Assignment 21

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course Mth 158

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.

021. `* 21

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Question: * 2.5.8 / 2.7.8 (was 2.6.6). y inv with sqrt(x), y = 4 when x = 9.

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Your solution:

y = k / sqrt(x)

4 = k / sqrt(9)

4 * 3 = k

k = 12

y = 12 / sqrt(x)

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confidence rating #$&*:

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Given Solution:

* * The inverse proportionality to the square root gives us y = k / sqrt(x).

y = 4 when x = 9 gives us

4 = k / sqrt(9) or

4 = k / 3 so that

k = 4 * 3 = 12.

The equation is therefore

y = 12 / sqrt(x). **

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Self-critique (if necessary):

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self-critique rating #$&*:

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Question: * 2.5.12 / 2.7.12 (was 2.6.10). z directly with sum of cube of x and square of y; z=1

and x=2 and y=3.

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Your solution:

z = k ( x^3 * y^2)

1 = k( 2^3 * 2^2)

1= k ( 8*9)

1 = k 72

k = 1/72

z = (1/72) (x^3 * y^2)

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confidence rating #$&*:

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Given Solution:

* * The proportionality is

z = k (x^3 + y^2).

If x = 2, y = 3 and z = 1 we have

1 = k ( 2^3 + 3^2) or

17 k = 1 so that

k = 1/17.

The proportionality is therefore

z = (x^3 + y^2) / 17. **

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Self-critique (if necessary): confused the term sum with product

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self-critique rating #$&*:

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Question: * 2.5.20 / 2.7.20 (was 2.6.20). Period varies directly with sqrt(length), const 2 pi

/ sqrt(32)

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Your solution:

T = k Sqrt(L)

T = ( 2 pi/ sqrt(32)) sqrt (L)

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confidence rating #$&*:

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Given Solution:

* * The equation is

T = k sqrt(L), with k = 2 pi / sqrt(32). So we have

T = 2 pi / sqrt(32) * sqrt(L). **

**** What equation relates period and length? ****

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Self-critique (if necessary):

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self-critique rating #$&*:

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Question: * 2.5.42 / 2.7.42 (was 2.7.34 (was 2.6.30). Resistance dir with lgth inversely with

sq of diam. 432 ft, 4 mm diam has res 1.24 ohms. **** What is the length of a wire with

resistance 1.44 ohms and diameter 3 mm? Give the details of your solution.

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Your solution:

R= kL/d^2

1.24 = (k 432) / 4^2

(16/432)*1.24 = k

k = 0.04592593

then sub in for K to find L

1.44 = (0.04592593 L)/ 3^2

(9/ 0.04592593) * 1.44= L

L = 282.19354839

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confidence rating #$&*:

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Given Solution:

* * We have

R = k * L / D^2. Substituting we obtain

1.24 = k * 432 / 4^2 so that

k = 1.24 * 4^2 / 432 = .046 approx.

Thus

R = .046 * L / D^2.

Now if R = 1.44 and d = 3 we find L as follows:

First solve the equation for L to get

L = R * D^2 / (.046). Then substitute to get

L = 1.44 * 3^2 / .046 = 280 approx.

The wire should be about 280 ft long. **

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Self-critique (if necessary):

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self-critique rating #$&*:

&#This looks very good. Let me know if you have any questions. &#

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