Problem 3 - The Calculus with Analytic Geometry

Question

Let \(r\) be the reciprocal of a number \(n .\) Find the instantaneous rate of change of \(r\) with respect to \(n\) and the relative rate of change of \(r\) per unit change in \(n\) when \(n\) is (a) 4 and (b) 10 .

Step-by-Step Solution

Answer

For \(n = 4\), instantaneous rate: -\(1/16\), relative rate: -\(1/4\). For \(n = 10\), instantaneous rate: -\(1/100\), relative rate: -\(1/10\).

Step 1 - Define the reciprocal

The reciprocal of a number is defined as the multiplicative inverse of that number. If the number is denoted as \(n\), then the reciprocal \(r\) is given by: \[ r = \frac{1}{n} \]

Step 2 - Differentiate the reciprocal

To find the instantaneous rate of change, compute the derivative of \(r\) with respect to \(n\). Using the power rule for differentiation: \[ \frac{dr}{dn} = \frac{d}{dn} \left( n^{-1} \right) = -n^{-2} = -\frac{1}{n^2} \]

Step 3 - Calculate the instantaneous rate of change at specific values

Substitute \(n = 4\) and \(n = 10\) into the derivative to find the rates: a) For \(n = 4\): \[ \left. \frac{dr}{dn} \right|_{n=4} = -\frac{1}{4^2} = -\frac{1}{16} \] b) For \(n = 10\): \[ \left. \frac{dr}{dn} \right|_{n=10} = -\frac{1}{10^2} = -\frac{1}{100} \]

Step 4 - Define the relative rate of change

The relative rate of change of \(r\) with respect to \(n\) per unit change in \(n\) is given by the ratio: \[ \text{Relative rate of change} = \frac{1}{r} \cdot \frac{dr}{dn} = n \cdot \left( -\frac{1}{n^2} \right) = -\frac{1}{n} \]

Step 5 - Calculate the relative rate of change at specific values

Substitute \(n = 4\) and \(n = 10\) into the relative rate of change formula: a) For \(n = 4\): \[ \text{Relative rate of change at } n = 4 = -\frac{1}{4} \] b) For \(n = 10\): \[ \text{Relative rate of change at } n = 10 = -\frac{1}{10} \]

Key Concepts

Reciprocal

A reciprocal represents the multiplicative inverse of a number. Simply put, for a given number denoted as \(n\), its reciprocal \(r\) is calculated by the formula: \( r = \frac{1}{n} \). For example, if \( n = 4 \), then \( r = \frac{1}{4} \). Reciprocals are quite useful in various mathematical operations and solving equations.

Power Rule for Differentiation

The power rule for differentiation is a fundamental technique in calculus used to find the derivative of functions of the form \( x^n \). When applying this rule, if you have a function \( f(x) = x^n \), the derivative \( f'(x) \) will be \( nx^{n-1} \). This rule simplifies the differentiation process significantly and is widely used. For instance, to differentiate \( n^{-1} \) (our reciprocal function), we get: \( \frac{d}{dn} \left( n^{-1} \right) = -n^{-2} = -\frac{1}{n^2} \).

Relative Rate of Change

The relative rate of change compares the rate of change of a function to the value of the function itself. For a function \( r \) dependent on \( n \), it is given by the formula: \( \frac{1}{r} \cdot \frac{dr}{dn} \). This concept helps in understanding how fast the function is changing relative to its current value. When applied to our reciprocal function \( r = \frac{1}{n} \) and its derivative \( -\frac{1}{n^2} \), it results in: \( n \cdot \left( -\frac{1}{n^2} \right) = -\frac{1}{n} \).

Derivative Computation

Derivative computation is crucial for finding the instantaneous rate of change of a function with respect to its variable. By differentiating the reciprocal function \( r = \frac{1}{n} \), we derived \( \frac{dr}{dn} = -\frac{1}{n^2} \). This gives us the rate at which \( r \) changes as \( n \) varies. By substituting specific values for \( n \), such as 4 and 10, we can find the instantaneous rates:
For \( n = 4 \): \( \left. \frac{dr}{dn} \right|_{n=4} = -\frac{1}{16} \)
For \( n = 10 \): \( \left. \frac{dr}{dn} \right|_{n=10} = -\frac{1}{100} \).