Problem 46 - Precalculus with Calculus Previews

Question

Power The electrical power \(P\) produced by a certain source is given by $$ P=\frac{E^{2} r}{R^{2}+2 R r+r^{2}} $$ where \(E\) is the voltage of the source, \(R\) is the resistance of the source, and \(r\) is the resistance in the circuit. Sketch the graph of \(P\) as a function of \(r\) using the values \(E=5\) volts and \(R=1\) ohm.

Step-by-Step Solution

Answer

The graph of \(P\) is zero at \(r=0\) and \(r=\infty\), with a peak at \(r=1\).

Step 1: Substitute Known Values

Substitute the given values of \(E=5\) volts and \(R=1\) ohm into the power formula. The formula becomes \(P = \frac{5^{2} r}{1^{2} + 2 \cdot 1 \cdot r + r^{2}}\). Simplifying further, it becomes \(P = \frac{25r}{r^2 + 2r + 1}\).

Step 2: Simplify Denominator

Simplify the expression in the denominator. Notice that \(r^2 + 2r + 1\) can be factored as \((r+1)^2\). Thus, the function simplifies to \(P = \frac{25r}{(r+1)^2}\).

Step 3: Analyze Behavior of Function

Consider the behavior of the function at extreme values of \(r\). As \(r\to 0\), \(P\to 0\). As \(r\to \infty\), \(P\to 0\) as well since the degree of the denominator is higher than that of the numerator.

Step 4: Find Critical Points

Find the derivative of \(P\) with respect to \(r\) and set it to zero to find critical points. Using the quotient rule, \(\frac{dP}{dr} = \frac{25\cdot (r+1)^2 - 25r\cdot 2(r+1)}{(r+1)^4}\). Setting the numerator equal to zero gives critical points, specifically when \(r = 1\).

Step 5: Determine Nature of Critical Points

Evaluate the second derivative or use the first derivative test to determine whether these points are maxima or minima. Calculating, we find \(r = 1\) is a maximum point.

Step 6: Plot Graph

Use the information gathered to sketch the plot. The graph has a peak at \(r = 1\), and it approaches zero as \(r\) goes to zero and infinity.

Key Concepts

Quotient Rule

The quotient rule is one of the most important tools for differentiation, especially when dealing with functions expressed as a ratio of two functions. It's like the chain rule but for fractions. When you have a function in the form of \( \frac{u}{v} \), where both \( u \) and \( v \) are differentiable, the derivative \( \frac{d}{dx}\left(\frac{u}{v}\right) \) is given by the formula: \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]
This formula tells you to differentiate \( u \) and \( v \), multiply them in a specific way, and finally divide by \( v^2 \).

• First, differentiate the numerator (\( u \)) and multiply by the denominator (\( v \)).
• Second, differentiate the denominator (\( v \)) and multiply by the numerator (\( u \)).
• Subtract these two results and then divide by \( v^2 \).

Applying the quotient rule in the context of the power function, \( P = \frac{25r}{(r+1)^2} \), helps us find the critical points where the function potentially has a maximum or minimum value. This understanding is crucial in analyzing the behavior of \( P \) around critical values of \( r \).

Critical Points

Critical points of a function occur where the first derivative equals zero or does not exist. These points are vital in understanding the peaks and troughs of the function.
For our power function \( P \), we find the critical points by setting the derivative of \( P \) with respect to \( r \) equal to zero. Using the quotient rule, we get: \[ \frac{dP}{dr} = \frac{25 \cdot (r+1)^2 - 25r \cdot 2(r+1)}{(r+1)^4} \] The critical points are determined by solving the equation resulting from setting the numerator to zero: \( 25 \cdot (r+1)^2 - 50r \cdot (r+1) = 0 \). Upon simplification, we find that \( r = 1 \) is a critical point.
Critical points help in identifying where the function's graph reaches significant turning points, like local maxima or minima, which is necessary for effective graph sketching.

Graph Sketching

Graph sketching is an essential visual approach in analyzing the behavior of functions. It involves considering several key aspects of the function such as critical points, intercepts, asymptotes, and behavior at infinity.
For the given power function \( P = \frac{25r}{(r+1)^2} \), we start by noting that it simplifies to showcase a key peak or a maximum point at \( r = 1 \). To sketch the graph:

• Identify the intercept: For \( P = \frac{25r}{(r+1)^2} \), the graph intercepts the origin \((0,0)\) since \( P = 0 \) when \( r = 0 \).
• Locate critical points: As calculated, there is a maximum at \( r = 1 \).
• Understand the limit behavior: As \( r \) approaches zero or infinity, \( P \) approaches zero due to the dominance of the \((r+1)^2\) term in growth compared to the numerator.

Using this information allows you to sketch the basic outline of the function, showing a rise to a peak at \( r = 1 \) and falling towards zero on either side, hence creating a parabolic arc-like shape.

Function Behavior

Understanding the behavior of a function is crucial in identifying how it acts over its domain. For our function, fatigue sets in as we assess the tendencies of \( P \) at various ranges of \( r \).
Analyzing \( P = \frac{25r}{(r+1)^2} \) gives insight into:

• The intercept at the origin \((0,0)\), indicating no power when resistance \( r \) is zero.
• As \( r \to 0 \) or \( r \to \infty \), \( P \to 0 \), demonstrating that power dissipates as resistance moves to these extremes, either due to the lack of resistance or overwhelming resistance.
• Critical point \( r = 1 \) represents the peak power output, a situation of balance where resistance is optimal for maximal power transfer.

Thus, function behavior analysis not only helps in mathematics but also gives insight into practical applications, such as optimal resistance settings in electrical circuits.