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anonymous
 one year ago
I need to find the tangent plane at (1,2,4), and I have a function z = 4x^2y^2+2y.
anonymous
 one year ago
I need to find the tangent plane at (1,2,4), and I have a function z = 4x^2y^2+2y.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Suppose f has a continuous partial derivatives. An equation of the tangent plane to the surface \[z=f(x,y)\] at the point \[P(x_0,y_0,z_0)\] is \[zz_0=f_x(x_0,y_0)(xx_0)+f_y(x_0,y_0)(yy_0)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So first you should find \[f_x(x_0,y_0) \] and \[f_y(x_0,y_0)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How do I find those?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You need to use partial derivatives, which means that you keep y constant when you are trying to find \[f_x(x_0,y_0)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay, is it correct that i've gotten f_x(x_0,y_0) = 8x and f_y(x_0,y_0) = 2y+2 ??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, that is correct. Now you just need to insert it in the function for the tangent plane.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you very much :)
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