# Math Help - Problem with integating for rocket boost phase, can anyone help?

1. ## Problem with integating for rocket boost phase, can anyone help?

I need to integrate this function once with respect to t to find an expression for h(t):
v(t) = -k ln(m-bt) - gt + k ln(m)

Can anyone show me how to do this? How would I resolve the constant?

I then need to show that, after substituting t into h(t) I get the expression:

h(t) = km/b [(1 - X) ln(1 - X) + X(1 - (g/2kb)Xm0)]

Sorry I wasn't sure how to put that into LaTeX, I hope it's clear enough in that form.

I should mention also that k, b, m and g are constants. This is all to find an expression for h(t) when the rocket reaches the end of its boost phase. Thanks.

2. You need an initial condition to resolve the constant.

In this case I would assume that h(0) = 0.
Substituting the constant in you can solve for zero.

The LaTeX commands you want are \ln{x} for $\ln{x}$
and \frac{1}{2} for $\frac{1}{2}$
and for subscripts a_n for $a_n$
to adjust bracket size use the following commands:
\left[ and \right] so $[\frac{1}{2}]$ becomes $\left[\frac{1}{2}\right]$
$v(t) = -k\ln(m-bt) - gt + k\ln(m)$
$h(t) = \frac{km}{b} \left[(1 - X)\ln(1 - X) + X(1 - \left(\frac{g}{2kb}\right)Xm_0)\right]$

3. Ok, thanks for your reply. However would you be able to show the steps to get the expression of h(t) from the v(t) expression?

Also how would I find t*?

4. $\int \ln(x) dx= x \ln(x)- x$ so to integrate $\int -k \ln(m-bt) - gt + k \ln(m)= -k\int \ln(m- bt) dt- g\int t dt+ k\ln(m)\int dt$, let u= m-bt in the first integral. Then du= -bdt so dt= -(1/b)dt and the integeral becomes $\frac{k}{b}\int \ln(u)du= \frac{k}{b}(u \ln(u)- u)= \frac{k}{b}((m-bt)\ln(m- bt)- (m- bt))$.

The entire integral is
$h= \frac{k}{b}((m-bt)\ln(m- bt)- (m- bt))-\frac{g}{2}t^2+ k\ln(m)t+ C$

As Haven said, set h(0) equal to the initial height to determine C.

You ask how to find t* but don't define t*. If you mean "at the end of the boost phase" then you will need some other condition to determine what t* itself is before putting it into that equation.

5. Thanks for your help! Yeah I did mean at the end of the boost stage. When the fuel runs out the equation is:

m(t*) = -bt + m = m - Xm

where m is the initial mass of the rocket and and the initial mass of fuel is Xm, where X is an unspecified fraction between 1 and 0.

How would I find an expression for t? Also how would I go about putting it into the expression for h(t) and v(t)?

6. Originally Posted by Kiche
Thanks for your help! Yeah I did mean at the end of the boost stage. When the fuel runs out the equation is:

m(t*) = -bt + m = m - Xm

where m is the initial mass of the rocket and and the initial mass of fuel is Xm, where X is an unspecified fraction between 1 and 0.

How would I find an expression for t? Also how would I go about putting it into the expression for h(t) and v(t)?
When you are doing problems involving integrals of logarithms, you are expected to know how to solve things like -bt*+ m= m-Xm!