Assignment 12b

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course mth 163

6/26 11am

012. `query 12*********************************************

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:

X= 30, y= 50, 50 = k 30^3, 50 = 27000k

50 / 27000 = .00185, so k = .00185 or 1/540

The function would be y = .00185x^3

confidence rating #$&*: 3

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

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Your solution:

Y = .00185 (100) ^3 = 1850, The storage capacity of a 100 cm box would be 1850 liters.

confidence rating #$&*: 3

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

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Your solution:

100 = .00185x ^3 or 100 = 1/540 x^3, we then need to solve the equation

- x^3 = 540 *100 = 54000, this means x = 54000 ^1/3 = 37.797

The box that will hold 100 liters is almost 38 cm.

confidence rating #$&*: 2

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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) = 38 approx.

The length of a box that will store 100 liters is thus about 38 cm. **

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Question: `qHow long would a geometrically similar 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:

450*1000 = 450000

X^3 = 540 * 450000 = 243000000

So x = 243000000 ^3 = 624, The length would be 624cm.

confidence rating #$&*: 3

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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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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:

.74 = .8^2k, .74 = .64k solve for k to get k = 1.16

The equation to solve for the time would be 1.16 (20) ^2 = 464 this is how long it would take to scrub the entire statue.

confidence rating #$&*: 3

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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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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:

The proportion would be y = k x^-2

5 = k 30^-2, 5 = .00111k solve for k = 4500

So the equation would be y =4500 x^-2

confidence rating #$&*: 2

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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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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:

It weighs more than 3 times a 1 unit cube because you have 3 layers of cubes with 9 cubes on each layer.

confidence rating #$&*: 3

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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 8 8 * 35 = 360

3 27 27 * 35 = 945

4 64 64 * 35 = 2240

5 125 125 * 35 = 4375

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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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:

For a 6 unit square there would be 36 cubes

For a 7 unit square there would be 49

For a 8 unit = 64

For a 9 unit = 81

For a 10 unit = 100

confidence rating #$&*: 3

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

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Your solution:

The volume ratio would be(5/3) ^3 = 4.629

The edge ratio would be 5/3 = 1.667

This means that the volume ratio is equal to the edge ratio ^3.

confidence rating #$&*: 3

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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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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:

When x1 = 3 we get y1 = a(3)^3

When x2 = 5 we get y2 = a(5)^3

confidence rating #$&*: 2

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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

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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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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:

When x1 = 3 we get y1 = a(3)^3

When x2 = 5 we get y2 = a(5)^3

confidence rating #$&*: 2

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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

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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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