Assignment 21

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course Mth 271

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. `query 21

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Question: `q **** problem 1 7th edition Query 2.8.4 dy/dt for (3,4) with x'=8; dx/dt for (4,3) with y'=-2 ****

What are your solutions?

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Your solution:

2x dx’ + -2y dy’ = 0

Let x = 3 and y = 4 and substitute

2(3) * 8 + 2 * -4 dy/dt = 0 so

-48/8 = -6

(4,3) y' is = -2,,,, substitute in

2 * 4 dx/dt + 2 * 3 * -2 = 0 so

8 dx/dt - 12 = 0 , solve for dx/dt and get 3/2

confidence rating #$&*:

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Given Solution:

`a At (3,4) you are given dx/dt as x ' = 8.

Since 2x dx/dt + 2y dy/dt = 0 we have

2(3) * 8 + 2 * 4 dy/dt = 0 so

dy/dt = -48/8 = -6.

At (4,3) you are given dy/dt as y' = -2. So you get

2 * 4 dx/dt + 2 * 3 * -2 = 0 so

8 dx/dt - 12 = 0 and therefore

8 dx/dt = 12. Solving for dx/dt we get

dx/dt = 12/8 = 3/2. **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `q **** problem 2 7th edition Query 2.8.6 roc of volume if r increases at rate 2 in/min, if r= 6 in and if r = 24 in **** What is the rate of volume change if r is increasing at 2 inches / minute?

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Your solution:

Volume of a sphere is determined by equation V= 4/3 pir^3

You would take the derivative o fthe equation and get V’=(4pir^2) dr/dt

rate = 2 in/min, or dr/dt=2

Plug values of dr/dt=2 and r=6 into the equation 4pir^2 * dr/dt and get

Plugging in dr/dt = 2 and r = 6 gives 4 pi (6^2) * 2 = 904.78

Plugging in dr/dt = 2 and r = 24 gives 4 pi (24^2) * 2 = 14,476

confidence rating #$&*:

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Given Solution:

`a The shape is a sphere. The volume of a sphere, in terms of its radius, is

V = 4/3 `pi r^3.

Taking the derivative with respect to t, noting that r is the only variable, we obtain

dV/dt = ( 4 `pi r^2) dr/dt

You know that r increases at a rate of 2 in / min, which means that dr/dt = 2.

Plugging in dr/dt = 2 and r = 6 gives 4 pi (6^2) * 2 = 288 pi = 904 approx.

Plugging in dr/dt = 2 and r = 24 gives 4 pi (24^2) * 2 = 4 pi (576)(2) = 4608 pi = 14,476 approx.. **

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Self-critique (if necessary):

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Self-critique Rating:

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

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