Integrating: mg/b(1 - e^(-bt/m)

[jmwachtel]

Homework Statement

I need help integrating: mg/b(1 - e^(-bt/m)

The Attempt at a Solution

I came up with mgt/b - mg/b(-b/m * e^(-bt/m)

It's been a while since calculus, can someone help me with this.

 

[G01]

If this is the integral you mean:

\frac{mg}{b}\int(1-e^{-bt/m})dt

then you are correct in your integration.

 

[jmwachtel]

That is it, but the numbers are coming out negative, they should be positive. Something is not right. I know I have to take it from 0 to t, but I'm not sure what to do with it. Please help.

 

[jmwachtel]

Anyone else know calculus?

 

[lavalamp]

Can you give more information on the problem? It could simply be that you've take a particular direction to be positive, and the answer in the back of the book assumes the opposite direction is positive.

 

[Cynapse]

I get the integral to be

= mg/b * {[t-(m/b)exp(-bt/m)]}
= {mgt/b + [(m^2)g/(b^2)]exp(-bt/m)}
= {mgt/b + mg/b(m/b * e^(-bt/m)}

you got the exponential integration wrong:
Integral of: exp(kx)dx = {1/(dkx/dx)exp(kx)} = {(1/k)exp(kx)}
in this case k = -b/m
so integral of exp[(-b/m)t] = (-m/b).exp[(-b/m)t]

YOUR exponential is a - to start with so it becomes a plus

 

[jmwachtel]

The answer should be around .3 positive when T is equal to .05. I get a negative -.02417. The integral is still not right even with the one above, what is going on?

 

[jmwachtel]

and: m = .001, g=-9.81, b=.025

 

[<---]

The integral you posted first is not correct as cynapse explained, even in corrected form it's an indefinite integral with the constant of integration missing. It sounds like you need a definite integral.
Did you get the right definite integral for the limits of integration 0 and T?
The answer for the equation as you described it is -0.0084.
This leads me to believe that you don't have the correct set up for the problem.
Are the units consistent?
Do you have the correct limits of integration?
If you post the full wording of the problem as well as units of known quantity's with a more complete description of what you are doing I can try to help you more.

 

[jmwachtel]

Everything is correct, we are setting up a spreadsheet in class. We use this for Analytical
approach. v is calculated using the equation above, so to get back to y, you have to integrate. I know what y should be which I stated above because that's through the numerical approach.

 

[Gear300]

It seems you're finding distance traveled while experiencing air resistance. Just do this:
\frac{mg}{b}\int(1-e^{-bt/m})dt
Then, integrate each term in the integral separately in respect to time t (find the integral of 1 and the integral of -e^(-bt/m) individually). For the integral of -e^(-bt/m), use (-bt/m) as a function u. Then simply use the chain rule for integration to solve the problem.

 

[jmwachtel]

Here is what I have got and it's still not working, this is driving me nuts!

=(((m*g/b)*t)+((m*g/b)*((m/b)*EXP(-b*t/m))))-(m^2*g/b^2)

What am I doing wrong?

 

[<---]

That's the correct integral for the limits 0 & T.
At least I think it is. I would recommend using the forums handy mathematical formatting so the equations are easier to read.
So I still think something else is wrong here.

 

[jmwachtel]

This still not solved. I can't figure out what I am doing wrong. Where is the equation formatting?

 

[jmwachtel]

[\frac{mgt}{b} + \frac{mg}{b} (\frac{m}{b} e^{\frac{-bt}{m}})] - [\frac{m^{2}g}{b^{2}}]

This is integration I have come up with for the equation listed above that is not working! IS my integration wrong? This from 0 to t.

 

[<---]

That is the right integral for the equation G01 posted for the lower limit of 0 and an upper limit of t.
Side note: your equations will look nicer and be easier to work with if you simplify them.

\frac{mg}{b}[T + \frac{m}{b}(e^\frac{-bT}{m} - 1)]

 

[jmwachtel]

Ok here is my teachers reply:

Your integration is still incorrect. See attachment for explanation. You MUST have correctly corresponding limits on the integrals. At time=0, y=yo; at time=t, y=y.

I don't understand. The attached says:

v = dy/dt

dy = vdt

\intdy (from yintial to y) = \intvdt (from 0 to t)

 

[jmwachtel]

Anyone understand this?