How Does Differential Calculus Apply to Filling a Jug with Liquid?

[eeriana]

Homework Statement

2 part question: part one is supposed to be a hint to part two.
1) find the general solution to the equation: dx/dt = x - x^2
2)A one gallon jug is half full of a liquid and half full of air. Additional liquid is being pumped in at a rate of 1/4 gallon per minute. If the rate at which the liquid flows into the container is proportional to both the volume of liquid in the jug and the volume of air in the jug, how long will it be before the jug is 90% full of liquid?

Homework Equations

dx/dt = r(i)c(i) - (r(o)/V)x

The Attempt at a Solution

If I have calculated the first part correctly, I came up with t = ln(x/1-x) - C

I know the rate in of the liquid is .25 gal/min. Rate out is 0. I am not seeing the connection to the first equation. If someone could point me in the right direction, I'd appreciate it.

Thanks

 

[HallsofIvy]

Homework Statement

2 part question: part one is supposed to be a hint to part two.
1) find the general solution to the equation: dx/dt = x - x^2
2)A one gallon jug is half full of a liquid and half full of air. Additional liquid is being pumped in at a rate of 1/4 gallon per minute. If the rate at which the liquid flows into the container is proportional to both the volume of liquid in the jug and the volume of air in the jug, how long will it be before the jug is 90% full of liquid?

Homework Equations

dx/dt = r(i)c(i) - (r(o)/V)x

This is meaningless without saying what the variables mean. I might assume that "x" is the amount of liquid in the jar, but, in that case, what is V? Are "r(i)" and "r(o)" "rate in" and "rate out"? But what is c(i)?

The Attempt at a Solution

If I have calculated the first part correctly, I came up with t = ln(x/1-x) - C

So what is x as a function of t?

[/quote]I know the rate in of the liquid is .25 gal/min. Rate out is 0. I am not seeing the connection to the first equation. If someone could point me in the right direction, I'd appreciate it.

Thanks
[/QUOTE]
It took me a while to make sense of it but I think that by "the rate at which the liquid flows into the container is proportional to both the volume of liquid in the jug" they mean the rate is jointly proportional: that is dV/dt= k(V)(A) where V is the amount of liquid in the bottle at time t and A is the amount of air in the bottle at time t, both measured in gallons. Since the bottle holds one gallon, A= 1- V so the equation is dV/dt= k(V)(1- V).

Unfortunately, now the "Additional liquid is being pumped in at a rate of 1/4 gallon per minute." makes no sense! I can only assume that is to tell you the "constant of proportionality", k, and guess that it is only at the beginning, at t= 0 that the rate is 1/4 gallon per minute. Since V(0)= 1/2, A(0)= 1/2, k(V)(A)= k(1/2)(1/2)= k/4= 1/4 so k= 1. If that was what was intended, they should have said "initially" the liquid was being pumped in at a rate of 1/4 gallon per minute.

 

[eeriana]

This is meaningless without saying what the variables mean. I might assume that "x" is the amount of liquid in the jar, but, in that case, what is V? Are "r(i)" and "r(o)" "rate in" and "rate out"? But what is c(i)?

c(i) is concentration in.



So what is x as a function of t?

I know the rate in of the liquid is .25 gal/min. Rate out is 0. I am not seeing the connection to the first equation. If someone could point me in the right direction, I'd appreciate it.

Thanks
[/QUOTE]
It took me a while to make sense of it but I think that by "the rate at which the liquid flows into the container is proportional to both the volume of liquid in the jug" they mean the rate is jointly proportional: that is dV/dt= k(V)(A) where V is the amount of liquid in the bottle at time t and A is the amount of air in the bottle at time t, both measured in gallons. Since the bottle holds one gallon, A= 1- V so the equation is dV/dt= k(V)(1- V).

Unfortunately, now the "Additional liquid is being pumped in at a rate of 1/4 gallon per minute." makes no sense! I can only assume that is to tell you the "constant of proportionality", k, and guess that it is only at the beginning, at t= 0 that the rate is 1/4 gallon per minute. Since V(0)= 1/2, A(0)= 1/2, k(V)(A)= k(1/2)(1/2)= k/4= 1/4 so k= 1. If that was what was intended, they should have said "initially" the liquid was being pumped in at a rate of 1/4 gallon per minute.[/QUOTE]

so k is the proportionality? I am not quite sure where k came from. I am also not getting the connection to the first equation, and my instructor said that the first equation is a hint to solving the other. Am I missing something?

Thank you for your help