A.P. Calculus 2 - The Glass
You will find from home a glass or vase that will hold liquid. The best type of glass is a wine
glass. For more than basic credit, you wish to stay away from a glass with straight edges. The
more “complicated” the glass, the better this assignment will work.
Simple Glass - too easy
Still Simple but better
The type of glass you want
Your job is to find 4 pieces of information:
1)
2)
3)
4)
Find the volume that the object will hold.
Find the inner surface area of the glass.
Find the amount of work need to empty the object by pumping it over the edge.
If water is being poured at the rate of 1 cubic inch per second, at what rate is the water
level rising when the glass contains half of its volume?
First, you may NOT use a measuring cup to determine the volume. You must generate
equations to model the shape of the glass. For the middle glass above, you can generate the side
of the glass using a line. To model the glass on the right, you will need a piecewise function.
I will not tell you how to generate equations for the graph. However, somehow, you need to
determine the radius of the glass at regularly spaced height intervals. You may use regression
on your calculator to determine the equation.
Once you generate your equations, you need to flip it about the x (easier) or y-axis to generate a
solid. Then calculate the volume in cubic inches. Then, you will use a measuring cup to
measure the volume and determine the difference between the actual volume and the volume
generated using your equations(s).
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Stu Schwartz
To find the surface area, you will use the inner surface area since some glasses have a fluted
outer edge which would complicate the outer surface calculations.
Use your work formulas to generate the work done to empty the glass.
To find the related rate, you need to determine the rate of change of the volume. Remember
what rate of change means.
The glass you should use should not be made of crystal. It should not be expensive. You may
want to go to a store in the mall and purchase a cheap plastic glass of your liking.
What you will hand in:
• The glass (which I will return to you). Make sure your name is taped to the inside of it.
• A paragraph describing how you generated the equations. Be sure you describe what you did.
If you generated a table of values, show that table and you you determined them.
• The equations you generated for one side of the glass.
• The formula you used for the volume and the answers you found.
• The true volume of the glass and the difference between the true area and the volume you
found with your equations.
• The formula for the work done and the answer.
• The analysis to the related rates problem and the answer.
How you will be graded:
The assignment will be worth 20 points times the difficulty level of the glass (which is
determined by me).
• A very simple glass has difficulty level 1 - maximum of 20 points
• A glass with straight sides (not vertical) has difficult level 1.25 - maximum of 25 points
• A glass with a piece-wise function sides has difficulty level 1.5 - maximum of 30 points
• A glass with more than 2 pieces has difficulty level 2 - maximum of 40 points
Due date: ________________________________________________
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Stu Schwartz
Volume of a Vase
Height (x)
0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
2.25
2.5
2.75
3
3.25
3.5
3.75
4
4.25
4.5
4.75
5
5.25
5.5
5.75
6
Circumference
9.75
10
12
13
13.5
14
14.5
15
15.5
15.75
16.5
16.75
17
17.25
17
16.5
16
15.5
15
13.5
10
9.5
10
10.5
11
Radius (y)
1.48926201
1.52905078
1.84736093
2.00651601
2.08609355
2.16567109
2.24524862
2.32482616
2.4044037
2.44419247
2.56355878
2.60334755
2.64313632
2.68292509
2.64313632
2.56355878
2.48398124
2.4044037
2.32482616
2.08609355
1.52905078
1.44947324
1.52905078
1.60862831
1.68820585
Graph of Vase
3
Radius of Vase
2.5
2
1.5
1
0.5
0
0
1
2
3
4
5
6
Height of Vase
Using a ruler, I measured the height of the vase from top to bottom. I neglected the solid bottom portion of the
vase in my measurement. Next, I measured the circumference of the glass in .25 increments. I divided the
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Stu Schwartz
circumference by 2π to find the radius. Then I subtracted the thickness of the vase (.0625 in.) to find the inner
radius of the vase. I designated the height as the x-value and the radius as the y-value. Finally, I fit three curves
to those points:
Y1 = ".129674x 2 + .779413x + 1.426891,
0 # x # 4.75
Y2 = 2.546479x 2 " 26.801692x + 71.912281,
Y3 = .318310x " .221655,
4.75 < x # 5.5
5.5 < x # 6
Volume of the vase
!
4.75
5.50
2
V = " # Y1 dx + "
0
V = 90.536 in
#Y
2
2
4.75
6
3
dx + " # Y1 dx
5.50
3
"
in 3 %
3
True Volume = 5.125 cups $14.4375
' = 73.992 in
cup &
#
Error = 16.544 in3
Percentage error = 18.271%
!
Surface Area!
b
$ # '2
Surface Area = 2" * r 1+ &Yi ) dx
% (
a
2
b
$ #'
2" * ri 1+ &Yi ) dx
% (
a
! #
Y1 = .259348x + .778413
#
Y2 = .2.092958x + 26.801692
#
Y3 = .318310
r1 = Y1
r2 = Y2
r3 = Y3
4.75
5.50
6
$ # '2
$ # '2
$ # '2
S.A. = 2" * Y1 1+ &Y1 ) dx + 2" * Y2 1+ &Y2 ) dx + 2" * Y3 1+ &Y3 ) dx
% (
% (
% (
0
4.75
5.50
S.A. = 90.231 in 2
Work
!
Weight Density = 62.4
4, 75
2
lb
lb
= .0361 3
3
ft
in
5.5
2
6.0
2
W = .0361 " Y1 (6 # x ) dx + .0361 " Y2 (6 # x ) dx + .0361 " Y3 (6 # x ) dx
0
4.75
5.5
W = 3.209
! in lbs.
!
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Stu Schwartz
Related Rates
V = 90.536 in3 so \F(1,2) V = 45.258 in3
h
45.268 =
"Y
2
1
dx
0
Solve graphically
h
(
h = 2.9157
)
2
V = " $ #.129674x 2 + .779413x + 1.426891 dx
0
!
(
2
dV d % h
= '" $ #.129674x 2 + .779413x + 1.426891 dx*
dt dt & 0
)
(
)
2 dh
dV
= " #.129674h 2 + .779413h + 1.426891
dt
dt
(
)
dh
1
=
2
dt
+
.2
"- #.129674(2.9157) + .779413(2.9157) + 41.4268910
,
/
dh
in
= .0472
dt
sec
!
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Stu Schwartz