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If$$f(x,y)=\begin{cases}9-x^2-y^2&\text{if }x^2+y^2\leq9\\0&\text{if }x^2+y^2>9\end{cases}$$study the continuity and existence of partial derivative with respect to $y$ at point $(3,0)$.

The graph of the domain of $f$ is:

Continuity study at $(3,0)$:

$f(3,0)=9-3^2-0=0$, but I do not know how to find $$\lim_{(x,y)\to(3,0)}f(x,y).$$ I tried the following: $$\lim_{(x,y)\to(3,0)}f(x,y)=\left\{\begin{array}{l}\displaystyle\underset{C_1\colon x^2+y^2=9}{\lim_{(x,y)\to(3,0)}}f(x,y)=\underset{C_1\colon x^2+y^2=9}{\lim_{(x,y)\to(3,0)}}(9-(x^2+y^2))=9-9=0\\\displaystyle\underset{C_2\colon x^2+y^2\neq9}{\lim_{(x,y)\to(3,0)}}f(x,y)=\underset{C_2\colon x^2+y^2\neq9}{\lim_{(x,y)\to(3,0)}}\text{??}=\text{??????}\end{array}\right.$$ but then I realized that the "curve" $C_2$ is actually NOT a curve but a set of infinite points, as shown in the previous image.

Existence of partial derivative with respect to $y$ at $(3,0)$:

I know that I need to study whether $$\frac\partial{\partial y}f(3,0)=f'((3,0);(0,1))=f_y(3,0)=\lim_{h\to0}\frac{f((3,0)+h(0,1))-f(3,0)}{h}=\lim_{h\to0}\frac{f(3,h)}{h}$$ exists or not, but I am not able to even find that limit.

Any help? Thanks!!

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  • $\begingroup$ Maybe we could use the following change of variables?: $$\left\{\begin{array}{c}x=\rho\cos\theta\\y=\rho\sin\theta\end{array}\right.\implies f(\rho\cos\theta,\rho\sin\theta)=\begin{cases}9-\rho^2&\text{if $\rho\leq3$},\\0&\text{if $\rho>3$}.\end{cases}$$ $\endgroup$ – manooooh Apr 23 at 6:08
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    $\begingroup$ Have you tried the one-variable situation? $f(x)=\begin{cases} 9-x^2, & |x|\le 3, \\ 0, & |x|\ge 3\end{cases}$? Your change of variables makes the same situation, but doesn't make the $y$-partial clear. $\endgroup$ – Ted Shifrin Apr 27 at 1:04
  • $\begingroup$ @TedShifrin thanks for your help. I did not tried the one-variable situation, but it seems that it has a lot of common with $2$-variables. I do not understand what you mean by "But doesn't make the $y$-partial clear". Could you explain it, please? $\endgroup$ – manooooh Apr 27 at 4:52
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    $\begingroup$ No, I take it back. My one-variable question was not relevant. I apologize. I will post a short answer to your original question. $\endgroup$ – Ted Shifrin May 1 at 22:26
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Regarding the partial derivative question, note that $f(3,y) = 0$ for all values of $y$. (The line $x=3$ touches the circle $x^2+y^2=9$ tangentially and is otherwise completely outside, where the function is identically $0$.) Thus, $\frac{\partial f}{\partial y}(3,0) = 0$.

Now, if $a^2+b^2>9$, $f$ is identically $0$ in a neighborhood of $(a,b)$ and so $\frac{\partial f}{\partial y}(a,b) = 0$. If $a^2+b^2<9$, then $\frac{\partial f}{\partial y}(a,b) = -2b$. What happens if $a^2+b^2=9$ but $b\ne 0$? Then $\frac{\partial f}{\partial y}(a,b)$ does not exist: \begin{multline*} \lim_{h\to 0+}\frac{f(a,b+h)-f(a,b)}h = 0, \quad\text{whereas}\\ \lim_{h\to 0-}\frac{f(a,b+h)-f(a,b)}h = -2b\ne 0. \end{multline*} Therefore, $\frac{\partial f}{\partial y}$ is not continuous at $(3,0)$, as it is not even defined at all points nearby. Perhaps the question wasn't asking about this; it's a bit ambiguous.

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  • $\begingroup$ One thing is clear: if the limit of the derivative wrt $y$ does not exists, then the partial derivative does not exist. (1) I am not sure why you impose the condition $b\neq0$ (what happens if $b=0$?). (2) I do not understand how you can study the lateral limits if the domain of the function, at point $(3,0)$, has infinite directions. $\endgroup$ – manooooh May 1 at 22:40
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    $\begingroup$ I dealt with the case $b=0$ in the first paragraph. In the second paragraph, I am interested in what happens as $(a,b)\to (3,0)$. If you want to discuss continuity of a function at $(3,0)$, this is what you need to consider. I'm discussing the three different cases for where $(a,b)$ happens to lie. $\endgroup$ – Ted Shifrin May 1 at 22:42
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Continuty at (3,0) : When $|(x,y)-(3,0)|=\sqrt{(x-3)^2+y^2}<\delta<1$ for some $\delta>0$ then $|f(x,y)|\leq 9-x^2-y^2=(3-x)(3+x)\leq 6 (3-x)\leq 6\sqrt{(x-3)^2+y^2}<6\delta.$ This implies that $f$ is continuous at $(3,0).$

Partial differential wrt $y$ : $f(3,y)=0$ for any $y$ and hence $f$ has a partial derivative at $(3,0)$ wrt $y$ and the partial derivative is $0.$

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  • $\begingroup$ Why do you say that $\delta<1$? $\endgroup$ – manooooh May 1 at 22:32
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    $\begingroup$ Just to make sure that $|x-3|$ and $|y|$ are small enough. In epsilon-delta definition of limit only small value of $\delta$ matters. $\endgroup$ – John_Wick May 1 at 23:41
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We have the piece-wise defined surface. The two surfaces are continuous and if the limits are equal at the join, then the surface is continuous.

And since the two surfaces are continuous we can just plug $(3,0)$ in for each.

$\lim_\limits{(x,y)\to (3,0)} 9-x^2-y^2 = 0$

$\lim_\limits{(x,y)\to (3,0)} 0 = 0$

Another way to do it would be to say:

$\forall \epsilon >0, \exists \delta > 0 : d((x,y),(3,0))< \delta \implies |f(x,y)|<\epsilon$

$d((x,y),(3,0))$ is the distance between $(x,y)$ and $(3,0)$

And what distance metric do we want to use? How about the taxicab metric?

Then the maximal distance some point $x_1,y_1$ can be from the point $(3,0)$ would be $(3+\pm \delta, \pm \delta)$

$f(x,y) < |6\delta| + |2\delta^2|$

Lets demand that $\delta < 1$ then $\delta^2 < \delta$

$\delta = \min (\frac {\epsilon}{8},1)$

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  • $\begingroup$ I do not think that evaluating the function to every piece is the correct method... Also, I did not study the taxicab metric :( :(. $\endgroup$ – manooooh May 1 at 22:33
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    $\begingroup$ The taxicab metric is $d((x_1,y_1),(x_2,y_2)) = |x_1-x_2| + |y_1-y_2|$ or if you lived in a city of rectangular blocks, how many blocks the taxi would have to travel to make it to the destination. And what is wrong with evaluating the pieces? $\endgroup$ – Doug M May 2 at 7:53

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