So, I can not find out what I'm doing wrong with this question, even if my life depended on it. I know instruction for doing it, but I can't seem to figure out what I'm doing wrong. Because I refuse to believe the normal is (-0.1292, -0.1292, 0) unless someone here confirms it. Everything seems to be running smoothly up until the creation of the tangent plane, that's where my suspicion lies at the moment.

This is the problem:

Given the function $$ f(x, y)=e^{-(x^2+y^2)} $$ a) Use $f\left(\frac{1}{2}, \frac{1}{2} \right)$ and it's partial derivatives to find a normal vector to $f(x, y)$ in the point $\left(\frac{1}{2}, \frac{1}{2} \right)$

Any help in helping me figure out is greatly appreciated. Thanks in advance.

(edit) A messy image of GeoGebra and what I've done https://i.imgur.com/Rc7tu0L.png

  • $\begingroup$ Can you please show what you did, so that someone might have a clue as to what you are doing wrong? $\endgroup$ – uniquesolution Apr 23 at 6:13
  • $\begingroup$ @uniquesolution Added an image, might make it easier to realize my mistake if you can understand the messy nature of it. $\endgroup$ – Viking Apr 23 at 6:19
  • $\begingroup$ A $z$-coordinate of $0$ is obviously not possible for the normal to the graph of a smooth function defined everywhere. It would be analogous to having a vertical tangent to the graph of a function of one variable. $\endgroup$ – amd Apr 23 at 6:46
  • $\begingroup$ Instead of adding a low-res image of some calculations and graphs, describe step-by-step what it is you’ve done to try to solve this problem. $\endgroup$ – amd Apr 23 at 6:49

A function $(x,y)\mapsto f(x,y)$ has no normal, but a curve in ${\mathbb R}^2$ or a surface in ${\mathbb R}^3$ have normals. Your link indicates that you are considering the surface $$S: \quad z=f(x,y)\tag{1}$$ in ${\mathbb R}^3$ and want to know the normal of $S$ at the point $P=\bigl({1\over2},{1\over2}, f\bigl({1\over2},{1\over2}\bigr)\bigr)\in S$.

In order to find this normal we need a parametric representation of $S$. To this end we use $(x,y)$ as parameters and convert $(1)$ into $${\bf f}:\quad {\mathbb R}^2\to{\mathbb R}^3,\qquad (x,y)\mapsto\bigl(x,y,f(x,y)\bigr)\ .$$ Given the parameter point $P'=\bigl({1\over2},{1\over2}\bigr)$ we obtain $P=\bigl({1\over2},{1\over2},e^{-1/2}\bigr)$. The vectors $${\bf f}_x(P')=\bigl(1,0,f_x(P')\bigr),\quad {\bf f}_y(P')=\bigl(0,1,f_y(P')\bigr)$$ are (linearly independent) tangent vectors to $S$ at $P$. Therefore the vector $${\bf n}:={\bf f}_x(P')\times{\bf f}_y(P')=\bigl(-f_x(P'),\>-f_y(P'),\>1\bigr)$$ is a normal vector to $S$ at $P$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.