Question

Let \(f: \mathbb{R}^{2} \rightarrow \mathbb{R}\) be differentiable. Define \(g: \mathbb{R} \rightarrow \mathbb{R}\) by \(g(s)=\) \(f(s, c-s)\), where \(c\) is a constant. Write down an expression for \(g^{\prime}(s) .\) If \(D_{1} f=D_{2} f\) on \(\mathbb{R}^{2}\), deduce that \(f(s, t)=h(s+t)\) for some \(h: \mathbb{R} \rightarrow \mathbb{R}\).

Short answer

Question: Find an expression for \(g'(s)\) and prove that \(f(s,t) = h(s+t)\) for some function \(h : \mathbb{R} \rightarrow \mathbb{R}\), given that \(g(s) = f(s, c-s)\) and the partial derivatives \(D_1 f = D_2 f\) on \(\mathbb{R}^2\). Answer: We have found that \(g'(s) = 0\), and proved that \(f(s,t) = h(s+t)\) for some function \(h : \mathbb{R} \rightarrow \mathbb{R}\).

Step-by-step solution

Calculate the gradient of \(f\)#

Since \(f\) is differentiable, we can compute its gradient, denoted by \(\nabla f\), which is a vector containing the partial derivatives of \(f\) with respect to its variables: $$ \nabla f = \begin{bmatrix} D_1 f \\ D_2 f \end{bmatrix}. $$ Keep in mind that we are given the condition \(D_1 f = D_2 f\).

Apply the chain rule to \(g(s)\)#

Now we will find the derivative of \(g\) with respect to \(s\). Apply the chain rule, which states that the derivative of the composition of two functions is the product of the gradient of the outer function and the derivative of the inner function: $$ g'(s) = \nabla f(s, c-s) \cdot \begin{bmatrix} \frac{d}{ds}(s) \\ \frac{d}{ds}(c-s) \end{bmatrix}. $$

Compute the derivative of the inner functions#

Compute the derivatives of the inner functions with respect to \(s\): $$ \begin{bmatrix} \frac{d}{ds}(s) \\ \frac{d}{ds}(c-s) \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}. $$

Calculate \(g'(s)\)#

Substitute the derivatives into our expression for \(g'(s)\) from step 2: $$ g'(s) = \begin{bmatrix} D_1 f(s, c-s) \\ D_2 f(s, c-s) \end{bmatrix} \cdot \begin{bmatrix} 1 \\ -1 \end{bmatrix} = D_1 f(s, c-s) - D_2 f(s, c-s). $$

Use the condition \(D_1 f = D_2 f\)#

Now, substitute the given condition \(D_1 f = D_2 f\) into our expression for \(g'(s)\): $$ g'(s) = D_1 f(s, c-s) - D_1 f(s, c-s) = 0. $$ This implies that \(g'(s)\) is always zero, meaning that \(g(s)\) is a constant function. Thus, we can write \(g(s) = k\), where \(k\) is a constant.

Rewrite \(f(s,t)\) as \(h(s+t)\)#

Since \(g(s) = f(s, c-s) = k\), we can interpret \(f(s,c-s)\) as a constant function of \(s+t\). Define a new function \(h : \mathbb{R} \rightarrow \mathbb{R}\) such that \(h(s+t) = f(s, c-s)\). Then, we can rewrite \(f(s,t)\) as \(h(s+t)\), and this function \(h\) satisfies the given conditions.

Key concepts

Gradient of a Function

In calculus and vector analysis, the gradient of a function is a crucial concept, particularly when dealing with functions of several variables. Imagine the gradient as an arrow pointing in the direction of the steepest incline of a multivariate function.

For a function like our differentiable function f that maps \( \mathbb{R}^{2} \) into \( \mathbb{R} \), the gradient is made up of the partial derivatives of \( f \) with respect to each of its variables, which can be visually represented as \( abla f = \begin{bmatrix} D_1 f \ D_2 f \end{bmatrix} \). The gradient demonstrates how much the function changes when we make infinitesimally small movements in the x or y directions, or in geometric terms, the slope of the tangent plane to the surface at any given point.

In the context of the given exercise, our prime interest is in the properties of these derivatives, especially since we are given the unique condition that \( D_1 f = D_2 f \), a rarity that suggests the function increases at the same rate in both the x and y directions. This ties into the broader topic of symmetric functions, and such a property guides us toward simplifying expressions for the function's behavior.

Chain Rule

Understanding the Chain Rule

The chain rule is a fundamental theorem in calculus that enables us to differentiate the composition of two or more functions. It provides a method to calculate the derivative of a complex function by breaking it down into simpler parts. As a powerful tool, the chain rule crosses over from simple one-variable calculus into the realm of multivariable calculus.

Given a function like \( g(s) = f(s, c-s) \), where \( f \) is a function of two variables, to find \( g'\), the derivative of \( g \) with respect to \( s \), we invoke the chain rule. The formula looks something like this: \( g'(s) = abla f(s, c-s) \cdot \begin{bmatrix} 1 \ -1 \end{bmatrix} \).

Here, you are seeing the gradient of \( f \) being dot-multiplied with the gradient vector of the inner functions, a concise way of saying that the change in \( g \) with respect to \( s \) is influenced by how \( f \) changes in both the \( s \) and \( c-s \) directions.

Partial Derivatives

Partial Derivatives and Their Role

Partial derivatives play a vital role in multivariable calculus by measuring the rate at which a function changes with respect to one variable while keeping other variables constant.

Take, for instance, our function \( f \) from the exercise: it depends on two variables, and the partial derivative \( D_1 f \) represents the rate of change of \( f \) with respect to the first variable, and \( D_2 f \) with respect to the second variable. A salient feature of partial derivatives is that they provide the slopes of the tangents to the curves obtained by intersecting the graph of the function with planes parallel to the coordinate axes.

When the problem states that \( D_1 f = D_2 f \), it simplifies the scenario significantly because it implies that the slopes of the function in both the x and y directions are identical, facilitating the deduction that the function \( f \) can be expressed in terms of another function \( h \) that depends on the sum of its variables, hence \( f(s, t) = h(s+t) \). This ties into the concept of level curves and contour lines of a function, resources that provide graphical representation for understanding such relationships between variables.