Let $f(x,y)=e^{-(x^2+y^2)}$ and $g(x,y)=\frac{e^{-1}}{x^2+y^2}$. Show that $f(a,b)=g(a,b)$ at any point $(a,b)$ where $a^2+b^2=1$.

Question

Let $f(x,y)=e^{-(x^2+y^2)}$ and $g(x,y)=\frac{e^{-1}}{x^2+y^2}$.

a) Show that $f(a,b)=g(a,b)$ at any point $(a,b)$ where $a^2+b^2=1$.

b) Show that the graph of $f$ is tangent to the graph of $g$ at every point $(a,\ b)$ where $a^2+b^2=1$.

For a), \begin{align*} f(a, b)&=g(a, b)\\ e^{-(x^2+y^2)}&=\frac{1}{e(x^2+y^2)}\\ \Rightarrow e(x^2+y^2)e^{-(x^2+y^2)}&=1 \\ x^2e^{-x^2-y^2+1}+y^2e^{-x^2-y^2+1}&=1 \end{align*}

I don't know how to continue with the calculation.

For b), I do not know if the following is useful: if two functions $f$ and $g$ have the same value $f(a, b)=g(a, b)$ and the same partial derivatives at $(a, b)$, then their tangent planes at $(a, b, f(a, b))$ are equal and we say that the graphs are tangent at that point.

Answer 1

Your work for (a) is correct! You just have to apply the condition that $a^2 + b^2 = 1$, and I recommend you do so on the second equality you obtained: $e^{-(a^2 + b^2)} = \frac{1}{e(a^2 + b^2)}$.

To show that the graph of $f$ is tangent to the graph of $g$, you need to compute the tangent plane to $g$ and verify that it matches $f$ at the points $(a, b)$ with $a^2 + b^2 = 1$. Recall that the tangent plane of $g$ at $(a, b)$ is given by \begin{equation*} T(a, b) = g(a, b) + \frac{\partial g}{\partial x}(a, b)(x - a) + \frac{\partial g}{\partial y}(a, b)(y - b). \end{equation*}

Answer 2

Where $a^2+b^2=1$: $f(a,b)=e^{-(a^2+b^2)}=e^{-1}$ and $g(a,b)=\frac{e^{-1}}{a^2+b^2}=e^{-1}$

Why did you post this???