A water tank initially holds 100L of water. It springs a leak so that water leaks out at a rate of𝐹(𝑡)=−3𝑒−0.2𝑡Litres per second, where 𝑡 is the number of seconds since the leak developed.1. Write an equation for the volume 𝑉 of water in the tank at time 𝑡. Simplify your answer.𝑉(𝑡)=

To find the volume $V(t)$ of water in the tank at time $t$, we need to integrate the rate of change of the volume of water over time. The rate of change of the volume of water is given by $F(t) = -3e^{-0.2t}$ liters per second.

1. Set up the integral:

   The volume of water at time $t$ can be found by integrating the rate of change of the volume from time $0$ to $t$. Since the tank initially holds 100 liters of water, we need to add this initial volume to the integral of the rate of change.

   $V(t) = V(0) + \int_0^t F(\tau) \, d\tau$

   Here, $V(0) = 100$ liters and $F(\tau) = -3e^{-0.2\tau}$.

2. Integrate the rate function:

   $\int_0^t -3e^{-0.2\tau} \, d\tau$

   To integrate $-3e^{-0.2\tau}$, we use the fact that the integral of $e^{ax}$ is $\frac{1}{a}e^{ax}$:

   $\int -3e^{-0.2\tau} \, d\tau = -3 \int e^{-0.2\tau} \, d\tau = -3 \left( \frac{1}{-0.2} e^{-0.2\tau} \right) = 15e^{-0.2\tau}$

3. Evaluate the definite integral:

   $\int_0^t -3e^{-0.2\tau} \, d\tau = \left[ 15e^{-0.2\tau} \right]_0^t = 15e^{-0.2t} - 15e^0 = 15e^{-0.2t} - 15$

4. Combine with the initial volume:

   $V(t) = 100 + \left( 15e^{-0.2t} - 15 \right)$

5. Simplify the expression:

   $V(t) = 100 + 15e^{-0.2t} - 15$

   $V(t) = 85 + 15e^{-0.2t}$

So, the equation for the volume $V$ of water in the tank at time $t$ is:

$V(t) = 85 + 15e^{-0.2t}$

To find the volume $V(t)$ of water in the tank at time $t$, we need to integrate the rate of change of the volume of water over time. The rate of change of the volume of water is given by $F(t) = -3e^{-0.2t}$ liters per second.

1. Set up the integral:

   The volume of water at time $t$ can be found by integrating the rate of change of the volume from 0 to $t$:

   $V(t) = V(0) + \int_0^t F(\tau) \, d\tau$

   Here, $V(0)$ is the initial volume of water in the tank, which is 100 liters.

2. Substitute the given rate function $F(t)$:

   $V(t) = 100 + \int_0^t -3e^{-0.2\tau} \, d\tau$

3. Integrate the function:

   To integrate $-3e^{-0.2\tau}$, we use the fact that the integral of $e^{ax}$ is $\frac{1}{a}e^{ax}$:

   $\int -3e^{-0.2\tau} \, d\tau = -3 \int e^{-0.2\tau} \, d\tau = -3 \left( \frac{e^{-0.2\tau}}{-0.2} \right) = 15e^{-0.2\tau}$

4. Evaluate the definite integral:

   $\int_0^t -3e^{-0.2\tau} \, d\tau = \left[ 15e^{-0.2\tau} \right]_0^t = 15e^{-0.2t} - 15e^0 = 15e^{-0.2t} - 15$

5. Combine the results:

   $V(t) = 100 + \left( 15e^{-0.2t} - 15 \right)$

6. Simplify the expression:

   $V(t) = 100 + 15e^{-0.2t} - 15$

   $V(t) = 85 + 15e^{-0.2t}$

So, the equation for the volume $V$ of water in the tank at time $t$ is:

$V(t) = 85 + 15e^{-0.2t}$