Download 6.7 Solution

Davis Buenger
Math 1172 6.7 Solutions
1. Find the mass(of the thin bar with density
1
0≤x≤2
function ρ(x) =
1 + x 2 < x ≤ 4.
Solution: As indicated by the box above,
to find the mass of a linear object we need
to integrate the density function over the
bounds of the object. We are given the density function and from the density function
we deduce that the object is bounded between 0 and 4.Thus
Z 4
M ass =
ρ(x) dx
0
Z 2
Z 4
=
ρ(x) dx +
ρ(x) dx
0
2
Z 2
Z 4
=
1 dx +
1 + x dx
0
2
4
x2
2
= x|0 + x +
2 2
22
42
− 2+
= (2 − 0) + 4 +
2
2
= 10.
September 8, 2015
Z
L
5e−x dx
0
L
e−x = 5
−1 0
=
= −5e−L − 5(−e−1·0 )
= 5 − 5e−L .
Similarly the mass of the second bar as a
function of L is
Z L
ρ2 (x) dx
M ass2 (L) =
0
Z L
7e−2x dx
=
0
L
e−2x = 7
−2 0
7 −2L
e−2·0
=
e
−7
−2
−2
7 7 −2L
− e .
=
2 2
Thus our problem has been reduced to comparing the values of these two functions. Let
us find the intersection pts of the two functions.
2. Two bars of length L have densities
7 7 −2L
5 − 5e−L =
− e
ρ1 (x) = 5e−x and ρ2 (x) = 7e−2x (Assume
2 2
that the ends of the bars are at x = 0 and
7 −L 2
3
(e ) − 5e−L +
= 0
x = L). For what values of L is the first bar
2
2
heavier? As the lengths increase, do their
masses increase without bound?
Solution: To solve the problem we need Thus by the quadratic formula,
q
to find the mass of each bar as a function
5
±
52 − 4 · 32 · 27
of the Rlength L. The mass of a linear ob−L
b
e
=
ject is a ρ(x)dx, where a and and b are the
2 · 72
√
bound of the linear object and ρ is the den3
5 ± 25 − 21
sity function. Thus the mass of the first bar
=
= ,1
7
7
as a function of L is
1
Z L
M ass1 (L) =
ρ1 (x) dx
0
So
7
L = 0, ln .
3
At ln(2)
M ass1 (ln(2)) = 5 − 5e− ln 2
5
= 5−
2
5
,
=
2
3. Find the mass of the thin bar over the
interval 0 ≤ x ≤ π with the density function
4x
. Give the exact value (not
ρ(x) =
3 + x2
a calculator approximation in decimal.)
Recall that the mass of a linear object lying
on the interval [a, b] with density function
ρ(x) is given by the formula
b
Z
ρ(x) dx
M ass =
a
and
7 7 −2·ln 2
− e
2 2
7 7
=
−
2 8
21
=
= 2.625.
8
M ass2 (ln(2)) =
Hence in the interval (0, ln 73 ), M ass2 (L) is
larger. At ln(3)
M ass1 (ln(3)) = 5 − 5e− ln 3
5
= 5−
3
10
=
= 3.3,
3
In our case the interval is [0, π] and our den4x
sity function is ρ(x) = 3+x
2 . Thus
π
Z
M ass =
0
Let u = 3+x2 . Then du = 2x dx or dx = du
.
2x
Our interval of integration changes with the
u substitution. At x = 0, u = 3 and when
x = π, u = 3 + π 2 . Thus
3+π 2
Z
M =
3
Z
4x du
·
u 2x
3+π 2
= 2
3
and
7 7 −2·ln 3
− e
2 2
7
7
−
=
2 18
56
28
=
=
= 3.1.
18
9
M ass2 (ln(3)) =
Thus in the interval (ln 73 , ∞), M ass1 (L) is
larger.
As L goes to infinity both e−2L and e−L go
to 0. Thus
lim M ass1 (L) = 5
L→∞
7
lim M ass2 (L) = .
L→∞
2
In both cases the mass has an upper bound 2
on how large it can be.
4x
dx
3 + x2
1
du
u
3+π2
= 2 ln(u)
3
= 2 ln(3 + π 2 ) − (2 ln(3))
3 + π2
= 2 ln
3
4. (Test Question) True or false: Given a
spring that obeys Hooke’s Law, the work required to stretch the spring from equilibrium
to 1 cm is the same as the work required
to stretch the spring from 1 cm to 2 cm.
False: Given a fixed spring constant k, the
work required to stretch the spring from
equilibrium to 1 cm is given by
Z 0.01
k
k(0.01)2
=
J.
kx =
2
20, 000
0
6. Calculate the work required to stretch a
spring 0.5 m from its equilibrium position if
it requires a force of 50N to hold the spring
in a stretched position 0.2 m from its equilibrium position.
Solution: By Hooke’s law, the work required to stretch a spring from equilibrium
to a position 0.5 m from equilibrium is
Z 0.5
k · x dx.
Meanwhile, the work required to stretch the
spring from 1 cm to 2 cm is given by
Z 0.02
k((0.02)2 − (0.01)2 )
3k
kx =
=
J.
2
20, 000
0.01
Here k is the spring constant. We will use
the second bit of information they provided
to calculate k and then integrate to find the
amount of work done. A second application
of Hooke’s Law at the position x = 0.2 shows
Thus the work streaching the spring from 1
cm to 2 cm is greater.
50N = k · 0.2m.
0
Thus k = 250 and
5. How much work is required to move an
object from x = 1 to x = 3 in the presence of
a force (in N) given by F (x) = 2/x2 acting
along the x-axis?
Solution:
Z 3
2
W ork =
dx
2
1 x
3
−2 =
x 1
−2 −2
4
−
= J
=
3
1
3
Z
0.5
250 · x dx
0.5
x2 = 250 2 0
(0.5)2
= 250
2
125
=
J.
4
W ork =
0
7. A water tank is shaped like an inverted cone with a height 6 m and a base radius 1.5. If
the tank is full, how much work is required to pump the water to the level of the top of the
tank? Second, is it true that it takes half as much work to pump the water out of the tank
when it is filled to half its depth as when it is full?
3
Solution:Place the bottom tip of the cone at
the origin and align the central axis with the
y−axis. As indicated in the above box we
find the work done emptying a tank by the
following integral
Z b
ρ · g · A(y) · D(y) dy.a
W ork =
a
triangle ABC and ADEb , where A, B, C, D
and E are the points in the diagram. Since
the triangles are similar the ratios of the sides
are equal. Thus
1.5
r(y)
=
y
6
or
The water starts at a height of y = 0 and
ends at a height of y = 6 thus our range of
integration is [0, 6].
r(y) =
y
4
2
Thus A(y) = π y16 and
W ork =
=
=
=
=
6
y2
1000 · 9.8 · π · (6 − y) dy
16
0
Z 6
1000 · 9.8 · π
6y 2 − y 3 dy
16
0
3
6
1000 · 9.8 · π y
y4
6 −
16
3
4 0
3
1000 · 9.8 · π
64
6
6 −
16
3
4
66150 · πJ
Z
If the tank is filled half way,
3
y2
1000 · 9.8 · π · (6 − y) dy
16
0
3
1000 · 9.8 · π
3
34
=
6 −
16
3
4
= 20671.875 · πJ
Z
W ork =
Our fluid is water, so ρ = 1000. The distance
the cross section at a height of y is 6 − y. The
cross sections in this case are circles. Thus to
find the area of a cross section it suffices to
find the radius of the circle as a function of y.
Fix a generic height y. Consider the similar
Thus by comparison one sees that the work
draining the tank when it is half full is not
half of the work draining the tank when the
tank is full.
a
Here a, b, ρ, g, A(y), and D(y) are as defined above. Thus to solve the problem we need to first identify
expressions for all the terms and then integrate.
b
They are similar since they share a common angle and both have other angles which are right
8. A water trough has a semicircular vertical cross section with a radius of .25 m and a length
of 3m. How much work is required to pump the water out when it is full?
4
Solution: Assume that the front face of the trough lies in the xy-plane and that the center
of the semicircle is the origin. Thus the equation for the boundary of the front face is given
2
by lower half of the circle y 2 + x2 = 14 . Horizontal cross sections are rectangles of length
3m with a width equal to the width of the semi circle at a height y.
p
After solving the equation of the semicircle for x we arrive at the equation
x
=
±
.252 − y 2 .
p
2
2
Thus
pat the height y the front edge of the cross section begins at (− .25 − y , y) and ends
atp( .252 − y 2 , y). The width of the horizontal cross section at a height y is width(y) =
2 .252 − y 2 m, and the area of the horizontal cross section at a height y is
p
Area(y) = 3 · 2 .252 − y 2 m2 .
During the draining process, each horizontal cross section starts at a height y and exits the
trough at a height of 0. Thus the distance the cross section at height y travels is (note here
that we are only considering negative heights y).
Distance = 0 − y = −y.
The density of water is 1000kg/m3 and we will approximate the force of gravity to be
9.8m/s2 . The water exists from a height of −.25m to a height of 0m. Thus
Z 0
Z 0
p
p
2
2
−2y .252 − y 2 dy.
W ork =
9.8 · 1000 · (−y) · 2 .25 − y dy = 9800
−.25
−.25
Let u = .252 − y 2 . Then du = −2ydy. At y = −.25 we have that u = 0, and at y = 0 we
have u = .252 . Thus
Z .252
√ du
W ork = 9800
−2y u
−2y
0
Z .252
√
= 9800
u du
0
.252
2 1225
= 9800u =
J
3 0
12
3
2
5
9. (Test Question) A large vase is constructed by revolving the portion of the graph
1
x = 2y 3 + 4y, 0 ≤ y ≤ 6 around the y-axis. (Assume that distances are measured in
meters.) Find the total volume that the vase can hold. Next find the work required to pump
all of the water in the vase over the top when the vase is filled to half its depth.
Solution: First let us find the volume using the washer/disk method. As we are revolving
a region around the y-axis to form our shape, we will integrate with respect to y it find the
volume. In this case the inner radius is x = 0 (i.e. we are considering disks not washers),
1
and the outer radius is x = 2y 3 + 4y. As indicated, our vase is limited to y values in the
interval [0, 6]. Thus
Z 6
1
(2y 3 + 4y)2 dy
V olume = π
Z0 6
4
2
= π
4y 3 + 16y 3 + 16y 2 dy
0
6
12 5 3 · 16 7 16 3
= π
y3 +
y3 + y
5
7
3
0
12 5 3 · 16 7 16 3
= π
63 +
63 + 6 .
5
7
3
Recall that the density of water is 1000kg/m3 and that the force of gravity is approximately
1
9.8m/s2 . The area of a cross-section at a height of y is given by π(2y 3 +4y)2 and the distance
the slice at a height of y travels to exit the vase is 6 − y. Finally as the bowl is only filled to
half its depth, the interval of integration is [0, 3]. Thus
Z 3
1
W ork =
1000 · 9.8(6 − y)π(2y 3 + 4y)2 dy
0
Z 3
1
4
= 9800π
12y 3 + 24y − 2y 3 − 4y 3 dy
0
3
4
6 7
2
4
3
3
= 9800π 9y + 12y − y − y
7
0
7
4
6
2
4
= 9800π 9 · 3 3 + 12 · 3 − 3 3 − 3 .
7
10. (Test Question) A tank is shaped like a parabolic bowl. It is formed by revolving
the graph of y = 4x2 for 0 ≤ x ≤ 3 (in meters) about the y-axis. The tank is filled
with water to a height of 30 meters. How much work is required to pump all of the water to an exit pipe at the top of the tank? [Note: The density of water is 1000 kg/m3 .]
6
Solution: Work for pumping a liquid out of a tank is given by the equation
Z b
W ork =
ρ · g · CSA(y) · D(y) dy.
a
Here [a, b] is the interval where the liquid exists. ρ is the density of the liquid. g is the
gravitational constant. CSA(y) is the area of the horizontal cross section at the height y.
D(y) is the distance the cross section at height y has to travel.
In our case the tank is filled up to 30m so our bounds of integration are a = 0 and b = 30.
ρ = 1000 and g ≈ 9.8. The tank is formed as a rotational solid with axis of rotation the y
axis. Thus the cross sections are circles and by solving the relation y = 4x2 for y we find the
radius as a function of height:
r
y
.
r(y) =
4
Thus
y
CSA(y) = π .
4
a
The height of the tank is 36 meters . Thus the distance the yth cross section has to travel
to pump out of the tank is 36 − y. Thus
Z 30
πy
(36 − y) dy
W ork ≈
1000 · 9.8 ·
4
0
Z
9800π 30
=
36y − y 2 dy
4
0
30
9800π
y3
2
=
18y −
4
3 0
9800π
303
2
=
18(30) −
4
3
≈ 69272118 J
a
Plug in x = 3 into the equation y = 4x2
11. A 60-m long, 9.4 mm diameter rope hangs freely from a ledge. The linear density of the
rope is 55g/m. How much work is needed to lift the entire rope to the ledge?
7
Solution:Before directly getting into the solution, note that the approach to this problem
will be similar to that of the tank draining problem with the exception that we need not
worry about cross-sectional area as we are given linear density instead of density.
Assign a coordinate system to the rope with the bottom tip of the rope at the origin and
the top end of the rope at the point (0,60). Our density is 0.055kg/m and the distance the
slice of the rope at height y travels is 60 − y. Thus
Z 60
W ork =
9.8(0.055)(60 − y) dy
0
60
1 2
= 9.8(0.055) 60y − y
2
0
1
2
= 9.8(0.055) 60(60) − (60)
2
1
=
9.8(0.055)(60)2 .
2
8