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The (Tsiolkovsky) rocket equation states that the velocity of a rocket can be calculated as

$$v(t) = v_0 \ln\frac{m_0}{m_0 - \dot m t}$$

where $m_0$ is the starting mass, $\dot m$ is the (constant) loss of mass per time due to the rocket engine which emits the exhaust gas with the velocity $v_0$. To get the distance travelled at a certain time after the launch, I have to integrate this formula. My approach is as follows:

$$ \begin{align} &\int \ln \frac{a}{a-bt}\text{d}t\\ = &\int \ln a \text{d}t - \int \ln (a-bt)\text{d}t\\ = & t\ln a - \left( -\frac{1}{b}((a-bt) \ln(a-bt) - (a-bt)\right) \end{align} $$

I know that there is a "nice" solution: $$\int \ln \frac{a}{a-bt}\text{d}t = \frac{a-bt}{b} \ln \frac{a-bt}{a} + t$$ but I can't see how to get there from my approach. Which steps are neccessary, or is there a different way to begin with?

PS: it seems to work if I remove the constant $\frac{a}{b}$ and add the constant term $-\frac{a}{b}\ln a$ in my antiderivative. But there has to be a better way!

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You may just use an integration by parts. $$ \begin{align} \int \ln \frac{a}{a-bt}\:\text{d}t&=t\:\ln \frac{a}{a-bt}-\int t\:\frac{\left(\frac{a}{a-bt}\right)'}{\frac{a}{a-bt}}\:\text{d}t\\\\ &=t\:\ln \frac{a}{a-bt}-\int t\:\frac{b}{a-bt}\:\text{d}t\\\\ &=t\:\ln \frac{a}{a-bt}+\int \frac{a-bt-a}{a-bt}\:\text{d}t\\\\ &=t\:\ln \frac{a}{a-bt}+t-a\int \frac{1}{a-bt}\:\text{d}t\\\\ &=t\:\ln \frac{a}{a-bt}+t-\frac ab\int \frac{b}{a-bt}\:\text{d}t\\\\ &=t\:\ln \frac{a}{a-bt}+t-\frac ab\ln \frac{a}{a-bt}\\\\ &=\left(t-\frac ab\right)\ln \frac{a}{a-bt}+t\\\\ &=\frac{a-bt}{b} \ln \frac{a-bt}{a} + t. \end{align} $$

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  • $\begingroup$ Could you please explain why you multiply with $b/b$, and more importantly, why do you add $\ln a$? Is this only to get to my proposed solution? $\endgroup$ – Jasper Feb 13 '15 at 22:04
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    $\begingroup$ @Jasper Yes, it is because you may notice that: $$\left(\ln \frac{a}{a-bt} \right)'=\frac{\left(\frac{a}{a-bt}\right)'}{\frac{a}{a-bt}}=\frac{b}{a-bt} $$ giving $$\int \frac{b}{a-bt}\:\text{d}t=\ln \frac{a}{a-bt}.$$ Thanks. $\endgroup$ – Olivier Oloa Feb 13 '15 at 23:21
  • $\begingroup$ But using just integration by substitution, one would get $\ln \frac {1}{a-bt}$, right? $\endgroup$ – Jasper Feb 14 '15 at 0:25
  • $\begingroup$ @Jasper Right. The general form is $$\ln \frac{1}{a-bt} +C $$ $C$ is any constant, if $C:=\ln a$ you get $$\ln \frac{1}{a-bt} +\ln a=\ln \frac{a}{a-bt} $$ as another primitive. $\endgroup$ – Olivier Oloa Feb 14 '15 at 1:33

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