Problem 15 - Precalculus: Functions and Graphs

Question

Electrical resistance The electrical resistance \(R\) of a wire varies directly as its length \(l\) and inversely as the square of its diameter \(d\) (a) Express \(R\) in terms of \(l, d,\) and a constant of variation \(k\) (b) A wire 100 feet long of diameter 0.01 inch has a resistance of 25 ohms. Find the value of \(k\) in part (a). (c) Sketch a graph of the relationship between \(R\) and \(d\) for \(I=100\) and \(d>0\) Find the resistance of a wire made of the same material that has a diameter of 0.015 inch and is 50 feet long.

Step-by-Step Solution

Answer

(a) \( R = k \frac{l}{d^2} \); (b) \( k = 2.5 \times 10^{-4} \); (c) New resistance is \( 0.5556 \) ohms.

Step 1: Define the equation for resistance

Since the resistance \( R \) varies directly with the length \( l \) and inversely with the square of the diameter \( d \), we can express the relationship as \( R = k \frac{l}{d^2} \), where \( k \) is the constant of variation.

Step 2: Determine the constant of variation \( k \)

Using the given values \( l = 100 \) feet, \( d = 0.01 \) inch, and \( R = 25 \) ohms, substitute into the equation \( R = k \frac{l}{d^2} \):\[ 25 = k \frac{100}{(0.01)^2} \]Solve for \( k \):\[ k = \frac{25 \times (0.01)^2}{100} = 2.5 \times 10^{-4} \].

Step 3: Construct the graph of \( R \) vs \( d \) for fixed \( l \)

For a fixed length of 100 feet, the equation becomes \( R = 2.5 \times 10^{-4} \times \frac{100}{d^2} = \frac{0.025}{d^2} \). This equation describes a hyperbola opening along the positive x-axis. As \( d \) increases, \( R \) decreases.

Step 4: Calculate the new resistance for a different wire

Given a wire 50 feet long with a diameter of 0.015 inch, we use the constant \( k = 2.5 \times 10^{-4} \) to find the new resistance:Substitute \( l = 50 \) feet and \( d = 0.015 \) inch into the equation:\[ R = 2.5 \times 10^{-4} \times \frac{50}{(0.015)^2} \]Calculating, we have:\[ R = 2.5 \times 10^{-4} \times \frac{50}{0.000225} \approx 0.5556 \] ohms.

Key Concepts

Direct Variation

In many mathematical relationships, one quantity varies directly with another. This means if one quantity increases, the other increases at a consistent rate, and likewise for decreases. For example, in the formula for electrical resistance of a wire, the resistance \( R \) varies directly with its length \( l \). This implies that if the length of the wire is doubled, the resistance will also double, assuming all other factors remain constant.

Direct variation is commonly expressed in the form \( y = kx \), where \( y \) and \( x \) are variables, and \( k \) is the constant of variation. In the case of resistance, our setup is a bit more complex as \( R = k \frac{l}{d^2} \), integrating both direct and inverse relationships. Regardless, the direct variation portion focuses solely on the relationship between resistance and length.

• An increase in length leads to a proportional increase in resistance.
• The relationship is linear as long as the constant is stable.

Inverse Variation

Inverse variation describes a situation where, as one value increases, the other value decreases proportionally. This is crucial in understanding the resistance of a wire based on its diameter. The way it works for this physics problem is that the resistance \( R \) varies inversely with the square of the diameter \( d \) of the wire.

In mathematical terms, this is represented as \( y = \frac{k}{x^2} \). Here, as the diameter of the wire increases, having a thicker wire, the electrical resistance decreases since it provides less opposition to the flow of electric current.

This part of our formula, \( R = k \frac{l}{d^2} \), shows that:

• Larger diameter equals lower resistance, assuming constant length and material.
• This relationship is not linear, but rather creates a hyperbolic shape in graph form.

Constant of Variation

The constant of variation, often denoted as \( k \), is a key factor in both direct and inverse variations. It provides a standardized element to express how one variable changes with respect to another. In our resistance equation, the constant \( k \) ensures the relationship adheres to real-world observations.

To find the constant of variation, we used known values of length, diameter, and resistance. By substituting these into the formula \( R = k \frac{l}{d^2} \), we solve for \( k \), ensuring our equation accurately models the relationship's real-world behavior.

The constant can be determined like this:

• Insert given values of \( R, l, \) and \( d \) into the formula.
• Rearrange to solve for \( k \).
• This provides a unique value that fits the specific material and conditions.

In our example, \( k = 2.5 \times 10^{-4} \), which unifies the relationship between length, diameter squared, and resistance for that specific material and condition.

Graphing Functions

Graphing the relationship between variables can reveal essential insights, especially when dealing with non-linear relationships. In our problem concerning resistance, depicting \( R \) vs \( d \) (diameter) with a fixed length provides a visual understanding.

The equation \( R = \frac{0.025}{d^2} \) represents this relationship. The graph of this function is a hyperbola:

• The graph opens along the positive x-axis, indicating a decrease in \( R \) as \( d \) increases.
• When \( d \) is small, \( R \) is large, reflecting high resistance in thin wires.
• Conversely, when \( d \) is large, \( R \) reduces, showing less resistance in thicker wires.

By plotting graph points, we can better predict how changes in diameter will affect resistance, allowing for effective material and design choices. Using graphical representations aids comprehension, bridging numerical relationships with visual understanding.