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 15.5 and 15.6 are consistent with the results in parts (a)-(d). $$f(x, y)=\left\\{\begin{array}{ll}\frac{2 x y^{2}}{x^{2}+y^{4}} & \text { if }(x, y) \neq(0,0) \\\0 & \text { if }(x, y)=(0,0)\end{array}\right.$$

Short answer

In conclusion, the function \(f(x, y) = \frac{2xy^2}{x^2 + y^4}\) is continuous and differentiable at the point (0, 0), and its partial derivatives \(f_x(0, 0)\) and \(f_y(0, 0)\) are also continuous at this point. These findings are consistent with Theorems 15.5 and 15.6, which establish the relationships between continuity, differentiability, and the existence of continuous partial derivatives.

Step-by-step solution

Part a - Continuity of f at (0,0)

To determine if \(f\) is continuous at (0, 0), we must check whether the limit of \(f(x, y)\) as \((x, y) \to (0, 0)\) exists and is equal to \(f(0,0) = 0\). Let's compute the limit: $$\lim_{(x, y) \to (0, 0)} f(x, y) = \lim_{(x, y) \to (0, 0)} \frac{2xy^2}{x^2 + y^4}$$ To compute this limit, we can use the polar coordinate system, by setting \(x = r\cos\theta\) and \(y= r\sin\theta\), where \(r > 0\) . Then, the limit becomes: $$\lim_{r\to 0} \frac{2(r\cos\theta)(r\sin\theta)^2}{(r\cos\theta)^2 + (r\sin\theta)^4}$$ Notice that the \(r\) term can be factored out from the numerator and denominator: $$\lim_{r\to 0} \frac{2r^3\cos\theta\sin^2\theta}{r^2\cos^2\theta + r^4\sin^4\theta}$$ Now, divide both numerator and denominator by \(r^2\): $$\lim_{r\to 0} \frac{2r\cos\theta\sin^2\theta}{\cos^2\theta + r^2\sin^4\theta}$$ As \(r \to 0\), the expression becomes: $$\frac{0}{\cos^2\theta + 0} = 0$$ Since the limit exists and equals \(f(0,0)\), the function is continuous at (0,0).

Part b - Differentiability of f at (0,0)

In order to determine if \(f\) is differentiable at (0,0), we need to compute the partial derivatives \(f_x\) and \(f_y\) at this point.

Part c - Evaluating \(f_x(0,0)\) and \(f_y(0,0)\)

Let's calculate the partial derivatives: $$f_x(x, y) = \frac{\partial}{\partial x}\left(\frac{2xy^2}{x^2 + y^4}\right) = \frac{2y^2(x^2 + y^4) - 2x^2y^2(2x)}{(x^2 + y^4)^2}$$ $$f_y(x, y) = \frac{\partial}{\partial y}\left(\frac{2xy^2}{x^2 + y^4}\right) = \frac{2x(x^2 + y^4) - 2xy^2(4y^3)}{(x^2 + y^4)^2}$$ Note that for both \(f_x(0,0)\) and \(f_y(0,0)\), the numerator becomes 0, so their values are: $$f_x(0,0) = 0$$ $$f_y(0,0) = 0$$

Part d - Continuity of \(f_x\) and \(f_y\) at (0,0)

Now, we need to determine if \(f_x\) and \(f_y\) are continuous at the point (0,0). We will apply the same method as in Part a: $$\lim_{(x, y) \to (0, 0)} f_x(x, y) = \lim_{(x, y) \to (0, 0)} \frac{2y^2(x^2 + y^4) - 2x^2y^2(2x)}{(x^2 + y^4)^2}$$ $$\lim_{(x, y) \to (0, 0)} f_y(x, y) = \lim_{(x, y) \to (0, 0)} \frac{2x(x^2 + y^4) - 2xy^2(4y^3)}{(x^2 + y^4)^2}$$ Converting to polar coordinates and considering the limit as \(r \to 0\), we get: $$\lim_{r\to 0} f_x(r\cos\theta, r\sin\theta) = 0$$ $$\lim_{r\to 0} f_y(r\cos\theta, r\sin\theta) = 0$$ Since both of these limits exist and equal the values of \(f_x(0,0)\) and \(f_y(0,0)\), the partial derivatives are continuous at (0,0).

Part e - Consistency with Theorems 15.5 and 15.6

Theorem 15.5 states that if all first-order partial derivatives are continuous in a neighborhood of \((a,b)\), then the function is differentiable at \((a,b)\). In our case, the values of \(f_x(0,0)\) and \(f_y(0,0)\) exist and are continuous, so according to Theorem 15.5, the function \(f\) is differentiable at (0,0). Theorem 15.6 states that if a function is differentiable at \((a,b)\), then it is also continuous at that point. Our results in Part a show the function is continuous at (0,0), and Parts b, c, and d show that \(f\) is differentiable and its partial derivatives are continuous at this point. Therefore, our results are consistent with both Theorems 15.5 and 15.6.

Key concepts

Continuity

Continuity in multivariable functions is akin to the unbroken flow of a single-variable function. For a function to be continuous at a point, the limit must exist and match the function's value there. Let's consider the specific function given in the exercise. To determine if it's continuous at (0,0), we check if \(\lim_{(x, y) \to (0, 0)} f(x, y) = f(0,0) \).
Here, this involves finding \(\lim_{(x, y) \to (0, 0)} \frac{2xy^2}{x^2 + y^4} \).Switching to polar coordinates simplifies evaluating the limit as follows: \(x = r\cos\theta\) and \(y = r\sin\theta\). The expression then becomes \(\lim_{r\to 0} \frac{2r^3\cos\theta\sin^2\theta}{r^2\cos^2\theta + r^4\sin^4\theta} \).Dividing through by \(r^2\) results in \(\lim_{r\to 0} \frac{2r\cos\theta\sin^2\theta}{\cos^2\theta + r^2\sin^4\theta} = 0\).Since this limit equals \(f(0,0) = 0\), the function \(f\) is continuous at (0,0).Using polar coordinates is particularly helpful here because it considers all possible paths simultaneously as \(r\to 0\).

Partial Derivatives

Partial derivatives are essential tools to grasp changes in multivariate functions. They represent the rate of change with respect to one variable while holding others constant. For our function, we find the partial derivatives \(f_x(x, y)\) and \(f_y(x, y)\).
The expression for \(f_x(x, y) = \frac{\partial}{\partial x}\left(\frac{2xy^2}{x^2 + y^4}\right)\) involves applying the quotient rule, resulting in:\[f_x(x, y) = \frac{2y^2(x^2 + y^4) - 2x^2y^2(2x)}{(x^2 + y^4)^2}\]Similarly, the expression for \(f_y(x, y) = \frac{\partial}{\partial y}\left(\frac{2xy^2}{x^2 + y^4}\right)\) yields:\[f_y(x, y) = \frac{2x(x^2 + y^4) - 2xy^2(4y^3)}{(x^2 + y^4)^2}\]At the point (0,0), both derivatives \(f_x(0,0)\) and \(f_y(0,0)\) compute to zero since their numerators become zero at this point.Partial derivatives help determine if the function is locally linear (differentiable) at the point in question.

Polar Coordinates

Polar coordinates offer a different perspective for analyzing multivariable functions, converting typical \((x, y)\) coordinates into \((r, \theta)\), where \(r\) represents the radial distance from the origin and \(\theta\) the angle.
This transformation is particularly valuable when evaluating limits at points like the origin. By expressing \(x\) as \(r\cos\theta\) and \(y\) as \(r\sin\theta\), we can rewrite our function's limit expression in terms without singularity issues at (0,0).Dividing by \(r^2\) in the limit process further simplifies, allowing the expression to retain dependency only on \(r\), which we can easily evaluate.Using polar coordinates, paths to zero are considered collectively, guaranteeing consistency regardless of approach, a crucial aspect for establishing continuity and differentiability within multivariable contexts.

Limits in Multivariable Calculus

Limits in multivariable calculus extend the concept from single variables by considering multiple inputs converging towards a point. Evaluating these limits can help us ascertain both continuity and differentiability.
The key challenge lies in ensuring the limit is the same no matter which path or direction we approach the point of interest.For the function \(f(x, y)\), the standard approach is to examine assorted paths to the origin, such as via the x-axis, y-axis, y = mx lines, or using polar coordinates.Polar coordinates are especially effective, as transforming into \(r\) and \(\theta\) consolidates potential pathway inconsistencies into a singular radially focused form.For \(f(x, y) = \frac{2xy^2}{x^2 + y^4}\), \(\lim_{(x,y) \to (0,0)} f(x,y)\) was analyzed, and carefully manipulating the expression led us to conclude it to be zero.This not only confirmed continuity but also opened doors to explore differentiable aspects through partial derivatives.Understanding limits across varied pathways ensures thorough comprehension of function behavior near specified points.