Problem 43 - Calculus an Applied Approach

Question

Find the rate of change of \(x\) with respect to \(p .\) \(p=\sqrt{\frac{200-x}{2 x}}, \quad 0

Step-by-Step Solution

Answer

The rate of change of \(x\) with respect to \(p\) is \(-\frac{(200-3x)*(200-x)}{(8x^3)}\).

Step 1: Rewriting the equation

Rewrite the equation \(p=\sqrt{\frac{200-x}{2 x}}\) as \(p^2 = \frac{200-x}{2 x}\), to simplify the differentiation process.

Step 2: Differentiate

Differentiate both sides of the equation with respect to \(x\). \[2p\frac{dp}{dx} = -\frac{200-3x}{2x^2}\] Remember the derivative of \(p\) with respect to \(x\) is denoted as \(\frac{dp}{dx}\) or \(p'\).

Step 3: Solve for dp/dx

We now need to find the derivative of \(p\) with respect to \(x\), \(dp/dx\), by rearranging the equation. So \[\frac{dp}{dx} = -\frac{(200-3x)*p}{4x^2}\].

Step 4: Substitute p value

Replace \(p\) in the equation by \(\sqrt{\frac{200-x}{2x}}\) to get \[\frac{dp}{dx} = -\frac{(200-3x)*\sqrt{\frac{200-x}{2x}}}{4x^2}.\]\] To simplify further, we can multiply both numerator and denominator by \(2x\) to get rid of the fractions.

Step 5: Simplification

After multiplying the top and bottom of the fraction by \(2x\), we get \[\frac{dp}{dx} = -\frac{(200-3x)*(200-x)}{(8x^3)}.\]

Key Concepts

Differentiation

Differentiation is a crucial concept in calculus that helps us understand how a function changes at any given point. In our exercise, we're interested in finding how the variable \( x \) changes with respect to \( p \), which means we are looking at the rate of change.
Differentiation involves taking derivatives, which are essentially the slopes of the tangent lines to the function at a point. This is a powerful tool because it tells us how steeply the function is changing:

• When the derivative is positive, the function is increasing at that point.
• If the derivative is negative, the function is decreasing.
• A zero derivative means the function is flat, which often indicates a local maximum or minimum.

In the context of our problem, to handle the differentiation on both sides of the transformed equation \( p^2 = \frac{200-x}{2x} \), we applied the chain rule, which is used when differentiating compositions of functions. Understanding how to apply the chain rule and differentiate such equations is a fundamental skill in calculus.

Derivative

A derivative is a measure of how a function changes as its input changes. It's the key output of the differentiation process. In mathematical terms, the derivative provides us with the rate of change.
For instance, in this exercise, differentiating both sides of the equation with respect to \( x \) involved recognizing how \( p \) and other terms varied with \( x \). The derivative \( \frac{dp}{dx} \) gives us the rate at which \( p \) changes with \( x \).

• The left side of the differential equation uses the product rule, involving both \( p \) and its derivative \( \frac{dp}{dx} \).
• The right side requires understanding of simple algebraic differentiation where we utilized the power rule and constant rule to obtain the expression \(-\frac{200-3x}{2x^2}\).

These processes highlight how derivatives play an important role in understanding changes and behaviors of functions in complex expressions. Taking derivatives step by step allows us to simplify and solve complex rate-of-change problems efficiently.

Implicit Differentiation

Implicit differentiation is useful when dealing with equations where one variable is not isolated on either side. This comes in handy when directly solving for one variable isn't feasible, as in complex expressions involving square roots or fractions.
In our example, the original equation \( p = \sqrt{\frac{200-x}{2x}} \) was rewritten as \( p^2 = \frac{200-x}{2x} \). By squaring both sides, we made differentiation more straightforward, as it avoided the square root.

• Here we used implicit differentiation on \( p^2 \), treating \( p \) as a function of \( x \), which let us deduce the equation \( 2p \frac{dp}{dx} \).
• Implicit differentiation helps break down functions where separating variables is impractical due to intertwining dependencies.
• Ultimately, complex rates and interactions are simplified to linear relationships or expressions easier to solve.

Implicit differentiation complements differentiation methods by allowing an indirect route to calculate derivatives without the need to untangle variables from highly couched forms.