class 051019
The edges of one cube are 2 feet long and the edges of another
are 3 feet long. If the cube was a die (a die would be one of a pair of
dice), the faces would have 1, 2, 3, 4, 5 and 6 dots on them, respectively.
Each cube has six square faces.
- How many solid 1-foot cubes would be required to
completely fill the cube with 2-foot edges?
- This would require 2 layers, each consisting of two rows
with two cubes in each row. That is, two layers of 4 cubes, or 8
cubes. If each weighs 150 lbs then the total weight would be 8 * 150
lbs = 1200 lbs.
- Assume that each 1-foot cube weighs 150 lb. How
much would the filled cube weigh?
- How many solid 1-foot cubes would be required to
completely fill the cube with 3-foot edges?
- This would require 3 layers, each consisting of three
rows with three cubes in each row. That is, three layers of 9 cubes,
or 27 cubes. If each weighs 150 lbs then the total weight would be 27
* 150 lbs = 4050 lbs.
- Assume again that each 1-foot cube weighs 150 lb.
How much would the filled cube weigh?
- What therefore would be the ratio of the weights of the
two cubes?
- The weights are 4050 lbs and 1200 lbs so the ratio is
4050 / 1200 = 3.375.
- What is the ratio of their edge lengths?
- The edge lengths are 3 and 2 so the ratio of edge lengths
is 3 / 2 = 1.5.
- How are these ratios related?
- i.e., how is 3.375 related to 1.5? The answer is
that 3.375 = 1.5 * 1.5 * 1.5 = 1.5^3 = 3.375.
- This illustrates one instance of the general fact that
volume ratio = (edge ratio)^3.
- What proportionality (i.e., y = k x, y = k x^2, y = k
x^-3, etc.) governs the relationship between the mass y and the edge length
x of a cube filled with 1-foot blocks?
- Since mass occupies volumes, which can be thought of as
filled with little cubes, the proportionality is y = k x^3.
- What is the value of the proportionality constant k?
- We know that when x = 2, the weight of the cube is y =
1200 lbs. We substitute to get
1200 = k * 2^3. Solving for k we get
k = 1200 / 2^3 = 150.
- What therefore is the specific proportionality
relationship between y and x (i.e., substitute your value of k into your
proportionality)?
- Since k = 150 the proportionality y = k x^3 becomes
y = 150 x^3.
This will then apply to any edge length x. For example
a cube with edge length 2.5 will have volume
y = 150 * 2.5^3 = 2400 lbs.
- How many 1-foot squares would be required to cover a
single face of the 2-foot cube? How many would be required to cover
the entire surface of that cube?
- We would require four 1-ft squares for each 2-ft-square
face.
- How many 1-foot squares would be required to cover a
single face of the 3-foot cube? How many would be required to cover
the entire surface of that cube?
- We would require nine 1-ft squares for each 3-ft-square
face. That would be a total of 9 * 6 = 54 squares for the entire
3-foot cube.
- What proportionality governs the relationship between y =
total surface area of a cube vs. x = edge length?
- y = k x^2 governs the relationship between surface area
and length for geometrically similar figures.
- What is the value of the proportionality constant k?
- When x = 3 we have y = 54. Plugging in to the
proportionality
54 = k * 3^2 so
k = 54 / 3^2 = 6.
Our proportionality is
y = 6 x^2.
- What therefore is the specific proportionality between
surface area and edge length for a cube?
Nutshell Summary
Mth163
I have a question about Swings per minute vs. pendulum length. I used fishing
line and a Christmas ordiament...is this ok?...My real question is what are you
asking me to do for question #4 For your number of swings vs. length data, use a
graphing cal. to obtain power function fits y=ax^p for p=-.3, -.4, -.5, -.6,
-.7, and plot the resulting functions with your data points. Which function fits
best? my data looks like this (1, 47) (2, 35) (3, 30) (4, 27) (5, 25)... Do you
want me to pluge in 1 for x and 47 for y and then p -.3 thru -.7...then slove
for a? I am really confused because if I did this then I would not have anything
to plot....Thanks.
I should have said to use DERIVE rather than your graphing calculator. You
would put the data into a line, using the syntax [ [1,47],[2,35],[3,30], etc. ].
Then for p = -.3 you would use the command FIT([x,ax^-.3], ##), where ## is the
number of the line containing the data.
Another way to see how the different functions work is to start with, say, y
= a x^-.3. Solving for a you would get a = y * x^.3. If we have the right model
we would expect a to remain nearly constant, so plugging in your x and y values
for the different points you would see if you get about the same value of a
every time. You would get
x y a = y x^.3
1 47 47
2 35 43.09005447
3 30 41.71167511
4 27 40.9243473
5 25 40.51641492
These values of a are not too much different, but they do show a definite
pattern of decreasing from one line to the next. So we try a different power.
Mth 163
There is a question in the homework which asks, What is the volRatio for a
pair of cubes with edges 12.7 and 2.3 (the ratio is [second volume]/[first
volume]?
Right. The volume of the first cube would be 12.7 * 12.7 * 12.7 = 2000 or
so, and the volume of the second would be 2.3 * 2.3 * 2.3, or about 11 or 12.
The ratio would be [second volume] / [ first volume].
Then it asks, What is the edge ratio^3? Does this mean the second volume
divided by the first volume then raise that answer to the power of three?
This means to find the edge ratio, and cube it. The edge ratio is 12.7 /
2.3. What is this number, and what is its cube?
I ask this because an earlier question led me to believe that we were to cube
the volumes first before dividing, or were we doing something different by
cubing the numbers first?
You cube the edge length to get the volume; we haven't encountered a case
where you cube the volume.
Using y = a x^-.6 we would get a = y * x^.6, and the following table would
result:
x y a = y x^.6
1 47 47
2 35 53.05007983
3 30 57.99546135
4 27 62.02971117
5 25 65.66319511
These values tend to increase from one x value to the next.
This would lead us to expect that p = .4 or p = .5 might give us more nearly
constant values of a.
Try y = a x^.4 and y = a x^.5 and see which gives you the most constant a
values.