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

1. richyw Group Title

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 $z-ln(5)=z_x(0,2)(x-0)+z_y(0,2)(y-2)$

2. richyw Group Title

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

3. richyw Group Title

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

4. richyw Group Title

nevermind. I see where I went wrong haha.

5. TuringTest Group Title

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.

6. TuringTest Group Title

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

7. richyw Group Title

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 $z-f(a,b)=f_x(a,b)(x-a)+f_y(a,b)(y-b)$

8. richyw Group Title

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

9. TuringTest Group Title

ah, gotchya thanks