Solving Variable Velocity Water Pouring Problem with Calculus

[hexlan]

I've been relearning c++ and my friend gave me a problem to try and solve programmatically. Here is the set up:

There is a pitcher with a gallon of water.
The water is being poured out beginning at a rate of 2^x milliliters per second.
x begins at 1 and increases by 1 over 5 seconds.

The goal is to find out how long it takes to pour out all the water.

I finished writing a program and got an answer, but unfortunately both me and my friend are quite rusty with our calculus and have no way of checking the answer.

Could someone possibly provide an answer, or better yet go through the proper way of solving it.

 

[mfb]

Does that mean x=1 + t/5?
WolframAlpha can integrate that (where u is the time in seconds), you just have to set the result equal to the number of milliliters in a gallon (I refuse to calculate with imperial units).

Or does the flow rate increase in steps (2, 4, 8, ... ml/s)? Then you need a sum.

 

[hexlan]

Does that mean x=1 + t/5?

That is correct.

 

[kbrookes]

dv/dt = 2^x, x=1+(t/5) therefore dv/dt = 2^(1+t/5). Integrate this to find v in terms of t, when t=0, v=0. Use this to find the constant of integration. From this you can sub the volume of a gallon in ml to find the value of t(seconds). I think this should work however the integration may be a little tricky. Thanks Kyle

 

[PJC01]

I appreciate your use of calculus in solving this variable velocity water pouring problem. Calculus is a powerful tool for analyzing and solving problems involving changing quantities, and it is commonly used in physics and engineering.

To solve this problem, we can use the concept of integration, which is a fundamental concept in calculus. Integration allows us to find the total amount of water poured out over a specific period of time by summing up the infinitesimal amounts poured out at each moment.

To begin, we can set up a function to represent the volume of water poured out at any given time t. Let's call this function V(t), where t is measured in seconds. Since the rate of pouring is given by 2x milliliters per second, we can express this function as V(t) = 2x(t) where x(t) represents the value of x at time t.

Next, we need to determine the limits of integration. In this case, we want to find the total amount of water poured out over a period of 5 seconds, so our limits of integration will be from 0 to 5 seconds.

Now, we can set up the integral to represent the total amount of water poured out over 5 seconds:

∫ 0 5 2x(t) dt

To solve this integral, we need to find an antiderivative of 2x(t). The antiderivative of 2x(t) is x^2(t), so our integral becomes:

∫ 0 5 x^2(t) dt

Evaluating this integral from 0 to 5 gives us the total amount of water poured out over 5 seconds, which is equal to x^2(5) - x^2(0) = x^2(5) - 1. Since x(0) = 1, we can simplify this to x^2(5) - 1 = (1 + 5)^2 - 1 = 36 - 1 = 35 milliliters.

Therefore, it takes 35 milliliters of water to be poured out over 5 seconds.

I hope this explanation helps you understand the proper way to solve this problem using calculus. Keep up the good work in your programming and problem-solving skills!