Q 53. - Calculus

Question

Determine the domains of the functions in Exercises 47–56, and find where the functions are continuous.  

$$\begin{gathered} f(x,y)=\left\{\frac{xy}{\sqrt{x^2+y^2}} if (x,y)\ne \left(0,0\right)\right\} \\ \left\{0 if (x,y)=(0,0) \right\} \end{gathered}$$

Step-by-Step Solution

Answer

The function $$\begin{gathered} f(x,y)=\left\{\frac{xy}{\sqrt{x^2+y^2}} if (x,y)\ne \left(0,0\right)\right\} \\ \left\{0 if (x,y)=(0,0) \right\} \end{gathered}$$ is continuous everywhere.

Step 1. Given information.

We have given expression: $$\begin{gathered} f(x,y)=\left\{\frac{xy}{\sqrt{x^2+y^2}} if (x,y)\ne \left(0,0\right)\right\} \\ \left\{0 if (x,y)=(0,0) \right\} \end{gathered}$$

Step 2: To find domain of the function.

The given function is a piece wise function, with a break point at $(x,y)=(0,0)$

The domain of a polynomial function is the entire set of real numbers.

Hence, it never constraints the domain of the function.

The radical term in the denominator makes sure that the term inside the radical sign is greater than or equal to 0.

Also, the rational expression means that the denominator cannot be equal to 0.

The left side of this inequality is sum of two squares. 

Hence, it is always positive. 

The only case where it is not so, when $x=y=0$ but that is not the case with the given function.

Hence, the function is defined for all the real values. 

Thus, the domain of the function is given by the set .

$Domain \left(f\right)=f^2 or \left\{\left(x,y\right) x,y\in f\right\}$

Step 3: To find continuous of the function.

Substitute the value of $x=r\cos \theta$ and $y=r \sin \theta$

$$\begin{gathered} \underset{(x,y)\to (0,0)}{\lim }\frac{xy}{\sqrt{x^2+y^2}}=\underset{r\to 0}{\lim }\frac{\left(r \cos \theta \right) \left(r \sin \theta \right)}{\sqrt{r^2 {\cos }^2\theta +r^2}{\sin }^2\theta } \\ =\underset{r\to 0}{\lim }\frac{r^2 \cos \theta \sin \theta }{\sqrt{r^2 ({\cos }^2\theta +}{\sin }^2\theta )} \\ =\underset{r\to 0}{\lim }\frac{r^2 \cos \theta \sin \theta }{\sqrt{r^2 \left(1\right)}} \\ =\underset{r\to 0}{\lim }\frac{r^2 \cos \theta \sin \theta }{r} \\ =\underset{r\to 0}{\lim }r\cos \theta \sin \theta \\ =0 \end{gathered}$$

Hence, there is no point of discontinuity for the given number.