91Ó°ÊÓ

Understanding the volume of a cone is critical in solving related rates problems involving conical shapes. A cone is a three-dimensional geometric shape that tapers smoothly from a flat circular base to a point called the apex. The formula for the volume of a cone is given by:

• \( V = \frac{1}{3} \pi r^2 h \)

Here, \(V\) represents the volume, \(r\) the radius of the base, and \(h\) the height of the cone. This formula accounts for the reduction in volume caused by the triangular cross-section as compared to a cylinder. Each slice from the base to the apex gets progressively smaller.
In our specific problem, the simplification is possible because the height \( h \) equals the radius \( r \). So, the formula becomes:

• \( V = \frac{1}{3} \pi r^3 \)

This relationship allows us to focus directly on how changes in the radius affect the volume, which is crucial when applying calculus-based methods like differentiation.

Differentiation is a key concept in calculus that deals with the rate of change. In the context of related rates problems, it helps us find how one variable changes with respect to another. When differentiating the cone's volume concerning time, we apply the formula for the derivative and use the chain rule effectively.
Given \( V = \frac{1}{3} \pi r^3 \), the differentiation based on the time \( t \) is:

• \( \frac{dV}{dt} = \pi r^2 \frac{dr}{dt} \)

This equation states that the rate of change of volume with respect to time, \( \frac{dV}{dt} \), depends on both the rate at which the radius changes \( \frac{dr}{dt} \) and the current size of the radius squared. In this exercise, we know that \( \frac{dV}{dt} \) is \(243 \pi\), enabling us to substitute this value to find the rate at which the radius changes over time.

Integration is essentially the reverse process of differentiation. It allows us to find a function when its rate of change is known. In our exercise, after finding the rate of change of the radius \( \frac{dr}{dt} = \frac{243}{r^2} \), we need to integrate to express the radius as a function of time:

• \( \int dr = \int \frac{243}{r^2} \, dt \)

Performing these integrations results in:

• \( r = 3\sqrt{81t} \)

This equation means that the radius \( r \) is directly related to the square root of \( t \). By applying the condition \( r(0) = 0 \), it aligns with the initial conditions given in our problem and formulates \( r(t) = 3t^{1/3} \). Integration ties the problem together by allowing us to move from the rate of change back to a specific mathematical function that describes the system precisely over time.