Problem 76 - College Algebra and Calculus: An Applied Approach

Question

The value \(V\) of a machine \(t\) years after it is purchased is inversely proportional to the square root of \(t+1\). The initial value of the machine is (a) Write \(V\) as a function of \(t\). (b) Find the rate of depreciation when \(t=1\). (c) Find the rate of depreciation when \(t=3\).

Step-by-Step Solution

Answer

The function representing the value of the machine is \(V(t) = \frac{V_0}{\sqrt{t+1}}\). The rate of depreciation after 1 year is \(- \frac{V_0}{2 \sqrt{2^3}}\), and after 3 years it is \(- \frac{V_0}{2 \sqrt{4^3}}\).

Step 1: Write the Function

Given that \(V\) is inversely proportional to the square root of \(t+1\), we can represent this as: \(V = \frac{k}{\sqrt{t+1}}\), where \(k\) is the constant of variation. Additionally, we know from the problem that when \(t=0\), \(V\) is the initial value of the machine. Thus the equation becomes: \(V_0 = \frac{k}{\sqrt{0+1}} = k\) which simplifies to just \(k = V_0\). So the function we get is \(V(t) = \frac{V_0}{\sqrt{t+1}}\).

Step 2: Find the Rate of Depreciation at \(t=1\)

The rate of depreciation of the machine at a certain time is represented by the derivative of the function. Use the chain rule to find the derivative: \(V'(t) = - \frac{V_0}{2 (t+1)^{(3/2)}}\). Substituting \(t=1\) into this function gives the rate of depreciation after 1 year.

Step 3: Find the Rate of Depreciation at \(t=3\)

Substitute \(t=3\) into the depreciation function \(V'(t) = - \frac{V_0}{2 (t+1)^{(3/2)}}\) to find the rate of depreciation after 3 years.

Key Concepts

Rate of Depreciation

The rate of depreciation refers to the speed at which an asset loses its value over time. For the machine in our exercise, this rate varies depending on how many years have passed since it was purchased, denoted by the variable \( t \). The key to understanding the depreciation of the machine lies in the derivative of its value function, \( V(t) = \frac{V_0}{\sqrt{t+1}} \).

In simple terms, depreciation is how much less the machine is worth after some time. To find the rate of this decrease when \( t=1 \) and when \( t=3 \), we need to compute the derivative of \( V(t) \) using calculus. The derivative you've learned about here, \( V'(t) = - \frac{V_0}{2 (t+1)^{(3/2)}} \), shows how fast the value decreases at any time \( t \).

• At \( t=1 \), substitute \( t \) with 1 in the derivative to find out how much the machine's value decreases after one year.
• At \( t=3 \), replace \( t \) with 3 to discover the depreciation rate after three years.

The derivative, being negative, shows that the machine decreases in value over time, indicating depreciation. This change in value is slower as time goes on since the denominator increases, reflecting the slowing loss of value in depreciation.

Constant of Variation

The constant of variation, often represented by \( k \), is a crucial component in functions that describe inverse variation. In our exercise, the value of the machine, \( V \), inversely relates to the square root of \( t+1 \). This relationship is expressed with the equation \( V = \frac{k}{\sqrt{t+1}} \).

Inverse variation means that as one thing increases, another thing decreases. For this case, as time increases, the value of the machine decreases because they are inversely related through the square root function.

• To find \( k \), assume \( t = 0 \) (this is when the machine is brand new).
• At \( t = 0 \), \( V \) equals the initial value \( V_0 \), making \( k = V_0 \).

This constant of variation helps define the unique characteristic of the specific machine since it represents how quickly the value drops in relation to the time passed. It essentially sets the starting point for the machine's value depreciation.

Chain Rule

The chain rule is a fundamental tool in calculus used to find derivatives of compositions of functions. In this exercise, we apply it to determine the rate of depreciation for the machine's value function, \( V(t) = \frac{V_0}{\sqrt{t+1}} \).

To use the chain rule effectively, we need to take into account both the function and its inner component. Here’s how the chain rule applies:

• First, identify the outer function, which is \( f(u) = \frac{1}{u} \).
• The inner function is \( u = \sqrt{t+1} \).

By applying the chain rule, we take the derivative of the outer function with respect to the inner function and multiply it by the derivative of the inner function. In simpler terms, we found \( V'(t) = -\frac{V_0}{2(t+1)^{3/2}} \).

The chain rule thus enables us to understand how different parts of a function affect the function as a whole, particularly useful in finding rates of change like depreciation in this scenario. Remember, the power of the chain rule lies in untangling complex functions into simpler parts to see how changes in one aspect cause changes in the entire function.