assignment 21

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course Mth 158

10/30 11

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:

start with y=k/sqrt(x) then you can plug in the numbers saying 4=k/sqrt(9) which is 4=k/3 multiply both sides by 3 and you get k=12 so we have 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) plug in the given numbers so you have 1 = k ( 2^3 + 3^2) which is 1=k(8+9) then 1=k(17) or k=1/17 so that leaves us with z = (1/17) (x^3 + y^2) or 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: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) and it says k=2 pi / sqrt(32) so I guess we just say

T=(2 pi / 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: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

plug in the given numbers to get 1.24 = (k * 432) / 4^2

1.24 = k * 432 / 16 multiply both sides by 16 to get 19.84=k*432 divide both sides by 432 and we have

k=.0459 approximately

so now given a wire resistance is 1.44 and diameter is 3 we plug the numbers in to get

1.44=(.0459*L)/3^2

1.44=(.0459*L)/9

12.96=. 0459*L

282.3529412=L

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):

close enough?

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Self-critique Rating:

@& Definitely close enough.

My solutions are ballpark estimates. *@

ok"

Self-critique (if necessary):

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Self-critique rating:

ok"

Self-critique (if necessary):

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Self-critique rating:

#*&!

&#Your work looks good. See my notes. Let me know if you have any questions. &#