# Piece of Pi

A while back, we deemed that it is utterly impractical to calculate the value of $\pi$ using the partial sum of Leibniz’s series due to its slow convergence (see “Pumpkin Pi“)

Fig. 1

As illustrated in Fig. 1, in order to determine each additional correct digit of $\pi$, the number of terms in the summation must increase by a factor of 10.

What we need is a fast converging series whose partial sum yields given number of correct digits with far fewer terms.

Looking back, we see that the origin of Leibniz’s series is the definite integral

$\frac{\pi}{4} = \int\limits_{0}^{1}\frac{1}{1+x^2} dx\quad\quad\quad(1)$

To find the needed new series, we consider a variation of (1), namely,

$\frac{\pi}{6} = \int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{1}{1+x^2} dx\quad\quad\quad(2)$

Given the fact (see “Pumpkin Pi“) that

$\frac{1}{1+x^2} = \sum\limits_{k=1}^{n}(-1)^{k+1}x^{2k-2}+\frac{(-1)^n x^{2n}}{1+x^2}\quad\quad\quad(3)$

We proceed to integrate (3) with respect to $x$ from $0$ to $\frac{1}{\sqrt{3}}$,

$\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{1}{1+x^2} dx=\int\limits_{0}^{\frac{1}{\sqrt{3}}}\sum\limits_{k=1}^{n}(-1)^{k+1}x^{2k-2} dx + \int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{(-1)^n x^{2n}}{1+x^2}dx$

$= \sum\limits_{k=1}^{n}(-1)^{k+1}\int\limits_{0}^{\frac{1}{\sqrt{3}}}x^{2k-2} dx + (-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx$

$= \sum\limits_{k=1}^{n}(-1)^{k+1}\frac{x^{2k-1}}{2k-1}\bigg|_{0}^{\frac{1}{\sqrt{3}}}+ (-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx$

As result, (2) becomes

$\frac{\pi}{6}=\sum\limits_{k=1}^{n}(-1)^{k+1}\frac{(\frac{1}{\sqrt{3}})^{2k-1}}{2k-1}+(-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx$

$= \sum\limits_{k=1}^{n}(-1)^{k+1}\frac{{(\frac{1}{3}})^k}{(2k-1)\frac{1}{\sqrt{3}}} + (-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx$

$= \sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)} + (-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx$

i.e.,

$\frac{\pi}{6} - \sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)} = (-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx\quad\quad\quad(4)$

By (4),

$|\frac{\pi}{6} - \sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)}| = |(-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx|=\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx<\int\limits_{0}^{\frac{1}{\sqrt{3}}}x^{2n}dx$

$=\frac{x^{2n+1}}{2n+1}\bigg|_0^{\frac{1}{\sqrt{3}}}$

$=\frac{1}{3^n \sqrt{3} (2n+1)}$

which gives

$|\sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)} - \frac{\pi}{6}| < \frac{1}{3^n \sqrt{3} (2n+1)}$.

And so

$-\frac{1}{3^n \sqrt{3} (2n+1)}<\sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)}-\frac{\pi}{6}<\frac{1}{3^n \sqrt{3} (2n+1)}\quad\quad\quad(5)$

Since $\lim\limits_{n\rightarrow \infty}\frac{1}{3^n \sqrt{3} (2n+1)}=0$, (5) implies

$\lim\limits_{n\rightarrow \infty} \sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)}-\frac{\pi}{6}= 0$.

Hence,

$\lim\limits_{n\rightarrow \infty} \sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)}=\frac{\pi}{6}$.

It follows that the value of $\pi$ can be approximated by the partial sum of a new series

$6\sqrt{3}\sum\limits_{k=1}^{\infty}\frac{(-1)^{k+1}}{3^k(2k-1)}$

Let’s compute it with Omega CAS Explorer (see Fig. 2, 3)

Fig. 2

Fig. 2 shows the series converges quickly. The sum of the first 10 terms yields the first 6 digits!

Fig. 3

Totaling the first 100 terms of the series gives the first 49 digits of $\pi$ (see Fig. 3)

Exercise 1. Show that $\lim\limits_{n\rightarrow \infty}\frac{1}{3^n \sqrt{3} (2n+1)}=0$.

Exercise 2. Can we use $\frac{\pi}{3} = \int\limits_{0}^{\sqrt{3}}\frac{1}{1+x^2}dx$ to compute the value of $\pi$ in a similar fashion? Explain.

# Viva Rocketry! Part 2

Fig. 1

A rocket with $n$ stages is a composition of $n$ single stage rocket (see Fig. 1) Each stage has its own casing, instruments and fuel. The $n$th stage houses the payload.

Fig. 2

The model is illustrated in Fig. 2, the $i^{th}$ stage having initial total mass $m_i$ and containing fuel $\epsilon_i m_i (0 < \epsilon_i <1, 1 \leq i \leq n)$. The exhaust speed of the $i^{th}$ stage is $c_i$.

The flight of multi-stage rocket starts with the $1^{st}$ stage fires its engine and the rocket is lifted. When all the fuel in the $1^{st}$ stage has been burnt, the $1^{st}$ stage’s casing and instruments are detached. The remaining stages of the rocket continue the flight with $2^{nd}$ stage’s engine ignited.

Generally, the rocket starts its $i^{th}$ stage of flight with final velocity achieved at the end of previous stage of flight. The entire rocket is propelled by the fuel in the $i^{th}$ casing of the rocket. When all the fuel for this stage has been burnt, the $i^{th}$ casing is separated  from the rest of the stages. The flight of the rocket is completed if $i=n$. Otherwise, it enters the next stage of flight.

When all external forces are omitted, the governing equation of rocket’s $i^{th}$ stage flight (see “Viva Rocketry! Part 1” or “An alternate derivation of ideal rocket’s flight equation (Viva Rocketry! Part 1.3)“) is

$0 = m_i(t) \frac{dv_i(t)}{dt} + c_i \frac{dm_i(t)}{dt}$

It can be written as

$\frac{dv_i(t)}{dt} = -\frac{c_i}{m_i(t)}\frac{dm_i(t)}{dt}\quad\quad\quad(1)$

Integrate (1) from $t_0$ to $t$,

$\int\limits_{t_0}^{t}\frac{dv_i(t)}{dt}\;dt = -c_i \int\limits_{t_0}^{t}\frac{1}{m_i(t)}\frac{dm_(t)}{dt}\;dt$

gives

$v_i(t) - v_i(t_0) = -c_i(\log(m_i(t))-\log(m_i(t_0)))=-c_i\log(\frac{m_i(t)}{m_i(t_0)})$.

At $t=t^*$ when $i^{th}$ stage’s fuel has been burnt, we have

$v_i(t^*) - v_i(t_0) = -c_i\log(\frac{m_i(t^*)}{m_i(t_0)})\quad\quad\quad(2)$

where

$m_i(t_0) = \sum\limits_{k=i}^{n} m_k +P$

and,

$m_i(t^*) = \sum\limits_{k=i}^n m_k - \epsilon_i m_i + P$.

Let $v^*_i = v_i(t^*), \; v^*_{i-1}$ the velocity of rocket at the end of ${i-1}^{th}$ stage of flight.

Since $v_i(t_0) = v^*_{i-1}$, (2) becomes

$v^*_i - v^*_{i-1} = -c_i \log(\frac{ \sum\limits_{k=i}^n m_k - \epsilon_i m_i+P}{\sum\limits_{k = i}^{n} m_k + P}) = -c_i\log(1-\frac{\epsilon_i m_i}{\sum\limits_{k = i}^{n} m_k + P})$

i.e.,

$v^*_i = v^*_{i-1}-c_i\log(1-\frac{\epsilon_i m_i}{\sum\limits_{k = i}^{n} m_k + P}), \quad\quad 1\leq i \leq n, v_0=0\quad\quad\quad(3)$

For a single stage rocket ($n=1$), (3) is

$v_1^*=-c_1\log(1-\frac{\epsilon_1}{1+\beta})\quad\quad\quad(4)$

In my previous post “Viva Rocketry! Part 1“, it shows that given $c_1=3.0\;km\;s^{-1}, \epsilon_1=0.8$ and $\beta=\frac{1}{100}$, (4) yields $4.7 \;km\;s^{-1}$, a value far below $7.8\;km\;s^{-1}$, the required speed of an earth orbiting  satellite.

But is there a value of $\beta$ that will enable the single stage rocket to produce the speed a satellite needs?

Let’s find out.

Differentiate (4) with respect to $\beta$ gives

$\frac{dv_1^*}{d\beta} = - \frac{c_1 \epsilon_1}{(\beta+1)^2 (1-\frac{\epsilon_1}{\beta+1})} = - \frac{c_1\epsilon_1}{(\beta+1)^2(\frac{1-\epsilon_1+\beta}{\beta+1})} < 0$

since $c_1, \beta$ are positive quantities and $0< \epsilon_1 < 1$.

It means $v_1^*$ is a monotonically decreasing function of $\beta$.

Moreover,

$\lim\limits_{\beta \rightarrow 0}v_1^*=\lim\limits_{\beta \rightarrow 0}-c_1\log(1-\frac{\epsilon_1}{1+\beta}) = - c_1 \log(1-\epsilon_1)\quad\quad\quad(5)$

Given $c_1=3.0\;km\;s^{-1}, \epsilon_1=0.8$, (5) yields approximately

$4.8\;km\;s^{-1}$

Fig. 3

This upper limit implies that for the given $c_1$ and $\epsilon_1$, no value of $\beta$ will produce a speed beyond (see Fig. 4)

Let’s now turn to a two stage rocket ($n=2$)

From (3), we have

$v_2^* = -c_1\log(1-\frac{\epsilon_1 m_1}{m_1+m_2+P}) - c_2\log(1-\frac{\epsilon_2 m_2}{m_2+P})\quad\quad\quad(6)$

If $c_1=c_2=c, \epsilon_1 = \epsilon_2 = \epsilon, m_1=m_2$ and $\frac{P}{m_1+m_2} = \beta$, then

$P = \beta (m_1+m_2) = 2m_1\beta = 2m_2\beta$.

Consequently,

$v_2^*=-c \log(1-\frac{\epsilon}{2(1+\beta)}) - c\log(1-\frac{\epsilon}{1+2\beta})\quad\quad\quad(7)$

When $c=3.0\;km\;s^{-1}, \epsilon=0.8$ and $\beta = \frac{1}{100}$

$v_2^* \approx 6.1\;km\;s^{-1}$

Fig. 5

This is a considerable improvement over the single stage rocket ($v^*=4.7\; km\;s^{-1}$). Nevertheless, it is still short of producing the orbiting speed a satellite needs.

In fact,

$\frac{dv_2^*}{d\beta} = -\frac{2c\epsilon}{(2\beta+2)^2(1-\frac{\epsilon}{2\epsilon+2})}-\frac{2c\epsilon}{(2\beta+1)^2(1-\frac{\epsilon}{2\beta+1})}= -\frac{2c\epsilon}{(2\beta+2)^2\frac{2\beta+2-\epsilon}{2\beta+2}}-\frac{2c\epsilon}{(2\beta+1)^2\frac{2\beta+1-\epsilon}{2\beta+1}} < 0$

indicates that $v_2^*$ is a monotonically decreasing function of $\beta$.

$\lim\limits_{\beta \rightarrow 0} v_2^*=\lim\limits_{\beta \rightarrow 0 }-c\log(1-\frac{\epsilon}{2+2\beta})-c\log(1-\frac{\epsilon}{1+2\beta})=-c\log(1-\frac{\epsilon}{2})-c\log(1-\epsilon)$.

Therefore, there is an upper limit to the speed a two stage rocket can produce. When $c=3.0\;km\;s^{-1}, \epsilon=0.8$, the limit is approximately

$6.4\;km\;s^{-1}$

Fig. 7

In the value used above, we have taken equal stage masses, $m_1 = m_2$. i.e., the ratio of $m_1 : m_2 = 1 : 1$.

Is there a better choice for the ratio of $m_1:m_2$ such that a better $v_2^*$ can be obtained?

To answer this question, let $\frac{m_1}{m_2} = \alpha$, we have

$m_1 = \alpha m_2\quad\quad\quad(8)$

Since $P = \beta (m_1+m_2)$, by (8),

$P = \beta (\alpha m_2 + m_2)\quad\quad\quad(9)$

Substituting (8), (9) into (6),

$v_2^* = -c\log(1-\frac{\epsilon \alpha m_2}{\alpha m_2+m_2+\beta(\alpha m_2+m_2))}) - c\log(1-\frac{\epsilon m_2}{m_2 + \beta(\alpha m_2 + m_2)})$

$= -c\log(1-\frac{\epsilon \alpha}{\alpha+1+\beta(\alpha+1)})-c\log(1-\frac{\epsilon}{1+\beta(\alpha+1)})$

$= c\log\frac{(\alpha+1)(\beta+1)((\alpha+1)\beta+1)}{(1-\epsilon+(\alpha+1)\beta)((1-\epsilon)\alpha + (\alpha+1)\beta+1)}\quad\quad\quad(10)$

a function of $\alpha$. It can be written as

$v_2^*(\alpha) = c\log(w(\alpha))$

where

$w(\alpha) = \frac{(\alpha+1)(\beta+1)((\alpha+1)\beta+1)}{(1-\epsilon+(\alpha+1)\beta)((1-\epsilon)\alpha + (\alpha+1)\beta+1)}$

This is a composite function of $\log$ and $w$.

Since $\log$ is a monotonic increasing function (see “Introducing Lady L“),

$(v_2^*)_{max} = c\log(w_{max})$

Here, $(v_2^*)_{max}, w_{max}$ denote the maximum of $v_2^*$ and $w$ respectively.

To find $w_{max}$, we differentiate $w$,

$\frac{dw}{d\alpha} = \frac{(\beta+1)(\alpha^2\beta-\beta-1)(\epsilon-1)\epsilon}{(\epsilon-\alpha \beta-\beta-1)^2(\alpha\epsilon-\alpha \beta-\beta-\alpha-1)^2}=\frac{(\beta+1)(\alpha^2\beta-\beta-1)(\epsilon-1)\epsilon}{(1-\epsilon+\alpha \beta+\beta)^2(\alpha(1-\epsilon)+\alpha \beta+\beta)^2}\quad\quad\quad(11)$

Solving $\frac{dw}{d\alpha} = 0$ for $\alpha$ gives

$\alpha = - \sqrt{1+\frac{1}{\beta}}$ or $\sqrt{1+\frac{1}{\beta}}$.

Fig. 8

By (8), the valid solution is

$\alpha = \sqrt{1+\frac{1}{\beta}}$.

It shows that $w$ attains an extreme value at $\sqrt{1+\frac{1}{\beta}}$.

Moreover, we observe from (11) that

$\alpha < \sqrt{1+\frac{1}{\beta}} \implies \frac{dw}{d\alpha} > 0$

and

$\alpha > \sqrt{1+\frac{1}{\beta}} \implies \frac{dw}{d\alpha} < 0$.

i.e., $w$ attains maximum at $\alpha=\sqrt{1+\frac{1}{\beta}}$.

It follows that

$v_2^*$ attains maximum at $\alpha = \sqrt{1+\frac{1}{\beta}}$.

Therefore, to maximize the final speed given to the satellite, we must choose the ratio

$\frac{m_1}{m_2} = \sqrt{1+\frac{1}{\beta}}$.

With $\beta =\frac{1}{100}$, the optimum ratio $\frac{m_1}{m_2}=10.05$, showing that the first stage must be about ten times large than the second.

Using this ratio and keep $\epsilon=0.8, c=3.0\;km\;s^{-1}$ as before, (10) now gives

$v_2 = 7.65\;km\;s^{-1}$

a value very close to the required one.

Fig. 9

Setting $\beta = \frac{1}{128}$, we reach the goal:

$v_2^* = 7.8\;km\;s^{-1}$

Fig. 10

Fig. 11

At last, it is shown mathematically that provided the stage mass ratios ($\frac{P}{m_1+m_2}$ and $\frac{m_1}{m_2}$)are suitably chosen, a two stage rocket can indeed launch satellites into earth orbit.

Exercise 1. Show that $\alpha < \sqrt{1+\frac{1}{\beta}} \implies \frac{dw}{d\alpha} > 0$ and $\alpha > \sqrt{1+\frac{1}{\beta}} \implies \frac{dw}{d\alpha} < 0$.

Exercise 2. Using the optimum $\frac{m_1}{m_2} = \sqrt{1+\frac{1}{\beta}}$ and $\epsilon=0.8, c=3.0\;km\;s^{-1}$, solving (10) numerically for $\beta$ such that $v_2^* = 7.8$.

# Does gravity matter ? (Viva Rocketry! Part 1.2)

A single rocket expels its propellant at a constant rate $k$.

Assuming constant gravity is the only external force, show that the equation of motion is

$(p+m_0-kt)\frac{dv}{dt}=ck-(p+m_0-kt)g$

where $v$ is the rocket’s speed, $c$ the speed of the propellant relative to the rocket, $p$ the payload mass, and $m_0$ the initial rocket mass.

If the rocket burn is continuous, show that the burn time is $\frac{\epsilon m_0}{k}$ and deduce that the final speed given to the payload is

$v=-c \log(1-\frac{\epsilon m_0}{m_0+p}) - \frac{g\epsilon m_0}{k}$

where $1-\epsilon$ is the structural factor of the rocket.

Estimate the percentage reduction in the predicted final speed due to the inclusion of the gravity term if

$\epsilon=0.8, \frac{p}{m_0}=\frac{1}{100}, c=3.0\;km\;s^{-1}, m_0=10^5\;kg$, and $k=5 \times 10^3\;kg\;s^{-1}$.

Find an expression for the height reached by the rocket during the burn and estimate its value using the data above.

Let’s recall the governing equation of rocket’s flight derived in “Viva Rocketry! Part 1“, namely,

$F = m\frac{dv}{dt} + u\frac{dm}{dt}$.

In the present context, $m = m_0 + p- k t$. It implies that

$\frac{dm}{dt}=-k$

and,

$F=-mg=-(m_0+p-kt)g$.

With $u = c$, we have

$-(p+m_0-kt)g = (p+m_0-kt)\frac{dv}{dt}+c\cdot(-k)$,

i.e.,

$(p+m_0-kt)\frac{dv}{dt}=ck-(p+m_0-kt)g$

or

$\frac{dv}{dt}=\frac{ck}{m_0+p-kt}-g$.

The structural factor $1-\epsilon$ indicates the amount of fuel is $\epsilon m_0$. Since the fuel is burnt at a constant rate $k$, it must be true that at burnt out time $T$,

$\epsilon m_0 = kT$.

Therefore,

$T=\frac{\epsilon m_0}{k}$.

The solution to initial-value problem

$\begin{cases} \frac{dv}{dt}=\frac{ck}{m_0+p-kt}-g\\ v(0) = 0 \end{cases}$

tells the speed of the rocket during its flight while fuel is burnt (see Fig. 1):

$v = -c \log(1-\frac{kt}{m_0+p})-gt\quad\quad\quad(1)$

Fig. 1

Evaluate (1) at burnt out time gives the final speed of the payload:

$v_1=-c \log(1-\frac{\epsilon m_0}{m_0+p}) - \frac{g\epsilon m_0}{k}\quad\quad\quad(2)$

Notice the first term of (2) is the burnt out velocity without gravity (see “Viva Rocketry! Part 1“)

It follows that the percentage reduction in the predicted final speed due to the inclusion of gravity is

$\frac{\frac{g \epsilon m_0} {k}}{-c \log(1-{\epsilon \over {1+\frac{p}{m_0}}})}\quad\quad\quad(3)$

Using the given values which are typical, the estimated value of (3) (see Fig. 2) is

$0.003\%$.

Fig. 2

This shows the results obtained without taking gravity into consideration can be regarded as a reasonable approximation and the characteristics of rocket flight indicated in “Viva Rocketry! Part 1” are valid.

Since $v = \frac{dy}{dt}$, (1) can be written as

$\frac{dy}{dt} = -c \log(1-\frac{kt}{m_0+p})-gt$

To find the distance travelled while the fuel is burnt, we solve yet another initial-value problem:

$\begin{cases}\frac{dy}{dt} = -c \log(1-\frac{kt}{m_0+p})-gt \\ y(0) = 0 \end{cases}$

Fig. 3

The solution (see Fig. 3) is

$y= -\frac{1}{2}g t^2 + ct - c\cdot (t-\frac{m_0+p}{k}) \cdot \log(1-\frac{kt}{m_0+p})$.

Hence, the height reached at the burnt out time $t=\frac{\epsilon m_0}{k}$ is

$h = -\frac{g\epsilon^2 m_0^2}{2k^2}+\frac{c\epsilon m_0}{k}+\frac{c}{k}\cdot (p+(1-\epsilon)m_0) \cdot \log(1-\frac{\epsilon m_0}{m_0+p})$.

Using the given values, we estimate that $h \approx 27 \; km$ (see Fig. 4)

Fig. 4

Exercise 1: Find the distance the rocket travelled while the fuel is burnt by solving the following initial-value problem:

$\begin{cases}\frac{d^2y}{dt^2} =\frac{ck}{p+m_0-kt}-g \\ y(0) = 0, y'(0)=0 \end{cases}$

# Thunderbolt (Viva Rocketry! Part 1.1)

Fig. 1

Shown in Fig. 1 is an experimental car propelled by a rocket motor. The drag force (air resistance) is given by $R = \beta v^2$. The initial mass of the car, which includes fuel of mass $m_f$, is $m_0$. The rocket motor is burning fuel at the rate of $q$ with an exhaust velocity of $u$ relative to the car.  The car is at rest at $t=0$. Show that the velocity of the car is given by, for $0 \leq t \leq T$,

$v(t) = \mu \cdot \frac{1-({m \over m_0})^{\frac{2\beta \mu}{q}}}{1+({m \over m_0})^{\frac{2\beta \mu}{q}}}$,

where $m=m_0-qt, \mu^2=\frac{qu}{\beta}$, and $T=\frac{m_f}{q}$ is the time when the fuel is burnt out.

We have derived the governing equation of rocket flight in “Viva Rocketry! Part 1“, namely,

$F = m \frac{dv}{dt} + u \frac{dm}{dt}\quad\quad\quad(1)$

From $m=m_0-qt$,  we have

$\frac{dm}{dt} = -q$.

Apply air resistance $R=\beta v^2$ as the external force, (1) becomes

$-\beta v^2 = (m_0 - q t) \frac{dv}{dt} - u q$.

And the car is at rest initially implies

$v(0)=0$.

It follows that the motion of the car can be modeled by an initial-value problem

$\begin{cases} -\beta v^2 = (m_0 - q t) \frac{dv}{dt} - u q \\ v(0) = 0 \end{cases}\quad\quad\quad(2)$

It suffices to show that the given $v(t)$ is the solution to this initial-value problem:

Fig. 2

An alternative is obtaining the stated $v(t)$ through solving (2).

Fig. 3

The fact that $m = m_0 -qt, (-1)^{2 \frac{\sqrt{b}\sqrt{u}}{\sqrt{q}}} = 1$ simplifies the result considerably,

$\frac{\sqrt q \sqrt u}{\sqrt \beta}\cdot \frac{m_0^{\frac{2 \sqrt \beta \sqrt u}{\sqrt q}}-m^{\frac{2 \sqrt \beta \sqrt u}{\sqrt q}}}{m_0^{\frac{2 \sqrt \beta \sqrt u}{\sqrt q}}-m^{\frac{2 \sqrt \beta \sqrt u}{\sqrt q}}}\quad\quad\quad(3)$

Divide both the numerator and denominator of (3) by $m_0^{\frac {2 \sqrt{\beta} \sqrt{u}}{\sqrt{q}}}$ then yields

$\frac{\sqrt q \sqrt u}{\sqrt \beta}\cdot \frac{1-(\frac{m}{m_0})^{\frac{2 \sqrt \beta \sqrt u}{\sqrt q}}}{1-(\frac{m}{m_0})^{\frac{2 \sqrt \beta \sqrt u}{\sqrt q}}}$

which is equivalent to the given $v(t)$ since $\mu^2=\frac{qu}{\beta}$.

At time $t=T$, the fuel is burnt out. It means

$m_0-m_f = m_0 - qT$.

Therefore,

$T = \frac{m_f}{q}$

Exercise 1: Solve (2) manually.

Hint: The differential equation of (2) can be written as $\frac{1}{uq - \beta v^2} \frac{dv}{dt} = \frac{1}{m_0 - q t}$.

Exercise 2: For $m_0=900\;kg, m_f=450\;kg, q=15\;kg/sec, u=500\;m/sec, \beta=0.3$, what is the burnout velocity of the car?

# Viva Rocketry! Part 1

In this post, we will first look at the main characteristics of rocket flight, and then examine the feasibility of launching a satellite as the payload of a rocket into an orbit above the earth.

A rocket accelerates itself by ejecting part of its mass with high velocity.

Fig. 1

Fig. 1 shows a moving rocket. At time $t+\Delta t$, the mass $\Delta m$ leaves the rocket in opposite direction. As a result, the rocket is being propelled away with an increased speed.

Let

$m(\square), m_{\square}$ – the mass of rocket at time ${\square}$

$\vec{v}_{\square}$ – the velocity of rocket at time $\square$

$v(\square), v_{\square}$ – the magnitude of $\vec{v}_{\square}$

$\vec{v}^*_{t+\Delta t}$ – the velocity of ejected mass $\Delta m$ at $t + \Delta t$

$v^*_{t+\Delta t}$ – the magnitude of $\vec{v}^*_{t+\Delta t}$

$u$ – the magnitude of $\Delta m$‘s velocity relative to the rocket when it is ejected. It is time invariant.

From Fig. 1, we have

$\Delta m = m_t - m_{t + \Delta t}$,

$\vec{v}_t = v_{t}$,

$\vec{v}_{t + \Delta t} = v_{t + \Delta t}$

and most notably, the relationship between $v^*_{t+\Delta t}, v_{t+\Delta t}$ and $u$ (see “A Thought Experiment on Velocities”):

$v^*_{t+\Delta t} = u - v_{t + \Delta t}$.

It follows that

$\vec{v}^*_{t+\Delta t} = -v^*_{t+\Delta t} = v_{t + \Delta t} - u$,

momentum at time $t$: $\vec{p}(t) = m_t \vec{v}_t = m_t v_t$

and,

momentum at time $t+\Delta t$$\vec{p}(t+\Delta t) = m_{t+\Delta t}\vec{v}_{t+\Delta t} + {\Delta m} \vec{v}^*_{t+\Delta t}=m_{t+\Delta t}\vec{v}_{t+\Delta t} + (m_t - m_{t+\Delta t}) \vec{v}^*_{t+\Delta t}$$= m_{t+\Delta t}v_{t+\Delta t} + (m_t - m_{t+\Delta t})(v_{t+\Delta t}-u)$.

Consequently, change of momentum in $\Delta t$ is $\vec{p}(t+\Delta t)- \vec{p}(t) = m_t (v_{t + \Delta t} - v_t) + u (m_{t + \Delta t} - m_t)$.

Apply Newton’s second law of motion to the whole system,

$\vec{F}= {d \over dt} \vec{p}(t)$

$= \lim\limits_{\Delta t \rightarrow 0} {{\vec{p}(t+\Delta t) - \vec{p}(t)} \over \Delta t}$

$= \lim\limits_{\Delta t \rightarrow 0} { {m_t (v_{t + \Delta t} - v_t) + u (m_{t + \Delta t} - m_t)} \over {\Delta t} }$

$= \lim\limits_{\Delta t \rightarrow 0} {m_t {{v_{t + \Delta t} - v_t} \over {\Delta t}} + {u {{m_{t + \Delta t} - m_t} \over {\Delta t}}}}$

$= m_t \lim\limits_{\Delta t \rightarrow 0}{(v_{t+\Delta t} - v_t) \over {\Delta t}} + u \lim\limits_{\Delta t \rightarrow 0}{(m_{t +\Delta t} - m_t) \over \Delta t}$

That is,

$\vec{F} = m(t) {d \over dt} v(t) + u {d \over dt} m(t)$

where $\vec{F}$ is the sum of external forces acting on the system.

To get an overall picture of the rocket flight, we will neglect all external forces.

Without any external force, $\vec{F} = 0$. Therefore

$0 = m(t) {d \over dt} v(t) + u {d \over dt} m(t)$

i.e.,

${d \over dt} v(t) = -{u \over m(t)} {d \over dt} m(t)\quad\quad\quad(1)$

That fact that $u, m(t)$ in (1) are positive quantities shows as the rocket loses mass (${d \over dt} m(t) < 0$), its velocity increases (${d \over dt} v(t) > 0$)

Integrate (1) with respect to $t$,

$\int {d \over dt} v(t)\;dt = -u \int {1 \over m(t)} {d \over dt} m(t)\;dt$

gives

$v(t) = -u \log(m(t)) + c$

where $c$ is the constant of integration.

At $t = 0, v(0)=0, m(0) = m_1 + P$ where $m_1$ is the initial rocket mass (liquid or solid fuel + casing and instruments, exclude payload) and $P$ the payload.

It means $c = u \log(m_1+P)$.

As a result,

$v(t) = -u \log(m(t)) + u \log(m_1+P)$

$= -u (\log(m(t) - \log(m_1+P))$

$= -u \log({m(t) \over m_1+P})$

i.e.,

$v(t) = -u \log({m(t) \over m_1+P})\quad\quad\quad(2)$

Since $m_1$ is divided into two parts, the initial fuel mass $\epsilon m_1 (0 < \epsilon < 1)$, and the casing and instruments of mass $(1-\epsilon)m_1$, $m(0)$ can be written as

$m(0) = \epsilon m_1 + ( 1 - \epsilon) m_1 + P$

When all the fuel has burnt out at $t_1$,

$m(t_1) = (1 - \epsilon)m_1 + P$

By (2), the rocket’s final speed at $t_1$

$v(t_1) = -u \log({m(t_1) \over {m_1+P}})$

$= -u \log({(1-\epsilon)m_1+P \over {m_1 + P}})$

$= -u \log({{m_1 + P -\epsilon m_1} \over {m_1+P}})$

$= -u \log(1-{{\epsilon m_1} \over {m_1+P}})$

$= -u \log(1-{\epsilon \over {1 + {P \over m_1}}})$

$= -u \log(1-{\epsilon \over {1 + \beta}})$

where $\beta = {P \over m_1}$.

In other words,

$v(t_1) =-u \log(1-{\epsilon \over {1 + \beta}})\quad\quad\quad(3)$

Hence, the final speed depends on three parameters

$u, \epsilon$ and $\beta$

Typically,

$u = 3.0\;km\;s^{-1}, \epsilon = 0.8$ and $\beta = 1/100$.

Using these values, (3) gives

$v_1 = 4.7\;km\;s^{-1}\quad\quad\quad(4)$

This is an upper estimate to the typical final speed a single stage rocket can give to its payload. Neglected external forces such as gravity and air resistance would have reduced this speed.

With (4) in mind, let’s find out whether a satellite can be put into earth’s orbit as the payload of a single stage rocket.

We need to determine the speed that a satellite needs to have in order to stay in a circular orbit of height $h$ above the earth, as illustrated in Fig. 2.

Fig. 2

By Newton’s inverse square law of attraction, The gravitational pull on satellite with mass $m_{s}$ is

${\gamma \; m_{s} M_{\oplus} \over (R_{\oplus} + h)^2}$

where universal gravitational constant $\gamma = 6.67 \times 10^{-11}$, the earth’s mass $M_{\oplus} = 5.9722 \times 10^{24}\; kg$, and the earth’s radius $R_{\oplus} = 6371\;km$.

For a satellite to circle around the earth with a velocity of magnitude $v$, it must be true that

${\gamma \; m_{s} M_{\oplus} \over (R_{\oplus} + h)^2} = {m_{s} v^2 \over (R_{\oplus}+h) }$

i.e,

$v = \sqrt{\gamma \; M_{\oplus} \over (R_{\oplus}+h)}$

On a typical orbit, $h = 100\;km$ above earth’s surface,

$v = 7.8\;km\cdot s^{-1}$

This is far in excess of (4), the value obtained from a single stage rocket.

The implication is that a typical single stage rocket cannot serve as the launching vehicle of satellite orbiting around earth.

We will turn to multi-stage rocket in “Viva Rocketry! Part 2”.

# What moves fast, will slow down, Part One

This post aims to explain mathematically how populations change.

Our first attempt is based on ideas put forward by Thomas Malthus’ article “An Essay on the Principle of Population” published in 1798.

Let $p(t)$ denotes total population at time $t$.

Assume in a small interval $\Delta t$, births and deaths are proportional to $p(t)$ and $\Delta t$. i.e.

births = $a \cdot p(t) \Delta t$

deaths = $b \cdot p(t) \Delta t$

where $a, b$ are constants.

It follows that the change of total population during time interval $\Delta t$ is

$p(t+\Delta t) - p(t) = a\cdot p(t)\Delta t - b \cdot p(t)\Delta t = r\cdot p(t)\Delta t$

where $r = a - b$.

Dividing by $\Delta t$ and taking the limit as $\Delta t \rightarrow 0$, we have

$\lim\limits_{\Delta \rightarrow 0} {p(t+\Delta t) - p(t) \over \Delta t} = r \cdot p(t)$

which is

${d \over dt} p(t) = r \cdot p(t)\quad\quad\quad(1)$

a first order differential equation.

Since (1) can be written as

${1 \over p(t)} {d \over dt} p(t) = r$,

integrate with respect to $t$; i.e.

$\int {1 \over p(t)}{d \over dt} p(t)dt = \int {r} dt$

$\log p(t) = r\cdot t + c$

where $c$ is the constant of integration.

If at $t=0, p(0) = p_0$, we have

$c = \log p_0$

and so

$p(t) = p_0 e^{r\cdot t}\quad\quad\quad(2)$

The result of our first attempt shows that the behavior of the population depends on the sign of constant $r$. We have exponential growth if $r > 0$, exponential decay if $r < 0$ and no change if $r = 0$.

The world population has been on a upward trend ever since such data is collected (see “World Population by Year“)

Qualitatively, our model (2) with $r>0$ indicates this trend. However, it also predicts the world population would grow exponentially without limit. And that, is most unlikely to occur, since there are so many limitation factors to growth: lack of food, insufficient energy, overcrowding, disease and war.

Therefore, it is doubtful that model (1) is The One.

Our second attempt makes a modification to (1). It takes the limitation factors into consideration by replacing constant $r$ in (1) with a function $r(t)$. Namely,

$r(t) = \gamma - \alpha \cdot p(t)\quad\quad\quad(3)$

where $\gamma$ and $\alpha$ are both positive constants.

Replace $r$ in (1) with (3),

${d \over dt} p(t) = (\gamma - \alpha \cdot p(t)) p(t) = \gamma (1 - {p(t) \over {\gamma \over \alpha}}) p(t)\quad\quad\quad(4)$

Since $r(t)$ is a monotonic decreasing function, it shows as population grows, the growth slows down due to the limitation factors.

Let $p_{\infty} = {\gamma \over \alpha}$,

${d \over dt} p(t) = \gamma (1- {p(t) \over p_{\infty}}) p(t)\quad\quad\quad(5)$

This is the Logistic Differential Equation.

Written differently as

${d \over dt} p(t) - \gamma \cdot p(t) = -{\gamma \over p_{\infty}} p(t)^2$,

the Logistic Differential Equation is also a Bernoulli’s equation (see “Meeting Mr. Bernoulli“)

Let’s understand (5) geometrically without solving it.

Two constant functions, $p(t) = 0$ or $p_{\infty}$ are solutions of (5), since

${d \over dt} 0 = \gamma (1-{0\over p_{\infty}}) 0 = 0$

and

${d \over dt} p_{\infty} = \gamma (1-{p_{\infty} \over {p_{\infty}}}) p_{\infty} = 0$.

Plot $p(t)$ vs. ${d \over dt} p(t)$ in Fig. 1, the points, $0$ and $p_{\infty}$, are where the curve of ${d \over dt} p(t)$ intersects the axis of $p(t)$.

Fig. 1

At point $A$ where $p(t) > p_{\infty}$, since ${d \over dt} p(t) < 0$, $p(t)$ will decrease; i.e., $A$ moves left toward $p_{\infty}$.

Similarly, at point $B$ where $p(t) < p_{\infty}, {d \over dt} p(t) > 0$ implies that $p(t)$ will increase and $B$ moves right toward $p_{\infty}$.

The model equation can also tell the manner in which $p(t)$ approaches $p_{\infty}$.

Let $p = p(t)$,

${d^2 \over dt^2} p(t) = {d \over dt}({d \over dt} p)$

$= {d \over dp} ({d \over dt}p) \cdot {d \over dt} p$

$= {d \over d p}(\gamma (1-{p \over p_{\infty}})p)\cdot {d \over dt }p$

$= \gamma(1 - {2 p\over p_{\infty}})\cdot \gamma (1-{p \over p_{\infty}})p$

$= \gamma^2 p ({{2 p} \over p_{\infty}} -1)({p \over p_{\infty}}-1)$

As an equation with unknown $p$, $\gamma^2 p ({{2 p} \over p_{\infty}} -1)({p \over p_{\infty}}-1)=0$ has three zeros:

$0, {p_{\infty} \over 2}$ and $p_{\infty}$.

Therefore,

${d^2 \over dt^2}p > 0$ if $p > p_{\infty}$,

${d^2 \over dt^2} p < 0$ if ${p_{\infty} \over 2} < p < p_{\infty}$

and

${d^2 \over dt^2} p > 0$ if $p < {p_{\infty} \over 2}$.

Consequently $p(t)$, the solution of initial-value problem

$\begin{cases} {d \over dt} p(t) = \gamma (1-{p(t) \over p_{\infty}}) p(t) \\ p(0)=p_0 \end{cases}\quad\quad(6)$

where $p_0 \neq 0, p_{\infty}$ behaves in the manner illustrated in Fig. 2.

Fig. 2

If  $p_0 > p_{\infty}, p(t)$ approaches $p_{\infty}$ on a concave curve. Otherwise, when ${p_{\infty} \over 2} \leq p_0 < p_{\infty}, p(t)$ moves along a convex curve.  For $p_0 < {p_{\infty} \over 2}$, the curve is concave first.  It turns convex after $p(t)$ reaches ${p_{\infty} \over 2}$.

Next, let’s solve the initial-value problem analytically for $p_0 \neq 0, p_{\infty}$.

Instead of using the result from “Meeting Mr. Bernoulli“, we will start from scratch.

At $t$ where $p(t) \neq 0, p_{\infty}$,  we re-write (5) as

${1 \over p(t)(1-{p(t) \over p_{\infty}}) }{d \over dt} p(t) = \gamma$.

Expressed in partial fraction,

$({1 \over p(t)} + {{1 \over p_{\infty}} \over {1-{p(t) \over p_{\infty}}}}) {d \over dt} p(t) = \gamma$.

Integrate it with respect to $t$,

$\int ({1 \over p(t)} + {{1 \over p_{\infty}} \over {1-{p(t) \over p_{\infty}}}}) {d \over dt} p(t) dt = \int \gamma dt$

gives

$\log p(t) - \log (1-{p(t) \over p_{\infty}}) = \gamma t + c$

where $c$ is the constant of integration.

i.e.,

$\log {p(t) \over {1-{p(t) \over p_{\infty}}}} = \gamma t + c$.

Since $p(0) = p_0$, we have

$c = {\log {p_{0} \over {1-{p_0 \over p_{\infty}}}}}$

and so

$\log ({{p(t) \over {1-{p(t) \over p_{\infty}}}} \cdot {{1-{p_0 \over p_{\infty}}}\over p_0}} )=\gamma t$.

Hence,

${{p(t) \over {1 - {p(t) \over p_\infty}}}= {{p_0 \cdot e^{\gamma t}} \over {1-{p_0 \over p_\infty}}}}$.

Solving for $p(t)$ gives

$p(t) = { p_{\infty} \over {1+({p_{\infty} \over p_0}-1)e^{-\gamma \cdot t}}}\quad\quad\quad(7)$

We proceed to show that (7) expresses the value of $p(t)$, the solution to (6) where $p_0 \neq 0, p_{\infty}$, for all $t$ ‘s (see Fig.3)

Fig. 3

From (7), we have

$\lim\limits_{t \rightarrow \infty} p(t) = p_{\infty}$.

It validates Fig. 1.

(7) also indicates that none of the curves in Fig. 2 touch horizontal line $p(t) = p_{\infty}$.

If this is not the case, then there exists at least one instance of $t$ where $p(t) = p_{\infty}$; i.e.,

${p_{\infty} \over {1+({p_{\infty} \over p_0}-1)e^{-\gamma \cdot t}}} = p_{\infty}$.

It follows that

${({p_{\infty} \over {p_0}} - 1) e^{-\gamma t}} = 0$

Since ${e^{-\gamma t}} > 0$ (see “Two Peas in a Pod, Part 2“), it must be true that

$p_0 = p_{\infty}$.

But this contradicts the fact that (7) is the solution of the initial-value problem (6) where $p_0 \neq 0,p_\infty$.

Reflected in Fig.1 is that $A$ and $B$ will not become $p_{\infty}$. They only move ever closer to it.

Last, but not the least,

${\lim \limits_{t \rightarrow \infty}} {d \over dt} p(t) = \gamma (1-{{ \lim\limits_{t \rightarrow \infty} p(t)} \over p_{\infty}}) {\lim\limits_{t \rightarrow \infty} p(t)} = \gamma (1 - {p_{\infty} \over p_{\infty}}) p_{\infty} = 0$.

Hence the title of this post.

# You say, “y” I say, “y[x]”

You see things; and you say “Why?”

But I dream things that never were; and I say “Why not?”

George Bernard Shaw in Back to Methuselah

The Wolfram Language function DSolve and NDSolve can solve differential equations.

Let’s look at a few examples.

Example 1 Solving an ODE symbolically. The solution, a function, is evaluated at a given point.

Example 2 Solving an ODE symbolically. Redefine a function and evaluate it at a given point.

Example 3 Solving an ODE initial-value problem symbolically. Get the value at a given point from the symbolic solution.

Example 4 Solving an ODE initial-value problem numerically. Get the value at a given point from the numerical solution.

Regarding whether to specify ‘y‘ or ‘y[x]‘ in DSolve, the only decent explanation I can find is in Stephen Wolfram’s book “The Mathematica Book”. This is straight from horse’s mouth:

When you ask DSolve to get you a solution for y[x], the rule it returns specify how to replace y[x] in any expression. However, these rules do not specify how to replace objects such as y'[x]. If you want to manipulate solutions that you get from DSolve, you will often find it better to ask for solutions for y, rather than y[x].

He then proceeds to give an illustration:

Had you started with DSolve[y'[x]==x+y[x], y[x], x], the result would be

As expected, only y[x] is replaced.