Problem 15 - Algebra and Trigonometry with Analytic Geometry

Question

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 \(l=100\) and \(d>0\). (d) 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) \(k = 0.000025\); (d) Resistance is 0.0056 ohms.

Step 1: Express Resistance in Terms of Given Variables

Since resistance \(R\) varies directly with length \(l\) and inversely with the square of the diameter \(d\), we can express it as \(R = k \frac{l}{d^2}\). This shows that resistance increases with length and decreases exponentially with diameter.

Step 2: Solve for the Constant of Variation

Plug the given values into the equation from step 1. The wire is 100 feet long with a diameter of 0.01 inch and has a resistance of 25 ohms. So, \(25 = k \frac{100}{(0.01)^2}\). Simplifying, \(25 = k \cdot 1000000\). Hence, \(k = \frac{25}{1000000} = 0.000025\).

Step 3: Sketch Graph of Resistance Versus Diameter

To graph \(R = k \frac{l}{d^2}\) for \(l = 100\) and \(d > 0\), substitute into the equation: \(R = 0.000025 \times \frac{100}{d^2}\). Plotting this indicates that as \(d\) increases, \(R\) decreases hyperbolically, approaching zero as \(d\) increases.

Step 4: Calculate Resistance for New Conditions

Use the equation from Step 1 with \(k = 0.000025\), \(l = 50\) feet, and \(d = 0.015\) inch: \(R = 0.000025 \frac{50}{(0.015)^2}\). Calculating, \(R = 0.000025 \frac{50}{0.000225} = 0.000025 \times 222.2222 = 0.0055555 = 0.0056\) ohms.

Key Concepts

Direct Variation

In mathematics, when we say that one quantity varies directly with another, it means that the two quantities increase or decrease together at a constant rate. In other words, the ratio of the two quantities remains constant.
For example, if we have two variables, such as electrical resistance \(R\) and wire length \(l\), and if \(R\) varies directly as \(l\), it implies:

• If the length of the wire increases, the resistance also increases.
• If the length of the wire decreases, the resistance decreases proportionally.

In the given exercise, the resistance \(R\) is expressed as \(R = k \cdot l/d^2\). This shows that as the wire length \(l\) gets longer, given the same diameter \(d\), the resistance \(R\) grows, maintaining a specific proportion dictated by the constant of variation \(k\).

Inverse Variation

Inverse variation describes a relationship where one quantity increases while another quantity decreases, or vice versa. Two quantities vary inversely if their product is a constant.
In the context of the exercise, the electrical resistance \(R\) varies inversely with the square of the wire's diameter \(d\). This can be represented as:

• If the diameter increases, the resistance decreases significantly.
• If the diameter decreases, the resistance increases.

In the formula \(R = k \frac{l}{d^2}\), the inverse relationship is evident, showing how the resistance reacts to changes in diameter. Doubling the diameter will quadruple the effect on resistance, decreasing it significantly due to this inverse square relationship.

Graphing Relationships

Graphing relationships allows us to visualize how changes in one variable affect another. In this problem, we are asked to graph the relationship between resistance \(R\) and diameter \(d\).
To do this, we use the equation: \(R = 0.000025 \times \frac{100}{d^2}\).
This relationship indicates a hyperbolic graph since we have inverse variation with the square of \(d\). If you graph \(R\) against \(d\), you will see:

• The graph curves downward as \(d\) increases, approaching zero but never touching the \(x\)-axis.
• The resistance decreases at a decreasing rate as the diameter increases.

This visual representation helps reinforce the concept of inverse variation and gives a clearer understanding of how diameter affects resistance.

Constant of Variation

The constant of variation \(k\) is a crucial component in expressions describing direct or inverse relationships. It represents a fixed multiplier that adjusts the formula to match real-world values.
In the exercise, we first expressed the resistance as \(R = k \cdot \frac{l}{d^2}\). Then, using given data (\(l = 100\) feet, \(d = 0.01\) inches, and \(R = 25\) ohms), we solved for \(k\). The calculations showed:

• \(25 = k \frac{100}{(0.01)^2}\)
• Simplifying, \( k = 0.000025\)

Understanding the constant \(k\) helps us apply the formula accurately across different scenarios. It ensures the calculated resistance reflects the material and physical context of the wire.