Easy math, but I'm dumb

Ok this is like rational applications or something, but I have no clue which way to do it.



Is it x/15 - x/20 = 1?

cause it fills at 1/15th per minute and drains at 1/20th per minute to change 1minute to t-minutes, set equation = one completed job?

rate of fill for faucets = Vtub/15min
rate of fill for drain = -Vtub/20min

rate of fill with both = Vtub/15min + (-Vtub/20min) = Vtub/60min

Where Vtub is the volume of the tub.

So it would take 60min.

thanx bro!

Dammit Shidoshi go to your Chemistry.

Is it x/15 - x/20 = 1?

cause it fills at 1/15th per minute and drains at 1/20th per minute to change 1minute to t-minutes, set equation = one completed job?

I'd love to see your image, because I'm having difficulty understanding your problem. However, since you mention "drain" and "rate", I'm going to assume that this problem has something to do with filling and draining a tank. Your conditions state that this tank fills at 1/15th per minute and drains at 1/20th per minute. I'm going to assume that this means that it's being filled at 1/15th of its volume per minute and being drained at 1/20th of its volume per minute. Therefore, you have:

dx/dt = x/15 - x/20

dx/dt = x/60

dx/x = dt/60

∫dx/x = 1/60*∫dt

ln(x) = t/60 + C1

x = e^(t/60 + C1)

x = C2*e^(t/60), where C2 = e^C1

At time t=0, x = C2, so C2 = x(0). Therefore:

x = x(0)*e^(t/60)

Since this equation returns the volume at any time, plugging in 1 minute will reveal that you have e^(1/60) times what you had at the beginning of the filling/draining session.

I have no idea how Shidoshi produced 60 minutes as the answer, because that would mean that the problem states that x(60)=x(0)*e, which seems like a rather unlikely choice as an initial value for this problem. *shrugs*

*Agrees with Shidoshi*

Why the hell are you using Calculus to answer a high school math problem... Superiority complex?

How else can you figure out the relation between volume and time? I don't see how it's even possible to guess that the solution has an exponential form. If it's possible, do tell how you can, without calculus, find this form based on the given problem? Also, how am I supposed to know that Tay is in high school? If he is, how am I supposed to know that he not currently taking calculus, a course that most high school students take in their junior or senior years? -.-"

While I can't prove this isn't a calculus question, I've taken high school math and did these kinds of problem solving questions. Tay's in high school; I think most people here know that.

You remind me of Sheldon from The Big Bang Theory (minus the humour) and I'll just leave it at that.

Yeah 2^32 I thought this was a related rate problem too lol. But then I realized it was just an example.

I'd love to see your image, because I'm having difficulty understanding your problem. However, since you mention "drain" and "rate", I'm going to assume that this problem has something to do with filling and draining a tank. Your conditions state that this tank fills at 1/15th per minute and drains at 1/20th per minute. I'm going to assume that this means that it's being filled at 1/15th of its volume per minute and being drained at 1/20th of its volume per minute. Therefore, you have:

dx/dt = x/15 - x/20

dx/dt = x/60

dx/x = dt/60

∫dx/x = 1/60*∫dt

ln(x) = t/60 + C1

x = e^(t/60 + C1)

x = C2*e^(t/60), where C2 = e^C1

At time t=0, x = C2, so C2 = x(0). Therefore:

x = x(0)*e^(t/60)

Since this equation returns the volume at any time, plugging in 1 minute will reveal that you have e^(1/60) times what you had at the beginning of the filling/draining session.

I have no idea how Shidoshi produced 60 minutes as the answer, because that would mean that the problem states that x(60)=x(0)*e, which seems like a rather unlikely choice as an initial value for this problem. *shrugs*

Because you are doing it wrong.

dV/dt = Vt/15 - Vt/20

dV = (Vt/60) dt

Integrating from (V,t)=(0,0) till (V,t)=(V,t)

V = (Vt/60)*t

Where Vt is the total volume.
The drain/filling rates are not dependant on the volume. They are constants.

The drain/filling rates are not dependant on the volume. They are constants.

Which is exactly why it is not a Calculus problem.

I don't know if this counts as Linear Algebra, but you sure can put the problem under the "vector view".
If you're going EAST 10 miles per hour, but something is pulling you back at 5 miles per hour, you'll end up with 5 miles per hour (or less due to fraction).
The same goes for Volume-filling, the filling rate of the pump ADDs water to the container, thus it is positive in algebraic expression (+ V1), the draining TAKES AWAY water from the container, thus it is negative (-V2). Find the combination between those two (V1 + -V2) and you can figure out where the thing is headed, and how fast.

Which is exactly why it is not a Calculus problem.

I don't know if this counts as Linear Algebra, but you sure can put the problem under the "vector view".
If you're going EAST 10 miles per hour, but something is pulling you back at 5 miles per hour, you'll end up with 5 miles per hour (or less due to fraction).
The same goes for Volume-filling, the filling rate of the pump ADDs water to the container, thus it is positive in algebraic expression (+ V1), the draining TAKES AWAY water from the container, thus it is negative (-V2). Find the combination between those two (V1 + -V2) and you can figure out where the thing is headed, and how fast.

I was just commenting why harrison's resolution was wrong. I know it's not a Calculus problem, I just solved it in the same way he did to better exemplify.

I was just commenting why harrison's resolution was wrong. I know it's not a Calculus problem, I just solved it in the same way he did to better exemplify.

Yeah, I just borrowed your line to follow up on someone above's comment on not being able to tell whether/why it's not a Calculus problem.

V = (Vt/60)*t

Where Vt is the total volume.
The drain/filling rates are not dependant on the volume. They are constants.

I did not know that the question was asking for 1/60th of the total volume (of the container) and not just its volume (the volume of the liquid that is already in the container). I assumed the latter because the former was probably not worth making a thread for.

I did not know that the question was asking for 1/60th of the total volume (of the container) and not just its volume (the volume of the liquid that is already in the container). I assumed the latter because the former was probably not worth making a thread for.

The question is what's the time needed to fill up the whole volume (which we don't have a way of knowing).