Question

(i) If \(f: \mathbb{R}^{2} \rightarrow \mathbb{R}\) is such that \(|f(s, t)| \leq|s t|\) for all \((s, t) \in \mathbb{R}^{2}\), prove that \(f\) is differentiable at \((0,0)\). Was it necessary to have the condition on \(f\) holding for all \((s, t) \in \mathbb{R}^{2}\) ? (ii) Let \(f: \mathbb{R}^{2} \rightarrow \mathbb{R}\) be defined by \(f(s, t)=s|t|\). At which points in \(\mathbb{R}^{2}\) is \(f\) differentiable?

Short answer

In conclusion: (i) We have shown that the function \(f\) is differentiable at the point \((0,0)\), even without the given condition for all \((s,t)\). The function is differentiable at \((0,0)\) by showing the existence of partial derivatives and the appropriate limit at this point. (ii) For the function \(f(s,t) = s|t|\), we have determined that it is differentiable at all points \((s,t) \in \mathbb{R}^2\) except for points of the form \((s,0)\) for \(s \neq 0\), i.e., along the \(s\)-axis (excluding the origin). The function is differentiable at \((0,0)\) as proven in part (i).

Step-by-step solution

Compute the partial derivatives at \((0, 0)\)

Firstly, we will need to calculate the partial derivatives of \(f(s,t)\) with respect to \(s\) and \(t\), at the point \((0, 0)\). Observe that: \(\lim_{s\to 0} \frac{f(s, 0) - f(0, 0)}{s} = \lim_{s\to 0} \frac{f(s, 0)}{s} \leq \lim_{s\to 0} \frac{|s\cdot0|}{s} = \lim_{s\to 0} 0 = 0\) \(\lim_{t\to 0} \frac{f(0, t) - f(0, 0)}{t} = \lim_{t\to 0} \frac{f(0, t)}{t} \leq \lim_{t\to 0} \frac{|0\cdot t| }{t} = \lim_{t\to 0} 0 = 0\) From the above calculations, we have that the partial derivatives of \(f(s,t)\) with respect to \(s\) and \(t\) at \((0, 0)\) are \(\frac{\partial{f}}{\partial{s}}(0,0) = 0\) and \(\frac{\partial{f}}{\partial{t}}(0,0) = 0\).

Evaluate the limit definition of the derivative

To prove that \(f\) is differentiable at \((0,0)\), we need to show that the following limit exists for \((h,k) \to (0,0)\): \(\lim_{(h,k)\rightarrow(0,0)} \frac{f(0+h, 0+k) - f(0,0) - \frac{\partial{f}}{\partial{s}}(0, 0)h - \frac{\partial{f}}{\partial{t}}(0, 0)k}{\sqrt{h^2 + k^2}}\) \(\lim_{(h,k)\rightarrow(0,0)} \frac{f(h,k)}{\sqrt{h^2 + k^2}}\) Since we know that \(|f(s,t)|\leq|s t|\) for all \((s,t)\in\mathbb{R}^2\), we get: \(|\frac{f(h,k)}{\sqrt{h^2 + k^2}}| \leq \frac{|hk|}{\sqrt{h^2 + k^2}}\) As \((h,k) \to (0, 0)\), then the limit exists and is equal to 0: \(\lim_{(h,k)\rightarrow(0,0)} \frac{f(h,k)}{\sqrt{h^2 + k^2}} = 0\) Thus, the function \(f\) is differentiable at \((0,0)\). For the given condition to hold for all \((s,t)\), then it would not be necessary. #For part (ii):#

Find the partial derivatives of the given function

We are given that \(f(s,t) = s|t|\). Let's find the partial derivatives of \(f\) with respect to \(s\) and \(t\): \(\frac{\partial{f}}{\partial{s}} = |t|\) $\frac{\partial{f}}{\partial{t}} = \begin{cases} s & \text{if } t > 0 \\ -s & \text{if } t < 0 \\ \text{undefined} & \text{if } t = 0 \end{cases}$

Determine where the partial derivatives are continuous

Observe that \(\frac{\partial{f}}{\partial{s}}\) is continuous everywhere. However, \(\frac{\partial{f}}{\partial{t}}\) is not continuous at \(t=0\). This means that for any fixed non-zero \(s\), \(f\) will be differentiable on the line \(t=0\), but for \(t=0\) and \(s\neq0\), the function \(f\) will not be differentiable. So, the function \(f(s,t) = s|t|\) is differentiable at all points \((s,t) \in \mathbb{R}^2\) except for points on the \(s\)-axis i.e., points of the form \((s,0)\) for \(s \neq 0\). Note that in part (i), we have already proven that \(f\) is differentiable at \((0,0)\).

Key concepts

Limit Definition of Derivative

The limit definition of derivative lies at the heart of understanding differentiability in multivariable calculus. This foundational concept is based on the idea that a function is differentiable at a point if the function closely approximates its tangent plane in the vicinity of that point. Formally, a two-variable function \( f(x, y) \) is differentiable at a point \( (a, b) \) if the limit

\[ \lim_{{(h,k)\rightarrow(0,0)}} \frac{f(a+h, b+k) - f(a,b) - f_x(a, b)h - f_y(a, b)k}{\sqrt{h^2 + k^2}} \]
exhibits a value of zero. Here, \( f_x(a, b) \) and \( f_y(a, b) \) are the partial derivatives of \( f \) with respect to \( x \) and \( y \) at \( (a, b) \), respectively. These derivatives represent the slopes of the tangent plane to the surface created by \( f \) at the given point. In the provided exercise, by showing that the limit equals zero, we verify the function is differentiable at the origin.

Partial Derivatives

Partial derivatives are a pivotal concept when discussing the differentiability of functions in higher dimensions. They measure the rate at which a function changes as one of its input variables is varied, while the others are held constant. If you have a function \( f(x, y) \), the partial derivative with respect to \( x \), denoted \( \frac{\partial f}{\partial x} \), tells you how \( f \) changes as \( x \) changes when \( y \) is fixed, and vice versa for \( \frac{\partial f}{\partial y} \).

In the exercise example, the partial derivatives at the origin capture how the function behaves in the immediate neighborhood of \( (0, 0) \). By calculating these derivatives, it was proven that they both equal zero, suggesting that at the origin, the surface is flat.

Continuity of Partial Derivatives

For a function to be smooth and differentiable, it is often necessary but not sufficient that its partial derivatives exist and are continuous in the region of interest. Continuous partial derivatives provide us with a smooth surface that does not have any sharp edges or abrupt changes in slope - often a prerequisite for differentiability. However, there are exceptions to this rule; a classic example is the function in part (i) of the exercise where differentiability at a point is proven without the need for continuous partial derivatives.

However, in most cases and as seen in part (ii) of the exercise, the discontinuity of \( \frac{\partial f}{\partial t} \) at \( t = 0 \) indicates that the function cannot have a tangent plane at points along the \( s \)-axis, thus \( f \) is not differentiable at those points, except at the origin where a different argument about differentiability was used.

Differentiable Functions

Differentiable functions in multivariable calculus exhibit a smoothness that allows for linear approximation near a point. This notion extends our understanding of derivatives from single-variable calculus into higher dimensions. When a function is differentiable at a point, it behaves nicely enough around that point such that it can be approximated by a plane—the tangent plane. This property is crucial for utilizing calculus to solve real-world problems where slight changes in input can be accurately predicted to result in changes in output.

In the context of the exercise, we saw that the function \( f \) met the criteria for differentiability at the origin, as its behavior could be akin to a flat plane at \( (0, 0) \). However, the function's differentiability failed at certain points along the \( s \)-axis due to a problem with the partial derivative with respect to \( t \), showing that differentiability is a local property and must be verified at each point individually.