We are tasked with finding the total volume of liquid that evaporates in the first 5 units of time, given the rate of evaporation function $f(t) = e^{-t/10}$. The problem also assumes that the initial volume of the liquid is equal to the volume of the cylindrical container, which has a height of 10 units and a radius of 2 units.

Step 1: Volume of the Cylindrical Container

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

$V = \pi r^2 h$

where:

• $r = 2$ is the radius,
• $h = 10$ is the height.

Thus, the initial volume of the liquid in the container is:

$V = \pi (2)^2 (10) = 40\pi \, \text{units}^3$

Step 2: Evaporation Model

The rate of evaporation is proportional to the surface area of the liquid, which is constant in this case, because the liquid is evaporating from the cylindrical top surface. However, the problem states that the evaporation rate as a function of time is $f(t) = e^{-t/10}$.

The total volume of liquid evaporated over a time interval $[0, T]$ is given by the integral of the evaporation rate function over that interval, multiplied by the initial volume of the container:

$V_{\text{evap}}(T) = V_0 \int_0^T f(t) \, dt$

where $V_0 = 40\pi$ is the initial volume and $f(t) = e^{-t/10}$.

Step 3: Set Up the Integral

We now compute the volume of liquid evaporated in the first 5 units of time. The total evaporated volume is:

$V_{\text{evap}}(5) = 40\pi \int_0^5 e^{-t/10} \, dt$

This is an exponential integral that can be solved as follows.

Step 4: Compute the Integral

The integral of $e^{-t/10}$ is straightforward. We have:

$\int e^{-t/10} \, dt = -10 e^{-t/10}$

Thus:

$\int_0^5 e^{-t/10} \, dt = \left[ -10 e^{-t/10} \right]_0^5 = -10 e^{-5/10} + 10 e^{-0} = -10 e^{-1/2} + 10$

Simplifying:

$\int_0^5 e^{-t/10} \, dt = 10(1 - e^{-1/2})$

Step 5: Total Volume Evaporated

The total volume evaporated in the first 5 units of time is:

$V_{\text{evap}}(5) = 40\pi \times 10 (1 - e^{-1/2}) = 400\pi (1 - e^{-1/2})$

Step 6: Approximation of the Infinite Series

We now approximate $e^{-1/2}$. Using the Taylor series expansion for $e^{-x}$, we have:

$e^{-1/2} = 1 - \frac{1}{2} + \frac{1}{2^2 \cdot 2!} - \frac{1}{2^3 \cdot 3!} + \cdots$

This gives:

$e^{-1/2} \approx 1 - 0.5 + 0.125 - 0.02083 = 0.60417$

Thus, the volume evaporated is approximately:

$V_{\text{evap}}(5) \approx 400\pi (1 - 0.60417) = 400\pi \times 0.39583 \approx 497.92 \, \text{units}^3$

Final Answer:

The total volume of liquid that evaporates in the first 5 units of time is approximately $497.92 \, \text{units}^3$, and the exact expression is:

$V_{\text{evap}}(5) = 400\pi (1 - e^{-1/2})$

Would you like more details or clarifications on any step?

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Additional Questions:

1. How does the rate of evaporation change as time progresses?
2. What is the total volume evaporated after 10 units of time?
3. How would the result change if the container had a different radius?
4. How can we derive the Taylor series for $e^{-x}$?
5. How would we calculate the evaporation if the rate $f(t)$ had a different functional form?

Tip:

When solving exponential integrals, always check if the base of the exponential involves time constants, as this impacts the integration result.