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.

 

volume ratio = (edge ratio)^3.

 

 

1200  = k * 2^3.  Solving for k we get

k = 1200 / 2^3 = 150.

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.

54 = k * 3^2 so

k = 54 / 3^2 = 6.

Our proportionality is

y = 6 x^2.

Nutshell Summary

which power function best fits data for swings per minute vs pendulum length

 

 

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.