Question

Geometric and Arithmetic Mean The geometric mean of \(u\) and \(v\) is \(G=\sqrt{u v}\) and the arithmetic mean is \(A=(u+v) / 2 .\) Show that if \(u=x, v=x+c, c\) a real number, then $$\frac{d G}{d x}=\frac{A}{G}$$

Short answer

The derivative of the geometric mean with respect to \(x\) is equal to the ratio of the geometric mean to the arithmetic mean.

Step-by-step solution

Calculate the Arithmetic Mean

Given the two numbers \(u = x\) and \(v = x + c\), where \(c\) is a real number, the arithmetic mean \(A\) is calculated as \(A = (u + v) / 2 = (x + (x + c)) / 2 = x + c/2\).

Calculate the Geometric Mean

The geometric mean \(G\) of the numbers \(u = x\) and \(v = x + c\) is \(G = \sqrt{u*v} = \sqrt{x * (x + c)}\)

Differentiate the Geometric Mean with respect to \(x\)

Using the chain rule for differentiation, (where the derivative of \(f(g(x))\) is \(f'(g(x))*g'(x)\)), we get \(\frac{d G}{d x} = \frac{1}{2\sqrt{x*(x+c)}} * (x+c+x)= \frac{G}{x+c/2} = \frac{G}{A} \) This involves simplification after differentiation.

Compare the derivatives

We can see from the above step that \( \frac{d G}{d x} = \frac{G}{A} \), which is what we were asked to show.

Key concepts

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It is a foundational tool in many sciences and engineering fields, as it provides a framework for modeling and understanding changes within various systems. There are two major branches of calculus: differential calculus, which concerns the rate of change and slopes of curves, and integral calculus, which deals with the accumulation of quantities and the areas under and between curves.

In the given exercise, calculus principles are being applied to find a relationship between the rates of change (derivatives) of the geometric and arithmetic means with respect to a variable. The process shows how calculus can be used to explore and prove mathematical properties and relationships.

Differentiation

Differentiation is a fundamental operation in calculus that measures how a function changes as its input changes. The derivative of a function at a certain point represents the slope of the tangent line to the function graph at that point, which is the instantaneous rate of change of the function with respect to one of its variables.

In the context of our problem, the differentiation of the geometric mean with respect to the variable, which is a part of the process, is crucial to establishing the relationship \(\frac{dG}{dx} = \frac{A}{G}\). Differentiation allows us to quantify the sensitivity of the geometric mean to changes in the variable \(x\).

Chain Rule

The chain rule is an essential differentiation rule in calculus that allows computation of the derivative of a composite function. If \(f(x)\) is a function and \(g(x)\) is another function, then the derivative of the composite function \(f(g(x))\) can be found using the chain rule: \(\frac{df}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx}\).

For our exercise, we apply the chain rule to differentiate the geometric mean \(G\). Since the geometric mean is a function of both \(x\) and \(x+c\), we must use the chain rule to differentiate \(G\) with respect to \(x\) accurately. This step is crucial in obtaining the final derivative that shows the relationship between \(\frac{dG}{dx}\) and the ratio of the arithmetic mean to the geometric mean.

Mean Value Theorem

The Mean Value Theorem is one of the most important theorems in calculus. It basically states that if a function \(f(x)\) is continuous on the closed interval \[a, b\] and differentiable on the open interval \(a, b\), then there exists at least one point \(c\) in the interval \(a, b\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\). This theorem guarantees that there is at least one point on a smooth curve where the instantaneous rate of change (the derivative) is the same as the average rate of change over the interval.

While the Mean Value Theorem is not directly applied in this exercise, understanding this theorem is vital because it tells us about the existence of specific points that quantify the idea of averaging the rates of change, analogous to finding means in a broader sense. It sets the stage for comprehending more complex relationships like the one shown between the geometric and arithmetic means' rates of change.