Solving for Initial Barrel Capacity: Wine Leaking Rate Problem"

[quantum_owl]

Homework Statement

Wine is leaking from a barrel at the rate of
r(t)= 10 - 0.1t litres per minute.

If the barrel is empty after 100 minutes, how much wine did the barrel initially hold?

The Attempt at a Solution

Well, I'm completely stuck. I feel like I'm missing something really, really obvious. Could anyone point me in the right direction?

Thanks everyone

 

[Borek]

How much wine leaked during first minute?

During second?

Third?

How much total during three minutes?

Can you calculate now how much wine leaked in 100 minutes?

 

[quantum_owl]

I don't know if this is the right way to go about it, but when t=0, the barrel is full, is that right? Which then means that when t=100, the barrel is empty.

Then, I can say that r(0)=10-0.1x0, which leaves 10. So the maximum capacity of the barrel is 10 litres, maybe.

I don't know, I'm not any good at these worded questions.

 

[JonF]

Sorry, but that's not right. Think of rates as “how fast”, at r(0) the rate wine is leaving the bottle is 10 lpm, this says nothing about how much is currently in the barrel. Compare it to speed, if ALL you know is that a car is currently going 10mph, you can't stay how far it's traveled. That number 10 you can up with is the rate that the problem STARTS at. Again compare it to speed, if I pushed a car really hard and it started at 10mph, at that very instant I pushed it how far has it moved? Sure it's going 10mph, but it hasn't moved yet, so it's gone 0 miles.


Let’s talk about p(t), where p(t) is how much wine has left the barrel.

We know p(0) = 0 (it started out with 0 line being leaked), we want to solve for p(100) (how much wine left it when it's empty).

I suspect your class gave you ways to convert r(t) into p(t). It might be put it in terms of position, acceleration, velocity, or some other rate words. If this is the case, ask yourself what is equivalent to position, velocity, and acceleration in this equation.

but it should be p(t) = 10t - 0.05t^2 + p(0)

If you are in calculus, you should recognize this as the integral of r(t)dt, with your +c term = p(0). If you aren’t they should have just given you this conversion or shown you how to get it by drawing a picture. If you don't understand how to get p(t), let me know, and let me know which method your class uses.

 

[quantum_owl]

First of all, thankyou to both of you for taking the time to reply. I really appreciate it.

Jon, I'm taking a course which is roughly equivalent to Australian maths C, or final year A-Level maths. I will be honest with you, and say that we have never been shown anything like what you have just shown me. Could you please explain to me how you got to the equation?

Many thanks

 

[BobG]

Integrating would be the obvious solution, as JonF mentioned.

You could also find the average rate and the multiply the average rate by the amount of time.