# Thread: Application: rocket being fired - how to integrate?

1. ## Application: rocket being fired - how to integrate?

I have these few lines that I need help with.

dv/dt = (ub/(Mo -bt)) - g

v(t) = -u ln(Mo -bt) - gt + C

1. I see that we integrated from the first to the next line, but what did we integrate?
2. How do you determine we could integrate from the first line?
3. Can you show me how to integrate whatever was integrated.

Thank you so much,
Yvonne

2. $v'(t) = \frac{ub}{M_{0} - bt} - g$

We can see that t is our variable so that is what we are integrating in respect to. $u, M_{0}, b, g$ are all constants so treat them as such.

$\underbrace{\int v'(t) dt}_{v(t)} = \int \frac{ub}{M_{0} - bt} \: dt - \int g \: dt$

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For the first integral, we have the s-sub: $s = M_{0} - bt \: \: \Rightarrow \: \: ds = -b \: dt \iff dt = -\frac{ds}{b}$

So we have: $\int \frac{ub \: dt}{M_{0} - bt} = \int \frac{ub\left(-\frac{ds}{b}\right)}{s} = -u \int \frac{ds}{s} = -u \ln |s| + C = -u \ln |M_{0} - bt| + C$
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As for the second integral, imagine: $\int g \: dt = \int g t^{0} \: dt$ and integrate accordingly.