Part Derivs: Minimizing the Weight of a Rocket

[kostoglotov]

Homework Statement

This is actually an Applied Project in the text, and overall is quite a large problem, so I won't post the entire thing, as there are lots of equations and steps where the text guides me by saying "show that...this thing...then...show that this other thing..."

What I need to know, is if my reasoning on one particular step is valid or not.

So, the overall objective is to find an expression for the minimal weights of each of 3 stages of a rocket based on various constant factors and a constraint that a given final velocity be reached.

3) Verify that \ln{\frac{M+A}{A}} is minimized at the same location as M.

Homework Equations

v_f: final velocity

M_1, M_2, M_3: masses of rocket stages

edit: A: a constant, mass of the payload

M = M_1+M_2+M_3

\frac{M+A}{A}=\frac{(1-S)^3N_1N_2N_3}{(1-SN_1)(1-SN_2)(1-SN_3)}

These N_is are functions of the three masses of the three stages.

The Attempt at a Solution

I went ahead and found the gradient vectors for both M and ln((M+A)/A).

\nabla M = \left< 1,1,1 \right>

\nabla \ln{\frac{M+A}{A}}=\nabla \left[\ln{(M+A)}-\ln{A}\right]=\left< \frac{1}{M+A},\frac{1}{M+A},\frac{1}{M+A} \right>

NB: these are grad vectors with respect to M_1,M_2,M_3

I have reasoned that these vectors are pointing in the same direction, so are therefore equivalent to each other, and so therefore could be substituted for one another in the Method of Lagrange Multipliers.

Or more formally, when using the Method of Lagrange Multipliers

\nabla M = k \nabla \ln{\frac{M+A}{A}} = \lambda \nabla constraint

But since k and lambda are constants, the k could just be absorbed into the lambda and the system solved in a similar, valid way.

Am I right? If not, why not? :)

 

[Ray Vickson]

Homework Statement

This is actually an Applied Project in the text, and overall is quite a large problem, so I won't post the entire thing, as there are lots of equations and steps where the text guides me by saying "show that...this thing...then...show that this other thing..."

What I need to know, is if my reasoning on one particular step is valid or not.

So, the overall objective is to find an expression for the minimal weights of each of 3 stages of a rocket based on various constant factors and a constraint that a given final velocity be reached.

3) Verify that \ln{\frac{M+A}{A}} is minimized at the same location as M.

Homework Equations

v_f: final velocity

M_1, M_2, M_3: masses of rocket stages

edit: A: a constant, mass of the payload

M = M_1+M_2+M_3

\frac{M+A}{A}=\frac{(1-S)^3N_1N_2N_3}{(1-SN_1)(1-SN_2)(1-SN_3)}

These N_is are functions of the three masses of the three stages.

The Attempt at a Solution

I went ahead and found the gradient vectors for both M and ln((M+A)/A).

\nabla M = \left< 1,1,1 \right>

\nabla \ln{\frac{M+A}{A}}=\nabla \left[\ln{(M+A)}-\ln{A}\right]=\left< \frac{1}{M+A},\frac{1}{M+A},\frac{1}{M+A} \right>

NB: these are grad vectors with respect to M_1,M_2,M_3

I have reasoned that these vectors are pointing in the same direction, so are therefore equivalent to each other, and so therefore could be substituted for one another in the Method of Lagrange Multipliers.

Or more formally, when using the Method of Lagrange Multipliers

\nabla M = k \nabla \ln{\frac{M+A}{A}} = \lambda \nabla constraint

But since k and lambda are constants, the k could just be absorbed into the lambda and the system solved in a similar, valid way.

Am I right? If not, why not? :)

No need to be so fancy; the result is automatically true, almost with no work. if $A$ is a positive constant, $M/A$ is minimized at the same point as $M$. Then, also, $1+ M/A$ is minimized at the same point. However, $1 + M/A = (M+A)/A$, so the latter is minimized at the same point as $M$. Finally, if $A,M > 0$, $\ln((A+M)/A)$ is a monotone strictly increasing function of $(A+M)/A$, so if we make $(M+A)/A$ as small (or large) as possible, we also make its logarithm as small (or large) as possible, and vice versa. That's all you need: it is not "rocket science", if you will forgive the term.