Find an equation for the tangent plane to the surface given by$$z=\ln{\left(1+x^2+y^2\right)}$$

Question

richyw · Answer

Sorry, at the point (0,2,ln5). why can't I just say$$z_x=\frac{2y}{\left(1+x^2+y^2\right)}$$$$z_y=\frac{2x}{\left(1+x^2+y^2\right)}$$And then that $$z-ln(5)=z_x(0,2)(x-0)+z_y(0,2)(y-2)$$

richyw · Answer

so $$z=\frac{4}{5}x+\ln(5)$$

richyw · Answer

I don't see where I am going wrong....

richyw · Answer

nevermind. I see where I went wrong haha.

turingtest · Answer

is it that you don't have the coefficient for z, since you didn't turn it into a function? I want to know because your way is different than mine.

turingtest · Answer

er, I mean didn't turn it into a 3-variable function...

richyw · Answer

my notation was sloppy. All I do to find the plane tangent to $z=f(x,y)$ at the point $P\left(a,b,f(a,b)\right)$ is use the formula $$z-f(a,b)=f_x(a,b)(x-a)+f_y(a,b)(y-b)$$

richyw · Answer

I just accidentally put f_y where f_x should have been...

turingtest · Answer

ah, gotchya
thanks