Calculate Water Volume In A Draining Tank After 2 Minutes

by Henrik Larsen 58 views

Introduction

In this article, we're diving into a classic differential equations problem that involves a tank draining water. Specifically, we have a tank initially containing 100 liters of water, and it's draining at a rate proportional to the remaining volume. We're given the proportionality constant, k = 0.05, and the initial volume, V(0) = 100 liters. Our main goal here is to figure out how much water is left in the tank after 2 minutes. So, let's break down the problem and solve it step by step. This kind of problem pops up in various fields, from chemical engineering to environmental science, whenever we need to model how quantities change over time, especially when the rate of change depends on the current amount. Think about it: it could be anything from the decay of a radioactive substance to the cooling of an object. So, understanding this type of problem can give you a solid foundation for tackling real-world scenarios. We'll start by setting up the differential equation, then we'll solve it using separation of variables, and finally, we'll plug in the values to get our answer. Stick with us, and you'll see how mathematical modeling can make complex situations much clearer. Trust me, this is something you'll want in your problem-solving toolkit!

Setting up the Differential Equation

To start, let's translate the problem statement into a mathematical form. The key phrase here is "drained at a rate proportional to the remaining volume." This tells us that the rate of change of the volume (dV/dt) is proportional to the volume itself (V). Since the tank is draining, the volume is decreasing, so we need to include a negative sign. This gives us the differential equation:

dV/dt = -kV

Where:

  • V(t) is the volume of water in the tank at time t (in liters).
  • t is the time (in minutes).
  • k is the proportionality constant, which is given as 0.05.

So, our equation becomes:

dV/dt = -0.05V

This is a separable differential equation, which means we can rearrange it so that all the V terms are on one side and all the t terms are on the other. This is a crucial step because it allows us to integrate both sides independently and find the general solution. Why is this important, guys? Because it's the foundation for understanding how the volume changes over time. Without this setup, we'd be stuck! Think of it like sorting your laundry before washing – you need to separate the colors from the whites before you can throw them in the machine. In math, we're separating the variables before we integrate. This step-by-step approach is what makes these kinds of problems manageable. We're not just throwing numbers into a formula; we're building a model that describes the physical situation, and that's what makes it powerful. Plus, once you get the hang of separating variables, you'll start seeing these kinds of equations everywhere, from physics to finance. It's a versatile tool, so let's make sure we nail it down!

Solving the Differential Equation

Now that we have our differential equation, dV/dt = -0.05V, let's solve it. The first step is to separate the variables. We'll divide both sides by V and multiply both sides by dt:

(1/V) dV = -0.05 dt

Now, we can integrate both sides. The integral of (1/V) dV is ln|V|, and the integral of -0.05 dt is -0.05t. Don't forget to add the constant of integration, C, on one side:

∫(1/V) dV = ∫-0.05 dt

ln|V| = -0.05t + C

To solve for V, we need to exponentiate both sides:

e^(ln|V|) = e^(-0.05t + C)

|V| = e^(-0.05t) * e^C

Since e^C is just another constant, let's call it A:

V = Ae^(-0.05t)

This is the general solution to our differential equation. It tells us how the volume changes over time, but we still need to find the value of A. That's where our initial condition comes in. Remember, we know that V(0) = 100 liters. So, we can plug in t = 0 and V = 100 into our general solution:

100 = Ae^(-0.05 * 0)

100 = A * e^0

100 = A

So, A = 100, and our particular solution is:

V(t) = 100e^(-0.05t)

Guys, this is a major milestone! We've gone from a differential equation describing a draining tank to an explicit formula that tells us the volume of water at any time t. This formula is like a magic key – it unlocks the answer to our question. But let's take a moment to appreciate what we've done here. We've used the power of calculus to model a real-world situation. We separated variables, integrated both sides, and used an initial condition to find a unique solution. This is the process that engineers and scientists use every day to solve complex problems. And now, we're just one step away from answering our original question: How much water is left after 2 minutes? Keep that momentum going; we're almost there!

Calculating the Volume After 2 Minutes

Now that we have the particular solution, V(t) = 100e^(-0.05t), we can easily find the volume of water after 2 minutes. We just need to plug in t = 2:

V(2) = 100e^(-0.05 * 2)

V(2) = 100e^(-0.1)

Using a calculator, we find that e^(-0.1) is approximately 0.9048:

V(2) = 100 * 0.9048

V(2) ≈ 90.48

So, after 2 minutes, there are approximately 90.48 liters of water left in the tank.

Isn't that cool? We started with a word problem, translated it into a differential equation, solved the equation, and now we have a precise answer. This is the power of mathematical modeling! We've essentially built a virtual model of the draining tank, and we can use it to predict the volume of water at any time. This is super useful in all sorts of applications. For example, engineers might use this kind of model to design drainage systems or to control the flow of liquids in a chemical plant. Environmental scientists might use it to study the rate at which pollutants are flushed out of a lake or river. The possibilities are endless. And remember, this whole process started with understanding the concept of proportionality and how to express it mathematically. So, next time you see a problem that involves rates of change, think about setting up a differential equation. You might be surprised at how powerful this technique can be. We've tackled this problem head-on, and now you've got a solid example of how to apply differential equations to real-world scenarios. Keep practicing, and you'll become a pro at this in no time!

Conclusion

In summary, we've solved a problem involving a tank draining water at a rate proportional to its volume. We set up a differential equation, solved it using separation of variables, and used the initial condition to find the particular solution. Finally, we calculated that after 2 minutes, there are approximately 90.48 liters of water left in the tank.

This problem illustrates a fundamental application of differential equations in modeling real-world phenomena. The key takeaways are:

  1. Translating word problems into mathematical equations.
  2. Solving separable differential equations.
  3. Using initial conditions to find particular solutions.
  4. Interpreting the results in the context of the problem.

Great job, guys! You've taken on a challenging problem and come out on top. Remember, the skills you've learned here – setting up equations, solving for variables, and interpreting results – are valuable in many different fields. Keep practicing, keep exploring, and keep pushing your mathematical boundaries. The world is full of problems waiting to be solved, and now you're one step closer to solving them!