The Gaussian Integral

 The Gaussian integral is not only important in mathematics but is also extremely important in studying physics, for example, quantum mechanics, quantum field theory, and statistical mechanics. Let us begin with the most simple Gaussian integral
$$G=\int_{-\infty}^\infty e^{-x^2}dx$$
This is not an easy integral to calculate because there is no closed form antiderivative of the function $e^{-x^2}$. However, it can be done easily if we extend it to an integral in two-dimensions as
\begin{align*}
G^2&=\int_{-\infty}^\infty e^{-x^2}dx\int_{-\infty}^\infty e^{-y^2}dy\\
&=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)}dxdy
\end{align*}
Now, in term of the polar coordinates, $x=r\cos\theta$, $y=r\sin\theta$ and the area element is $dA=rdrd\theta$, so $G^2$ can be written as
\begin{align*}
G^2&=\int_0^{2\pi}\int_0^\infty re^{-r^2}drd\theta\\
&=\pi
\end{align*}
The integral $\int_0^\infty re^{-r^2}dr$ can be easily evaluated using the substitution $u=-r^2$ and its value is $\frac{1}{2}$. Now, we obtain
$$G=\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi}$$
From this you can evaluate more complicated Gaussian integrals using a simple substitution and/or some algebra. For example, $\int_{-\infty}^\infty e^{-ax^2}dx$ for a positive real number $a$ can be evaluated easily by a simple substitution $u=\sqrt{a}x$:
\begin{align*}
\int_{-\infty}^\infty e^{-ax^2}dx&=\frac{1}{\sqrt{a}}\int_{-\infty}^\infty e^{-u^2}du\\
&=\sqrt{\frac{\pi}{a}} \tag{1}
\end{align*}
If $a=\frac{1}{2}$, then $\int_{-\infty}^\infty e^{-\frac{1}{2}x^2}dx=\sqrt{2\pi}$. Normally one will have to use the integration by parts to calculate $\int_{-\infty}^\infty x^2e^{-x^2}dx$ or $\int_{-\infty}^\infty x^4e^{-x^2}dx$. There is a neat trick of evaluating such integrals by differentiating (1) with respect to $a$:
$$\frac{d}{da}\int_{-\infty}^\infty e^{-ax^2}dx=-\int_{-\infty}^\infty x^2e^{-ax^2}dx$$
and
$$\frac{d}{da}\sqrt{\frac{\pi}{a}}=-\frac{1}{2a}\sqrt{\frac{\pi}{a}}$$
Thus, we have
$$
\int_{-\infty}^\infty x^2e^{-ax^2}dx=\frac{1}{2a}\sqrt{\frac{\pi}{a}} \tag{2}
$$
For $a=1$, we obtain
$$\int_{-\infty}^\infty x^2e^{-x^2}dx=\frac{\sqrt{\pi}}{2}$$
Differentiating (2) with respect to $a$ leads to
$$
\int_{-\infty}^\infty x^4e^{-ax^2}dx=\frac{3}{4a^2}\sqrt{\frac{\pi}{a}} \tag{3}
$$
For $a=1$, we obtain
$$
 \int_{-\infty}^\infty x^4e^{-x^2}dx=\frac{3}{4}\sqrt{\pi} \tag{4}
$$
I learned this trick from the book Quantum Field Theory in a Nutshell by Anthony Zee. Another important variation is
$$
\int_{-\infty}^\infty e^{-ax^2+bx}dx \tag{5}
$$
This integral can be evaluated from (1) with a little bit of algebra. First, by completing the square, we write
\begin{align*}
-ax^2+bx&=-a\left(x^2-\frac{b}{a}x\right)\\
&=-a\left(x^2-\frac{b}{a}x+\left(-\frac{b}{2a}\right)^2-\left(-\frac{b}{2a}\right)^2\right)\\
&=-a\left(x-\frac{b}{2a}\right)^2+\frac{b^2}{4a}
\end{align*}
Now, (5) is evaluated as
\begin{align*}
\int_{-\infty}^\infty e^{-ax^2+bx}dx&=\int_{-\infty}^\infty e^{-a\left(x-\frac{b}{2a}\right)^2+\frac{b^2}{4a}}dx\\
&=e^{\frac{b^2}{4a}}\int_{-\infty}^\infty e^{-a\left(x-\frac{b}{2a}\right)^2}dx\\
&=e^{\frac{b^2}{4a}}\int_{-\infty}^\infty e^{-au^2}du\ \left[u=x-\frac{b}{2a}\right]\\
&=e^{\frac{b^2}{4a}}\sqrt{\frac{\pi}{a}}
\end{align*}

This time let us try to calculate
$$
\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty (x^2+y^2+z^2)e^{-(x^2+y^2+z^2)}dxdydz \tag{6}
$$
You stumble upon this type of integral, for example, in a derivation of the Maxwell-Boltzmann statistics in the form of
$$\int d^3p\frac{p^2}{2m}\exp\left(-\beta\frac{p^2}{2m}\right)$$
The integral (6) can be easily evaluated using the spherical coordinates:
\begin{align*}
x&=r\sin\theta\cos\phi\\
y&=r\sin\theta\sin\phi\\
z&=r\cos\theta\\
\end{align*}
where
$$0\leq r<\infty,\   0\leq\theta\leq\pi,\ 0\leq\theta\leq 2\pi$$
The volume element in the spherical coordinates is $dV=r^2dr\sin\theta d\theta d\phi$, so
\begin{align*}
\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty (x^2+y^2+z^2)e^{-(x^2+y^2+z^2)}dxdydz&=\int_0^{2\pi}\int_0^\pi\int_0^\infty r^4e^{-r^2}dr\sin\theta d\theta d\phi\\
&=\frac{3}{2}\pi\sqrt{\pi}
\end{align*}
We obtain
$$\int_0^\infty r^4e^{-r^2}dr=\frac{3}{8}\sqrt{\pi}$$
using (4).

Marginal Functions

 Let $C(x)$, $R(x)$, and $P(x)$ denote, respectively, the total cost function, the total revenue function, and the total profit function.

Definition.

• $C(x+1)-C(x)$, the cost of producing the $x+1$st item,  is called the marginal cost at production level $x$.
• $R(x+1)-R(x)$, the revenue derived by the producer when $x+1$ items are sold, is called the marginal revenue at $x$.
• $P(x+1)-P(x)$, the producer's profit due to the $x+1$st item, is called marginal profit at $x$.
  

Let $E(x)$ be an economic function which represents any of the total cost, revenue, and profit functions and we assume that it is differentiable. In many real applications, the approximation
$$E'(x)\approx E(x+1)-E(x)$$
or
$$E'(x)\approx E(x)-E(x-1)$$
is valid. That is, $E'(x)$ is approximately the marginal economic function at level $x$ or level $x-1$.

Example. Given that the cost, in dollar, of producing $x$ short wave radios is $C(x)=x^2+80x+3500$,

1. Find the cost of producing; (a) the 100th radio; (b) the 101st radio.
2. Find $C'(100)$ and interpret this.
   

Solution.

1.  (a) It is the marginal cost at level 99, $C(100)-C(99)=279$ dollars. (b) It is the marginal cost at level 100, $C(101)-C(100)=281$ dollars. 
2. $C'(x)=2x+80$, so $C'(100)=280$ dollars. This can be used to approximate the marginal cost at level 99 or at 100. In either case, the error is 1 dollar and the percentage error in the approximation is 0.36.
   

Definition. The average cost function $\bar C(x)$ is defined by
$$\bar C(x)=\frac{C(x)}{x}$$

Example. Suppose that the relationship between price of and demand for a certain type of large color television set is given by the demand equation $10p+x=10000$, where $p$ is the per unit price in dollars, and $x$ is the number of sets demanded. If the producer's cost is $C(x)=60x+3000$,

1. Determine the revenue function.
2. Determine the marginal revenue function.
3. Determine the profit function.
4. Determine the marginal profit function.
5. What is the price for each color television when the marginal profit is zero?
6. Sketch the profit function, drawing the tangent line when the marginal profit is zero. What does this profit represent?
7. What can you conclude about the profit at the price obtained in 5?
   

Solution.

1. Solving the demand equation for price, we find $$p=-0.1x+1000$$ and so the revenue function is given by $$R(x)=px=-0.1x^2+1000x$$

2. The marginal revenue function at $x$ is $R(x+1)-R(x)$ but for practical purposes, it can be replaced by the approximation $R'(x)=-0.2x+1000$.

3. \begin{align*}P(x)&=R(x)-C(x)\\&=-0.1x^2+400x-3000\end{align*}

4. $P'(x)=-0.2x+400$.

5. When $P'(x)=-0.2x+400=0$, $x=2000$ i.e. the production level when the marginal profit is zero is 2000. The corresponding price per color television unit is $p=-0.1(2000)+1000=800$.

6.

 

 7. At $p=800$ ($x=2000$), the profit is maximized and the maximum value is $P(2000)=397000$ dollars.

In the preceding example, the profit is maximized when the marginal profit is zero. This is not a coincidence. The marginal profit is
$$P'(x)=R'(x)-C'(x)$$
If $P'(x)=0$, $R'(x)=C'(x)$ i.e. the marginal revenue equals the marginal cost. This means that the cost of producing one more item equals the revenue obtained by producing it, so there is no gain to be made by making the next item. As long as the marginal cost is less than the marginal revenue, it pays to keep producing more items. When the marginal cost becomes equal to the marginal revenue, it is time to stop producing more items.

Calculus 29: Newton's Method

 Quadratic equations can be easily solved using the quadratic formula. For cubic and quartic equations there are also such formulas for solutions, but they are not easy to use in general. For polynomials of higher-order degree of 5 or higher there are not even such formulas for roots. Newton's Method allows us to find an approximate solution to such equations. Newton's method is not limited to solving polynomial equations. It can solve any type of equations of the form $f(x)=0$ as long as $f(x)$ is differentiable. I will use a simple example to explain how it works and then formulate Newton's method in general. Let us consider the function $f(x)=x^4-2$. Newton's method begins with by guessing the first solution. In order for Newton's method to work, one needs to come up with the first guess close enough to the actual solution, otherwise Newton's method may return an undesirable result. (I will show you an example of such case later on.) We can come up with a reasonable first guess say $x_0$ using the graph of the function. However, even without knowing the graph, one  can come up with a reasonable first guess using the Intermediate Value Theorem as discussed here. For example, $f(1)=-1<0$ and $f(3)=79>0$, so by the IVT, there is a root of $f(x)$ in the interval $[-1,3]$ and we may choose $x_0=2$. Since $f(2)=14>0$, we can narrow the interval further down to $[1,2]$ in which case we may choose $x_0=1.5$.

Figure 1. The graph of $f(x)=x^4-2$.

From the graph, we choose $x_0=2$. Of course, one can choose even a closer point, for example $x_0=1.5$. The tangent line to the graph of $f(x)$ at $x_0=2$ is
$$y-f(x_0)=f'(x_0)(x-x_0)$$
Setting $y=0$, we find the $x$-intercept $x_1$
$$x_1=x_0-\frac{f(x_0)}{f'(x_0)}=1.562500000$$

Figure 2. The first iteration of Newton's method with $x_0=2$.

In Figure 2, we see that $x_1$ is closer to the actual solution than $x_0$. This time we find the $x$-intercept $x_2$ of the tangent line to the graph of $f(x)$ at $x_1=1.562500000$.
$$x_2=x_1-\frac{f(x_1)}{f'(x_1)}=1.302947000$$

Figure 3. The second iteration of Newton's method with $x_1=1.562500000$.

In Figure 3, we see that $x_2$ is closer to the actual solution than $x_1$. Similarly, we can find the next approximate solution $x_3=1.203252569$ which is closer to the actual solution than $x_2$ as shown in Figure 4.

Figure 4. The third iteration of Newton's method with $x_2=1.302947000$.

Continuing this process, the 6th approximate solution is given by $x_6=1.189207115$ which is correct to 9 decimal places. The exact solution is $\root 4\of{2}=1.189207115002721$.

In general, Newton's Method is given by
\begin{align*}x_0&=\mbox{initial approximate}\\x_{n+1}&=x_n-\frac{f(x_n)}{f'(x_n)}\end{align*}
for $n=0,1,2,\cdots$. Here, the assumption is that $f'(x_n)\ne 0$ for $n=0,1,2,\cdots$.

Earlier, I mentioned that if we don't choose the initial approximate $x_0$ close enough to the actual solution, Newton's method may return an undesirable result. Let me show you an example. Let us consider the function $f(x)=x^3-2x-5$. Figure 5 shows its graph.

Figure 5. The graph of $f(x)=x^3-2x-5$.

If we choose $x_0=-4$ and run Newton's method, we obtain the following approximates.

                  X[1] = -2.673913043
                  X[2] = -1.708838801
                  X[3] = -0.7366532045
                  X[4] = -11.29086856
                  X[5] = -7.553673519
                  X[6] = -5.065760748
                  X[7] = -3.400569565
                  X[8] = -2.252794796
                  X[9] = -1.350919123
                 X[10] = 0.01991182580
                 X[11] = -2.501495587
                 X[12] = -1.568413258
                 X[13] = -0.5049189040

As we can see, the numbers do not not appear to be converging to somewhere which indicates that Newton's method is not working well for this case. In certain cases when we choose $x_0$ too far from the actual solution, we may end up getting $f'(x_n)=0$ for some $n$ in which case Newton's method fails. For $x_0=4$, we obtain
 
                   X[1] = 2.891304348
                   X[2] = 2.311222795
                   X[3] = 2.117035157
                   X[4] = 2.094830999
                   X[5] = 2.094551526

The fifth approximate $x_5=2.094551526$ is correct to 6 decimal places.

Newton's method is not suitable to be carried out by hand. An open source computer algebra system Maxima has a built-in package mnewton for Newton's method. If you want to install Maxima on your computer, you can find an instruction here under Number Theory and Cryptography. Let us redo the above example using mnewton with initial approximate $x_0=4$.

(%i1) load("mnewton") (Note: put a dollar sign at the end of the command before executing it)
(%i2) mnewton([x^3-2*x-5], [x], [4]);
(%o2) [[x = 2.094551481542326]]

What I find interesting about mnewton is that even if you use an initial approximate that didn't work out for the standard Newton's method such as $x_0=-4$ in the above example, it instantly returns the answer. (Try it yourself.)

Newton's method can be used to calculate Internal Rate of Return (IRR) in finance. It is the discount rate at which Net Present Value (NPV) is equal to zero. NPV is the sum of the present values of all cash flows, or alternatively, NPV can be defined as the difference between the present value of the benefits (cash inflows) and the present value of the costs (cash outflows). Here is an example.

Example. If we invest 100 dollars today and receive 110 dollars in one year, then NPV can be expressed as
$$\mathrm{NPV}=-100+\frac{110}{1+\mathrm{IRR}}$$
Setting $\mathrm{NPV}=0$, we have
$$\mathrm{IRR}=\frac{110}{100}-1=0.1=10\ \mbox{percent}$$
If we have multiple future cash inflows 90 dollars, 50, dollars and 30 dollars at the end of each year for the next three years, NPV is given by
$$\mathrm{NPV}=-100+\frac{90}{1+\mathrm{IRR}}+\frac{50}{(1+\mathrm{IRR})^2}+\frac{30}{(1+\mathrm{IRR})^3}$$
Setting $\mathrm{NPV}=0$, we obtain a cubic equation
$$100x^3-90x^2-50x-30=0$$
where $x=1+\mathrm{IRR}$. Using Newton's method, we find $x=1.41$, so $\mathrm{IRR}=0.41=41\ \mbox{percent}$.

(%i1) load("mnewton") (Note: put a dollar sign at the end of the command before executing it)
(%i2) mnewton([100x^3-90x^2-50*x-30], [x], [1]);
(%o2) [[x = 1.406937359155343]]

I wrote a simple Maple script that runs Newton's method. If you have Maplesoft, you are more than welcome to download the Maple worksheet here and use it.

Is Starship Earth Possible?

 The Wandering Earth is a novella written by a Chinese Sci-Fi writer Liu Cixin. He is also well-known for the trilogy Remembrance of Earth's past (the original Chinese title is 地球往事 whose literal translation is Earth’s past). The English translation of the trilogy won a Hugo Award for Best Novel and it was the firsst novel by an Asian author to win a Hugo Award. The first novel of the trilogy is 三体 meaning three-body and its English title is The Three-Body Problem. There is a Chinese TV series (Title: 三体) as well as an American TV series (Title: 3 Body Problem, available on Netflix) adapted from the first novel.

The premise of the Wandering Earth is that the Sun will soon become a supernova. Facing the ultimate cataclysmic extinction event, people on Earth turns their entire planet into a spaceship and attempt to relocate it to Proxima Centauri. It is going to be a long 2,500 years journey. So, at what speed the starship Earth must travel? Assuming that it makes no stops, on average, it will be 510 km/sec which is 0.19% of the speed of light. This is quite a fast speed. So far the fastest object that has ever been built is Parker Solar Probe. By 2025, it is expected to travel as fast as 191 km/sec which is 0.064% of the speed of light. Is it physically possible for the starship Earth to travel at 510 km/sec? The energy required for Earth to travel at 510 km/sec can be easily calculated using $E=\frac{1}{2}mv^2$. With $m=5.9722\times 10^{24}$ kg, the Earth's mass and $v=510$ km/sec, the energy $E$ is calculated to be $7.767\times 10^{29}$ J=$1.856\times 10^{14}$ megatons. This is 2,000 times the energy output of the Sun per second which is $9.1\times 10^{10}$ J. The most powerful nuclear weapon that has ever been created and tested is Tsar bomb (Царь-бомба) by the Soviet Union. Interestingly, the project was overseen by the famed physicist Andrei Sakharov. Its yield was about 50 megatons. In terms of Tsar bomb, the energy is equivalent to detonating $3.712\times 10^{12}$, i.e. almost 4 trillion Tsar bombs! It seems such an enormous amount of energy is beyond our reach even in a distant future. Ultimately, the answer to the question about whether we can put giant thrusters on Earth to make it travel at 510 km/sec may be determined by the famous Tsiolkovsky rocket equation
$$
\Delta v=v_e\ln\frac{m_0}{m_f}=I_{\mathrm{sp}}g_0\ln\frac{m_0}{m_f} \tag{1}
$$
where

• $\Delta v$ is the maximum change of velocity of the vehicle;
• $v_e=I_{\mathrm{sp}}g_0$ is the effective exhaust velocity;
• $I_{\mathrm{sp}}$ is the specific impulse in dimension of time;
• $g_0=9.8\ \mathrm{m}/\mathrm{sec}^2$ is the gravitational acceleration of an object in a vacuum near the surface of the Earth;
• $m_0$, called wet mass, is the initial mass, including propellant;
• $m_f$, called dry mass, is the final total mass without propellant.
  

Tsiolkovsky rocket equation is named after the Russian rocket scientist Konstantin Eduardovich Tsiolkovsky (September 5, 1857 - September 19, 1935). He is dubbed the father of Russian rocket science. For a derivation of the rocket equation, see here. From (1), we obtain
$$
\frac{m_0-m_f}{m_0}=1-\frac{m_f}{m_0}=1-e^{-\frac{\Delta v}{v_e}} \tag{2}
$$
(2) gives rise to the percentage of the initial total mass which has to be propellant. This tell us how efficient the rocket engine is.

I don't see why Tsiolkovsky Rocket Equation wouldn't apply beyond Earth. The most realistic propulsion method for the starship Earth would be a nuclear-thermal rocket. As far as I know, the best performing nuclear-thermal rocket engine that has ever been built and tested was the USSR made nuclear-thermal rocket engine RD0410. It was developed in 1965-94. Its specific impulse is $I_{\mathrm{sp}}=910$ sec (see [1]). We need effective exhaust velocity using the gravitational acceleration for the Sun which is  $g_0=274\ \mathrm{m}/\mathrm{sec}^2$. The resulting effective exhaust velocity is $v_e=249$ km/sec. The orbiting speed of the Earth around the Sun can be calculated using the formula $v=\sqrt{\frac{GM}{r}}$. With $G=6.67\times 10^{-11}\ \mathrm{N}\mathrm{m}^2/\mathrm{kg}^2$, $r=1.5\times 10^{11}$ m and $M=1.99\times 10^{30}$ kg, we have $v=29.7$ km/sec. Since the escape velocity is $v_{\mathrm{escape}}=\sqrt{\frac{2GM}{r}}$, for an orbiting object its escape velocity is just $\sqrt{2}$ times its orbiting speed. So, the minimum velocity required for the Earth to break away from its orbit around the Sun is 42 km/sec. With $\Delta v=42$ km/sec and $v_e=249$ km/sec, (2) is evaluated to be $$1-e^{\frac{-\Delta v}{v_e}}=0.155$$ This means that about 16% of the mass of Earth has to be propellant just to break away from the orbit. Since the mass of Earth is $6\times 10^{24}$ kg=$6\times 10^{21}$ tons, 16% would be $10^{21}$ tons. For a nuclear-thermal rocket, the usual propellant is liquid hydrogen. (RD0410's propellant is also liquid hydrogen.) The basic principle is that liquid hydrogen is heated to a high temperature in a nuclear (fission) reactor and then expands through a rocket nozzle to create thrust. Earth does not even remotely have that much amount of hydrogen. While hydrogen is the most abundant element in the universe, Earth does not have it a lot. This decisively concludes that the starship Earth can't even break away from its orbit around the Sun let alone travel at the speed of 150 km/sec. Regardless, for the sake of completion, let us calculate how much propellant the starship Earth would need just to reach the speed of 150 km/sec. Now, with $\Delta v=150$ km/sec, we calculate (2) to be $$1-e^{\frac{-\Delta v}{v_e}}=0.453$$ That is, 45% of the mass of Earth has to be propellant!

By the way, Moving Earth is not just a science fiction. It is nothing like relocating Earth to a distant star system but scientists have been pondering how to shift Earth's orbit farther away from the Sun in order to mitigate rising temperatures on Earth. In my opinion, such an extreme meddling in Mother Nature must not be attempted even if possible as it likely results in unintended catastrophic consequences.

References:

1. Ядерный ракетный двигатель РД0410 (РД0411), Воронежский центр ракетного двигателестроения

Tsiolkovsky Rocket Equation

In this note, we derive the so called Tsiolkovsky rocket equation or simply rocket equation. It is given by
$$\Delta v=v_e\ln\frac{m_0}{m_f}=I_{\mathrm{sp}}g_0\ln\frac{m_0}{m_f} \tag{1}$$
where

• $\Delta v$ is the maximum change of velocity of the vehicle;
• $v_e=I_{\mathrm{sp}}g_0$ is the effective exhaust velocity;
• $g_0=9.8\ \mathrm{m}/\mathrm{s}^2$ is the gravitational acceleration of an object in a vacuum near the surface of the Earth;
• $m_0$, called wet mass, is the initial mass, including propellant;
• $m_f$, called dry mass, is the final total mass without propellant.
  

The equation (1) is named after the Russian scientist Konstantin Eduardovich Tsiolkovsky (September 5, 1857 - September 19, 1935). He is dubbed the father of Russian rocket science. It is also called fuel equation.

By the Newton's second law of motion, the net external force $\vec{F}$ to the change in linear momentum $\vec{P}$ of the whole system (including rocket and exhaust) is
$$\vec{F}=\frac{d\vec{P}}{dt}=\lim_{\Delta t\to 0}\frac{\Delta\vec{P}}{\Delta t}$$
$\Delta\vec{P}=\vec{P}_2-\vec{P}_1$, where $\vec{P}_1=m\vec{V}$ is the momentum of the rocket at time $t=0$ and $\vec{P}_2=(m-\Delta m)(\vec{V}+\Delta\vec{V})+\Delta m\vec{V}_e$ is the momentum of the rocket and exhausted mass at $t=\Delta t$. Here, with respect to the observer, $\vec{V}$ is the velocity of the rocket at time $t=0$, $\vec{V}$ is the velocity of the rocket at time $t=\Delta t$, $\vec{V}_e$ is the velocity of the mass added to the exhaust and lost by the rocket during tim $\Delta t$, $m$ is the mass of the rocket at time $t=0$, and $m-\Delta m$ is the mass of the rocket at time $t=\Delta t$. The velocity of the exhaust $\vec{V}_e$ in the observer frame is related to the velocity of the exhaust in the rocket $\vec{v}_e$ by
$$\vec{v}_e=\vec{V}_e-\vec{V}$$
or
$$\vec{V}_e=\vec{V}+\vec{v}_e$$
Now, $\Delta\vec{P}$ can be written as
$$\Delta\vec{P}=m\Delta\vec{V}+\vec{v}_e\Delta m-\Delta m\Delta\vec{V}$$
Since $\Delta m\to 0$ as $\Delta t\to 0$, we have
$$
\vec{F}=m\frac{d\vec{V}}{dt}+\vec{v}_e\frac{dm}{dt} \tag{2}
$$
If there are no external forces, then $\vec{F}=0$ i.e. $\frac{d\vec{P}}{dt}=0$ (conservation of linear momentum).
(2) then becomes the separable differential equation
$$
-m\frac{d\vec{V}}{dt}=\vec{v}_e\frac{dm}{dt} \tag{3}
$$
Assuming that $\vec{v}_e$ is constant (Tsiolkovsky's hypothesis) $v_e$, and integrating (3) we have
$$\int_v^{v+\Delta v}dv=-v_e\int_{m_0}^{m_f}\frac{dm}{m}$$
where $v=|\vec{V}|$, $\Delta v=|\Delta\vec{V}|$, $m_0$ is the initial total mass and $m_f$ is the final mass. Finally,  evaluating the integral yields the rocket equation (1).

From (1), we obtain
$$
\frac{m_0-m_f}{m_0}=1-\frac{m_f}{m_0}=1-e^{-\frac{\Delta v}{v_e}} \tag{4}
$$
The equation (4) gives rise to the percentage of the initial total mass which has to be propellant. This tells us how efficient the rocket engine is as shown in the following example.

Example. Let us consider an SSTO (Single-Stage-To-Orbit) rocket. (Most rockets we are seeing are two-stage-to-orbit or three-stage-to-orbit ones.) The rocket uses liquid hydrogen/liquid oxygen for its propellant, so specific impulse is about $I_{\mathrm{sp}}=350$ s. The exhaust velocity is then given by $v_e=3.43$ km/s. $\Delta v$ needed to get the rocket to a 322 km high LEO (Low Earth Orbit) is 8 km/s. With these values (4) is evaluated to be
$$1-e^{-\frac{\Delta v}{v_e}}=0.9$$
This means that 90% of the initial total mass has to be propellant. The remaining 10% is for the engines, the fuel tank, and the payload. The payload would account for only about 1% of the initial total mass. This kind of rocket is obviously very inefficient and expensive.

Update: In the Sci-Fi novella The Wandering Earth by Liu Cixin (there is also a movie of the same title on Netflix), the Sun will soon become a supernova and facing the ultimate cataclysmic extinction event, people on Earth turns their entire planet into a spaceship and attempt to relocate it to Proxima Centauri which is the closest star to the Sun (about 4.2 light-years). This is an extremely bold idea even in Chinese scale. (Well, they built the Great Wall!) Disappointingly though, in here, I showed using the rocket equation that it is not even possible for startship Earth to break away from its orbit around the Sun.

Homology: Cycle Groups and Boundary Groups

 Let us use $\langle\cdots\rangle$ for an unoriented simplex and $(\cdots)$ for an oriented simplex.

Example. $(p_0p_1)=-(p_1p_0)$. See figure 1.

 

Figure 1. An Oriented Simplex $(p_0p_1)$

Example.
\begin{align*}
\sigma_2&=(p_0p_1p_2)=(p_2p_0p_1)=(p_1p_2p_0)\\
&=-(p_0p_2p_1)=-(p_2p_1p_0)=-(p_1p_0p_2).
\end{align*}
See figure 2.

Figure 2. An Oriented Simplex $(p_0p_1p_2)$

Let $K=\{\sigma_\alpha\}$ be an $n$-dimensional simplicial complex of oriented simplexes.

Definition. The $r$-chain group $C_r(K)$ of a simplicial complex $K$ is a free abelian group generated by $r$-simplexes of $K$. If $r>\dim K$, $C_r(K):=0$. An element of $C_r(K)$ is called an $r$-chain.

Let there be $N_r$ $r$-simplexes in $K$. Denote them by $\sigma_{r,i}$ ($1\leq i\leq N$). Then $c\in C_r(K)$ is expressed as
$$c=\sum_{i=1}^{N_r}c_i\sigma_{r,i},\ c_i\in\mathbb Z.$$
The integers $c_i$ are called the coefficients of $c$. The addition of two $r$-chains $c=\sum_ic_i\sigma_{r,i}$ and $c'=\sum_ic_i'\sigma_{r,i}$ is
$$c+c'=\sum_i(c_i+c_i')\sigma_{r,i}.$$
The unit element is $0=\sum_i0\cdot\sigma_{r,i}$. The inverse element of $c$ is $-c=\sum_i(-c_i)\sigma_{r,i}$. Hence we see that $C_r(K)$ is a free abelian group of rank $N_r$
$$C_r(K)\cong\stackrel{N_r}{\overbrace{\mathbb Z\oplus\mathbb Z\oplus\cdots\oplus\mathbb Z}}.$$

Denote the boundary of an $r$-simplex $\sigma_r$ by $\partial_r\sigma_r$. Since a 0-simplex has no boundary,
$$\partial_0p_0=0.$$
For a 1-simplex $(p_0p_1)$,
$$\partial_1(p_0p_1):=p_1-p_0.$$
Let $\sigma_r=(p_0\cdots p_r)$ ($r>0$) be an oriented $r$-simplex. The boundary $\partial_r\sigma_r$ of $\sigma_r$ is an $(r-1)$-chain defined by
$$\partial_r\sigma_r:=\sum_{i=0}^r(-1)^i(p_0p_1\cdots\hat{p}_i\cdots p_r)$$
where the point $p_i$ under $\hat{}$ is omitted. For example,
\begin{align*}
\partial_2(p_0p_1p_2)&=(p_1p_2)-(p_0p_2)+(p_0p_1),\\
\partial_3(p_0p_1p_2p_3)&=(p_1p_2p_3)-(p_0p_2p_3)+(p_0p_1p_3)-(p_0p_1p_2).
\end{align*}
The boundary $\sigma_r$ defines a homomorphism called the boundary operator
$$\partial_r: C_r(K)\longrightarrow C_{r-1}(K);\ c=\sum_i c_i\sigma_{r,i}\longmapsto\partial_rc=\sum_ic_i\partial_r\sigma_{r,i}.$$

Let $K$ be an $n$-dimensional simplicial complex. Then there exists a sequence of free abelian groups and homomorphisms
$$0\stackrel{i}{\hookrightarrow}C_n(K)\stackrel{\partial_n}{\longrightarrow}C_{n-1}(K)\stackrel{\partial_{n-1}}{\longrightarrow}\cdots\stackrel{\partial_2}{\longrightarrow}C_1(K)\stackrel{\partial_1}{\longrightarrow}C_0(K)\stackrel{\partial_0}{\longrightarrow}0.$$
This sequence is called the chain complex associated with $K$ and is denoted by $C(K)$.

Definition. $Z_r(K):=\ker\partial_r\subset C_r(K)$ is called the $r$-cycle group. The elements of $Z_r(K)$ are called $r$-cycles. If $c\in Z_r(K)$, i.e. if $c$ is an $r$-cycle, $\partial_rc=0$. If $r=0$, $\partial_rc=0$ for all $c\in C_0(K)$, so $C_0(K)=Z_0(K)$.

Definition. Let us consider $C_{r+1}(K)\stackrel{\partial_{r+1}}{\longrightarrow}C_r(K)$ and let $c\in C_r(K)$. If there exists $d\in C_{r+1}(K)$ such that $c=\partial_{r+1}d$, then $c$ is called an $r$-boundary. The set of $r$-boundaries $B_r(K)$ ($=\partial_{r+1}C_{r+1}(K)={\rm Im}\partial_{r+1}$) is a subgroup of $C_r(K)$ called the $r$-boundary group. If $K$ is an $n$-dimensional simplicial complex, $B_n(K)=0$.

Consider $C_{r+1}(K)\stackrel{\partial_{r+1}}{\longrightarrow}C_r(K)\stackrel{\partial_r}{\longrightarrow}C_{r-1}(K)$. Then the following lemma holds.

Lemma. The composite map $\partial_r\partial_{r+1}:C_{r+1}(K)\longrightarrow C_{r-1}(K)$ is a zero map.

Proof. Since $\partial_r$ is a linear operator on $C_r(K)$, it suffices to prove the identity $\partial_r\partial_{r+1}=0$ for the generators of $C_{r+1}(K)$. If $r=0$, $\partial_0\partial_1=0$ since $\partial_0$ is a zero operator. Let us assume that $r>0$. Take $\sigma=(p_0\cdots p_rp_{r+1})\in C_{r+1}(K)$.
\begin{align*}
\partial_r\partial_{r+1}\sigma=&\partial_r\sum_{i=0}^{r+1}(-1)^i(p_0\cdots \hat{p}_i\cdots p_{r+1})\\
=&\sum_{i=0}^{r+1}(-1)^i\partial_r(p_0\cdots \hat{p}_i\cdots p_{r+1})\\
=&\sum_{i=0}^{r+1}(-1)^i[\sum_{j=0}^{i-1}(-1)^j(p_0\cdots \hat{p}_j\cdots\hat{p}_i\cdots p_{r+1})+\\
&\sum_{j=i+1}^{r+1}(-1)^{j-1}(p_0\cdots \hat{p}_i\cdots\hat{p}_j\cdots p_{r+1})]\\
=&\sum_{j<i}(-1)^{i+j}(p_0\cdots \hat{p}_j\cdots\hat{p}_i\cdots p_{r+1})+\\
&\sum_{j>i}(-1)^{i+j-1}(p_0\cdots \hat{p}_i\cdots\hat{p}_j\cdots p_{r+1})\\
=&0.
\end{align*}

Theorem. $B_r(K)\subset Z_r(K)$ or equivalently $\mathrm{Im}\partial_{r+1}\subset\ker\partial_r$.

Proof. Let $c\in B_r(K)$. Then $c=\partial_{r+1}d$ for some $d\in C_{r+1}(K)$. By the preceding lemma, $\partial_rc=\partial_r\partial_{r+1}d=0$. Hence, $c\in\ker\partial_r=Z_r(K)$.

Homology: Simplexes and Simplicial Complexes

 Definition. $0$-simplex $\langle p_0\rangle$ is a point or a vertex. A $1$-simplex $\langle p_0p_1\rangle$ is a line or an edge. A $2$-simplex $\langle p_0p_1p_2\rangle$ is a triangle with its interior included. $3$-simplex $\langle p_0p_1p_2p_3\rangle$ is a solid tetrahedron. A 0-simplex $\langle p_0\rangle$ may be simply written as $p_0$. See figure 1.

Figure 1. Simplicies

 Note that in order for an $r$-simplex to represent an $r$-dimensional object, the vertices $p_i$ must be geometrically independent, i.e. no $(r-1)$-dimensional hyperplane contains all the $r+1$ points. Let $p_0,\cdots,p_r$ be points geometrically independent in $\mathbb R^m$ ($m\geq r$). The $r$-simplex
$$\sigma_r=\{x\in\mathbb R^m: x=\sum_{i=0}^r c_ip_i,\ c_i\geq 0,\ \sum_{i=0}^r c_i=1\}$$
has the points $p_0,\cdots,p_r$ as its vertices. The ordered $(r+1)$-tuple $(c_0,c_1,\cdots,c_r)$ is called the barycentric coordinate of $x$. The 3-simplex $\langle p_0p_1p_2p_3\rangle$ has four 0-faces (vertices) $p_0,p_1,p_2,p_3$; six 1-faces (edges) $\langle p_0p_1\rangle$, $\langle p_0p_2\rangle$, $\langle p_0p_3\rangle$, $\langle p_1p_2\rangle$, $\langle p_1p_3\rangle$, $\langle p_2p_3\rangle$; four 2-faces (faces) $\langle p_0p_1p_2\rangle$, $\langle p_0p_2p_3\rangle$, $\langle p_0p_1p_3\rangle$, $\langle p_1p_2p_3\rangle$.

Let $K$ be a set of finite number of simplexes in $\mathbb R^m$. If these simplexes are nicely fitted together, $K$ is called a simplicial complex. By nicely fitted together we mean that:

1. An arbitrary face of a simplex of $K$ belongs to $K$.
2. If $\sigma$ and $\sigma'$ are two simplexes of $K$, $\sigma\cap\sigma'$ is either empty or a face of $\sigma$ and $\sigma'$.
   

See figure 2. The dimension of a simplicial complex is defined to be the maximum dimension of simplexes in $K$.

Figure 2. A Simplicial Complex

 Let $\sigma_r$ be an $r$-simplex and $K$ be the set of faces of $\sigma_r$. Then $K$ is an $r$-dimensional simplicial complex. For example, take $\sigma_3=\langle p_0p_1p_2p_3\rangle$. Then
$$\begin{array}{c}K=\{p_0,p_1,p_2,p_3,\langle p_0p_1\rangle,\langle p_0p_2\rangle,\langle p_0p_3\rangle,\langle p_1p_2\rangle,\langle p_1p_3\rangle,\langle p_2p_3\rangle,\\\langle p_0p_1p_2\rangle,\langle p_0p_1p_3\rangle,\langle p_0p_2p_3\rangle,\langle p_1p_2p_3\rangle,\langle p_0p_1p_2p_3\rangle\}.\end{array}$$

Definition. Let $K$ be a simplicial complex of simplexes in $\mathbb R^m$. The union of all the simplexes of $K$ is a subset of $\mathbb R^m$ called the polyhedron $|K|$ of a simplicial complex $K$. Note that $\dim |K|=\dim K$.

Let $X$ be a topological space. If there is a simplicial complex $K$ and a homeomorphism $f:|K|\longrightarrow X$, $X$ is said to be triangulable and the pair $(K,f)$ is called a triangulation of $X$.

Example. Figure 3 shows a triangulation of $S^1\times [0,1]$.

Figure 3. A Triangulation of $S^1\times[0,1]$

 Example. Figure 4 shows an example that is not a triangulation of $S^1\times [0,1]$.

Figure 4. Not a Triangulation of $S^1\times[0,1]$

The reason is that for $\sigma_2=\langle p_0p_1p_2\rangle$ and $\sigma_2'=\langle p_2p_3p_0\rangle$ $\sigma_2\cap\sigma_2'=\langle p_0\rangle\cup\langle p_2\rangle$. This is neither $\emptyset$ nor a simplex.

Homology: Free Abelian Groups

 Before we study homology groups, we review some basics of abelian group theory in this note.

The group operation for an abelian group is denoted by $+$. The unit element is denoted by $0$

Let $G_1$ and $G_2$ be abalian groups. A map $f: G_1\longrightarrow G_2$ is said to be a homomorphism if $$f(x+y)=f(x)+f(y),\ x,y\in G_1.$$ If $f$ is also a bijection (i.e one-to-one and onto), $f$ is called an isomorphism. If there is an isomorphism $f: G_1\longrightarrow G_2$, $G_1$ is said to be isomorphic to $G_2$ and we write $G_1\stackrel{f}{\cong} G_2$ or simply $G_1\cong G_2$.

Example. Define a map $f: \mathbb{Z}\longrightarrow\mathbb{Z}_2=\{0,1\}$ by $$f(2n)=0\ \mbox{and}\ f(2n+1)=1.$$ Then $f$ is a homomorphism.

A subset $H\subset G$ is a subgroup if it is a group with respect to the group operation of $G$. If $H$ is a subgroup of $G$, we write $H\leq G$.

Example. For any $k\in\mathbb N$, $k\mathbb{Z}=\{kn: n\in\mathbb{Z}\}\leq\mathbb{Z}$.

Example. $\mathbb{Z}_2=\{0,1\}\not\leq\mathbb{Z}$.

Let $H\leq G$. Define a relation on $G$ by
$$\forall x,y\in G,\ x\sim y\ \mbox{if}\ x-y\in H.$$
Then $\sim$ is an equivalence relation on $G$. The equivalence class of $x\in G$ is denoted by $[x]$, i.e. \begin{align*}
[x]&=\{y\in G: y\sim x\}\\
&=\{y\in G: y-x\in H\}.
\end{align*}
Let $G/H$ be the quotient set
$$G/H=\{[x]: x\in G\}.$$
Define an operation $+$ on $G/H$ by
$$[x]+[y]=[x+y],\ \forall [x],[y]\in G/H.$$
Then $G/H$ becomes an abelian group with this operation.

Example. $\mathbb{Z}/2\mathbb{Z}=\{[0],[1]\}$. Define $\varphi: \mathbb{Z}/2\mathbb{Z}\longrightarrow\mathbb{Z}_2$ by
$$\varphi([0])=0\ \mbox{and}\ \varphi([1])=1.$$
Then $\mathbb{Z}/2\mathbb{Z}\cong\mathbb{Z}_2$. In general, for every $k\in\mathbb N$, $\mathbb{Z}/k\mathbb{Z}\cong\mathbb{Z}_k$.

Lemma. Let $f: G_1\longrightarrow G_2$ be a homomorphism. Then

1. $\ker f=\{x\in G_1: f(x)=0\}=f^{-1}(0)$ is a subgroup of $G_1$.
2. $\mathrm{im}f=\{f(x): x\in G_1\}$ is a subgroup of $G_2$.
   

Theorem. [Fundamental Theorem of Homomorphism]
Let $f: G_1\longrightarrow G_2$ be a homomorphism. Then
$$G_1/\ker f\cong\mathrm{im}f.$$

Example. Let $f: \mathbb{Z}\longrightarrow\mathbb{Z}_2$ be defined by
$$f(2n)=0,\ f(2n+1)=1.$$
Then $\ker f=2\mathbb{Z}$ and $\mathrm{im}f=\mathbb{Z}_2$. By Fundamental Theorem of Homomorphism,
$$\mathbb{Z}/2\mathbb{Z}\cong\mathbb{Z}_2.$$

Take $r$ elements $x_1,x_2,\cdots,x_r$ of $G$. The elements of $G$ of the form
$$n_1x_1+n_2x_2+\cdots+n_rx_r\ (n_i\in\mathbb{Z},\ 1\leq i\leq r)$$
form a subgroup of $G$, which we denote $\langle x_1,\cdots,x_r\rangle$. $\langle x_1,\cdots,x_r\rangle$ is called a subgroup of $G$ generated by the generators $x_1,\cdots,x_r$. If $G$ itself is generated by finite elements, $G$ is said to be finitely generated. If $n_1x_1+\cdots+n_rx_r=0$ is satisfied only when $n_1=\cdots=n_r=0$, $x_1,\cdots,x_r$ are said to be linearly independent.

Definition. If $G$ is finitely generated by $r$ linearly independent elements, $G$ is called a free abelian group of rank $r$.

Example. $\mathbb{Z}$ is a free abelian group of rank 1 generated by 1 (or $-1$).

Example. Let $\mathbb{Z}\oplus\mathbb{Z}=\{(m,n):m,n\in\mathbb{Z}\}$. The $\mathbb{Z}\oplus\mathbb{Z}$ is a free abelian group of rank 2 generated by $(1,0)$ and $(0,1)$. More generally,
$$\stackrel{r\ \mbox{copies}}{\overbrace{\mathbb{Z}\oplus\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}}}$$
is a free abelian group of rank $r$.

Example. $\mathbb{Z}_2=\{0,1\}$ is finitely generated by 1 but is not free. $1+1=0$ so 1 is not linearly independent.

If $G=\langle x\rangle=\{0,\pm x,\pm 2x,\cdots\}$, $G$ is called a cyclic group. If $nx\ne 0$ $\forall n\in\mathbb{Z}\setminus\{0\}$, it is an infinite cyclic group. If $nx=0$ for some $n\in\mathbb{Z}\setminus\{0\}$, it is a finite cyclic group. Let $G=\langle x\rangle$ and let $f:\mathbb{Z}\longrightarrow G$ be a homomorphism defined by $f(k)=kx$, $k\in\mathbb{Z}$. $f$ is an epimorphism (i.e. onto homomorphism), so by Fundamental Theorem of Homomorphism,
$$G\cong\mathbb{Z}/\ker f.$$
If $G$ is a finite group, then there exists the smallest positive integer $N$ such that $Nx=0$. Thus
$$\ker f=\{0,\pm N,\pm 2N,\cdots\}=N\mathbb{Z}.$$
Hence
$$G\cong\mathbb{Z}/N\mathbb{Z}\cong\mathbb{Z}_N.$$
If $G$ is an infinite cyclic group, $\ker f=\{0\}$. Hence,
$$G\cong\mathbb{Z}/\{0\}\cong\mathbb{Z}.$$

Lemma. Let $G$ be a free abelian group of rank $r$, and let $H\leq G$. Then one may always choose $p$ generators $x_1,\cdots,x_p$ out of $r$ generators of $G$ so that $k_1x_1,\cdots,k_px_p$ generate $H$. Hence,
$$H\cong k_1\mathbb{Z}\oplus\cdots\oplus k_p\mathbb{Z}$$
and $H$ is of rank $p$.

Theorem. [Fundamental Theorem of Finitely Generated Abelian Groups]
Let $G$ be a finitely generated abelian group with $m$ generators. Then
$$G\cong\stackrel{r}{\overbrace{\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}}}\oplus \mathbb{Z}_{k_1}\oplus\cdots\oplus\mathbb{Z}_{k_p}$$
where $m=r+p$. The number $r$ is called the \emph{rank} of $G$.

Proof. Let $G=\langle x_1, \cdots,x_m\rangle$ and let $f: \mathbb{Z}\oplus\cdots\oplus\mathbb{Z}\longrightarrow G$ be the onto homomorphism
$$f(n_1,\cdots,n_m)=n_1x_1+\cdots +n_mx_m.$$
Then by Fundamental Theorem of Homomorphism
$$\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}/\ker f\cong G.$$
$\stackrel{m}{\overbrace{\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}}}$ is a free abelian group of rank $m$ and $\ker f$ is a subgroup of $\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}$, so by the preceding lemma
$$\ker f\cong k_1\mathbb{Z}\oplus\cdots\oplus k_p\mathbb{Z}.$$
Define $\varphi:\stackrel{p}{\overbrace{\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}}}/k_1\mathbb{Z}\oplus \cdots\oplus k_p\mathbb{Z}\longrightarrow\mathbb{Z}/k_1\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}/k_p\mathbb{Z}$ by
$$\varphi((n_1,\cdots,n_p)+k_1\mathbb{Z}\oplus\cdots\oplus k_p\mathbb{Z})=(n_1+k_1\mathbb{Z},\cdots,n_p+k_p\mathbb{Z}).$$
Then
$$\stackrel{p}{\overbrace{\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}}}/k_1\mathbb{Z}\oplus\cdots\oplus k_p\mathbb{Z}\stackrel{\varphi}{\cong}\mathbb{Z}/k_1\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}/k_p\mathbb{Z}.$$
Hence,
\begin{align*}
G&\cong\stackrel{m}{\overbrace{\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}}}/\ker f\\
&\cong\stackrel{m}{\overbrace{\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}}}/k_1\mathbb{Z}\oplus\cdots\oplus k_p\mathbb{Z}\\
&\cong\stackrel{m-p}{\overbrace{\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}}}\oplus\mathbb{Z}/k_1\mathbb{Z}\oplus\cdots\oplus Z/k_p\mathbb{Z}\\
&\cong\stackrel{m-p}{\overbrace{\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}}}\oplus\mathbb{Z}_{k_1}\oplus\cdots\oplus\mathbb{Z}_{k_p}.
\end{align*}

Calculus 28: L'Hôpital's Rule

 If two functions $f(x)$ and $g(x)$ both approach zero as $x\to a$, the function $\frac{f(x)}{g(x)}$ is said to assume the indeterminate form $\frac{0}{0}$ at $x=a$. Regardless of the term, $\frac{f(x)}{g(x)}$ may approach a limit as $x$ approaches $a$. The following theorem named after the French mathematician G. F. A. de L'Hôpital (1661-1704) is useful for the process of determining this limit if it exists.

Theorem (L'Hôpital's Rule). If the functions $f(x)$ and $g(x)$ are continuous in an interval containing $a$ and if $f'(x)$ and $g'(x)$ exist such that $g'(x)\ne 0$ in this interval (except possibly at $x=a$), then when $\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0$ (or equivalently $f(a)=g(a)=0$), we have $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$$ provided the limit on the right exists.

Example. Evaluate $\lim_{x\to 0}\frac{\tan x}{x}$.

Solution. The limit is an indeterminate form $\frac{0}{0}$. By applying L'Hôpital's Rule, we obtain $$\lim_{x\to 0}\frac{\tan x}{x}=\lim_{x\to 0}\frac{\sec^2x}{1}=1$$

The Indeterminate form $\frac{\infty}{\infty}$

If $f(x)\to\infty$ and  $g(x)\to \infty$ as $x\to a$ (or $x\to\pm\infty$), the function $\frac{f(x)}{g(x)}$ is said to assume the indeterminate form $\frac{\infty}{\infty}$ at $x=a$ (or at $x=\pm\infty$). The limit of $\frac{\infty}{\infty}$ as $x\to a$ (or $x\to\pm\infty$) may still be found by L'Hôpital's Rule if it exists.

Example. $$\lim_{x\to\infty}\frac{x^2}{e^x}=\lim_{x\to\infty}\frac{2x}{e^2}=\lim_{x\to\infty}\frac{2}{e^x}=0$$

Example. Evaluate $\lim_{x\to 0+}\frac{e^{-\frac{1}{x}}}{x}$.

Solution. Applying L'Hôpital's Rule, we have $$\lim_{x\to 0+}\frac{e^{-\frac{1}{x}}}{x}=\lim_{x\to 0+}\frac{e^{-\frac{1}{x}}}{x^2}$$ While L'Hôpital's rule can still be applied, we would accomplish nothing by doing so. This example shows that L'Hôpital's rule may not necessarily leads to a desirable result. For the above example, the limit can be found by $$\lim_{x\to 0+}\frac{e^{-\frac{1}{x}}}{x}=\lim_{x\to 0+}\frac{\frac{1}{x}}{e^{\frac{1}{x}}}=\lim_{z\to\infty}\frac{z}{e^z}=\lim_{z\to\infty}\frac{1}{e^z}=0$$

The Indeterminate form $0\cdot\infty$

If $f(x)\to 0$ and $g(x)\to\infty$ as $x\to a$ (or $x\to\pm\infty$), the function $f(x)g(x)$ is said to assume the indeterminate form $0\cdot\infty$ at $x=a$ (or at $x=\pm\infty$). If the limit of $f(x)g(x)$ as $x\to a$ (or $x\to\pm\infty$) exists, it may be found by writing $f(x)g(x)$ as $$\frac{f(x)}{\frac{1}{g(x)}}\ \mbox{or}\ \frac{g(x)}{\frac{1}{f(x)}}$$ and applying L'Hôpital's rule.

Example.

1. $\lim_{x\to\infty}xe^{-x}=\lim_{x\to\infty}\frac{x}{e^x}=\lim_{x\to\infty}\frac{1}{e^x}=0$.
2. $\lim_{x\to 0}\sin 3x\cot 2x=\lim_{x\to 0}\frac{\sin 3x}{\tan 2x}=\lim_{x\to 0}\frac{3\cos 3x}{2\sec^22x}=\frac{3}{2}$.
   

The Indeterminate Form $\infty-\infty$

If $f(x)\to\infty$ and $g(x)\to\infty$ as $x\to a$ (or $x\to\pm\infty$), the difference $f(x)-g(x)$ is said to assume the indeterminate form $\infty-\infty$  at $x=a$ (or at $x=\pm\infty$). If the limit of $f(x)-g(x)$ as $x\to a$ (or as $x\to\pm\infty$) exists, it may be found by transforming the difference into a fraction by algebraic means and applying L'Hôpital's rule.

Example. $\lim_{x\to 0}(\csc x-\cot x)=\lim_{x\to 0}\frac{1-\cos x}{\sin x}=\lim_{x\to 0}\frac{\sin x}{\cos x}=0$

Example. Evaluate $\lim_{x\to\infty}(x-\ln x)$.

Solution. While $x-\ln x$ assumes the indeterminate form $\infty-\infty$ at $x=\infty$, there is no algebraic means to transform it to a fraction. An indeterminate of the form $\infty-\infty$ such as the one in consideration may be evaluated by finding the limit of its exponential. Let $y=x-\ln x$. Then $$e^y=e^{x-\ln x}=\frac{e^x}{e^{\ln x}}=\frac{e^x}{x}$$ Hence, $$\lim_{x\to\infty}e^y=\lim_{x\to\infty}\frac{e^x}{x}=\lim_{x\to\infty}\frac{e^x}{1}=\infty$$ Since $y\to\infty$ when $e^y\to\infty$, $$\lim_{x\to\infty}(x-\ln x)=\infty$$

The Indeterminate Forms $0^0$, $\infty^0$, $1^\infty$

If $f(x)\to 0$ and $g(x)\to 0$, or $f(x)\to\infty$ and $g(x)\to 0$, or $f(x)\to 1$ and $g(x)\to\infty$ as $x\to a$ (or $x\to\pm\infty$), $f(x)^{g(x)}$ is said to assume the indeterminate form $0^0$, $\infty^0$, or $1^\infty$, respectively at $x=a$ (or at $x=\pm\infty$). If the limit of $f(x)^{g(x)}$ exists as $x\to a$ (or $x\to\pm\infty$), it may be found by denoting $f(x)^{g(x)}$ by $y$ and investigating the limit approached by the logarithm $$\ln y=g(x)\ln f(x)$$ If $\lim_{x\to a}\ln y=k$, then $\lim_{x\to a}y=e^k$.

Example. Evaluate $\lim_{x\to 0+}x^x$.

Solution. $x^x$ assumes the indeterminate form $0^0$ at $x=0$. Let $y=x^x$. Then $\ln y=x\ln x$ and $$\lim_{x\to 0+}\ln y=\lim_{x\to 0+}\frac{\ln x}{\frac{1}{x}}=\lim_{x\to 0+}\frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim_{x\to 0+}(-x)=0$$ Hence $\lim_{x\to 0+}y=\lim_{x\to 0+}x^x=e^0=1$.

Example. Evaluate $\lim_{x\to 0}(1-\sin x)^{\frac{1}{x}}$.

Solution. The function assumes the indeterminate form $1^\infty$ at $x=0$. Let $y=(1-\sin x)^{\frac{1}{x}}$. Then $\ln y=\frac{\ln(1-\sin x)}{x}$ and $$\lim_{x\to 0+}\ln y=\lim_{x\to 0+}\frac{\ln(1-\sin x)}{x}=\lim_{x\to 0+}\frac{\frac{-\cos x}{1-\sin x}}{1}=-1$$ Hence, $$\lim_{x\to 0+}(1-\sin x)^{\frac{1}{x}}=e^{-1}=\frac{1}{e}$$ Similarly, we also find $$\lim_{x\to 0-}(1-\sin x)^{\frac{1}{x}}=e^{-1}=\frac{1}{e}$$

Example. Use L'Hôpital's rule to show the limit $$\lim_{x\to 0}(1+x)^{\frac{1}{x}}=e$$

Proof. Let $y=(1+x)^{\frac{1}{x}}$. Then $y$ assumes the indeterminate form $1^\infty$ at $x=0$. $\ln y=\frac{\ln(1+x)}{x}$ and $$\lim_{x\to 0+}\ln y=\lim_{x\to 0+}\frac{\ln(1+x)}{x}=\lim_{x\to 0+}\frac{1}{1+x}=1$$ Hence, $$\lim_{x\to 0+}(1+x)^{\frac{1}{x}}=e$$ Similarly, we also find $$\lim_{x\to 0-}(1+x)^{\frac{1}{x}}=e$$

Calculus 27: Inverse Trigonometric Functions

 $y=\sin x$ is not a one-to-one function so there cannot be an inverse function of $y=\sin x$. However if we restrict its domain to the closed interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, it becomes a one-to-one function as shown in Figure 1.

Figure 1. The graph of $y=\sin(x)$ on $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$

 

This means that we can consider $y=\sin^{-1}x$, the inverse function of $y=\sin x$. That is to say, $y=\sin^{-1}x$ is the value in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ for which $x=\sin y$. $y=\sin^{-1}x$ is also denoted by $y=\arcsin x$. Similarly,

1. $y=\cos^{-1}x$ (or $y=\arccos x$) is the value in $[0,\pi]$ for which $x=\cos y$.
2. $y=\tan^{-1}x$ (or $y=\arctan x$) is the value in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ for which $x=\tan y$.
3. $y=\cot^{-1}x$ (or $y=\mathrm{arccot} x$) is the value in $(0,\pi)$ for which $x=\cot y$.
   

Remark. In general, $y=\sin^{-1}x$ is an inverse relation of $y=\sin x$ which is multiple-valued. In order to consider differentiation, we require it to be single-valued and when $-\frac{\pi}{2}\leq\sin^{-1}x\leq\frac{\pi}{2}$ we call it the principal value of $\sin^{-1}x$ and denote it by $\mathrm{Sin}^{-1}x$. Throughout this note we will only consider principal values so we won't be using the traditional notation like $y=\mathrm{Sin}^{-1}x$. 

Recall that the graph of $y=f(x)$ and the graph of its inverse function $y=f^{-1}(x)$ are symmetric about the line $y=x$. Using this symmetry one can obtain the graph of an inverse trigonometric function. For example, Figure 2 shows the graph of $y=\sin x$ and the graph of $y=\sin^{-1} x$.

Figure 2. The graphs of $y=\sin(x)$ (in red), $y=\arcsin(x)$ (in blue) and $y=x$ (in black).

Figure 3. The graph of $y=\arcsin(x)$

Figure 4. The graph of $y=\arccos(x)$

Figure 5. The graph of $y=\arctan(x)$ on [-20,20].

Figure 6. The graph of $y=\mathrm{arccot}(x)$ on [-20,20].

In addition, $y=\sec^{-1}x$ has domain $|x|\geq 1$ and range $\left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right]$.

Figure 7. The graph of $y=\mathrm{arcsec}(x)$. The horizontal asymptote is $y=\frac{\pi}{2}$.

$y=\csc^{-1}x$ has domain $|x|\geq 1$ and range $\left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right]$.

Figure 8. The graph of $y=\mathrm{arccsc}(x)$.

Inverse Functions and Their Derivatives

Let $y=f^{-1}(x)$ denote the inverse function of $f(x)$. Then
$$
x=f(y) \tag{1}
$$
Differentiate (1) with respect to $x$.
$$
1=f'(y)\frac{dy}{dx} \tag{2}
$$
Solving (2) for $\frac{dy}{dx}$ we have
$$
\frac{dy}{dx}=\frac{1}{f'(f^{-1}(x))} \tag{3}
$$

Example. Let $f(x)=\ln x$. Knowing $f'(x)=\frac{1}{x}$ find the derivative of $f^{-1}(x)=e^x$.

Solution. Using (3) $$\frac{d}{dx}e^x=\frac{1}{\frac{1}{e^x}}=e^x$$

Although knowing the formula (3) is convenient, you can always find the derivative of an inverse function by following the same process of deriving (3).  It's not actually anymore difficult or complicated than using (3).

The Derivatives of Inverse Trigonometric Functions

Let $f(x)=\sin x$. Then $f^{-1}(x)=\sin^{-1}x$. Using (3)
\begin{align*}
\frac{d}{dx}\sin^{-1}x&=\frac{1}{\cos(\sin^{-1}x)}\\
&=\frac{1}{\sqrt{1-\sin^2(\sin^{-1}x)}}\\
&=\frac{1}{\sqrt{1-x^2}}
\end{align*}
where $|x|<1$. The reason the sign in front of $\sqrt{}$ is positive is that $-\frac{\pi}{2}<\sin^{-1}x<\frac{\pi}{2}$ so $\cos(\sin^{-1}x)> 0$.

Alternative derivation: Let $y=\sin^{-1}x$. Then  $x=\sin y$. By implicit differentiation
$$1=\cos y\frac{dy}{dx}$$
Hence
$$\frac{dy}{dx}=\frac{1}{\cos y}=\frac{1}{\cos(\sin^{-1}x)}=\frac{1}{\sqrt{1-x^2}}$$

If $u$ is a functions of $x$, then
$$\frac{d}{dx}\sin^{-1}u=\frac{1}{\sqrt{1-u^2}}\frac{du}{dx},\ |u|<1$$
Similarly we obtain the rest of derivative formulas.
\begin{align*}
\frac{d}{dx}\cos^{-1}u&=-\frac{1}{\sqrt{1-u^2}}\frac{du}{dx},\ |u|<1\\
\frac{d}{dx}\tan^{-1}x&=\frac{1}{1+u^2}\frac{du}{dx}\\
\frac{d}{dx}\cot^{-1}x&=-\frac{1}{1+u^2}\frac{du}{dx}\\
\frac{d}{dx}\sec^{-1}u&=\frac{1}{|u|\sqrt{u^2-1}}\frac{du}{dx},\ |u|>1\\
\frac{d}{dx}\csc^{-1}x&=-\frac{1}{|u|\sqrt{u^2-1}}\frac{du}{dx},\ |u|>1
\end{align*}
In turns out we don't really have to calculate all these formulas. We just need to calculate for example $\frac{d}{dx}\sin^{-1}x$, $\frac{d}{dx}\tan^{-1}x$, $\frac{d}{dx}\sec^{-1}x$ the rest can be obtained by inverse function-inverse cofunction identities
\begin{align*}
\cos^{-1}x&=\frac{\pi}{2}-\sin^{-1}x\\
\cot^{-1}x&=\frac{\pi}{2}-\tan^{-1}x\\
\csc^{-1}x&=\frac{\pi}{2}-\sec^{-1}x
\end{align*}
In case you haven't seen this identities before they can be easily obtained from cofunction identities. For example sine and cosine are cofunctions of each other as you learned in trigonometry, namely
$$\sin\left(\frac{\pi}{2}-x\right)=\cos x,\ \cos\left(\frac{\pi}{2}-x\right)=\sin x$$
Let
$$\cos\left(\frac{\pi}{2}-y\right)=\sin y=x$$
Then
$$\frac{\pi}{2}-y=\cos^{-1}x$$
i.e. the first identity above
$$\cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x$$

Integration Formulas

For any constant $a\ne 0$,

1. $\int\frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\left(\frac{u}{a}\right)+C$, valid for $u^2<a^2$
2. $\int\frac{du}{a^2+u^2}=\frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right)+C$
3. $\int\frac{du}{u\sqrt{u^2-a^2}}=\frac{1}{a}\sec^{-1}\left|\frac{u}{a}\right|+C$, valid for $|u|>a>0$
   

Example.
\begin{align*}
\int_{\sqrt{2}/2}^{\sqrt{3}/2}\frac{dx}{\sqrt{1-x^2}}&=\left.\sin^{-1}x\right|_{\sqrt{2}/2}^{\sqrt{3}/2}\\
&=\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)-\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)\\
&=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}
\end{align*}

Example.
\begin{align*}
\int\frac{dx}{\sqrt{3-4x^2}}&=\frac{1}{2}\int\frac{du}{\sqrt{3-u^2}} (u=2x)\\
&=\frac{1}{2}\sin^{-1}\left(\frac{u}{\sqrt{3}}\right)+C\\
&=\frac{1}{2}\sin^{-1}\left(\frac{2x}{\sqrt{3}}\right)+C
\end{align*}

Example.
\begin{align*}\int\frac{dx}{4x-x^2}&=\int\frac{dx}{4-(x-2)^2}\\
&=\int\frac{du}{4-u^2} (u=x-2)\\
&=\sin^{-1}\left(\frac{u}{2}\right)+C\\
&=\sin^{-1}\left(\frac{x-2}{2}\right)+C\end{align*}

Example.
\begin{align*}\int\frac{dx}{4x^2+4x+2}&=\int\frac{dx}{4\left(x+\frac{1}{2}\right)^2+1}\\
&=\frac{1}{2}\int\frac{du}{u^2+1}  (u=2x+1)\\
&=\frac{1}{2}\tan^{-1}u+C\\
&=\frac{1}{2}\tan^{-1}(2x+1)+C
\end{align*}

Example.
\begin{align*}
\int\frac{dx}{e^{2x}-6}&=\int\frac{du}{u\sqrt{u^2-6}}\ (u=e^x)\\
&=\frac{1}{\sqrt{6}}\sec^{-1}\left(\frac{e^x}{\sqrt{6}}\right)+C
\end{align*}
where $e^x>\sqrt{6}$ or equivalently $x>\ln\sqrt{6}\approx 0.8959$.

Integrating Inverses of Functions

Let $y=f^{-1}(x)$. Then $x=f(y)$ and $dx=f'(y)dy$. So with integration by parts, we have
\begin{align*}
\int f^{-1}(x)dx&=\int yf'(y)dy\\&=yf(y)-\int f(y)dy \tag{4}\\
&=xf^{-1}(x)-\int f(y)dy
\end{align*}
Using (3) we can also rewrite (4) as
$$
\int f^{-1}(x)dx=xf^{-1}(x)-\int\frac{x}{f'(f^{-1}(x))}dx \tag{5}
$$

Using (4)
\begin{align*}
\int\cos^{-1}xdx&=x\cos^{-1}x-\sin y+C\\
&=x\cos^{-1}x-\sin(\cos^{-1}x)+C
\end{align*}
Since $x=\cos y$, $$\sin(\cos^{-1}x)=\sin y=\sqrt{1-x^2}$$ (Recall that $0\leq\cos^{-1}x\leq\pi$ so $\sin^{-1}x\geq 0$.) Hence,
$$
\int\cos^{-1}xdx=x\cos^{-1}x-\sqrt{1-x^2}+C \tag{6}
$$
Of course one can obtain (6) using (5) though the required calculation is a bit longer. The integral (6) can be also found without using (4) or (5). Using integration by parts
\begin{align*}
\int\cos^{-1}xdx&=x\cos^{-1}x+\int\frac{x}{\sqrt{1-x^2}}dx\\
&=x\cos^{-1}x-\sqrt{1-x^2}+C
\end{align*}
The rest of the integrals of inverse trigonometric functions are given by
\begin{align*}
\int\sin^{-1}xdx&=x\sin^{-1}x+\sqrt{1-x^2}+C\\
\int\tan^{-1}xdx&=x\tan^{-1}x-\frac{1}{2}\ln(1+x^2)+C\\
\int\cot^{-1}xdx&=x\cot^{-1}x+\frac{1}{2}\ln(1+x^2)+C\\
\int\sec^{-1}xdx&=x\sec^{-1}x-\frac{x}{|x|}\ln|x+\sqrt{x^2-1}|+C\\
\int\csc^{-1}xdx&=x\csc^{-1}x+\frac{x}{|x|}\ln|x+\sqrt{x^2-1}|+C
\end{align*}

Calculus 26: Optimization Problems II: Business and Economic Optimization Problems

In here, we discussed several examples of optimizations problems, mostly geometric optimization problems. In this note, we study business and economic optimization problems. Let us begin with the following example.

Example. Suppose that the price and demand for a particular luxury automobile are related by the demand equation $p+10x=200,000$, where $p$ is the price per car in dollars and $x$ is the number of cars that will be purchased at that price. What price should be charged per car if the total revenue is to be maximized?

Solution. Recall that the total revenue $R$ is given by $R=xp$, where $p$ is the price per item and $x$ is the number of items sold. In terms of $x$, the revenue is written as
$$R(x)=200,000x-10x^2$$
$R(x)=200,000-20x$ and set it equal to 0, we find the critical point $x=10,000$. Since $R^{\prime\prime}(x)=-20<0$, $R(10,000)=100,000,0000$, i.e. a billion dollars. We don't actually need calculus to see this. $R(x)$ is a quadratic function with negative leading coefficient, so it assumes as maximum at the $x$-coordinate of its vertex, $x=-\frac{b}{2a}=-\frac{200,000}{2\cdot 10}=10,000$. To answer the question, the price at which the total revenue $R$ is maximized is
$$p=-10(10,000)+200,000=100,000\ \mathrm{dollars}$$
As you might have noticed by now, it could have been shorter if we wrote $R$ in terms of the price $p$ since the question is about the price at which the revenue is maximized. In terms of $p$, $R$ is written as
$$R(p)=-\frac{p^2}{10}+20,000p$$
Either by solving $R'(p)=-\frac{p}{10}+20,000=0$ or by $p=-\frac{b}{2a}=\frac{20,000}{\frac{2}{10}}$, we find $p=\$ 100,000$.

Example. Suppose that the cost, in dollars, of producing $x$ hundred bicycles is given by $C(x)=x^2-2x+4900$. What is the minimum cost?

Solution. Either by solving $C'(x)=2x-2=0$ or by $x=-\frac{b}{2a}=-\frac{-2}{2\cdot 1}$, we find that $C(x)$ assumes the minimum at $x=1$, i.e. the cost of production is the minimum when 100 bicycles are produced. The minimum cost is $C(1)=4899$ dollars.

Example. In the preceding example, find the minimum average cost.

Solution. Recall that the average cost $\bar C(x)$ is given by $\bar C(x)=\frac{C(x)}{x}$, so we have
$$\bar C(x)=x-2+\frac{4900}{x}$$
Setting $C'(x)=1-\frac{4900}{x^2}$ equal to 0, we find $x=70$. Since $C^{\prime\prime}(x)=\frac{9800}{x^3}>0$ when $x=70$, $\bar C(70)=138$ (in dollars) is the minimum average cost.

Here is an interesting theorem from economics.

Theorem. The average cost is minimized at a level of production at which marginal cost equals average cost, i.e. when
$$C'(x)=\bar C(x)$$

Proof. Since $\bar C(x)=\frac{C(x)}{x}$, we obtain by the quotient rule
$$\bar C'(x)=\frac{xC'(x)-C(x)}{x^2}$$
Setting $\bar C'(x)=0$, we have
$$xC'(x)-C(x)=0,$$
that is
$$C'(x)=\frac{C(x)}{x}=\bar C(x)$$

The preceding example can be quickly answered using this Theorem. Setting $C'(x)=\bar C(x)$, we have
$$2x-2=x-2+\frac{4900}{x}$$
Simplifying this we obtain
$$x^2=4900$$
Hence, $x=70$ as we found earlier.

Example. The cost in dollars of producing $x$ stereos is given by $C(x)=70x+800$. The demand equation is $20p+x=18000$. (a) What level of production maximizes profit? (b) What is the price per stereo when profit is maximized? (c) What is the maximum profit?

Solution. From the demand equation, we obtain $p=-0.05x+900$. Recall that the profit function $P(x)$ is given by
\begin{align*} P(x)&=R(x)-C(x)\\ &=xp-C(x)\\ &=-0.05x^2+900x-(70x+800)\\ &=0.05x^2+830x-800 \end{align*}
Setting $P'(x)=-0.1 x+830$ equal to 0, we find the critical point $x=8300$. $P^{\prime\prime}(x)=-0.1<0$, so the profit has the maximum at $x=8300$.

(a) The level of production that maximizes profit is $x=8300$.

(b) The price per stereo at which profit is maximized is
$$p=-0.05(8300)+900=485\ \mathrm{dollars}$$

(c) The maximum profit is
$$P(8300)=-0.05(8300)^2+830(8300)-800=3,443,700\ \mathrm{dollars}$$

Here is another interesting theorem from economics.

Theorem. The profit is maximized when the marginal revenue equals the marginal cost, that is, when $R'(x)=C'(x)$.

Proof. Differentiating revenue function $P(x)=R(x)-C(x)$, we have
$$P'(x)=R'(x)-C'(x)$$
The critical point is obtained from $P'(x)=0$, i.e. when $R'(x)-C'(x)=0$ or $R'(x)=C'(x)$. That is, when the marginal revenue equals the marginal cost. This completes the proof.

The preceding example can be answered quickly using this theorem. Setting $R'(x)=C'(x)$, we have
$$-0.1x+900=70$$
or
$$x=8300$$

Example. A theater has 204 seats. The manager finds that he can fill all the seats if he charges $4.00 per ticket. For each ten cents that he raises the ticket price he will sell three fewer seats. What ticket price should he charge to maximize the ticket revenue?

Solution. When the manager increases $n$ cents per ticket price, the number $x(n)$ of tickets sold is $x(n)=204-3n$ and the price $p(n)$ per ticket is $p(n)=4+0.1n$. Then the total revenue is given by
\begin{align*}
R(n)&=x(n)p(n)\\
&=(204-3n)(4+0.1n)\\
&=-0.3n^2+8.4n+816
\end{align*}
Setting $R'(n)=-0.6n+8.4$ equal to 0, we find $n=14$. Since $R^{\prime\prime}(n)=-0.6<0$, the ticket revenue is maximized when $n=14$, i.e. when the ticket price is 4.00 dollars+1.40 dollars=5.40 dollars.

Alternatively, one can easily find the demand equation which is linear in this case. The equation of line through two points $(204,4)$ and $(201,4.1)$ is given by
$$p=-\frac{1}{30}x+\frac{54}{5}$$
and so we obtain the revenue function
$$R(x)=-\frac{1}{30}x^2+\frac{54}{5}x$$
Setting $R'(x)=-\frac{x}{!5}+\frac{54}{5}$ equal to 0, we find $x=162$ and plugging this into the demand equation for $x$ gives $p=5.40$.

Let us consider a demand equation given as $x=D(p)$. If one were to consider $\frac{dx}{dp}$, the rate of change of demand with respect to price, often it would be convenient to have it as a dimensionless quantity, i.e. one that does not depend on particular units. For that we define a new quantity by dividing $\frac{dx}{dp}$ by $\frac{x}{p}$. The resulting ratio is called the elasticity of demand and is denoted by $\epsilon_D$.

Definition. The elasticity of demand $\epsilon_D$ is defined by
$$\epsilon_D=\frac{\frac{dx}{dp}}{\frac{x}{p}}=\frac{p}{x}\frac{dx}{dp}$$

In economics, demand decreases as price increases, so demand function is a decreasing function, i.e. $\frac{dx}{dp}<0$. Since both $x$ and $p$ are positive, the elasticity $\epsilon_D$ is always negative. The demand is said to be elastic if $|\epsilon_D|>1$, inelastic if $|\epsilon_D|<1$, and unitary if $|\epsilon_D|=1$.

Definition. The relative change of a function whose equation is $p=f(a)$ as $q$ changes from $q_1$ to $q_2$ is
$$\frac{f(q_2)-f(q_1)}{f(q_1)}$$
The percentage change is defined as
$$100\times\frac{f(q_2)-f(q_1)}{f(q_1)}$$

Example. (a) If the demand equation is $x=100-3p$ , find the elasticity of demand when $p=1$.

(b) Show that the elasticity equals the ratio of the relative change in demand to the relative change in price when $p$ changes from 1 to 2.

Solution. (a) $x=100-3p$, $\frac{dx}{dp}=-3$, and when $p=1$, $x=97$, so
$$\epsilon_D=\frac{p}{x}\frac{dx}{dp}=-\frac{3}{97}$$
Since $|\epsilon_D|<1$, the demand is inelastic.

(b) As $p$ changes from 1 to 2, the relative change in price is $\frac{2-1}{1}=1$. When $p$ changes from 1 to 2, $x$ changes from 97 to 94, so the relative change in demand is $-\frac{3}{97}$. Hence, the ration of the relative change in demand to the relative change in price is
$$\frac{-\frac{3}{97}}{1}=-\frac{3}{97}=\epsilon_D$$

Example. Given the demand equation $x=\sqrt{100-2p}$, find the elasticity of demand when $p=18$. Is the demand elastic or inelastic at $p=18$?

Solution. The elasticity $\epsilon_D$ at $p=18$ is
\begin{align*} \epsilon_D&=\frac{p}{x}\frac{dx}{dp}\\ &=\frac{18}{\sqrt{100-2(18)}}\left(-\frac{1}{\sqrt{100-2p}}\right)_{p=18}\\ &=\frac{18}{8}\left(-\frac{1}{8}\right)\\ &=-\frac{9}{32} \end{align*}
Since $|\epsilon_D|=\frac{9}{32}<1$, the demand at $p=18$ is inelastic.

Example. Show that when the revenue is maximized, $|\epsilon_D|=1$.

Solution. The total revenue is $R=xp$, so
\begin{align*} \frac{dR}{dp}&=\frac{dx}{dp}p+x\\ &=x\left(1+\frac{p}{x}\frac{dx}{dp}\right)\\ &=x(1+\epsilon_D) \end{align*}
Since $x>0$, $\frac{dR}{dp}=0$ if and only if $\epsilon_D=-1$. If $\epsilon_D<-1$, then $|\epsilon_D|>1$ i.e. the demand is elastic, and $\frac{dR}{dp}<0$ (the revenue is decreasing). If $-1<\epsilon_D<0$, then $|\epsilon_D|<1$ i.e. the demand is inelastic, and $\frac{dR}{dp}>0$ (the revenue is increasing). So, the revenue is maximized when $|\epsilon_D|=1$.

When demand is a function of price, the elasticity can be written as
$$\epsilon_D=\frac{\frac{p}{x}}{\frac{dp}{dx}}$$

Calculus 25: Optimization Problems

 In mathematics and also in applications, we often encounter problems that require to maximize or minimize the value of a certain quantity. The general procedure can be summarized as:

1. Express the quantity to be maximized or minimized in terms of a single variable. The quantity may be described in terms of two variables however with given constraint it could be reduced to a single variable.
2. Differentiate the function obtained in step 1 and set the derivative equal to 0.
3. Solve the equation from step 2 to obtain critical values and determine whether they maximize or minimize the given quantity. Usually the first or second derivative test is a convenient tool for the required inspection.
   

Example. A farmer has 2400 ft of fencing and wants to fence off a rectangular field that borders a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area?

Solution. Let $x$ and $y$ denote the length and the width of the rectangular field. Suppose that the side along the river has the length $x$. Then the area is $A=xy$ and the required fencing in terms $x$ and $y$ is $x+2y=2400$. This fencing is a constraint and solve it for $y$ to obtain $y=1200-\frac{x}{2}$. Plugging this into $A$ for $y$, the area can be written as a function of a single variable $x$: $$A(x)=1200x-\frac{x^2}{2}$$ $A'(x)=1200-x$ and setting this equal to 0, we find $x=1200$. Since $A^{\prime\prime}(x)=-1<0$, by the second derivative test $x=1200$ gives rise to the absolute maximum of $A(x)$. The required dimensions are $1200\ \mbox{ft}\times 600\ \mbox{ft}$ where the side that borders the river is 1200 ft and the resulting largest area is 720,000 $\mbox{ft}^2$.

Example. A box with a square base and open top must have a volume of 32,000 $\mbox{cm}^3$. Find the dimensions of the box that minimize the amount of material used.

Solution. Let $x$ and $h$ be the length and the height of the box, respectively. Then $x^2h=32000$ and we want to minimize the surface area $A=x^2+4xh$. Solve the volume constraint for $h$ to obtain $h=\frac{32000}{x^2}$. Plugging this into $A$ for $h$, we write $A$ as a function of a single variable $x$: $$A(x)=x^2+\frac{128000}{x}$$ $A'(x)=2x-\frac{128000}{x^2}$ and setting it equal to 0, we find $x=40$. Since $A^{\prime\prime}(x)=2+\frac{256000}{x^3}>0$ for all $x>0$, $A(40)$ is the absolute minimum. Therefore the required dimensions are $40\ \mbox{cm}\times 40\ \mbox{cm}\times 20\ \mbox{cm}$.

Example. If 1200 $\mbox{cm}^2$ of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

Solution. Let $x$ and $h$ be the length and the height of the box, respectively. Then $x^2+4xh=1200$ and we want to maximize $V=x^2h$. Solve the area for $h$ to obtain $h=\frac{1200-x^2}{4x}$. Plugging this into $V$, we write the volume as a function of a single variable $x$: $$V(x)=300x-\frac{1}{4}x^3$$ $V'(x)=300-\frac{3}{4}x^2$ and setting it equal to 0, we find $x=20$. Since $V'(x)$ is a quadratic polynomial with a negative leading coefficient, $V(20)=4000\ \mbox{cm}^3$ is the largest possible volume of the box.

Example. Find the point on the parabola $y^2=2x$ that is the closest to the point $(1,4)$.

Solution. Let $(x,y)$ denote a point on the parabola $y^2=2x$. The distance between $(x,y)$ and $(1,4)$ is $d=\sqrt{(x-1)^2+(y-4)^2}$ and we want to minimize this. Note minimizing $d$ is equivalent to minimizing $d^2=(x-1)^2+(y-4)^2$. Solve the equation of parabola for $x$ to obtain $x=\frac{y^2}{2}$. Plugging this into $d^2$, we can write it as a function of a single variable $y$: $$f(y)=\left(\frac{y^2}{2}-1\right)^2+(y-4)^2=\frac{y^4}{4}-8y+17$$ $f'(y)=y^3-8$ and setting it equal to 0, we find $y=2$. Since $f^{\prime\prime}(y)=3y^2>0$ for all $y\ne 0$, $(x,y)=(2,2)$ is the point on the parabola $y^2=2x$ that is the closest to $(1,4)$.

The shortest distance from $(1,4)$ to the parabola $y^2=2x$.

Remark. The above problem also can be solved using a simple geometric fact that the shortest path from $(1,4)$ to the parabola $y^2=2x$ would be normal to the tangent line (i.e. the path is perpendicular to the tangent line). Let $(a,b)$ be the point on the parabola that is closest to $(1,4)$. By implicit differentiation we find $\frac{dy}{dx}=\frac{1}{y}$ and so the normal line at $(a,b)$ has the slope $-b$. The equation of the normal line is then $y-4=-b(x-1)$. Since this line is passing through $(a,b)$, $b-a=-b(a-1)$ or $ab=4$. $(a,b)$ is also on the parabola so we have $b^2=2a$. Solve the two equations simultaneously to obtain $b=2$ and hence $a=2$. Therefore, $(a,b)=(2,2)$.

Example. An open box is to be made out of a 6-inch by 18-inch piece of cardboard by cutting out squares of equal size from the four corners and bending up the sides. Find the dimensions of the resulting box that has the largest volume.

Solution. When you tackle this type of problems, it is very important to draw a picture that properly depicts the description of the problem as shown in the following figure.

Box

In the figure, let $y$ represent the length and $z$ the width of the box. (It doesn't matter if you choose $y$ to be the width and $z$ the length of the box. You will end up having the same answer.) Since the dimensions of the cardboard are $6\times 18$, the length $y$ and the width $z$ are, respectively, $y=6-2x$ and $z=18-2x$. Thus, the volume $V$ is $$V(x)=(6-2x)(18-2x)x=4x^3-48x^2+108x$$ To find the critical points, set $$V'(x)=12x^2-96x+108=0$$ or equivalently $x^2-8x+9=0$. This quadratic equation has two solutions $x=4\pm\sqrt{7}$. Recall that in order for a critical point $c$ to maximize the volume, it is required that $$V^{\prime\prime}(c)=24c-96<0,$$ i.e. $c<4$. Thus, $x=4-\sqrt{7}=1.35425$ maximizes the volume $V(x)$. The dimensions of the box that has the largest volume is then $$3.2915\times 15.2915\times 1.35425$$

Example.  A fence 4 feet tall runs parallel to a tall building at a distance of 2 feet from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building?

Solution. The following figure depicts the description of the problem.

The big and small right triangles are similar triangles, so we have $$\frac{x+2}{y}=\frac{x}{\sqrt{x^2+16}}$$ which is equal to $\cos\theta$. Solving the equation for $y$, we obtain $$y=\sqrt{x^2+16}+\frac{2\sqrt{x^2+16}}{x}$$ To minimize $y$, we find $$y'=\frac{x^3-32}{x^2\sqrt{x^2+16}}$$ and there is only one critical point $x=\root 3\of{32}=2\root 3\of{4}$. The resulting $y$ value, i.e. the length of the shortest ladder is then $8.32388$. By the second derivative, one can confirm that the critical point indeed minimizes the length of the ladder. But even without using the second derivative, one can make the length of such ladder as long as one likes, so there is no maximum length of such ladder.