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Math Help - Differential calculus- first order partial derviatives

  1. #1
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    Differential calculus- first order partial derviatives

    Hi
    I need help with the followoing question:

    Find the slopes of the curves of intersection of the surface z= f(x,y) with the planes perpendicular to the x axis and y axis respectively at a given point.

    z= sin (4x+y) at (0, pi/2)

    I did fx(x,y)= 4 cos(4x+y) and at this point i get 0

    To find the perpendicular planes i think you do -1/0 but that is not defined.

    is this the correct method?

    kind regards
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  2. #2
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    You need to review the notion of a directional derivative and, in particular, the partial derivative. For the plane perpendicular to the y-axis, the slope of the curve will be f_x(0,\pi/2)=0. That's it. I have no idea where you're getting -1/0 from.
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  3. #3
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    This can be done without partial derivatives:

    The "plane perpendicular to the x-axis" at (0, \pi/2) is the y-z plane, x= 0. Set x= 0 in the formula: z= sin(y) and the derivative of that is cos(y) which is cos(\pi/2)= 0 at (0, \pi/2).

    The "plane perpendicular to the y-axis" at (0, \pi/2) is the plane y= \pi/2: z= sin(4x+\pi/2) and the derivative of that is 4 cos(4x+ \pi/2) which is 4 sin(\pi/2)= 4 at (0, \pi/2).
    Last edited by HallsofIvy; May 31st 2010 at 04:00 AM. Reason: Thanks to ojones
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  4. #4
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    HallsofIvy: If your goal is to get through your course on multivariable calculus without ever learning what a partial derivative is, you're not going to make it. By the way, I think you're second answer is incorrect (you didn't differentiate).

    One more thing, the derivatives you're doing ARE partial derivatives.
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  5. #5
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    thanks halls of ivy and ojones! but ojones- there really isn't any need to get so worked up over hall of ivy's post- any help is good and we all make mistakes
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