assignment 21

course MTH 158

7/25/09 5:30 P.M.

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)

y=12/sqrt(x)

4=k/3

k=12

y=12/3

y = 4

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):

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 + (3)^3)

1 = K (8+9)

1 = K(17)

1/17 = k

z = 1/17(x^3 + y^2)

confidence rating: 3

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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):

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= (2pi/sqrt(32))(sqrt(l))

confidence rating: 3

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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):

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=K(l/d^2)

1.24=K(432/16)

16(1.24)=K(432)

19.84/432 = K

K = .046

1.44 = .046(L/9)

(9)1.44 = .046L

12.96 = .046L

12.96/.046 = L

L = 281.74'

confidence rating: 3

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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):

Self-critique Rating:

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