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Math Help - partial derivatives: Find an equation for the plane

  1. #1
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    partial derivatives: Find an equation for the plane

    Find an equation for the plane tangent to the graph of z = 6 cos(xy) at the point (2, π/6, 3).

    i went on to find
    partial derivative x = -6 y sin(xy)
    partial derivative y = -6 x sin(xy)

    then i used the equation for tangent plane
    z=f(a,b)+fx(a,b)(x-a)+fy(a,b)(y-b)

    z = 6cos(xy) + (-6)ysin(xy)(x-2)+(-6)xsin(xy)(y-\pi/6)

    z = 6(.5) + (-6)(\frac{\pi}{6})sin(2 * \frac{\pi}{6})(x - 2) + (-6)(2)sin(2 * \frac{\pi}{6})(y - \frac{\pi}{6})

    z = 3 + (-\pi)*\frac{\sqrt{3}}{2} * (x-2) + (-12)\frac{\sqrt{3}}{2}(y - \frac{\pi}{6})


    The thing is, i have no idea if i did this problem correctly or not. I haven't been taught how to deal with problems like this and trying to learn on my own. Can anyone help? Thanks!
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  2. #2
    MHF Contributor

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    You should learn how to check it yourself. If [tex]x= 2[/math and y= \pi/6 in your formula for the plane give z= 3? Is the "x-slope", the coefficient of x in the plane formula, equal to -6(3) sin(2(\pi/6))? Is the coefficient of y in theplane formula equal to -6(\pi/6)sin(2(\pi/6))?

    Here, by the way is what I consider to be a simpler way to find the tangent plane to z= f(x,y). The function g(x, y, z)= z- f(x,y) has z- f(x,y)= 0 or z= f(x, y) as a "level surface". The gradient of g, \nabla g is perpendicular to that level surface and so perpendicular to its tangent plane, the tangent plane to z= f(x,y).

    If a normal to a plane at (x_0, y_0, z_0) is [tex]\nabla g= <-f_x, -f_y, 1>[tex], then the plane is given by -f_x(x_0, y_0)(x- x_0)- f_y(x_0, y_0)(y- y_0)+ (z- z_0)= 0
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  3. #3
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    so does that mean

    -6ysin(xy)(x-2)-6xsin(xy)(y-\frac{\pi}{6})-(z-3) = 0
    -6(\frac{\pi}{6})sin(2*\frac{\pi}{6})(x-2)-6(2)sin(2*\frac{\pi}{6})(y-\frac{\pi}{6})-(z-3) = 0
    -\pi*sin(\frac{\pi}{3})(x-2)-12sin(\frac{\pi}{3})(y-\frac{\pi}{6})-(z-3) = 0
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