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.