course MTH 272 05/12, 2300 assignment #028028.
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20:20:54 Query Problem 7.4.8 fy for xy / (x^2+y^2)
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RESPONSE --> xy / (x^2+y^2) (xy) (x^2 + y^2)^-1 fy=(1)(x)(-1)(2y)(x^2 + y^2)^-2 =(-2xy)/(x^2 + y^2)^2 confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:21:24 What is the requested partial derivative?
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RESPONSE --> (-2xy)/(x^2 + y^2)^2 confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:27:26 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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RESPONSE --> w = 1/sqrt(1-x^2-y^2-z^2) (1-x^2-y^2-z^2)^(-1/2) wx=(-1/2)(-2x)(1-x^2-y^2-z^2)^(-3/2) =x / (1-x^2-y^2-z^2)^(3/2) wy=(-1/2)(-2y)(1-x^2-y^2-z^2)^(-3/2) = y / (1-x^2-y^2-z^2)^(3/2) wz=(-1/2)(-2z)(1-x^2-y^2-z^2)^(-3/2) =z /(1-x^2-y^2-z^2)^(3/2) confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:27:59 What are the values of the three requested partial derivatives at the specified point?
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RESPONSE --> Specified point: confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:29:15 What are the three partial derivative functions?
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RESPONSE --> wx=x/ (1-x^2-y^2-z^2)^(3/2) wy=y /(1-x^2-y^2-z^2)^(3/2) wz=z / (1-x^2-y^2-z^2)^(3/2) confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:41:04 Query problem 7.4.40 (was 7.4.36) fx = fy = 0 for 3x^3-12xy+y^3
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RESPONSE --> 3x^3-12xy+y^3 fx=9x^2 - 12y=0 fy=-12x + 3y^2=0 -12x=-3y^2 x=3y^2 / 12 x=y^2 / 4 9(y^2 / 4) - 12y=0 (9/4) y^2 - 12y=0 y ((9/4)y -12)=0 y=0 x=0^2/4=0 (0,0) (9/4)y -12=0 (9/4)y=12 y=(16/3) x=(16/3)^2 / 4=7.1 (7.1, 5.3) confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:41:36 What are the coordinates of the point or points where the two partial derivatives are both zero?
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RESPONSE --> (0,0) and (7.1,5.3) confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:42:51 What system of simultaneous equations did you solve to get your result?
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RESPONSE --> fx=9x^2 - 12y fy=-12x + 3y^2 confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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