Question

\( \text { Rate of Change } \) You are given the dollar value of a product in 2004 and the rate at which the value of the product is expected to change during the next 5 years. Write a linear equation that gives the dollar value \(V\) of the product in terms of the year \(t\). (Let \(t=0\) represent 2000.) $$ \frac{2004 \text { Value }}{\$ 2540} $$ $$ \frac{\text { Rate }}{\$ 125 \text { increase per year }} $$

Short answer

The linear equation that gives the dollar value of the product in terms of the year is \(V = 125t + 2540\).

Step-by-step solution

- Identify the initial value

The dollar value of the product in 2004 is given as \$2540. Since the year 2000 is considered as 't=0', then the year 2004 will correspond to 't=4'. So, the initial value (also known as the y-intercept), or \(V_0\), is therefore \$2540.

- Identify the rate

The rate at which the value of the product is expected to change each year is given as \$125 increase per year. This rate is the slope of the linear equation, m.

- Formulate the linear equation

The general formula for a linear equation is \(y = mx + b\), where m is the slope and b is the y-intercept. Here, the value of the product \(V\) depends on the time \(t\). Hence, the formula becomes \(V = mt + V_0\). Plug in the slope value (m=\$125/year) and y-intercept value (V_0=\$2540), the linear equation becomes \(V = 125t + 2540\).

Key concepts

Rate of Change

Understanding the concept of rate of change is like recognizing the speed at which something happens over time. In a mathematical sense, especially when dealing with linear equations, the rate of change describes how much one variable changes on average with respect to a change in another variable. For the given problem, the product's value increases by \( \$125 \) each year. This constant increase represents the rate of change of the value, V, in terms of time, t.

In your daily life, you can observe rates of change quite frequently. For instance, when you drive a car and observe the speedometer, you're looking at a rate of change - your speed - which tells you how many miles you travel per hour. Similarly, in our exercise, the rate of change helps us predict the future value of a product, which is crucial for planning budgets or making investments.

Y-Intercept

The y-intercept is essentially the starting point of a linear equation on a graph. It's where the line crosses the y-axis, and it represents the value of the dependent variable when the independent variable is zero. In the context of the problem we're solving, the y-intercept is the value of the product in the year 2000, which has been defined as \( t= 0 \).

This concept is particularly handy when interpreting graphs and understanding the initial conditions of a scenario. Imagine you're planting a tree; the y-intercept would be the height of the tree when you plant it. Then, as years pass (your independent variable), the tree grows (the dependent variable increases), and the height changes at a rate which would be the slope of the line if you charted the growth on a graph.

Slope

The slope of a line in a linear equation basically tells us how steep the line is. Mathematically, it's the ratio that expresses the amount of vertical change (rise) for each unit of horizontal change (run) as you move along the line. In the exercise, the slope is how much the product's value goes up or down for each year that passes, which in this case, is a \( \$125 \) increase per year.

In real life, you can think of slope like an actual slope or hill: the steeper the hill, the higher the slope value. Likewise, in the exercise, the steeper the slope of our value line, the greater the change in the dollar value of the product per year. The concept of slope is crucial in understanding trends over time, such as economic growth, changes in population, or, like our example, the increase in a product's value.