Assignment 21

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course Mth 158

3/14/13 Around 8:30AM

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 = k / 3

 k = 12

 y = 12 / sqrt(x)

 

 

confidence rating #$&*: 3

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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): OK

 

 

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Self-critique Rating: OK

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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 + 3^2)

 1 = k(17)

 k = 1/17

 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): OK

 

 

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Self-critique Rating: OK

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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(L)) / sqrt 32

 

 

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): OK

 

 

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Self-critique Rating: OK

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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) / 4^2

 1.24 = k(432) / 16

 19.84 = k(432)

 k = 0.04592593

1.44 = (0.04592593*L) / 3^2

 12.96 = 0.04592593*L

 L = 282.2, approximately

 

 

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): OK

 

 

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Self-critique Rating:OK

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