The volume $V$ of a right circular cylinder is given by the formula:

$V = \pi r^2 h$

where $r$ is the radius and $h$ is the height of the cylinder.

Step 1: Express the radius and height in terms of time $t$

• Radius $r = t + 8$
• Height $h = \frac{1}{8}t$

Step 2: Substitute the radius and height into the volume formula

$V = \pi (t + 8)^2 \left( \frac{1}{8}t \right)$

Step 3: Simplify the volume expression

$V = \frac{\pi}{8} t (t + 8)^2$

Step 4: Differentiate the volume with respect to time $t$

We will use the product rule and chain rule to differentiate:

1. First, let $u = (t + 8)^2$, so the derivative of $u$ is: $\frac{d}{dt} (t + 8)^2 = 2(t + 8)$

2. Now apply the product rule to $V = \frac{\pi}{8} t (t + 8)^2$: $\frac{dV}{dt} = \frac{\pi}{8} \left[ (t + 8)^2 + t \cdot 2(t + 8) \right]$

Step 5: Simplify the derivative

$\frac{dV}{dt} = \frac{\pi}{8} \left[ (t + 8)^2 + 2t(t + 8) \right]$ $\frac{dV}{dt} = \frac{\pi}{8} \left[ (t + 8)(t + 8) + 2t(t + 8) \right]$ $\frac{dV}{dt} = \frac{\pi}{8} \left[ (t + 8)(t + 8 + 2t) \right]$ $\frac{dV}{dt} = \frac{\pi}{8} (t + 8)(3t + 8)$

Final Answer:

The rate of change of the volume with respect to time is: $\frac{dV}{dt} = \frac{\pi}{8} (t + 8)(3t + 8) \ \text{in}^3/\text{sec}$

Would you like further details on any of these steps?

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Here are 5 related questions to expand on this:

1. How does the rate of change of volume behave as time $t$ increases?
2. What happens to the volume if the height or radius were constant instead of changing with time?
3. Can we find the rate of change of surface area of this cylinder with respect to time?
4. How do changes in radius and height affect the instantaneous volume growth?
5. What is the volume at $t = 0$ and how does it change at $t = 1$?

Tip: When dealing with composite functions like this, always apply the chain rule and product rule carefully during differentiation.