable t.
Include a plot of each solution.
•
dy
dt
3t
e
= sin(y)
, y(0) = π2 .
This equation is separable, so we separate, writing the equation as an equation of differentials:
sin(y)dy = e3t dt.
We solve by integrating both sides:
Z
Z
sin(y)dy = e3t dt,
1
− cos(y) = e3t + C.
3
π
When t = 0, we have y = 2 , so − cos( π2 ) = 0 = 13 + C, so C = − 13 .
So the solution is cos(y) = 13 (1 − e3t ), or y = arccos( 31 (1 − e3t )).
The graph of the solution is a concave up curve, with a horizontal asymptote
y = arccos( 13 ) = 1.2309594173, which the curve approaches from above
as t → −∞.
For positive t, the solution curve terminates, when 13 (1 − e3t ) = 1, so when
t = 31 ln(4) = 0.4620981204, for then y = π and the curve has infinite
slope.
•
dy
dt
+ 4ty = 20t,
y(0) = 4.
This is in standard linear form, with P (t) = 4t and Q(t) = 20t and the
integrating factor is:
R
J(t) = e
P (t)dt
R
=e
4tdt
Then the equation is rewritten:
d
(yJ(t)) = Q(t)J(t),
dt
d
2
2
(ye2t ) = 20te2t .
dt
4
2
= e2t .
Integrating both sides, we get:
ye
2t2
Z
=
2
20te2t dt.
We put u = 2t2 and du = 4tdt:
Z
2
2t2
ye = 5eu du = 5eu + C = 5e2t + C,
2
2
2
y = e−2t (5e2t + C) = 5 + Ce−2t .
When t = 0, y = 4, which gives: 4 = 5 + C, so C = −1.
2
So the required solution is: y = 5 − e−2t .
The plot is an upside down bell-shaped curve, symmetrical about the y-axis,
which is a global minimum of y.
2
We have y 0 = 4te−2t , which is positive for t > 0 and negative for t < 0, so
t = 0 is the only critical point.
The curve is increasing for t > 0 and decreasing for t < 0.
2
We have y 00 = 4e−t (1 − 4t2 ), so for |t| < 21 , we have y 00 > 0 and the
solution is concave up, whereas for |t| > 21 , the solution is concave down.
So there are inflection points at ( 12 , 5 − √1e ) and (− 12 , 5 − √1e ).
As t → ±∞, the solution approaches the horizontal asymptote y = 5 from
below.
5
Question 3
A circuit has an inductance 20 henry and a resistance of 40 ohms connected in
series.
If the current is initially zero and a voltage source 80 sin(10t) volts is switched
on, find and plot the current as a function of time.
The differential equation for the current I amperes at time t seconds is:
L
dI
+ RI = E.
dt
Here L is the inductance, R is the resistance and E is the voltage source.
Putting in the data, our equation is:
20
dI
+ 40I = 80 sin(10t),
dt
dI
+ 2I = 4 sin(10t),
dt
The integrating factor is:
R
J(t) = e
2dt
= e2t .
So the equation becomes:
d
(IJ(t)) = e2t J(t),
dt
d
(Ie2t ) = e2t sin(10t),
dt
Z
2t
Ie = e2t sin(10t)dt.
This integral is awkward, done by using a table or by two integrations by parts
(ignoring the constant of integration):
Z
A = e2t sin(10t)dt
Z
1
u = sin(10t), du = 10 cos(10t)dt, dv = e dt, v = dv = e2t .
2
Z
Z
Z
1
1 2t
A = udv = uv − vdu = e2t sin(2t) −
e 10 cos(2t)dt
2
2
2t
6
1
= e2t sin(10t) − 5B,
2 Z
B = e2t cos(10t)dt.
Z
1
u = cos(10t), du = −10 sin(10t)dt, dv = e dt, v = dv = e2t .
2
Z
Z
Z
1
1 2t
B = udv = uv − vdu = e2t cos(10t) +
e 10 sin(10t)dt
2
2
1
= e2t cos(10t) + 5A.
2
So we now have two equations for A and B:
2t
1
A + 5B = e2t sin(10t),
2
1
B − 5A = e2t cos(10t).
2
Multiplying the second equation by 5 and subtracting from the first equation eliminates B and gives:
e2t
A=
(sin(10t) − 5 cos(10t)).
52
So now we have:
Ie2t =
e2t
(sin(10t) − 5 cos(10t)) + C,
52
I = e−2t (
e2t
(sin(10t) − 5 cos(10t)) + C),
52
1
(sin(10t) − 5 cos(10t)) + Ce−2t .
52
When t = 0, we have I = 0, so we get:
=
0=
1
(−5) + C,
52
C=
5
,
52
1
(5e−2t + sin(10t) − 5 cos(10t)).
52
If we plot the solution curve, we see a wave pattern.
As t increases, the exponential term dies off quickly, so after about t = 2, the
1
graph becomes almost identical to the graph of the function 52
(sin(10t)−5 cos(10t)),
1
π
√
which gives a sinusoidal curve of period 5 and amplitude 104 = .09806.
I=
7
Question 4
Consider the following integral:
6
Z
J=
1
1
dx.
x
• Estimate J with the approximation M10 , the midpoint rule with 10 intervals.
With 10 intervals, the interval width is 6−1
= 12 .
10
The data for M10 are as follows:
x
f (x) =
1
x
5
4
4
5
7
4
4
7
9
4
4
9
11
4
4
11
13
4
4
13
15
4
4
15
17
4
4
17
We add the data and multiply by the interval width
19
4
4
19
1
2
21
4
4
21
23
4
4
23
to give M10 :
596340296
1 4 4 4 4 4 4 4 4 4 4
= 1.7820390106.
M10 = ( + + + + + + + + + ) =
2 5 7 9 11 13 15 17 19 21 23
334639305
• Is the estimate M10 too high or too low?
Explain your answer and illustrate with a sketch.
The graph of the function y = x1 is a standard rectangular hyperbola, with
the axes as asymptotes.
We have the following derivatives:
f (x) =
1
1
, f 0 (x) = − 2 ,
x
x
f 00 (x) =
2
,
x3
f 000 (x) = −
6
,
x4
f 0000 (x) =
24
.
x5
In the given interval [1, 6], the graph is concave up, since f 00 (x) = x23 > 0,
so the tangent trapezoids used in the midpoint rule lie below the curve, so
M10 is an underestimate of the true result.
• Use the error estimate to estimate the error in M10 .
We have f 00 (x) = x23 , which is positive and decreasing on the interval [1, 6],
so its maximum size K2 on the interval [1, 6] is its value at x = 1, which is
2.
Then the error estimate EM for M10 is:
EM =
K2 (b − a)3
2(6 − 1)3
250
5
=
=
=
.
24n2
24(102 )
2400
48
8
• Estimate J with the approximation T10 , the trapezoidal rule with 10 intervals. The data for T10 are as follows:
x
f (x) = x1
weightw
wf (x)
1 32 2 52 3 72 4 92 5
1 23 12 25 13 27 14 29 15
1 2 2 2 2 2 2 2 2
1 43 1 45 23 47 12 49 25
11
2
2
11
6
2
1
4
11
1
6
1
6
We add the weighted data and multiply by the one half of the interval width
1
to give T10 :
2
1
4
4 2 4 1 4 2 4 1
6277
T10 = (1+ +1+ + + + + + + + ) =
= 1.8115440115.
4
3
5 3 7 2 9 5 11 6
3465
• Is the estimate T10 too high or too low?
Explain your answer and illustrate with a sketch.
In the given interval [1, 6], the graph is concave up, since f 00 (x) = x23 > 0,
so the trapezoids used in the trapezoidal rule lie above the curve, so T10 is
an overestimate of the true result.
• Use the error estimate to estimate the error in T10 .
The error estimate ET for T10 is:
2(6 − 1)3
250
5
K2 (b − a)3
ET =
=
=
= .
2
2
12n
12(10 )
1200
24
• Estimate J with the approximation S10 , Simpson’s rule with 10 intervals.
The data for S10 are as follows:
x
f (x) = x1
weightw
wf (x)
1 32 2 52 3 72 4 92 5
1 23 12 25 13 27 14 29 15
1 4 2 4 2 4 2 4 2
1 83 1 85 23 87 12 89 25
11
2
2
11
6
4
1
8
11
1
6
1
6
We add the weighted data and multiply by the one third of the interval
width 12 to give S10 :
1
8
8 2 8 1 8 2 8 1
3728
S10 = (1+ +1+ + + + + + + + ) =
= 1.7931697932.
6
3
5 3 7 2 9 5 11 6
2079
9
• Use the error estimate to estimate the error in S10 . We have f 0000 (x) = x245 ,
which is positive and decreasing on the interval [1, 6], so its maximum size
K4 on the interval [1, 6] is its value at x = 1, which is 24.
Then the error estimate ES for S10 is:
ES =
24(6 − 1)5
K4 (b − a)5
120
1
=
=
= .
4
4
180n
180(10 )
180(16)
24
The exact value of J is [ln(x)]61 = ln(6) = 1.7917594692.
The actual percentage error for each rule is as follows:
• For M10 : 0.5425 percent.
• For T10 : 1.1042 percent.
• For S10 : 0.0787 percent.
So for this example with ten intervals Simpson’s rule is considerably more accurate.
10
Question 5
A region R in the x-y plane is bounded by the curves y = x2 − 6x + 9 and the
line y = 2x + 9.
• Sketch the region R.
Put g(x) = x2 − 6x + 9 = (x − 3)2 and f (x) = 2x + 9.
The curves meet where f (x) = g(x), so where x2 − 6x + 9 = 2x + 9, or
x2 − 8x = 0, so where x = 0 or x = 8, so at the points (0, 9) and (8, 25).
The graph y = g(x) is a concave up parabola, touching the x-axis at (3, 0).
The line y = f (x) is a straight line of slope 2 meeting the parabola at (0, 9)
and (8, 25).
We have f − g = 8x − x2 and x(f − g) = 8x2 − x3 .
• Find the area of the region R.
The required area A is:
Z 8
Z 8
x3
1 1
83
256
A=
(f −g)dx =
(8x−x2 )dx = [4x2 − ]80 = 83 ( − ) =
=
.
3
2 3
6
3
0
0
• Find the volume of the solid of revolution obtained by rotating the region R
about the y-axis.
By the cylinder method, the required volume is:
Z 8
Z 8
2π
x(f − g)dx = 2π
(8x2 − x3 )dx
0
0
3
8x
x 8
1
84
2048π
4 1
= 2π[
− ]0 = 8 ( − ) = 2π =
.
3
4
3 4
12
3
• Find the volume of the solid of revolution obtained by rotating the region R
about the x-axis.
By the washer method, the required volume is:
Z 8
Z 8
2
2
π(f − g )dx = π
((2x + 9)2 − ((x − 3)2 )2 )dx
0
Z
4
0
8
1
1
((2x + 9)2 − (x − 3)4 )dx = π[ (2x + 9)3 − (x − 3)5 ]80
6
5
0
1
1
1
243
= π( (253 − 93 ) − (55 − (−3)5 )) = π( (15625 − 729 − 3750) −
)
6
5
6
5
π
243
π
π
= (11146) −
= (27865 − 729) = (27136).
6
5
15
15
=π
11
Question 6
A plate of density 2 kilograms per square meter is bounded by the curves:
y = x2 − 6x and y = 36 − x2 .
Units are metric.
• Sketch the plate.
Put g(x) = x2 − 6x and f (x) = 36 − x2 .
Then y = f (x) is a concave down parabola, whereas y = g(x) is a concave
up parabola.
The curves meet where 0 = f (x) − g(x) = 36 + 6x − 2x2 , so where
x2 − 3x − 18 = 0 or (x − 6)(x + 3) = 0, so at the points (−3, 27) and (6, 0).
We have f − g = 36 + 6x − 2x2 .
Also x(f − g) = 36x + 6x2 − 2x3 .
Also 12 (f 2 − g 2 ) = 12 (1296 − 72x2 + x4 − (x4 − 12x3 + 36x2 )) = 648 −
54x2 + 6x3 .
• Find the mass of the plate.
The area of the plate A is:
Z 6
Z 6
A=
(f − g)dx =
(36 + 6x − 2x2 )dx
−3
−3
2
= [36x + 3x2 − x3 ]6−3
3
2
= 36(6 − (−3)) + 3(36 − 9) − (216 + 27)
3
2
= 36(9) + 3(27) − (243) = 405 − 162 = 243.
3
Then, since the density ρ = 2 the mass M of the plate is M = (243)ρ =
243(2) = 486.
• Find the center of mass of the plate.
We have:
Z
1 6
1
[x(f − g), (f 2 − g 2 )]dx
[x, y] =
A −3
2
Z 6
1
=
[36x + 6x2 − 2x3 , 648 − 54x2 + 6x3 ]dx
243 −3
12
1
3
1
[18x2 + 2x3 − x4 , 648x − 18x3 + x4 ]6−3
243
2
2
1 3
1
([0, 648][x]6−3 + [18, 0][x2 ]6−3 + [2, −18][x3 ]6−3 + [− , ][x4 ]6−3 )
=
243
2 2
1
1 3
([0, 648](9) + [18, 0](27) + [2, −18](243) + [− , ](15)(81)
=
243
2 2
5 15
= [0, 24] + [2, 0] + [2, −18] + [− ,
]
2 2
3 27
=[ ,
].
2 2
=
• Find the moments of the plate about the axes.
The moment about the x-axis is M y = 486( 27
) = 6561.
2
3
The moment about the y-axis is M x = 486( 2 ) = 729.
• The plate is rotated about the line x = 6.
What volume is traced out by the plate?
By Pappus’ Theorem, the required volume is:
2π(6 − x)(243) = 9π(243) = 2187.
13
Question 7
Consider the following parametrized curve:
x = 2t3 − 2,
y = 3t2 − 3.
• Sketch the part AB of the curve, between the points A = (0, 0) and B =
(14, 9) on the curve.
The point A has 2t3 − 2 = 0, so t3 = 1, so t = 1 and the point B has
2t3 − 2 = 14, so t3 = 8, so t = 2 .
Plotting the curve, we see that it is increasing, and slightly concave down
from A to B.
• Find the length LAB of the curve AB.
We have:
dx dy
V = [ , ] = [6t2 , 6t] = 6t[t, 1],
dt dt
√
|V = 6|t| t2 + 1,
√
ds = |V |dt = 6|t| t2 + 1dt,
Z B
Z 2 √
LAB =
ds =
6t t2 + 1dt.
A
1
2
We make the substitution: u = t + 1, du = 2tdt.
When t = 1, u = 2 and when t = 2, u = 5.
Z 5
1
LAB =
3u 2 du
2
3
= 2[u 2 ]52
3
3
= 2(5 2 − 2 2 )
√
√
= 10 5 − 4 2.
14
• Write an integral for the surface area of the curved surface, obtained by
rotating the curve about the x-axis.
The surface area S is:
Z B
S=
2πyds
A
2
Z
= 2π
√
(3t2 − 3)6t t2 + 1dt
1
Z
= 36π
2
√
(t2 − 1)t t2 + 1dt
1
We again make the substitution: u = t2 + 1, du = 2tdt.
When t = 1, u = 2 and when t = 2, u = 5.
So now we have:
Z 5
1
S = 18π
(u − 2)u 2 du
2
Z
5
3
1
(u 2 − 2u 2 )du
= 18π
2
2 5 4 3
= 18π[ u 2 − u 2 ]52
5
3
5
5
3
3
12π
(3(5 2 − 2 2 ) − 10(5 2 − 2 2 ))
5
√
√
√
√
12π
=
(3(25 5 − 4 2) − 10(5 5 − 2 2))
5
√
96 √
= π(60 5 +
2).
5
=
15
Question 8
A dam is in the shape of an inverted isosceles triangle ABC (with AB = AC).
The lower vertex A is at a depth of 100 meters.
The upper vertices B and C are level with the surface of the water and are 150
meters apart.
• Assuming the dam wall is vertical, write and evaluate an integral for the
total force on the dam wall.
Consider the force on a slice of the dam of thickness dx at height x above
the vertex A.
Since the depth of the slice is 100 − x meters, the pressure on the slice is
p = ρg(100 − x).
y
x
If the width of the slice is y, then by similar triangles, we have BC
= 100
,
x
3
or y = 100 (150) = 2 x.
So the area of the slice is dA = ydx = 32 xdx.
So the force dF on the slice is:
3
dF = pdA = ρg(100 − x) xdx = k(100x − x2 )dx,
2
3
k = ρg.
2
Then the total force F on the dam, in Newtons is:
Z x=100
Z 100
F =
dF =
k(100x − x2 )dx
x=0
= k[100
=
0
x2 x3 ‘00
1 1
500000k
− ]0 = k(100)3 ( − ) =
2
3
2 3
3
500000 3
( )ρg = 250000ρg = (2.5)(105 )(103 )(9.81) = 2.4525(109 ).
3
2
16
• If instead the dam makes an angle of 5 degrees with the vertical, what then
is the total force on the dam wall?
π
The geometry of the slice is the same, but now its depth is (100−x) cos( 36
),
π
so the force is the same as before multipled by a factor of cos( 36 ).
π
π
So the total force in Newtons is: F cos( 36
) = 250000ρg cos( 36
) = 2.443(109 ).
• If the water in the vertical dam floods to a depth of five meters above BC,
how much extra force acts on the dam wall?
Here the depth is increased by 5, so the pressure at height x above A is
ρg(105 − x).
So the extra force on the slice is dG = ρg(5)( 32 )x = 15
ρgx.
2
The total extra force G, in Newtons, is then:
Z x=100
Z 100
15ρg 2 100
15
ρgxdx =
[x ]0
G=
dG =
2
4
x=0
0
= 37500ρg = 37500(103 )(9.81) = 367875000 = 3.67875(108 ).
17
Question 9
An object consists of two semi-circular metal discs A and B, joined at their common point.
• Disc A is the upper half of the circular disc of radius 6 cm centered at (0, 6)
(units in centimeters).
• Disc B is the right hand half of the circular disc of radius 6 cm centered at
(6, 0).
• If the discs have equal density where is their center of mass?
By example 3, page 557, the center of mass of a semi-circular disc of radius
4r
r is on the axis of symmetry at the point 3π
above the center.
8
So the centroid of disc A is at (0, 6 + π ).
Also the centroid of disc B is at (6 + π8 , 0).
If the discs have equal density, their masses are the same and the center of
mass is at the midpoint between their centroids, so at the point :
1
8
8
4
4
((0, 6 + ) + (6 + , 0)) = (3 + , 3 + ).
2
π
π
π
π
• If disc A has density 20 grams per square centimeter and disc B has density
40 grams per square centimeter, where now is their center of mass?
Here disc B counts twice as much as does disc A, so the center of mass is
at:
1
8
8
16
8
(1(0, 6 + ) + 2(6 + , 0)) = (4 +
,2 +
).
(1 + 2)
π
π
3π
3π
• If the geometrical region defined by the discs is rotated about the line x+y =
6, what is the volume of the solid of revolution traced out by the region?
By symmetry, the nearest point to the centroid on the line x + y = 6 is the
point (3, 3).
By Pythagoras its distance from the centroid is:
r
√
4
4
4 2
2
2
(3 + − 3) + (3 + − 3) =
.
π
π
π
Then by Pappus, the required volume in cubic centimeters is:
√
√
2 4 2
2π(π6 )(
) = 288π 2 = 1279.5502861896
π
18
Question 10
A water tank 10 meters long has a cross-section that is an isosceles triangle ABC,
with AB = BC.
The edge AB is level with water surface and is four meters long.
The vertex C is three meters below the surface.
How much work is done in pumping all the water out of the tank through a pipe
that exits the tank level with the edge AB?
Consider the work dW in removing a horizontal sliver of width dx, x meters above
the base.
We have dW = (3 − x)(ρgdV ), where dV is the volume of the sliver, ρ is the
density of water and g is the acceleration due to gravity, since the cross-section
moves up a distance 3 − x meters against gravity.
Then dV = Adx, where A is the area of the cross-section at level x.
If the width of the cross-section is y, we have A = 10y, since the length of the
tank is 10 meters.
By similar triangles, we have: xy = 43 , so y = 4x
, so A = 40x
, so dV = 40x
dx and
3
3
3
40
40
2
dW = ρg 3 x(3 − x)dx = k(3x − x )dx, where k = 3 ρg.
So the required work done in Joules is:
Z x=3
Z 3
3x2 x3 3
dW = k
(3x − x2 )dx = k[
− ]0
2
3
x=0
0
1 1
9
9 40
= 27k( − ) = k = ( ρg) = 60ρg
2 3
2
2 3
= 60(1000)(9.81) = 578600.
19
Question 11
Consider the differential equation
dy
dx
= y − 2x.
• Plot the slope field of the differential equation, including all points where
the slope is (a) 0, (b) 1, (c)−1, (d) 12 , (e) − 12 .
We have:
– (a) slope 0 for the slope field all along the line of slope 2, through the
origin, y = 2x.
– (b) slope 1 for the slope field all along the line of slope 2, through the
point (0, 1), y = 2x + 1.
– (c) slope −1 for the slope field all along the line of slope 2, through
the point (0, −1), y = 2x − 1.
– (d) slope 21 for the slope field all along the line of slope 2, through the
point (0, 12 ), y = 2x + 12 .
– (e) slope − 12 for the slope field all along the line of slope 2, through
the point (0, − 21 ), y = 2x − 12 .
• Discuss the behaviors of the various solutions of the differential equation.
In general we may construct the slope field by plotting a slope of c at the
point on the y-axis (0, c) and then copying that slope at every point along
the line through (0, c) with slope 2.
In particular, if we start at the point (0, 2), then the slope points directly
along the line, so we get a linear solution of the differential equation, namely
y = 2x + 2.
If we start above the line y = 2x + 2, then the solution approaches the line
for large negative x and goes to infinity for large positive x.
If we start below the line y = 2x + 2, then the solution approaches the line
for large negative x and goes to minus infinity for large positive x.
• Solve the differential equation with the initial condition y(0) = 2 and plot
the solution on your graph of the slope field.
From the slope field, we see that the required solution is just y = 2x + 2, a
straight line of slope 2.
Alternatively, we first find the general solution of the differential equation:
dy
= y − 2x,
dx
20
dy
− y = −2x.
dx
Here we have the equation in standard linear form, with P (x) = −1 and
Q(x) R= −2x.
R
R
Then P (x)dx = −1dx = −x, so the integrating factor J(x) = e P (x)dx =
e−x .
Then the differential equation is rewritten:
d
(yJ(x)) = Q(x)J(x),
dx
d
(ye−x ) = −2xe−x .
dx
Integrating both sides, we get:
Z
−x
ye = −2 xe−x dx.
We integrate by parts:
Z
u = x, du = dx, dv = e−xdx, v =
Z
Z
Z
dv =
−x
e−x dx = −e−x ,
Z
= udv = uv − vdu = −xe − (−e−x )dx
Z
−x
= −xe + e−x dx = −xe−x − e−x + A = −e−x (x + 1) + A.
Z
−x
ye = −2 xe−x dx = 2e−x (x + 1) + C,
xe
−x
Z
y = ex (2e−x (x + 1) + C) = 2(x + 1) + Cex .
If now y = 2, when x = 0, we get: 2 = 2(1) + Ce0 , so C = 0 and the
required solution is y = 2x + 2.
Any other solution, for which C 6= 0, has the exponential term Cex , which
goes to zero as x → −∞ and which dominates the linear term in y, blowing
up to infinity if C > 0 and to minus infinity if C < 0.
So the solutions with C 6= 0 have the behavior that they approach closely
to the line y = 2x + 2 (called a slant asymptote), for large negative x and
diverge to either infinity or minus infinity, as x → ∞, in agreement with the
picture derived from the slope field.
21