Question

Consider the following functions \(f.\) a. Is \(f\) continuous at (0,0)\(?\) b. Is \(f\) differentiable at (0,0)\(?\) c. If possible, evaluate \(f_{x}(0,0)\) and \(f_{y}(0,0)\) d. Determine whether \(f_{x}\) and \(f_{y}\) are continuous at \((0,0)\) e. Explain why Theorems 5 and 6 are consistent with the results in parts \((a)-(d).\) $$f(x, y)=\left\\{\begin{array}{cc}-\frac{x y}{x^{2}+y^{2}} & \text { if }(x, y) \neq(0,0) \\ 0 & \text { if }(x, y)=(0,0)\end{array}\right.$$

Short answer

If not, determine the partial derivatives at (0,0) and their continuity. Answer: The given function is not continuous at (0,0) and hence, not differentiable at that point. The partial derivatives exist and are equal to 0 at (0, imageNamed0), but they are not continuous at this point.

Step-by-step solution

Check Continuity at (0,0)

To check whether the function \(f(x, y)\) is continuous at (0,0), we need to show that the limit exists and equals the value of the function at (0,0), which is 0. \(\lim_{(x, y) \to (0,0)} f(x, y) = \lim_{(x, y) \to (0,0)} \frac{-xy}{x^2 + y^2}\) Using polar coordinates: \(x=r\cos(\theta)\) and \(y=r\sin(\theta)\), we have, \(\lim_{r \to 0} \frac{-r\cos(\theta)r\sin(\theta)}{r^2\cos^2(\theta)+r^2\sin^2(\theta)} = \lim_{r \to 0} \frac{-r^2\sin(\theta)\cos(\theta)}{r^2}=-\sin(\theta)\cos(\theta)\) Since the limit depends on the angle \(\theta\) and is not equal to \(0\) (the value of the function at (0,0)), the function is not continuous at (0,0).

Checking differentiability at (0, 0)

Since the function is not continuous at (0,0), it cannot be differentiable at that point.

Evaluate partial derivatives at (0, 0)

Although the function is not differentiable at (0,0), we can go ahead and compute its partial derivatives at (0,0) for the sake of completeness. Using the definition of partial derivatives: \(f_x(0,0) = \lim_{x \to 0} \frac{f(0,0)-f(0,0)}{x-0} = 0\) \(f_y(0,0) = \lim_{y \to 0} \frac{f(0,0)-f(0,0)}{y-0} = 0\) So, the partial derivatives both exist and are equal to 0 at (0,0).

Check the continuity of the partial derivatives at (0,0)

To see if the partial derivatives are continuous at (0,0), we'll take their limits as \((x, y) \to (0,0)\): \(f_x = \frac{-y}{x^2 + y^2}\) and \(f_y = \frac{x}{x^2 + y^2}\) Using polar coordinates again, \(\lim_{r \to 0} \frac{-r\sin(\theta)}{r^2}=-\sin(\theta)\) and \(\lim_{r \to 0} \frac{r\cos(\theta)}{r^2}=\cos(\theta)\) Since the limits depend on angle \(\theta\), neither partial derivative is continuous at (0,0).

Interpret the results using the theorems

Theorems 5 and 6 state that if a function has continuous partial derivatives, it is differentiable. In our case, both partial derivatives are not continuous, and we have already proved that the function is not differentiable at (0,0). This result is consistent with theorems 5 and 6.

Key concepts

Continuity

In mathematics, continuity of a function at a point means that the function does not "jump" or "break" at that point. For a function \( f(x, y) \), it is continuous at a point \((x_0, y_0)\) if the limit as \((x, y)\) approaches \((x_0, y_0)\) is equal to the function’s value at that point.

In the given exercise, we examine the continuity of the function \( f(x, y) = \frac{-xy}{x^2 + y^2} \), when \((x, y) eq (0,0)\), and \( f(x, y) = 0 \) at \((0, 0)\). By using polar coordinates to convert \(x\) and \(y\) to \(r\cos(\theta)\) and \(r\sin(\theta)\), we find the limit depends on \(\theta\), the angle.

This means that different approaches towards \((0, 0)\) can yield different results, proving that the function is not continuous at this point.

Differentiability

Differentiability extends the idea of continuity by considering how a function behaves as you zoom in on a point. A differentiable function has a well-defined tangent plane at that point. If a function is not continuous at a point, it cannot be differentiable there, because differentiability implies continuity.

In the problem, since the function \( f(x, y) \) is not continuous at \((0,0)\), it's immediately impossible for it to be differentiable at that point. Differentiability would require not only continuous behavior but also consistent, smooth changes without sudden shifts or undefined behavior.

As such, for \( f(x,y) \), the discontinuity indicates the absence of a derivative or tangent plane at \((0,0)\). This aligns with the mathematical rule that differentiability implies continuity.

Polar Coordinates

Polar coordinates present a different way of describing points in a plane using a radius and angle, \(r\) and \(\theta\), respectively. They are highly useful in analyzing limits and continuity, especially when dealing with circles or rotational symmetry.

When applying these coordinates in the given problem, we redefine \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\). This transformation simplifies the limit analysis by consolidating variables under a single parameter \(r\).

In the problem, polar coordinates reveal that the limit of our function does not settle on a single value as \(r \to 0\). Instead, it varies with \(\theta\), exposing the function's lack of continuity. This technique is invaluable when dealing with multi-variable limits.

Limit Analysis

Limit analysis is foundational in understanding behavior near a specific point, such as determining continuity or differentiability. This involves evaluating the function’s behavior as it "approaches" the point in question.

The exercise uses limit analysis to ascertain the behavior of \( f(x, y) \) near \((0, 0)\). We compute \( \lim_{(x, y) \to (0,0)} \frac{-xy}{x^2 + y^2} \) and analyze if the result is consistent from all directions using polar coordinates.

For our function, the limit \( -\sin(\theta)\cos(\theta) \) shows dependability on angle \(\theta\), meaning the function's path to a point changes its outcome. As such, limit analysis unveils that the function \( f(x, y) \) is discontinuous at \((0, 0)\), and emphasizes how an inconsistent limit leads to discontinuity. This practice is crucial for functions with complex behaviors near the origin.