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The idea of the cross-sectional area is pivotal in understanding how certain physical processes work, such as the time it takes to empty a tank through a pipe. Picture the cross-sectional area as the size of the opening through which fluid flows. The larger the area, the more fluid can pass through at any given moment. This is like comparing a water bottle opening to a garden hose. The wider the opening (or larger the cross-sectional area), the quicker more water can flow through.

In the context of the given problem, the cross-sectional area, denoted as \(A\), directly affects the rate at which the tank empties. When you have a bigger drainage pipe, indicated by a larger \(A\), the tank can empty faster due to the increased passage for water flow. This relationship is important to grasp because it factors into the whole concept of inverse variation, telling us that increasing \(A\) will decrease the time \(t\) needed as \(t = \frac{k}{A}\).

Variation problems often deal with relationships where one quantity changes relative to another. Inverse variation, specifically, occurs when one value decreases as another increases. This is opposite of direct variation, where both increase or decrease together.

In inverse variation, as seen in the exercise, the relationship is expressed as \( t = \frac{k}{A} \). Here, \(t\) is time, \(A\) is the cross-sectional area, and \(k\) is a constant. This means that as the area \(A\) increases, the time \(t\) must decrease, assuming \(k\) stays the same. Understanding how these variables interact is crucial for solving problems related to physical processes that rely on varying rates of flow.

The key to solving such problems involves:

• Identifying the type of variation (inverse, direct, joint, or partial)
• Formulating an equation based on the variation type
• Solving for the constant if it's unknown
• Using the equation to find unknown values

In this problem, the question clearly describes an inverse relationship between time \(t\) and area \(A\).

The constant of variation, \(k\), is a central element in both direct and inverse variation problems. It represents a fixed value that relates the variables involved. Once you know \(k\), you can predict how changes in one quantity will affect another.

In this problem, we are given a known set of values: - Time \(t = 2.0\) hours- Area \(A = 48\) in\(^2\)Using the relationship \( t = \frac{k}{A} \), we calculate \(k\) by substituting these values: \[2.0 = \frac{k}{48}\]Solving for \(k\), we find \(k = 96\).

This constant tells us that for any cross-sectional area \(A\), the corresponding time \(t\) to empty the tank can be found by rearranging the formula to solve for \(t\) given the new \(A\). For instance, when the area changes to \(68\) in\(^2\), we compute:\[t = \frac{96}{68} \approx 1.41 \, \text{hours}\]
This process illustrates how powerful the constant is for predicting and understanding the dynamics of the underlying variation problem.