Assignment 21 query

course Mth 271

021. `query 21*********************************************

Question: `q **** Query 2.8.4 dy/dt for (3,4) with x'=8; dx/dt for (4,3) with y'=-2 **** What is dy/dt if x=3, y=4 and dx/dt = 8?

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Your solution:

dy/dt for (3,4) is x’ = 8

dy/dt for (4,3) is x’ = -2

confidence rating:

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I am not confident about this problem.

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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 Rating:

Where did the 2x come from in the given solution? You stated that 2x dx/dy + 2y dy/dt = 0, but how do I know that?

The function, as should have been given in your text, is x^2 + y^2 = 25.

Taking the derivative of this function with respect to t (remembering to apply the chain rule) we get

2 x dx/dt + 2 y dy/dt = 0.

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Question: `q **** 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:

V = 4/3 pi r^3

dV/dr = 4 pi r^2 dr/dt

dV/dr = 4 pi (6) (2) = 904

dV/dr = 4 pi (24) (2) = 14,476

confidence rating:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

I feel fairly confident about this problem.

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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 Rating:

I had to look ahead to the answer to understand how to go about the problem. How do I know that the shape is a square?

I believe that's the way the problem is stated in the text.

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