91Ó°ÊÓ

Understanding the volume of a cone is crucial when addressing related rates problems involving geometric shapes. The formula for the volume of a cone is given by \[V = \frac{1}{3} \pi r^2 h\] where:

• \(V\) is the volume,
• \(r\) is the radius of the base, and
• \(h\) is the height of the cone.

In this exercise, the radius \(r\) of the waffle cone is expressed as a function of the height \(h\) using the relationship \(r = \frac{h}{6}\). This expression allows us to substitute \(r\) into the volume formula. As a result, we focus solely on height \(h\) when examining how volume changes over time. These geometric insights lay the groundwork for further mathematical analysis.

The chain rule is a fundamental principle in calculus for differentiating composite functions. In the context of our cone problem, it’s used to determine the rate at which the yogurt's height \(h\) changes inside the cone over time. If we have a function \(V(h(t))\), where both \(V\) and \(h\) depend on time \(t\), we employ the chain rule:\[\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}\]Here:

• \(\frac{dV}{dt}\) is the rate of change of volume with respect to time.
• \(\frac{dV}{dh}\) is the derivative of volume with respect to height.
• \(\frac{dh}{dt}\) is the rate of change of height with respect to time.

We apply this rule after expressing the volume in terms of height to link how fast the frozen yogurt is dispensed to the rise of height in the cone.

Differentiation is the process of finding the derivative of a function to determine rates of change. In this exercise, we use differentiation to find how the volume \(V\) changes with respect to height \(h\). Initially, the volume formula is expressed in terms of height as \[V = \frac{\pi}{108}h^3\] To differentiate, we find \[\frac{dV}{dh} = \frac{\pi}{108} \times 3h^2 = \frac{\pi}{36}h^2\] This step uncovers the relationship between the change in volume and height. To find the instantaneous rate at which the yogurt height is increasing within the cone, we equate the rate of volume change with respect to time to the expression derived from differentiating the volume with respect to height and apply the chain rule.

Calculus, particularly using related rates, enables us to solve real-world problems where multiple quantities change simultaneously. In this frozen yogurt cone problem, calculus lets us compute how quickly the yogurt fills up the cone in height rather than volume. Given that the volume changes at a constant rate (expressed after conversion to cubic centimeters per second), we apply all previously derived expressions and the chain rule as follows:

• Substitute known values into the derived equation \(\frac{dV}{dt} = \frac{\pi}{36}h^2 \cdot \frac{dh}{dt}\).
• Replace \(\frac{dV}{dt}\) with \(16.67\) and \(h\) with \(6\) (since that is the height at the moment of interest).
• Find \(\frac{dh}{dt}\), the rate at which the yogurt's height increases.

Following these steps systematically, calculus applications demonstrate their power by solving the rate of change in a shape's geometric dimensions concisely and efficiently.