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A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at liftoff (including its fuel) is $m$, the fuel is consumed at rate $r$, and the exhaust gases are ejected with constant velocity $v_e$ (relative to the rocket). A model for the velocity of the rocket at time t is given by the equation

$$v(t) = - gt - v_e \ln\frac{m-rt}m$$

where g is the acceleration due to gravity and t is not too large. If $g = 9.8\, \mathrm{m}/\mathrm{s}^2$, $m = 29,000\, \mathrm{kg}$, $r = 170\, \mathrm{kg}/\mathrm{s}$, and $v_e = 2,900\, \mathrm{m}/\mathrm{s}$. What would the height of the rocket be one minute after liftoff?

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Any insights of your own to add? –  Ron Gordon Feb 13 '13 at 9:25

1 Answer 1

The height of the rocket at time $t$ is

$$h(t) = \int_0^t dt' \: v(t') = -\frac{1}{2} g t^2 - v_e \int_0^t dt' \: \ln{\left (1-\frac{r}{m} t' \right )}$$

(The value of $g$ should be $-9.8 \, \mathrm{m}/\mathrm{s}^2$.)

Your job is to evaluate the integral on the right. Substitute $u=1-(r/m)t'$ and use the fact that

$$\int du \log{u} = u \log{u} - u + C$$

to get your result.

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Out of curiosity, why is it that you put your variable of integration before the integrand? I've never seen it done in that order before and it intrigues me. –  tesc Feb 13 '13 at 9:53
    
It's a habit I got from optics (my field of study). I found it easier to keep track of multiple integrals that way. In physics, it tuns out that it's easier to treat the integral as an operator this way as well. –  Ron Gordon Feb 13 '13 at 10:06
    
Interesting. I quite like how close the variable is to the integration sign itself and thereby making it easier to keep track of, as you say, but at the same time my institutionalised my cannot help but feel you're solving $\int 1\,du\;(= u + C)$. Thanks for the answer! –  tesc Feb 13 '13 at 10:11
    
Yeah, I've had that thrown at me a few times. My response is that they can read it however way they want and be completely baffled as to how I got the right answer, or be a little flexible in their notation and see what I have clearly done. I was trained the standard way, but every time I see a multiple integral, I tire of having to figure out which variable goes with which integral. In my notation, that is never an issue. –  Ron Gordon Feb 13 '13 at 10:14
    
I certainly think it's healthy to question notation. There's nothing to lose in trying to make the world a little bit more pedagogical! More out of curiosity: would you stylise an iterated integral as $\int dy \int dx f(x, y)$? –  tesc Feb 13 '13 at 10:22

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