Problem 84 - College Algebra with Modeling and Visualization

Question

Let the number of gallons \(G\) of water in a pool after \(t\) hours be given by \(G(t)=4000-100 t\) for \(0 \leq t \leq 40\) (a) Find \(G(t+h)\) (b) Find the difference quotient. Interpret your result.

Step-by-Step Solution

Answer

(a) \(G(t+h) = 4000 - 100t - 100h\). (b) The difference quotient is \(-100\), meaning the water decreases by 100 gallons per hour.

Step 1: Substitute into Function

To find \(G(t+h)\), substitute \(t+h\) into the function \(G(t)\). This yields:\[G(t+h) = 4000 - 100(t + h)\] distribute the \(-100\):\[G(t+h) = 4000 - 100t - 100h\].

Step 2: Find the Difference Quotient

The difference quotient is given by \(\frac{G(t+h) - G(t)}{h}\). Substitute the expressions for \(G(t+h)\) and \(G(t)\):\[ \frac{4000 - 100t - 100h - (4000 - 100t)}{h} \].

Step 3: Simplify the Expression

In the expression \(\frac{4000 - 100t - 100h - (4000 - 100t)}{h}\), cancel out like terms. The 4000 and \(-100t\) terms cancel out, leaving:\[ \frac{-100h}{h} \].

Step 4: Simplify the Quotient

Cancel \(h\) in the numerator and denominator:\[ -100 \]. This represents the rate of change of gallons of water per hour.

Key Concepts

Rate of Change

The rate of change is a fundamental concept that helps us understand how a quantity changes with respect to another. In the context of the original exercise, we have a pool where the number of gallons of water decreases over time. This change is represented by the function \(G(t) = 4000 - 100t\), where \(t\) is the time in hours.

The aim here is to determine how fast the water level is dropping each hour, which is defined by the rate of change. In mathematical terms, this is captured using the difference quotient. The difference quotient for the function \(G(t)\) is expressed as \(\frac{G(t+h) - G(t)}{h}\). This essentially gives us the average rate of change over a small interval from \(t\) to \(t+h\).

For the exercise, when you simplify this expression, as shown in the step-by-step solution, you get \(-100\). This tells us that the water level is decreasing by 100 gallons every hour. Thus, the rate of change in this scenario is constant.

Substitution in Functions

Substitution is a straightforward but crucial step in handling functions. It refers to plugging in specific values for the variables in a function, which helps us understand how outputs change based on different inputs. In the exercise, the function given is \(G(t) = 4000 - 100t\).

To find \(G(t+h)\), where \(h\) is a small increment in time, we substitute \(t+h\) into the function instead of \(t\). This substitution looks like this:

• Start with the function: \(G(t) = 4000 - 100t\)
• Substitute \(t+h\) for \(t\): \(G(t+h) = 4000 - 100(t+h)\)

After substituting, simplify the function to get \(G(t+h) = 4000 - 100t - 100h\). This formula now represents the number of gallons of water in the pool after \(t+h\) hours.

Simplification of Expressions

Simplifying expressions is an important skill that makes it easier to see what's really happening mathematically. After substitution and finding the difference quotient, we often end up with complex expressions. Simplification helps us reduce these down to their most basic terms.

In the step-by-step solution, once you replace the expressions \(G(t+h)\) and \(G(t)\) into the difference quotient \(\frac{G(t+h) - G(t)}{h}\), you get:

\[ \frac{4000 - 100t - 100h - (4000 - 100t)}{h} \]

To simplify, start by canceling out matching terms, like \(4000\) and \(-100t\). You are then left with:

• \(\frac{-100h}{h}\)

Further simplification involves canceling the \(h\) present in both the numerator and the denominator. This leaves us with \(-100\), clearly showing the constant rate of change.