Assignment - 12

course Mth163

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

012.

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Self-critique (if necessary):

ok

Self-critique Rating:

*********************************************

Question: `q001. Note that this assignment has 3 questions

If we know that y = k x^2, then if (x2/x1) = 7, what is (y2/y1)?

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Your solution:

y is equal to the squaring of x so y1 would be equal to the squaring of x1 and y2 would be equal to the squaring of x2 giving you the function:

y1 = kx1^2 and y2 = kx2^2 giving you:

y2 / y1 = (kx2^2) / (kx1^2) k divides out giving you:

y2 / y1 = x2^2 / x1^2 or (x2 / x1)^2

knowing that (x2 / x1) = 7 we have:

y2 / y1 = 7^2 = 49

Confidence rating: mostly

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Given Solution:

If y2 = k x2^2 and y1 = k x1^2, then y2 / y1 = (k x2^2) / ( k x1^2). Since k / k = 1 this is the same as

y2 / y1 = x2^2 / x1^2, which is the same as

y2 / y1 = (x2 / x1)^2.

In words this tells us if y to is proportional to the square of x, then the ratio of y2 to y1 is the same as the square of the ratio of x2 to x1.

Now if (x2 / x1) = 7, we see that y2 / y1 = (x2 / x1)^2 = 7^2 = 49.

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Self-critique (if necessary):

ok

Self-critique Rating:

*********************************************

Question: `q002. If we know that y = k x^3, then if (x2/x1) = 7, what is (y2/y1)?

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Your solution:

Doing the problem same as before except we are cubing the x’s:

y2 / y1 = (kx2^3) / ( kx1^3) = (x2 / x1)^3

given (x2 / x1) = 7 then:

y2 / y1 = 7^3 = 343

Confidence rating: positive

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

If y2 = k x2^3 and y1 = k x1^3, then y2 / y1 = (k x2^3) / ( k x1^3). Since k / k = 1 this is the same as

y2 / y1 = x2^3 / x1^3, which is the same as

y2 / y1 = (x2 / x1)^3.

In words this tells us if y to is proportional to the cube of x, then the ratio of y2 to y1 is the same as the cube of the ratio of x2 to x1.

Now if (x2 / x1) = 7, we see that y2 / y1 = (x2 / x1)^3 = 7^3 = 343.

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Self-critique (if necessary):

ok

Self-critique Rating:

*********************************************

Question: `q003. If we know that y = k x^-2, then if (x2/x1) = 64, what is (y2/y1)?

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Your solution:

giving the same problem as the first on except x^-2 we would have:

y2 / y1 = (kx2^-2) / ( kx1^-2) = (x2 / x1)^-2 if(x2 / x1) = 64 we have:

y2 / y1 = 64 ^ -2 = 1/64^2 = 1 / 4096

Confidence rating: positive

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

If y2 = k x2^-2 and y1 = k x1^-2, then y2 / y1 = (k x2^-2) / ( k x1^-2). Since k / k = 1 this is the same as

y2 / y1 = x2^-2 / x1^-2, which is the same as

y2 / y1 = (x2 / x1)^-2, which is the same as

1 / (x2 / x1)^2, which gives us

(x1 / x2)^2.

So if y = k x^-2, then (y2 / y1) = (x1 / x2)^2.(

In words this tells us if y to is inversely proportional to the square of x, then the ratio of y2 to y1 is the same as the square of the ratio of x1 to x2 (note that this is a reciprocal ratio).

Now if (x2 / x1) = 64, we see that y2 / y1 = (x1 / x2)^2 = (1/64)^2 = 1/ 4096.

012. `query 12

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Self-critique (if necessary):

ok

Self-critique Rating:

*********************************************

Question: `qproblem 1. box of length 30 centimeters capacity 50 liters .

What is the proportionality for this situation, what is the proportionality constant and what is the specific equation that relates capacity y to length x?

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Your solution:

the function for proportionality is v = kx^3 ( y = v and y = capacity) then:

50 = k (30)^3 ……50 = k(27000)…….k = 50 / 27000……k = .00185 approx.

giving you a function for y: y = (.00185)x^3

Confidence rating: positive

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Given Solution:

** The proportionality for volume is y = k x^3, where y is capacity in liters when x is length in cm.

Since y = 50 when x = 30 we have

50 = k * 30^3 so that

k = 50 / (30^3) = 50 / 27,000 = 1/540 = .0019 approx.

Thus y = (1/540) * x^3. **

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Self-critique (if necessary):

ok, I understand were you got (1/540) but why would you use that instead of using the solved part of that and does it matter which one you use?

In nearly any practical application 1/540 is no different from .00185. However if the numbers are considered to be exact, 1/540 would be more accurate than .00185.

Self-critique Rating:

*********************************************

Question: `qWhat is the storage capacity of a box of length 100 centimeters?

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Your solution:

using the equation I have received above:

y = .00185 * x^3 and x = length

y = .00185 * 100^3…..y = .00185 * 1,000,000……y = 1850

Confidence rating: positive

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Given Solution:

** The proportionality is y = 1/540 * x^3 so if x = 100 we have

y = 1/540 * 100^3 = 1900 approx.

A 100 cm box geometrically similar to the first will therefore contain about 1900 liters. **

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Self-critique (if necessary):

ok, it should be in liters

Self-critique Rating:

*********************************************

Question: `qWhat length is required to obtain a storage capacity of 100 liters?

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Your solution:

y = .00185 x^3 with y being the capacity

100 = .0085 x^3…….x^3 = .0085 / 100…..x^3 = I get an error on my calculator.

going back to the last problem and using your y = 1/540*x^3:

100 = 1 / 540 * x^3……x^3 = 54000^ (1/3) ……x = 37.8

Confidence rating: unsure

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Given Solution:

** If y = 100 then we have

100 = (1/540) * x^3 so that

x^3 = 540 * 100 = 54,000.

Thus x = (54,000)^(1/3) = 185 approx.

The length of a box that will store 100 liters is thus about 185 cm. **

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Self-critique (if necessary):

By using your function I have everything right except the last step which I don’t understand how you received that answer. My calculator gives me 37.8 rounded

Your calculation is correct. It appears I somehow took the 1/3 power of 54 000 000. That in itself isn't bad, but I should certainly have recognized that the result didn't make sense.

Self-critique Rating:

*********************************************

Question: `qHow long would a box have to be in order to store all the water in a swimming pool which contains 450 metric tons of water? A metric ton contains 1000 liters of water.

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Your solution:

First you have to calculate how many liters you have: 450 * 1000 = 450,000 liters of water

using your function y = 1/540 * x^3, with y = capacity of 450,000

450,000 = 1/540 * x^3……x^3 = 243,000,000…….x = 624

Confidence rating: positive

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Given Solution:

** 450 metric tons is 450 * 1000 liters = 450,000 liters. Thus y = 450,000 so we have the equation

540,000 = (1/540) x^3

which we solve in a manner similar to the preceding question to obtain

x = 624, so that the length of the box is 624 cm. **

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Self-critique (if necessary):

ok. length should be recorded in cm.

Self-critique Rating:

*********************************************

Question: `qproblem 2. cleaning service scrub the surface of the Statute of width of finger .8 centimeter vs. 20-centimeter width actual model takes .74 hours.

How long will it take to scrub the entire statue?

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Your solution:

using t = k * x^2 t = time which is .74 and x is cubed because of it being surface and being the width as .8 cm

.74 = k * .8 ^2 …… .74 = k * .64…..k = .74 / .64 = 1.16 rounded

now to solve for the whole statue y = k * x^2 x = 20 cm…the entire length of statue.

y = 1.16 * 20^2…… y = 464 hours

Confidence rating:

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Given Solution:

** y = k x^2 so

.74 = k * .8^2. Solving for k we obtain

k = 1.16 approx. so

y = 1.16 x^2.

The time to scrub the actual statue will be

y = 1.16 x^2 with x = 20.

We get

y = 1.16 * 20^2 = 460 approx..

It should take 460 hrs to scrub the entire statue. **

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Self-critique (if necessary):

ok. but I went back and calculated the last step again and still received 464, unless you are rounding it to 460.

Self-critique Rating:

*********************************************

Question: `qproblem 3. illumination 30 meters is 5 foot-candles. What is the proportionality for this situation, what is the value of the proportionality constant and what equation relates the illumination y to the distance x?

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Your solution:

y = k * x^-2 where y = illumination in ft – candles and x = distance in meters.

5 = k * 30^-2 ……..5 = k * (1/30^2)……k = 5 * 900…..k = 4500

we get y = 4500* x^2

Confidence rating: mostly

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Given Solution:

** The proportionality should be

y = k x^-2,

where y is illumination in ft candles and x the distance in meters.

We get

5 = k * 30^-2, or

5 = k / 30^2 so that

k = 5 * 30^2 = 4500.

Thus y = 4500 x^-2.

We get an illumination of 10 ft candles when y = 10. To find x we solve the equation

10 = 4500 / x^2. Multiplying both sides by x^2 we get

10 x^2 = 4500. Dividing both sides by 10 we have

x^2 = 4500 / 10 = 450 and

x = sqrt(450) = 21 approx..

For illumination 1000 ft candles we solve

1000 = 4500 / x^2,

obtaining solution x = 2.1 approx..

We therefore conclude that the comfortable range is from about x = 2.1 meters to x = 21 meters. **

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Self-critique (if necessary):

ok, to show a constant proportionality I need to add some y values into this equation.

Self-critique Rating:

*********************************************

Question: `qproblem 5.

Does a 3-unit cube weigh more or less than 3 times a 1-unit cube? Why is this?

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Your solution:

a 1 unit cube consist of 9 cubes and three layers of that will be 27 cubes (9 + 9 + 9 = 27)same as if you had three times 1 unit cube ( 3 * 9 = 27) which would give you the same weight.

Confidence rating: positive

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Given Solution:

** A 3-unit cube is equivalent to 3 layers of 1-unit cubes, each layer consisting of three rows with 3 cubes in each row.

Thus a 3-unit cube is equivalent to 27 1-unit cubes.

If the weight of a 1-unit cube is 35 lbs then we have the following:

Edge equiv. # of weight

Length 1-unit cubes

1 1 35

2 4 4 * 35 = 140

3 9 9 * 35 = 315

4 16 16 * 35 = 560

5 25 25 * 35 = 875.

Each weight is obtained by multiplying the equivalent number of 1-unit cubes by the 35-lb weight of such a cube. **

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Self-critique (if necessary):

ok, even though I was right you wanted me to show the proportionality and show it was equivalent to 1 unit cubes by the 35 lb weight of that cube.

Self-critique Rating:

*********************************************

Question: `qproblem 6. Give the numbers of 1-unit squares required to cover 6-, 7-, 8-, 9- and 10-unit square, and also an n-unit square.

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Your solution:

6 unit square has 6 squares per one unit with 6 rows each giving you 36 one square units.

7 unit square has 7 squares per units with 7 rows each giving you 49 one square units.

8 units squared has 8 squares per one unit with 8 rows each giving you 64 one square units.

9 units squared has 9 squares per one unit with 9 rows each giving you 81 one square units.

10 units squared has 10 squares per one unit and 10 rows each giving you 100 one squared units.

with n squared units you would have n squares per unit with n rows each giving you n * n.

Confidence rating: positive

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Given Solution:

**

To cover a 6-unit square requires 6 rows each containing 6 1-unit squares for a total of 36 one-unit squares.

To cover a 7-unit square requires 7 rows each containing 7 1-unit squares for a total of 49 one-unit squares.

To cover a 8-unit square requires 8 rows each containing 8 1-unit squares for a total of 64 one-unit squares.

To cover a 9-unit square requires 9 rows each containing 9 1-unit squares for a total of 81 one-unit squares.

To cover a 10-unit square requires 10 rows each containing 10 1-unit squares for a total of 100 one-unit squares.

To cover an n-unit square requires n rows each containing n 1-unit squares for a total of n*n=n^2 one-unit squares. **

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Self-critique (if necessary):

ok, could have wrote n * n as n^2 on unit squares

Self-critique Rating:

*********************************************

Question: `qproblem 8. Relating volume ratio to ratio of edges.

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Your solution:

a volume ratio of a cube with an edge of 5 units and one of 3 units will give you (5/3)^3 = 4.629 or approx.4.6

the ratio of the edge is 5/3 = 1.666

to compare the two we need to have:

volume ratio = edge ratio^3….. 4.6 = 1.666 ^ 3 ….4.6 = 4.6

Confidence rating: positive

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Given Solution:

** right idea but you have the ratio upside down.

The volume ratio of a 5-unit cube to a 3-unit cube is (5/3)^3 = 125 / 27 = 4.7 approx..

The edge ratio is 5/3 = 1.67 approx.

VOlume ratio = edgeRatio^3 = 1.678^3 = 4.7 approx..

From this example we see how volume ratio = edgeRatio^3.

If two cubes have edges 12.7 and 2.3 then their edge ratio is 12.7 / 2.3 = 5.5 approx..

The corresponding volume ratio would therefore be 5.5^3 = 160 approx..

If edges are x1 and x2 then edgeRatio = x2 / x1. This results in volume ratio

volRatio = edgeRatioo^3 = (x2 / x1)^3. **

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Self-critique (if necessary):

so you take the edge ratio 5/3 = 1.67 and cube that for your volume ratio; 1.67^3 = 4.77 approx.

Self-critique Rating:

*********************************************

Question: `qproblem 9. Relating y and x ratios for a cubic proportionality.

What is the y value corresponding to x = 3 and what is is the y value corresponding to x = 5?

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Your solution:

if you have y = k x^3 and you want the y value corresponding with x values of 3 & 5.

we can say 3 = x1 and 5 = x2, then you would have:

y1 = k * x1^3 and y2 = k * x2^3

to show the proportionality of y and x :

y2 / y1 = (k * x2 ^3) / (k * x1^3) the k cancels itself out or is equal to one giving you:

y2 / y1 = x2^3 / x1 ^3 or (x2 / x1) ^3

Confidence rating: positive

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Given Solution:

** If y = a x^3 then

if x1 = 3 we have y1 = a * 3^3 and

if x2 = 5 we have y2 = a * 5^3.

This gives us ratio y2 / y1 = (a * 5^3) / (a * 3^3) = (a / a) * (5^3 / 3^3) = 1 * 125 / 27 = 125 / 27.

In general if y1 = a * x1^3 and y2 = a * x2^3 we have

}

y2 / y1 = (a x2^3) / (a x1^3) = (a / a) * (x2^3 / x1^3) = (x2/x1)^3.

This tells you that to get the ratio of y values you just cube the ratio of the x values. **

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Self-critique (if necessary):

ok

Self-critique Rating:

*********************************************

Question: `qproblem 10. Generalizing to y = x^p.

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Your solution:

If we have the power = 2 function then we have y = f(x) = x^-2 then we can perform a similar analysis as the problem right before:

x1 = 3 and x2 = 5 and y1 =k* x1^-2 and y2 =k* x2^-2 giving you the proportionality of:

y2 / y1 = ( k*5^-2) / (k*3 ^ -2) or (5/3)^-2 = 1 / 2.79

Confidence rating: positive

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** If y = a x^2 then

y2 / y1 = (a x2^2) / (a x1^2) = (a / a) * (x2^2 / x1^2) = (x2/x1)^2.

This tells you that to get the ratio of y values you just square the ratio of the x values.

If y = f(x) = a x^p then

y1 = f(x1) = a x1^p and

y2 = f(x1) = a x2^p so that

y2 / y1 = f(x2) / f(x1) = (a x2^p) / (a x1^p) = (a / a) ( x2^p / x1^p ) = x2^p / x1^p = (x2 / x1)^p. **

Add comments on any surprises or insights you experienced as a result of this assignment.

this was a pretty easy assignment to comprehend, I did like the ratio stuff looks like it will come in handy ** this stuff is very important in most areas of study **

****ok, I used the x values given in our notes from problem 10 but derived the same concept.

"

&#This looks good. See my notes. Let me know if you have any questions. &#

Assignment - 12

course Mth163

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

012.

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Self-critique (if necessary):

ok

Self-critique Rating:

*********************************************

Question: `q001. Note that this assignment has 3 questions

If we know that y = k x^2, then if (x2/x1) = 7, what is (y2/y1)?

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Your solution:

y is equal to the squaring of x so y1 would be equal to the squaring of x1 and y2 would be equal to the squaring of x2 giving you the function:

y1 = kx1^2 and y2 = kx2^2 giving you:

y2 / y1 = (kx2^2) / (kx1^2) k divides out giving you:

y2 / y1 = x2^2 / x1^2 or (x2 / x1)^2

knowing that (x2 / x1) = 7 we have:

y2 / y1 = 7^2 = 49

Confidence rating: mostly

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

If y2 = k x2^2 and y1 = k x1^2, then y2 / y1 = (k x2^2) / ( k x1^2). Since k / k = 1 this is the same as

y2 / y1 = x2^2 / x1^2, which is the same as

y2 / y1 = (x2 / x1)^2.

In words this tells us if y to is proportional to the square of x, then the ratio of y2 to y1 is the same as the square of the ratio of x2 to x1.

Now if (x2 / x1) = 7, we see that y2 / y1 = (x2 / x1)^2 = 7^2 = 49.

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Self-critique (if necessary):

ok

Self-critique Rating:

*********************************************

Question: `q002. If we know that y = k x^3, then if (x2/x1) = 7, what is (y2/y1)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Doing the problem same as before except we are cubing the x’s:

y2 / y1 = (kx2^3) / ( kx1^3) = (x2 / x1)^3

given (x2 / x1) = 7 then:

y2 / y1 = 7^3 = 343

Confidence rating: positive

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

If y2 = k x2^3 and y1 = k x1^3, then y2 / y1 = (k x2^3) / ( k x1^3). Since k / k = 1 this is the same as

y2 / y1 = x2^3 / x1^3, which is the same as

y2 / y1 = (x2 / x1)^3.

In words this tells us if y to is proportional to the cube of x, then the ratio of y2 to y1 is the same as the cube of the ratio of x2 to x1.

Now if (x2 / x1) = 7, we see that y2 / y1 = (x2 / x1)^3 = 7^3 = 343.

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Self-critique (if necessary):

ok

Self-critique Rating:

*********************************************

Question: `q003. If we know that y = k x^-2, then if (x2/x1) = 64, what is (y2/y1)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

giving the same problem as the first on except x^-2 we would have:

y2 / y1 = (kx2^-2) / ( kx1^-2) = (x2 / x1)^-2 if(x2 / x1) = 64 we have:

y2 / y1 = 64 ^ -2 = 1/64^2 = 1 / 4096

Confidence rating: positive

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

If y2 = k x2^-2 and y1 = k x1^-2, then y2 / y1 = (k x2^-2) / ( k x1^-2). Since k / k = 1 this is the same as

y2 / y1 = x2^-2 / x1^-2, which is the same as

y2 / y1 = (x2 / x1)^-2, which is the same as

1 / (x2 / x1)^2, which gives us

(x1 / x2)^2.

So if y = k x^-2, then (y2 / y1) = (x1 / x2)^2.(

In words this tells us if y to is inversely proportional to the square of x, then the ratio of y2 to y1 is the same as the square of the ratio of x1 to x2 (note that this is a reciprocal ratio).

Now if (x2 / x1) = 64, we see that y2 / y1 = (x1 / x2)^2 = (1/64)^2 = 1/ 4096.

012. `query 12

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Self-critique (if necessary):

ok

Self-critique Rating:

*********************************************

Question: `qproblem 1. box of length 30 centimeters capacity 50 liters .

What is the proportionality for this situation, what is the proportionality constant and what is the specific equation that relates capacity y to length x?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

the function for proportionality is v = kx^3 ( y = v and y = capacity) then:

50 = k (30)^3 ……50 = k(27000)…….k = 50 / 27000……k = .00185 approx.

giving you a function for y: y = (.00185)x^3

Confidence rating: positive

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** The proportionality for volume is y = k x^3, where y is capacity in liters when x is length in cm.

Since y = 50 when x = 30 we have

50 = k * 30^3 so that

k = 50 / (30^3) = 50 / 27,000 = 1/540 = .0019 approx.

Thus y = (1/540) * x^3. **

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Self-critique (if necessary):

ok, I understand were you got (1/540) but why would you use that instead of using the solved part of that and does it matter which one you use?

In nearly any practical application 1/540 is no different from .00185. However if the numbers are considered to be exact, 1/540 would be more accurate than .00185.

Self-critique Rating:

*********************************************

Question: `qWhat is the storage capacity of a box of length 100 centimeters?

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Your solution:

using the equation I have received above:

y = .00185 * x^3 and x = length

y = .00185 * 100^3…..y = .00185 * 1,000,000……y = 1850

Confidence rating: positive

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Given Solution:

** The proportionality is y = 1/540 * x^3 so if x = 100 we have

y = 1/540 * 100^3 = 1900 approx.

A 100 cm box geometrically similar to the first will therefore contain about 1900 liters. **

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Self-critique (if necessary):

ok, it should be in liters

Self-critique Rating:

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Question: `qWhat length is required to obtain a storage capacity of 100 liters?

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Your solution:

y = .00185 x^3 with y being the capacity

100 = .0085 x^3…….x^3 = .0085 / 100…..x^3 = I get an error on my calculator.

going back to the last problem and using your y = 1/540*x^3:

100 = 1 / 540 * x^3……x^3 = 54000^ (1/3) ……x = 37.8

Confidence rating: unsure

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Given Solution:

** If y = 100 then we have

100 = (1/540) * x^3 so that

x^3 = 540 * 100 = 54,000.

Thus x = (54,000)^(1/3) = 185 approx.

The length of a box that will store 100 liters is thus about 185 cm. **

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Self-critique (if necessary):

By using your function I have everything right except the last step which I don’t understand how you received that answer. My calculator gives me 37.8 rounded

Your calculation is correct. It appears I somehow took the 1/3 power of 54 000 000. That in itself isn't bad, but I should certainly have recognized that the result didn't make sense.

Self-critique Rating:

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Question: `qHow long would a box have to be in order to store all the water in a swimming pool which contains 450 metric tons of water? A metric ton contains 1000 liters of water.

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Your solution:

First you have to calculate how many liters you have: 450 * 1000 = 450,000 liters of water

using your function y = 1/540 * x^3, with y = capacity of 450,000

450,000 = 1/540 * x^3……x^3 = 243,000,000…….x = 624

Confidence rating: positive

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Given Solution:

** 450 metric tons is 450 * 1000 liters = 450,000 liters. Thus y = 450,000 so we have the equation

540,000 = (1/540) x^3

which we solve in a manner similar to the preceding question to obtain

x = 624, so that the length of the box is 624 cm. **

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Self-critique (if necessary):

ok. length should be recorded in cm.

Self-critique Rating:

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Question: `qproblem 2. cleaning service scrub the surface of the Statute of width of finger .8 centimeter vs. 20-centimeter width actual model takes .74 hours.

How long will it take to scrub the entire statue?

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Your solution:

using t = k * x^2 t = time which is .74 and x is cubed because of it being surface and being the width as .8 cm

.74 = k * .8 ^2 …… .74 = k * .64…..k = .74 / .64 = 1.16 rounded

now to solve for the whole statue y = k * x^2 x = 20 cm…the entire length of statue.

y = 1.16 * 20^2…… y = 464 hours

Confidence rating:

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Given Solution:

** y = k x^2 so

.74 = k * .8^2. Solving for k we obtain

k = 1.16 approx. so

y = 1.16 x^2.

The time to scrub the actual statue will be

y = 1.16 x^2 with x = 20.

We get

y = 1.16 * 20^2 = 460 approx..

It should take 460 hrs to scrub the entire statue. **

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Self-critique (if necessary):

ok. but I went back and calculated the last step again and still received 464, unless you are rounding it to 460.

Self-critique Rating:

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Question: `qproblem 3. illumination 30 meters is 5 foot-candles. What is the proportionality for this situation, what is the value of the proportionality constant and what equation relates the illumination y to the distance x?

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Your solution:

y = k * x^-2 where y = illumination in ft – candles and x = distance in meters.

5 = k * 30^-2 ……..5 = k * (1/30^2)……k = 5 * 900…..k = 4500

we get y = 4500* x^2

Confidence rating: mostly

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Given Solution:

** The proportionality should be

y = k x^-2,

where y is illumination in ft candles and x the distance in meters.

We get

5 = k * 30^-2, or

5 = k / 30^2 so that

k = 5 * 30^2 = 4500.

Thus y = 4500 x^-2.

We get an illumination of 10 ft candles when y = 10. To find x we solve the equation

10 = 4500 / x^2. Multiplying both sides by x^2 we get

10 x^2 = 4500. Dividing both sides by 10 we have

x^2 = 4500 / 10 = 450 and

x = sqrt(450) = 21 approx..

For illumination 1000 ft candles we solve

1000 = 4500 / x^2,

obtaining solution x = 2.1 approx..

We therefore conclude that the comfortable range is from about x = 2.1 meters to x = 21 meters. **

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Self-critique (if necessary):

ok, to show a constant proportionality I need to add some y values into this equation.

Self-critique Rating:

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Question: `qproblem 5.

Does a 3-unit cube weigh more or less than 3 times a 1-unit cube? Why is this?

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Your solution:

a 1 unit cube consist of 9 cubes and three layers of that will be 27 cubes (9 + 9 + 9 = 27)same as if you had three times 1 unit cube ( 3 * 9 = 27) which would give you the same weight.

Confidence rating: positive

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Given Solution:

** A 3-unit cube is equivalent to 3 layers of 1-unit cubes, each layer consisting of three rows with 3 cubes in each row.

Thus a 3-unit cube is equivalent to 27 1-unit cubes.

If the weight of a 1-unit cube is 35 lbs then we have the following:

Edge equiv. # of weight

Length 1-unit cubes

1 1 35

2 4 4 * 35 = 140

3 9 9 * 35 = 315

4 16 16 * 35 = 560

5 25 25 * 35 = 875.

Each weight is obtained by multiplying the equivalent number of 1-unit cubes by the 35-lb weight of such a cube. **

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Self-critique (if necessary):

ok, even though I was right you wanted me to show the proportionality and show it was equivalent to 1 unit cubes by the 35 lb weight of that cube.

Self-critique Rating:

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Question: `qproblem 6. Give the numbers of 1-unit squares required to cover 6-, 7-, 8-, 9- and 10-unit square, and also an n-unit square.

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Your solution:

6 unit square has 6 squares per one unit with 6 rows each giving you 36 one square units.

7 unit square has 7 squares per units with 7 rows each giving you 49 one square units.

8 units squared has 8 squares per one unit with 8 rows each giving you 64 one square units.

9 units squared has 9 squares per one unit with 9 rows each giving you 81 one square units.

10 units squared has 10 squares per one unit and 10 rows each giving you 100 one squared units.

with n squared units you would have n squares per unit with n rows each giving you n * n.

Confidence rating: positive

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Given Solution:

**

To cover a 6-unit square requires 6 rows each containing 6 1-unit squares for a total of 36 one-unit squares.

To cover a 7-unit square requires 7 rows each containing 7 1-unit squares for a total of 49 one-unit squares.

To cover a 8-unit square requires 8 rows each containing 8 1-unit squares for a total of 64 one-unit squares.

To cover a 9-unit square requires 9 rows each containing 9 1-unit squares for a total of 81 one-unit squares.

To cover a 10-unit square requires 10 rows each containing 10 1-unit squares for a total of 100 one-unit squares.

To cover an n-unit square requires n rows each containing n 1-unit squares for a total of n*n=n^2 one-unit squares. **

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Self-critique (if necessary):

ok, could have wrote n * n as n^2 on unit squares

Self-critique Rating:

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Question: `qproblem 8. Relating volume ratio to ratio of edges.

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Your solution:

a volume ratio of a cube with an edge of 5 units and one of 3 units will give you (5/3)^3 = 4.629 or approx.4.6

the ratio of the edge is 5/3 = 1.666

to compare the two we need to have:

volume ratio = edge ratio^3….. 4.6 = 1.666 ^ 3 ….4.6 = 4.6

Confidence rating: positive

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Given Solution:

** right idea but you have the ratio upside down.

The volume ratio of a 5-unit cube to a 3-unit cube is (5/3)^3 = 125 / 27 = 4.7 approx..

The edge ratio is 5/3 = 1.67 approx.

VOlume ratio = edgeRatio^3 = 1.678^3 = 4.7 approx..

From this example we see how volume ratio = edgeRatio^3.

If two cubes have edges 12.7 and 2.3 then their edge ratio is 12.7 / 2.3 = 5.5 approx..

The corresponding volume ratio would therefore be 5.5^3 = 160 approx..

If edges are x1 and x2 then edgeRatio = x2 / x1. This results in volume ratio

volRatio = edgeRatioo^3 = (x2 / x1)^3. **

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Self-critique (if necessary):

so you take the edge ratio 5/3 = 1.67 and cube that for your volume ratio; 1.67^3 = 4.77 approx.

Self-critique Rating:

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Question: `qproblem 9. Relating y and x ratios for a cubic proportionality.

What is the y value corresponding to x = 3 and what is is the y value corresponding to x = 5?

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Your solution:

if you have y = k x^3 and you want the y value corresponding with x values of 3 & 5.

we can say 3 = x1 and 5 = x2, then you would have:

y1 = k * x1^3 and y2 = k * x2^3

to show the proportionality of y and x :

y2 / y1 = (k * x2 ^3) / (k * x1^3) the k cancels itself out or is equal to one giving you:

y2 / y1 = x2^3 / x1 ^3 or (x2 / x1) ^3

Confidence rating: positive

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Given Solution:

** If y = a x^3 then

if x1 = 3 we have y1 = a * 3^3 and

if x2 = 5 we have y2 = a * 5^3.

This gives us ratio y2 / y1 = (a * 5^3) / (a * 3^3) = (a / a) * (5^3 / 3^3) = 1 * 125 / 27 = 125 / 27.

In general if y1 = a * x1^3 and y2 = a * x2^3 we have

}

y2 / y1 = (a x2^3) / (a x1^3) = (a / a) * (x2^3 / x1^3) = (x2/x1)^3.

This tells you that to get the ratio of y values you just cube the ratio of the x values. **

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Self-critique (if necessary):

ok

Self-critique Rating:

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Question: `qproblem 10. Generalizing to y = x^p.

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Your solution:

If we have the power = 2 function then we have y = f(x) = x^-2 then we can perform a similar analysis as the problem right before:

x1 = 3 and x2 = 5 and y1 =k* x1^-2 and y2 =k* x2^-2 giving you the proportionality of:

y2 / y1 = ( k*5^-2) / (k*3 ^ -2) or (5/3)^-2 = 1 / 2.79

Confidence rating: positive

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Given Solution:

** If y = a x^2 then

y2 / y1 = (a x2^2) / (a x1^2) = (a / a) * (x2^2 / x1^2) = (x2/x1)^2.

This tells you that to get the ratio of y values you just square the ratio of the x values.

If y = f(x) = a x^p then

y1 = f(x1) = a x1^p and

y2 = f(x1) = a x2^p so that

y2 / y1 = f(x2) / f(x1) = (a x2^p) / (a x1^p) = (a / a) ( x2^p / x1^p ) = x2^p / x1^p = (x2 / x1)^p. **

Add comments on any surprises or insights you experienced as a result of this assignment.

this was a pretty easy assignment to comprehend, I did like the ratio stuff looks like it will come in handy ** this stuff is very important in most areas of study **

****ok, I used the x values given in our notes from problem 10 but derived the same concept.

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&#This looks good. See my notes. Let me know if you have any questions. &#