Problem 49 - Calculus

Question

The functions \(r=f(t)\) and \(V=g(r)\) give the radius and the volume of a commercial hot air balloon being inflated for testing. The variable \(t\) is in minutes, \(r\) is in feet, and \(V\) is in cubic feet. The inflation begins at \(t=0 .\) In each case, give a mathematical expression that represents the given statement. The volume of the balloon if its radius were twice as big.

Step-by-Step Solution

Answer

The volume expression is \(V = g(2f(t))\).

Step 1: Determine the expression for the radius when doubled

The radius of the balloon as a function of time is given by the function \(r = f(t)\). If the radius is doubled, the expression for the new radius \(r'\) would be \(r' = 2f(t)\).

Step 2: Substitute the new radius into the volume formula

The volume of the balloon is given by the function \(V = g(r)\). To find the volume when the radius is doubled, substitute \(r' = 2f(t)\) into the volume function. The new volume expression is \(V' = g(2f(t))\).

Key Concepts

Function Composition in Calculus

Function composition is a way of combining two functions where the output of one function becomes the input of another. In this exercise, we have two functions: one that gives the radius, \(r = f(t)\), and another that gives the volume, \(V = g(r)\). When we compose these functions, what we are really doing is creating a function that combines both steps.

• First, \(f(t)\) takes a time \(t\) and outputs a radius \(r\).
• Then, \(g(r)\) takes that radius \(r\) and outputs the volume \(V\).

This step-by-step process is captured in the expression \(V = g(f(t))\). Thus, function composition allows us to express complex relationships in a simplified manner. It paints a clearer picture of how changes in one variable affect another across interconnected steps.

Volume Calculation Formula

Volume calculation often involves understanding three-dimensional geometrical figures. For a balloon, which resembles a sphere, the formula for volume in terms of its radius is given by:\[V = \frac{4}{3} \pi r^3\]In the context of our problem, the volume function \(V = g(r)\) can represent such a formula with \(r\) being the radius of the balloon. When applying calculus, knowing how the volume changes as the radius changes is crucial.By substituting different values into the volume function, such as a doubled radius, we use the equation:

• With original radius: \(V = g(r) = \frac{4}{3} \pi r^3\)
• With doubled radius: \(V' = g(2r) = \frac{4}{3} \pi (2r)^3 = \frac{4}{3} \pi (8r^3) = 8g(r)\)

This transformation underscores how small changes in radius can lead to significant changes in volume for spherical objects like balloons.

Geometry of Spheres

Understanding the geometry of spheres, such as the hot air balloon in this exercise, is central to solving problems involving volume and surface area. A sphere is a perfectly round three-dimensional shape, where all points on its surface are equidistant from its center.Key properties of spheres include:

• Radius (\(r\)): the distance from the center to any point on the surface.
• Diameter: twice the radius, meaning \(d = 2r\).
• Surface area: calculated by \(4\pi r^2\).
• Volume: \(\frac{4}{3} \pi r^3\), as discussed in the volume calculation section.

Recognizing these properties helps bridge geometry and calculus, allowing us to apply formulas universally across different disciplines and solve real-world problems.

Derivatives in Calculus

Derivatives are a cornerstone in calculus, providing a way to understand rates of change. In the context of our balloon problem, we can think about derivatives with respect to the radius and how they might affect the volume.When seeking to find the rate at which the volume changes with respect to time \(t\), we can use the chain rule. This is because the volume \(V\) is a function of the radius \(r\), which itself is a function of time \(t\) (i.e., \(r = f(t)\)).
The chain rule states:\[\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}\]Here:

• \(\frac{dV}{dr}\) measures how the volume changes as the radius changes.
• \(\frac{dr}{dt}\) measures how the radius changes over time.

Through this chain rule, calculus allows us to analyze and predict dynamic changes in the volume of the balloon, making it a powerful tool in understanding real-world phenomena.