an oil tank contains some oil. the oil is leaking from the tank. it is found that the volume of oil V (in cm3) in the tank can be estimated by v=32(3+e^(1/5t))^(-1). and t(t>=0) is the number of hours elasped since the leaking begins. find the value of V when dV/dT attains its least value



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.

An oil tank contains some oil. The oil is leaking from the tank. It is found that the volume of oil V (in cm³) in the tank can be estimated by V = 32(3 + e^(1/5t))^(-1), and t (t >= 0) is the number of hours elapsed since the leaking begins. Find the value of V when dV/dt attains its least value.

Explore how to find the volume of oil in a tank as it leaks over time. The volume is given by V(t) = 32(3 + e^(1/5t))^(-1), and we find the volume when the rate of change dV/dt is minimized. This problem covers differentiation using the chain and quotient rules. Suitable for advanced calculus students in Grades 11-12.

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