assignment 28

course mth 272

......!!!!!!!!...................................Applied Calculus II

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Asst # 28

07-31-2006

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18:19:07

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

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18:40:43

for x as constant: x+xy(x^2+2y)/ x^2+y^2

for y as constant: y + xy(2x+y^2)/ x^2+y^2

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18:40:44

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**** What is the requested partial derivative?

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18:40:49

for x as constant: x+xy(x^2+2y)/ x^2+y^2

for y as constant: y + xy(2x+y^2)/ x^2+y^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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18:40:49

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**** 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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18:53:24

2x / (25+x^2)^(3/2)

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18:53:25

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

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19:14:45

# 32:

2x / (25+x^2)^(3/2)

The expression 1-x^2-y^2-z^2 is an 'inner function' and won't change. This result should be

2x / (1-x^2-y^2-z^2)^(3/2).

Every partial derivative involves the chain rule with inner function 1 = x^2 - y^2 - z^2 and outer function 1/ sqrt(z) = z^(-1/2).

The partial derivatives with respect to x, y and z of the 'inner function' (1 - x^2 - y^2 - z^2) are respectively -2x, -2y and -2z.

The derivative of the 'outer function' z^-(1/2) is -1/2 z^(-3/2).

So the partial derivatives of the entire function are

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

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

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

At the origin we have x = y = z = 0 so that the three partial derivatives are all of form 0 / 1 = 0.*&*&

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19:14:46

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**** What are the three partial derivative functions?

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19:15:20

i rewrote the expression and then applied the general power rule

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19:15:21

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

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19:56:33

* i did number 32 in a wrong section* :

x: x(1-x^2-y^2-z^2)^(1/2)

y: y(1-x^2-y^2-z^2)^(1/2)

z: z(1-x^2-y^2-z^2)^(1/2)

this question: not sure. do you find the partial derivative and try to equal it to each other and solve.

** 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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19:56:33

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

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19:56:47

?

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19:56:48

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

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19:57:18

if my intuition is correct than the partial derivatives with respect to x and y

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Applied Calculus II

Asst # 29

07-31-2006

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19:57:31

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Be sure to pay careful attention to my notes, and let me know if you have questions.