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Chapter 2: Problem 84

Volume The radius of a right circular cylinder is given by \(\sqrt{t+2}\) and its height is \(\frac{1}{2} \sqrt{t},\) where \(t\) is time in seconds and the dimensions are in inches. Find the rate of change of the volume with respect to time.

Short Answer

Expert verified
The rate of change of the volume with respect to time is given by \( \frac{dV}{dt} = \pi \frac{1}{2} (3t+2)\).

Step by step solution

01

Calculate the volume of the cylinder

The volume, \(V\), of a cylinder can be calculated by \(V = \pi r^2h\). Here, the radius, \(r\), and the height, \(h\), are given in terms of \(t\). So, substituting the given radius and height the formula becomes \(V(t) = \pi (\sqrt{t+2})^2 (\frac{1}{2}\sqrt{t})\).
02

Simplification of the volume function

The equation \(V(t) = \pi (\sqrt{t+2})^2 (\frac{1}{2}\sqrt{t})\) simplifies to \(V(t) = \pi \frac{t}{2} (t+2)\).
03

Differentiate the volume function with respect to time

Find \(\frac{dV}{dt}\) by applying the product rule to the simplified volume function. After differentiation, we get \( \frac{dV}{dt} = \pi \frac{1}{2} (3t+2) \).
04

Interpretation of the result

The equation \( \frac{dV}{dt} = \pi \frac{1}{2} (3t+2) \) represents the rate at which the volume of the cylinder expands for each value of \(t\).

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