We are given the volume of oil $V(t)$ in a tank as a function of time $t$:

$V(t) = 32 \left( 3 + e^{\frac{1}{5}t} \right)^{-1}$

We are tasked with finding the value of $V$ when $\frac{dV}{dt}$ (the rate of change of the volume) reaches its minimum value.

Step 1: Find $\frac{dV}{dt}$

To begin, we differentiate $V(t)$ with respect to $t$.

$V(t) = 32 \left( 3 + e^{\frac{1}{5}t} \right)^{-1}$

We apply the chain rule. Let $u = 3 + e^{\frac{1}{5}t}$, so:

$V(t) = \frac{32}{u}$

The derivative of $V(t)$ with respect to $t$ is:

$\frac{dV}{dt} = -32 \left( 3 + e^{\frac{1}{5}t} \right)^{-2} \cdot \frac{d}{dt} \left( 3 + e^{\frac{1}{5}t} \right)$

Now differentiate $3 + e^{\frac{1}{5}t}$:

$\frac{d}{dt} \left( 3 + e^{\frac{1}{5}t} \right) = \frac{1}{5} e^{\frac{1}{5}t}$

Thus, the derivative of $V(t)$ becomes:

$\frac{dV}{dt} = -32 \left( 3 + e^{\frac{1}{5}t} \right)^{-2} \cdot \frac{1}{5} e^{\frac{1}{5}t}$

Simplifying this expression:

$\frac{dV}{dt} = -\frac{32}{5} \cdot \frac{e^{\frac{1}{5}t}}{\left( 3 + e^{\frac{1}{5}t} \right)^2}$

Step 2: Minimize $\frac{dV}{dt}$

To find when $\frac{dV}{dt}$ attains its least value, we need to find the critical points by differentiating $\frac{dV}{dt}$ again and setting it equal to zero.

Let $\frac{dV}{dt} = f(t)$. We now differentiate $f(t)$ with respect to $t$:

$f(t) = -\frac{32}{5} \cdot \frac{e^{\frac{1}{5}t}}{\left( 3 + e^{\frac{1}{5}t} \right)^2}$

This step involves applying the quotient rule, and it may become lengthy, so I will calculate this explicitly to find the value of $t$ when $\frac{dV}{dt}$ is minimized.The critical point occurs when $t = 5 \ln(3)$. Now, we will substitute this value into the original equation for $V(t)$ to find the corresponding volume $V$ when $\frac{dV}{dt}$ is minimized.

Step 3: Calculate $V(t)$ at $t = 5 \ln(3)$

We substitute $t = 5 \ln(3)$ into the expression for $V(t)$:

$V(t) = 32 \left( 3 + e^{\frac{1}{5}t} \right)^{-1}$

Let me calculate this value for you.The value of the volume $V$ when $\frac{dV}{dt}$ attains its least value is approximately $5.33 \, \text{cm}^3$.

Would you like more details on the process, or do you have any further questions?

Here are 5 related questions:

1. How do you apply the quotient rule when differentiating functions like $\frac{dV}{dt}$?
2. What is the significance of critical points in minimizing or maximizing a function?
3. How does exponential decay affect the rate of change in real-world problems like oil leakage?
4. Can we explore the behavior of $V(t)$ over different time intervals?
5. How do logarithmic functions help in finding the critical points?

Tip: Always check the second derivative when minimizing or maximizing a function to confirm whether it's a minimum or maximum point.