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The surface area of a shape is the total area that covers its exterior. For a sphere or spherical object like a hot-air balloon, the formula for the surface area is given by \(S = 4 \pi r^2\). Here, \(r\) represents the radius of the sphere in meters, and \(\pi\) (Pi) is a mathematical constant approximately equal to 3.14159.

If we imagine inflating a balloon, as the radius \(r\) increases, the surface area \(S\) also increases. Understanding the relationship between surface area and radius is crucial for this exercise. The given formula \(S(r) = 4 \pi r^2\) allows us to calculate the surface area if we know the radius.

Understanding that the surface area depends on the square of the radius helps you see how changes in the radius significantly affect the surface area.

In this problem, the radius of the balloon changes over time. The function provided for the radius is \(r(t) = \frac{2}{3} t^3\), where \(t\) is the time in seconds.

This means that as time passes, the radius of the balloon grows following a cubic relationship with time, i.e., the radius increases rapidly as time progresses. For example, if you know the time \((t)\), you can substitute it into the radius function to find the radius at that point in time.

Understanding the radius function is important because it lays the foundation for determining how other aspects, like the surface area, change over time.

Substitution is a key step in solving this problem. We have two functions: the surface area function \(S(r) = 4 \pi r^2\) and the radius function \(r(t) = \frac{2}{3} t^3\). To find the surface area as a function of time, we need to substitute the radius function into the surface area function.

First, we rewrite the surface area formula by replacing \(r\) with \(\frac{2}{3} t^3\). This gives us \(S(t) = 4 \pi \left( \frac{2}{3} t^3 \right)^2 \). Substitution helps us express one variable in terms of another, which is crucial for solving composite functions.

The final step in solving the problem involves simplifying the substituted function to make it more manageable. After substitution, we have \(S(t) = 4 \pi \left( \frac{2}{3} t^3 \right)^2\).

First, we simplify the expression inside the parentheses: \( \left( \frac{2}{3} t^3 \right)^2 = \left( \frac{2}{3} \right)^2 t^6 = \frac{4}{9} t^6\).

Next, we multiply by the constant: \(S(t) = 4 \pi \frac{4}{9} t^6 = \frac{16 \pi}{9} t^6\).

Simplification allows us to reduce complex expressions to simpler forms, making calculation and interpretation easier. Now, we have the surface area of the balloon as a function of time: \(S(t) = \frac{16 \pi}{9} t^6\).