Problem 41 - Calculus an Applied Approach

Question

Find the rate of change of \(x\) with respect to \(p .\) \(p=\frac{2}{0.00001 x^{3}+0.1 x} \quad x \geq 0\)

Step-by-Step Solution

Answer

The rate of change of \(x\) with respect to \(p\) is \( \frac{dx}{dp} = - \frac{1}{ p \frac{ (0.00003 x^{2} + 0.1)}{0.00001 x^{3}+0.1 x}} \)

Step 1: Rewrite the Function

The given function is \( p = \frac{2}{0.00001 x^{3} + 0.1 x} \). We will rewrite it to a form more suitable for implicit differentiation. Multiply both sides by \( 0.00001 x^{3} + 0.1 x \) to get rid of the fraction: \( p(0.00001 x^{3} + 0.1 x) = 2 \).

Step 2: Take the Derivative

Apply the derivative to both sides to implicitly differentiate the function. Remember that \(p\) is a function of \(x\) and the chain rule is applied to differentiate \(p\). The derivative is: \( p'(0.00001 x^{3} + 0.1 x) + p(0.00003 x^{2} + 0.1) = 0 \). Therefore, \( p' = -p \frac{ (0.00003 x^{2} + 0.100)}{(0.00001 x^{3} + 0.1 x)} \) . Note that \(p'\) represents the derivative of \(p\) with respect to \(x\). We are interested in the opposite \(\frac{dx}{dp}\), so we need to find the reciprocal.

Step 3: Find the Reciprocal

The rate of change of \(x\) with respect to \(p\) is the reciprocal of the derivative found in step 2. The result is: \( \frac{dx}{dp} = - \frac{1}{ p \frac{ (0.00003 x^{2} + 0.1)}{0.00001 x^{3}+0.1 x}} \)

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus, used to differentiate composite functions. Imagine you have a function that is composed of two or more functions, like when you nest functions inside each other. The chain rule helps you find the derivative of such a function by breaking it down into simpler parts.
To apply the chain rule, follow these steps:

• Differentiation of the outer function first.
• Then multiply by the derivative of the inner function.

This method is especially useful when dealing with functions that depend on other variables. In our problem, the chain rule helps differentiate the function with respect to another variable, making complex differentiations manageable.

Reciprocal

A reciprocal is simply the inverse of a number or expression. For example, the reciprocal of a number \(a\) is \(\frac{1}{a}\). In calculus, finding the reciprocal of a derivative is essential when you want to switch from the rate of change of one quantity with respect to another.
In the problem, after finding the derivative \(p'\), which is \(\frac{dp}{dx}\), we are actually interested in \(\frac{dx}{dp}\).

• Taking the reciprocal lets us switch the focus from how \(p\) changes with \(x\), to how \(x\) changes with \(p\).

This transformation is crucial for solving certain types of calculus problems.

Rate of Change

The rate of change is a measure of how one quantity changes in relation to another. In most practical scenarios, it tells us how fast or slow a particular variable, like speed, volume, or temperature, is changing.
In mathematical terms, rate of change is often represented as a derivative. So when we say \(\frac{dp}{dx}\), we mean the rate at which \(p\) changes with respect to \(x\).
In our problem:

• We aim to determine how \(x\) alters when \(p\) shifts, represented as \(\frac{dx}{dp}\).

This helps us understand the relationship dynamics between two variables.

Derivative

A derivative represents an instantaneous rate of change or the slope of a function at any given point. It's a core concept in calculus that lets us find how a function changes as its input changes.
To calculate the derivative, we apply important rules like the power rule, product rule, and our previously discussed chain rule.
In the context of the given exercise, the derivative \(p'\) we calculated shows how \(p\) changes with \(x\).

• Once the derivative was found, we transformed it to focus on \(\frac{dx}{dp}\).

Thus, derivatives serve as the mathematical tool to explore and analyze changes within functions.