Assignment 21

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course MTH 158

11:38 July 17

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=4 when x=9

4=k/sqrt(9)

4=k/3

k=4*3=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:3

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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) x=2, y=3 and z=1

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

17k=1

k=1/17

The proportionality is z=(x^3+y^2)/17

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:3

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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:

Our equation will be 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):ok

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Self-critique Rating:3

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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:

When we substitute we get

1.24=k*432/4^2

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

r=.046*l/d^2

l=r*d^2/(.46)

l=1.44*3^2/(.046)=280

The wire should be 280 ft long

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:3"

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