Answer to Question #292861 in Calculus for jaya

Question

15. Let

f(x,y) { 0, x² < y < 2x² 1. otherwise

Verify the existence of partial derivatives fr(0,0) and f(0, 0), and the differentiability of fat (0,0).

Answer 1

From the given definition, we have that;

$\displaystyle f(x,y)=\begin{cases} 0 &\text{if } x²<y<2x²\\ 1&\text{if otherwise} \end{cases}$

That is,

$\displaystyle f(x,y)=\begin{cases} 0 &\text{if } y\in(x^2,2x^2)\\ 1&\text{if }y\in(-\infty,x^2]\cup[2x^2,\infty) \end{cases}$

Now, $\displaystyle f(0,0)=1 \text{ since }0\in(-\infty,x^2]\cup[2x^2,\infty)$

Thus,

1] $\displaystyle f_x(0,0)=\lim_{h\rightarrow0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\rightarrow0}\frac{f(h,0)-1}{h}$

But,$\displaystyle \,\lim_{h\rightarrow0}\frac{f(h,0)-1}{h}\text{ exists iff }\lim_{h\rightarrow0^-}\frac{f(h,0)-1}{h}\text{ and }\lim_{h\rightarrow0^+}\frac{f(h,0)-1}{h}\text{ exists and are}\\ \text{equal. }$

Now, $\displaystyle \lim_{h\rightarrow0^-}\frac{f(h,0)-1}{h}=\lim_{h\rightarrow0^-}\frac{1-1}{h}=0, \text{since }f(h,0) \rightarrow1\text{ as }h\rightarrow0^- \text{ because}\\$

$\displaystyle 0\in(-\infty,h^2]\cup[2h^2,\infty)$

Also, $\displaystyle \lim_{h\rightarrow0^+}\frac{f(h,0)-1}{h}=\lim_{h\rightarrow0^+}\frac{1-1}{h}=0, \text{since }f(h,0) \rightarrow1\text{ as }h\rightarrow0^+ \text{ because}\\$

$\displaystyle 0\in(-\infty,h^2]\cup[2h^2,\infty)$

Since both left and right limit exists, we have;

$\displaystyle f_x(0,0)=\lim_{h\rightarrow0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\rightarrow0}\frac{f(h,0)-1}{h}=0$

2] $\displaystyle f_y(0,0)=\lim_{h\rightarrow0}\frac{f(0,h)-f(0,0)}{h}=\lim_{h\rightarrow0}\frac{f(0,h)-1}{h}$

But,

$\displaystyle \,\lim_{h\rightarrow0}\frac{f(0,h)-1}{h}\text{ exists iff }\lim_{h\rightarrow0^-}\frac{f(0,h)-1}{h}\text{ and }\lim_{h\rightarrow0^+}\frac{f(0,h)-1}{h}\text{ exists and are}\\ \text{equal. }$

Now, $\displaystyle \lim_{h\rightarrow0^-}\frac{f(0,h)-1}{h}=\lim_{h\rightarrow0^-}\frac{1-1}{h}=0, \text{since }f(0,h) \rightarrow1\text{ as }h\rightarrow0^- \text{ because}\\$

$\displaystyle h\in(-\infty,0]\cup[0,\infty)$

Also, $\displaystyle \lim_{h\rightarrow0^+}\frac{f(0,h)-1}{h}=\lim_{h\rightarrow0^+}\frac{1-1}{h}=0, \text{since }f(0,h) \rightarrow1\text{ as }h\rightarrow0^+ \text{ because}\\$

$\displaystyle h\in(-\infty,0]\cup[0\infty)$

Since both left and right limit exists, we have;

$\displaystyle f_y(0,0)=\lim_{h\rightarrow0}\frac{f(0,h)-f(0,0)}{h}=\lim_{h\rightarrow0}\frac{f(0,h)-1}{h}=0$.

3] We need to check the continuity of the function at $\displaystyle (0,0).$ That is, we need to show that $\displaystyle \lim_{(x,y)\rightarrow(0,0)}f(x,y)=f(0,0)$.

first off, recall that $\displaystyle f(0,0)=1$.

Now, to evaluate $\displaystyle \lim_{(x,y)\rightarrow(0,0)}f(x,y)$, consider the limit along the power curve $\displaystyle x=t$,

$\displaystyle y=at^2$ as $\displaystyle t\rightarrow 0^+$.

Then,

$\displaystyle \lim_{(x,y)\rightarrow(0,0)}f(x,y)=\lim_{t\rightarrow 0^+}f(t,at^2)=\begin{cases}0 &\text{if }a\in(1,2)\\1&\text{if }a\in(-\infty,1]\cup[2,\infty)\end{cases}$

Since $\lim_{(x,y)\rightarrow(0,0)}f(x,y)=\lim_{t\rightarrow 0^+}f(t,at^2)=\begin{cases}0 &\text{if }a\in(1,2)\\1&\text{if }a\in(-\infty,1]\cup[2,\infty)\end{cases}$

depends on a, thus we have that $\displaystyle \lim_{(x,y)\rightarrow(0,0)}f(x,y)$ does not exist.

$\Rightarrow f(x,y)$ is not continuous at $\displaystyle (0,0).$

$\Rightarrow f(x,y)$ is not differentiable at $\displaystyle (0,0).$