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anonymous
 one year ago
In session 27, it is mentioned that the tangent plane should contain the tangent lines of the partial functions at point P.
Does that not meant that the normal of the tangent plane is equal to the cross product of the tangent lines? If so, does it not contradict
grad f = normal of tangent plane?
anonymous
 one year ago
In session 27, it is mentioned that the tangent plane should contain the tangent lines of the partial functions at point P. Does that not meant that the normal of the tangent plane is equal to the cross product of the tangent lines? If so, does it not contradict grad f = normal of tangent plane?

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1there should be no contradiction as \( \vec r = <x,y,f(x,y)>\) then \( \vec {r_x} = <1,0,f_x>\) and \( \vec {r_x} = <0, 1, f_y>\) and \(\vec {r_x} \ \times \vec {r_y}\) is \[\left\begin{matrix}\hat x & \hat y & \hat z\\ 1 & 0 & f_x \\ 0 & 1 & f_y\end{matrix}\right\] which is \(<f_x, f_y, 1>\) in terms of the gradient, it is important to note that it is NOT grad f that gives the normal but grad \(\Phi\) where \(\Phi = z  f(x,y) = 0\) thus \(\nabla \Phi = <f_x, f_y, 1>\), which is the same thing IOW \(\nabla \Phi \ne \nabla z\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1oops, typo, that second \( \vec{r_x}\) should *obviously* be \( \vec{r_y}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks! I was being careless :(
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