Problem 44 - Calculus an Applied Approach

Question

Find the rate of change of \(x\) with respect to \(p .\) \(p=\sqrt{\frac{500-x}{2 x}}, \quad 0

Step-by-Step Solution

Answer

The rate of change of \(x\) with respect to \(p\) is \(-\frac{4x^3p}{x-500}\).

Step 1: Rewrite the expression

We start by rewriting the equation as: \( p^2 = \frac{500-x}{2x} \). This will make it suitable for applying derivative rules.

Step 2: Implicit Differentiation

Next, we will differentiate both sides with respect to \(x\). Treat \(p\) as a function of \(x\) and apply the chain rule: \( 2p\frac{dp}{dx} = -\frac{1}{2x^2} *(500-x) - \frac{1}{2x}.\) This equation gives us the expression for \(\frac{dp}{dx}\).

Step 3: Solving for rate of change

Finally, we want to solve for \(\frac{dx}{dp}\), which is the reciprocal of \(\frac{dp}{dx}\). Therefore, \(\frac{dx}{dp} = \frac{1}{\frac{dp}{dx}} = -\frac{4x^3p}{x-500}\).

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used when a function is not explicitly solved for one variable in terms of another. In such cases, it's not possible to directly differentiate the function as you would in standard differentiation. Instead, you assume that the variables are implicitly connected through some function.

For example, when given an equation like \(p^2 = \frac{500-x}{2x}\), where \(p\) is a function of \(x\), we differentiate both sides with respect to \(x\) while treating \(p\) as a function that implicitly depends on \(x\). This involves using derivative rules and the chain rule to take the derivative of each term. The challenging part is to differentiate terms involving \(p\) correctly, as one needs to multiply by \(\frac{dp}{dx}\) due to the chain rule. Through this process, we can find the relationship between the rates of change of the variables involved, such as \(\frac{dp}{dx}\) or \(\frac{dx}{dp}\). When solving for these rates, you are deciphering how one variable changes in relation to another within an implicitly defined function.

Derivative Rules

Understanding derivative rules is essential for solving calculus problems effectively. These rules, including the power rule, product rule, quotient rule, and chain rule, allow us to take derivatives of various functions systematically. For instance, the power rule states that for any algebraic expression of the form \(x^n\), the derivative is \(nx^{n-1}\).

Applying these rules to the given problem, we use the constant multiple rule to bring out the constants, and the difference rule for the expression \(500-x\) inside the square root. By using these rules, we systematically take the derivative of complex expressions. As you can see in the step-by-step solution, the derivative of \(p^2\) is simple by the power rule, but the right side of the equation requires careful application of the quotient rule and the power rule to differentiate correctly. Knowing when and how to apply each rule is key to finding the derivative efficiently and accurately.

Chain Rule

The chain rule is indispensable when you need to differentiate a function that is composed of other functions — a function inside a function. It helps you take derivatives of composite functions where one function is nested within another. The rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

In our exercise, \(p\) being the outer function and \(\sqrt{\frac{500-x}{2x}}\), the inner function, necessitates the use of the chain rule to find \(\frac{dp}{dx}\). When performing implicit differentiation on the equation \(p^2 = \frac{500-x}{2x}\), we must apply the chain rule to account for the fact that \(p\) is itself a function of \(x\), which is not directly stated but implied by the equation. Therefore, after taking the derivative of the outer function \(p^2\), we multiply by \(\frac{dp}{dx}\), which is the rate of change of \(p\) with respect to \(x\), to complete the differentiation process according to the chain rule.

Solving for Derivatives

Solving for derivatives is the process of finding the rate at which one variable changes with respect to another. It involves differentiating an equation and possibly rearranging it to isolate the desired derivative. Specifically, in our exercise, after differentiating the equation using implicit differentiation, we obtain an expression for \(\frac{dp}{dx}\).

However, we are interested in the rate of change of \(x\) with respect to \(p\), which is the inverse rate \(\frac{dx}{dp}\). To find this, we take the reciprocal of \(\frac{dp}{dx}\), which yields \(\frac{dx}{dp} = \frac{1}{\frac{dp}{dx}}\). The final step is to simplify the expression to make the answer clearer. This might involve multiplying by a conjugate, factoring, or simplifying fractions. For students looking to fully grasp how changes in one variable affect another, becoming adept at solving for derivatives is a powerful tool in calculus.