# Calculus integration problem

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

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

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

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
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