can you please help me solve this?

Question

f(x,y)=2−ln[1+(x−2)^2 +(y−2)^2]

find and classify its stationary points. Then, find the gradient at (1;1). Finally, find the equation of the tangent plane to the graph of the function at the point (2; 2).

I found the stationary points at (2;2)

and found the second derivative but I don't know if it's correct , if you can please solve it so I can compare it, my professor didn't provide the solutions...

Answer 1

Hello Lily,

I see that you are having trouble computing the second derivative of the function. I understand that this can be a bit tricky and I can definitely provide some help. Unfortunately, we cannot just compute one second order derivative to use the Second Order Derivative Test. We must use all four possible derivatives, specifically the two second order unmixed derivatives and the two second order mixed derivatives. We then put these into a matrix called the Hessian Matrix and then compute the determinant of the matrix to get a special function with the unoriginal name the Hessian.

The answers you are looking for are in the LaTeX code below. You can paste this code into the LaTeX image generator (latexeditor.lagrida.com) to get neat looking images, so you don't have to decipher anything too complicated.

\begin{align}

f(x,y) &= 2−\ln(1+(x−2)^2 +(y−2)^2)\\

f_x(x,y) &= -\frac{2(x-2)}{1+(x−2)^2 +(y−2)^2}\\

f_y(x,y) &= -\frac{2(y-2)}{1+(x−2)^2 +(y−2)^2}\\

f_{xx}(x,y) &= -\frac{2[1+(x−2)^2 +(y−2)^2] - 4(x-2)^2}{[1+(x−2)^2 +(y−2)^2]^2}\\

&= -\frac{2-2(x−2)^2 +2(y−2)^2] }{[1+(x−2)^2 +(y−2)^2]^2}\\

f_{yy}(x,y) &= -\frac{2[1+(x−2)^2 +(y−2)^2] - 4(y-2)^2}{[1+(x−2)^2 +(y−2)^2]^2}\\

&= -\frac{2+2(x−2)^2 -2(y−2)^2] }{[1+(x−2)^2 +(y−2)^2]^2}\\

f_{xy}(x,y)=f_{yx}(x,y)= &= \frac{4(x-2)(y-2)}{1+(x−2)^2 +(y−2)^2]^2}\\

\end{align}

Here is the Hessian Matrix

\mathbf{H}_f=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\\\end{bmatrix}

Here the Hessian is simply

H_f = f_{xx}f_{yy}-(f_{xy})^2

because the second mixed derivatives are the same.

The second derivatives are f_{xx} = -2, f_{yy} = -2, and f_{xy} = 0. So the Hessian is H(x,y)=(-2)(-2)-(0)^2=4. This tells us that we have a local minimum of f(x,y) at (x,y)=(2,2).