Answer to Question #335783 in Calculus for Dhanush

Question

Let f(x,y)={y^3/x^2+y^2 , 0 otherwise if (x,y) is not =(0,0) show that f is continuous but not differentiable at (0,0)

Answer 1

Since it is not clear how the function looks like, consider two cases:

1. The function $f(x,y)$ has the form: $f(x,y)=\frac{y^3}{x^2}+y^2$. Set $y=t$, $x=t^2$. We receive $f(t^2,t)=\frac{t^3}{t^4}+t^2$ . This function is not continuous at $(0,0)$, since it approaches $\infty$ as $(x,y)\rightarrow(0,0)$ along the curve $x=t^2,y=t$. Thus, it is not differentiable at $(0,0).$
2. The function $f(x,y)$ is: $f(x,y)=\frac{y^3}{x^2+y^2}$. Consider polar coordinates: $x=r\cos\phi$, $y=r\sin\phi$. We get: $f(r\cos\phi,r\sin\phi)=\frac{r^3\sin^3\phi}{r^2}=r\sin^3\phi$. The latter approaches as $(r,\phi)\rightarrow(0,0)$. It corresponds to $(x,y)\rightarrow(0,0)$. Thus, we can put $f(0,0)=0$ and the function will be continuous at $(0,0)$. By definition, the function $f$ is differentiable at $(0,0)$, in case there are numbers $a$ and $b$ such that $\lim_{(x,y)\rightarrow(0,0)}\frac{f(x,y)-ax-by}{\sqrt{x^2+y^2}}=0$.

Consider polar coordinates: $x=r\cos\phi,y=r\sin\phi$. Then, $\frac{f(x,y)-ax-by}{\sqrt{x^2+y^2}}=\sin^3\phi-a\cos\phi-b\sin\phi$. Thus, for $a=0$ we receive: $\lim_{(x,y)\rightarrow(0,0)}\frac{f(x,y)-ax-by}{\sqrt{x^2+y^2}}=0$. Therefore, the function is differentiable at $(0,0).$