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

#### Kiche

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.

#### Haven

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

#### Kiche

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*?

#### HallsofIvy

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

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

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.

Last edited:
• Kiche

#### 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)?

#### HallsofIvy

MHF Helper
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!