Assignment 28

course Mth 272

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16:46:03

Query Problem 7.4.8 fy for xy / (x^2+y^2)

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RESPONSE -->

confidence assessment:

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16:46:15

What is the requested partial derivative?

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RESPONSE -->

fx(x,y) = -2x^2y/(x^2+y^2)^2

fy(x,y) = -2xy^2/(x^2+y^2)^2

confidence assessment: 2

** You have to use the quotient rule. The derivative is taken with respect to y, so the ' stands for the derivative with respect to y. You get

[ (xy)' (x^2 + y^2) - xy ( x^2 + y^2)' ] / (x^2 + y^2) ^ 2. Remembering that ' represents derivative with respect to y we get

[ x ( x^2 + y^2) - xy ( 2y ) ] / (x^2 + y^2 ) ^ 2 or

[ x^3 + x y^2 - 2 x y^2 ] / (x^2 + y^2) ^ 2, which simplifies to

[ x^3 - x y^2 ] / (x^2 + y^2) ^ 2 or

x [ x^2 - y^2 ] / (x^2 + y^2) ^ 2 . **

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16:46:18

Query problem 7.4.32 wx, wy, wz at origin for w = 1/sqrt(1-x^2-y^2-z^2)

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confidence assessment:

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16:46:29

What are the values of the three requested partial derivatives at the specified point?

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RESPONSE -->

Point (0,0,0) all values are 0

confidence assessment: 2

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16:46:40

What are the three partial derivative functions?

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RESPONSE -->

fx = -1/2(1-x^2-y^2-z^2)^-3/2(-2x) = x/(1-x^2-y^2-z^2)^3/2

fy = -1/2(1-x^2-y^2-z^2)^-3/2(-2y) = y/(1-x^2-y^2-z^2)^3/2

fz = -1/2(1-x^2-y^2-z^2)^-3/2(-2z) = z/(1-x^2-y^2-z^2)^3/2

confidence assessment: 2

Good.

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16:46:43

Query problem 7.4.40 (was 7.4.36) fx = fy = 0 for 3x^3-12xy+y^3

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confidence assessment:

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16:46:56

What are the coordinates of the point or points where the two partial derivatives are both zero?

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RESPONSE -->

points would be (0,0)

confidence assessment: 2

I think you've got it. Details for your reference and comparison:

** The partial derivatives are

fx = 9x^2 - 12y

fy = -12x + 3y^2.

If the two partial derivatives are zero we get the equations

9x^2 - 12 y = 0

-12x + 3 y^2 = 0.

Solving the first equation for y we get y = 3x^2 / 4.

Substituting this expression for y in the second we have

-12 x + 3 ( 3x^2 / 4)^2 = 0 so

-12 x + 27 x^4 / 16 = 0. Factoring out -3x:

-3x ( 4 - 9 x^3 / 16) = 0. This is so if

-3x = 0 or 4 - 9 x^3 / 16 = 0.

-3x = 0 gives solution x = 0.

4 - 9 x^3 / 16 = 0 if x^3 = 4 * 16 / 9, which happens when

x = 64^(1/3) / 9^(1/3) = 4 * 3^(-2/3), which is expressed in standard form as 4 * 3 ^(1/3) / 3.

If x = 0 then since 9 x^2 - 12 y = 0 we have y = 0.

If x = 4 * 3^(1/3)/3 then 9 x^2 - 12 y = 0 gives us

9 [ 4 * 3^(1/3)/3 ] ^2 - 12 y = 0 so

y = 9 [ 4 * 3^(1/3)/3 ] ^2 / 12 = 3/4 * 16 * 3^(2/3)/9 = 4 * 3^(2/3) / 3.

So the two partials are both zero at (0,0) and at ( 4 * 3^(1/3)/3, 4 * 3^(2/3)/3. **

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16:47:09

What system of simultaneous equations did you solve to get your result?

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RESPONSE -->

9x^2 - 12y = 0

-12x + 3y^2 = 0

confidence assessment: 2

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