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Find a point P(a,b,c) on the graph z=ln(4x^2+y^2) such that the tangent plane at that point is parallel to the plane 2x+2y-z=3
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I already found the gradient to be <8x/(4x^2+y^2) , 2y/(4x^2+y^2), -1> And I figured that you would have to set the gradient equal to a scalar times the normal of the plane (which is n=<2,2,-1> so my thoughts are that both the partial derivatives of z have to equal 2... But I can't find any values that would make that work nicely... Did I make a mistake or is the question just mean?
may have just found it...
there we go. ended up trying a different method and it worked out
Just in time for me to solve it myself :D thanks anyways haha
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