91Ó°ÊÓ

Derivatives play a pivotal role in calculus as they help us understand how a function changes in response to variations in its variables. When considering a cylinder's dimensions as functions of time, we need derivatives to determine the rates at which the radius and height are changing. In this exercise, the radius \(x\) and height \(y\) of the cylinder are described by the functions \(x(t) = \frac{1}{2}t\) and \(y(t) = \frac{1}{3}t\), respectively. To find the rate at which \(x\) and \(y\) are changing with respect to time \(t\), we differentiate these functions, obtaining \(\frac{dx}{dt} = \frac{1}{2}\) and \(\frac{dy}{dt} = \frac{1}{3}\).

• The derivative \(\frac{dx}{dt} = \frac{1}{2}\) indicates that for every unit increase in time \(t\), the radius \(x\) increases by half a unit.
• Similarly, \(\frac{dy}{dt} = \frac{1}{3}\) shows that for each unit increment in \(t\), the height \(y\) grows by one-third unit.

Understanding these derivatives is crucial because they provide the basis for calculating how other related quantities, like volume, evolve over time.

The formula for the volume of a cylinder is derived from its geometrical properties, specifically the area of its base and its height. For a right circular cylinder, the volume \(V\) is given by \(V = \pi x^2 y\), where \(x\) is the radius, and \(y\) is the height of the cylinder. In our exercise, both \(x\) and \(y\) are increasing functions of time.To express the volume as a function of time, substitute \(x(t) = \frac{1}{2}t\) and \(y(t) = \frac{1}{3}t\) into the volume formula:\[V(t) = \pi \left(\frac{1}{2}t\right)^2 \left(\frac{1}{3}t\right) = \pi \cdot \frac{1}{4}t^2 \cdot \frac{1}{3}t = \frac{\pi}{12} t^3\]This transformation helps us understand how the cylinder's volume evolves as time progresses, allowing us to link the dimensions' growth directly with the overall expansion in volume. Recognizing how each factor contributes to the total volume is essential when analyzing how the volume will change given certain rates of dimensional growth.

Differentiation with respect to time is a technique that allows us to quantify how variables that change over time affect other time-dependent quantities. This technique is especially useful in physics and engineering problems, where modeling and predicting the behavior of dynamic systems are crucial.In the provided problem, the derivative of the volume function \(V(t) = \frac{\pi}{12} t^3\) is taken with respect to time \(t\) to find how fast the volume changes:\[\frac{dV}{dt} = \frac{d}{dt} \left( \frac{\pi}{12} t^3 \right) = \frac{\pi}{12} \, 3t^2 = \frac{\pi}{4} t^2\]This expression \(\frac{\pi}{4} t^2\) represents the rate of change of volume over time for the cylinder. When calculating this rate at a specific moment, such as when \(x = 2\) and \(y = 5\), and under the consistent time value \(t = 6\), we substitute \(t\) into the derivative:\[\frac{dV}{dt} = \frac{\pi}{4} (6)^2 = 36 \times \frac{\pi}{4} = 9\pi\]This final step reveals how quickly the volume of the cylinder is increasing at that specific moment, illustrating the practical application of differentiation in tracking dynamic changes.