Wormhole Physics: The Einstein-Rosen Bridge Part 1/2

 A wormhole is a hypothetical celestial object which is a shortcut to a distant part of the universe. It can be used as a pathway for interstellar travel and even as a possible time machine. Wormholes are not some creation of science fiction. Like black holes, wormholes are also solutions of Einstein's field equations. However, unlike black holes, there is no known physical mechanism of how or reason why they can be created or they must exist. The only way, I know of, that a wormhole can be created naturally is by a black hole. But such a wormhole is not traversable. If you attempt to cross it, you die. It is like jumping into a black hole. In [2], the authors claim that traversable wormholes may be built by a highly advanced civilization.

The first paper on wormholes with serious calculations was the 1935 paper by Einstein and Rosen [1]. Einstein and Rosen used the term "bridge". The word "wormhole" was coined by John Archibald Wheeler much later. Einstein and Rosen didn't study the bridge by means of interstellar travel but to build a geometrical model of an elementary particle. Einstein and Rosen considered two types of bridges: neutral and quasicharged. In this note, we discuss the neutral bridge.

Let us consider the ordinary Schwarzschild metric with $c=G=1$
$$ds^2=-\left(1-\frac{2M}{r}\right)dt^2+\frac{dr^2}{1-\frac{2M}{r}}+r^2d\Omega^2$$
where $d\Omega^2=d\theta^2+\sin^2\theta d\phi^2$. Clearly, $r=0$ is a singularity. It is not a physical singularity but a coordinate singularity. What this means is that one can remove the singularity by a suitable coordinate transformation. Einstein and Rosen introduced the coordinate transformation $u^2=r-2M$ and this results in the Einstein-Rosen form
$$
ds^2=-\frac{u^2}{u^2+2M}dt^2+4(u^2+2M)du^2+(u^2+2M)^2d\Omega^2 \tag{1}
$$
where $-\infty<u<\infty$. The coordinate change removes the region $0\leq r<2M$ containing the singularity and twice covers the asymptotically flat region, $2M\leq r<\infty$. The region near $u=0$ is interpreted as a bridge connecting the asymptotically flat region $u=\infty$ with asymptotically flat region near $u=-\infty$. Such an interpretation makes sense. If we consider a spherical surface defined by taking $u=\mbox{constant}$, the area of this surface is $A(u)=4\pi(u^2+2M)^2$. The minimum area is $A(0)=16\pi M^2$. The narrowest part of this geometry with the minimum area is defined to be the throat and the region nearby is called the bridge. Note that $u=0$ ($r=2M$) corresponds to the event horizon, i.e. the throat is the event horizon. For the creation of a wormhole, the existence of a physical singularity (curvature singularity) surrounded by a horizon is essential. A physical singularity which is not surrounded by a horizon is called a naked singularity. No neutral Einstein-Rosen bridge construction is possible without a horizon. The neutral Einstein-Rosen bridge (1) is also called the Schwarzschild wormhole. The Schwarzschild wormhole is actually a black hole so it is not traversable. This rules out the possibility of using naturally formed wormholes for interstellar travel.

References:

1. Albert Einstein and N. Rosen, The particle problem in general theory of relativity, Phys. Rev., 48:73-77, 1935
2. Michael S. Morri, Kip S. Thorne, and Ulvi Yurtsever, Wormholes, Time Machines, and Weak Energy Condition, Physical Review Letters, Volume 61, Number 13, pp. 1446-1449, 26 September 1988

Calculus 23: Mean Value Theorem

The following theorem something that can be easily understood by intuition.

Theorem. [Rolle's Theorem]
Let $f$ be continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. If $f(a)=f(b)$, then there exists a number $c$ in $(a,b)$ such that $f'(c)=0$.

Example. Show that the equation $x^3+x-1=0$ has exactly only one real root.

Solution. Let $f(x)=x^3+x-1$. Note that $f(0)=-1$ and $f(1)=1$. So by the Intermediate Value Theorem, we see that there exists at least a root of the equation $x^3+x-1=0$ in the interval $(0,1)$. Now suppose that there are two different roots $a$ and $b$ of the equation $x^3+x-1=0$. Without loss of generality, we may assume that $a<b$. Then $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$. By Rolle's Theorem then, there exist a number $c$ in $(a,b)$ such that $f'(c)=0$. However, $f'(x)=3x^2+1\geq 1$ for all real number $x$. This is a contradiction. Therefore, there should be only one root of the equation. 

Figure 1. The graph of $f(x)=x^3+x-1$

Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Define $g(x)$ to be the distance between $f(x)$ and the line segment from $(a,f(a))$ to $(b,f(b))$, i.e.
$$g(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a)-f(a).$$ Then $g(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Since $g(a)=g(b)=0$, by Rolle's theorem there exists a number $c$ in $(a,b)$ such that $g'(c)=f'(c)-\frac{f(b)-f(a)}{b-a}=0$. Therefore, we proved the following theorem.

Figure 2. Mean Value Theorem

Theorem. [Mean Value Theorem]
Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Then there exists a number $c$ in $(a,b)$ such that
$$f'(c)=\frac{f(b)-f(a)}{b-a}.$$

Remark. Why the name Mean Value Theorem? The average $\mathrm{av}(f)$ of a continuous function $f$ on a closed interval $[a,b]$ can be defined by $$\mathrm{av}(f)=\frac{1}{b-a}\int_a^b f(x)dx$$ If $f'(x)$ is continuous on $[a,b]$, its average on $[a,b]$ is given by $$\frac{1}{b-a}\int_a^bf'(x)dx=\frac{f(b)-f(a)}{b-a}$$ That is, Mean Value Theorem states that one of the values of $f'(x)$ on $(a,b)$ becomes the average of $f'(x)$ on $[a,b]$.

The following examples are applications of the Mean Value Theorem.

Example. Suppose that $f(0)=-3$ and $f'(x)\leq 5$ for all values of $x$. How large can $f(2)$ possibly be?

Solution. By the Mean Value Theorem, there exists a number $c$ in $(0,2)$ such that
$$f'(c)=\frac{f(2)-f(0)}{2-0}=\frac{f(2)+3}{2}.$$
Since $f'(c)\leq 5$,
\begin{align*}
f(2)&=2f'(c)-3\\
&\leq 2\cdot 5-3=7.
\end{align*}
Hence, $7$ is the largest possible value of $f(2)$.

Example. A trucker handed in a ticket at a toll booth showing in 2 hours she had covered 159 mi on a toll road with speed limit of 65 mph. The trucker was cited for speeding. Why?

Solution. The average speed was $\frac{159}{2}=79.5$ mph. By the Mean Value Theorem the trucker was driving at the speed 79.5 mph at some point.

Using Mean Value Theorem, one can prove the following theorem.

Theorem. If $f'(x)=0$ for all $x$ in the open interval $(a,b)$, then $f$ is constant on $(a,b)$.

Proof. Let $x,y$ be any two numbers in $(a,b)$. Without loss of generality, we may assume that $x<y$. Then $f(x)$ is continuous on $[x,y]$ and is differentiable on $(x,y)$. So, by the Mean Value Theorem, there exists $x<z<y$ such that
$$f'(z)=\frac{f(y)-f(x)}{y-x}$$
Since $f'(z)=0$, $f(x)=f(y)$. This completes the proof.

Calculus 22: Maximum and Minimum

 Maximum and Minimum

 There are two different types of extremum (maximum or minimum) values of a function $y=f(x)$. We may consider a value of $y$ that is an extremum globally on the domain or we may also consider a value of $y$ that is an extremum locally around an $x$ value.

A function $f$ has an absolute maximum at $c$ if $f(c)\geq f(x)$ for all $x$ in the domain of $f$. Similarly, $f$ has an absolute minimum at $c$ if $f(c)\leq f(x)$ for all $x$ in the domain of $f$.

A function $f$ has a local maximum (or relative maximum) at $c$ if $f(c)\geq f(x)$ in some neighborhood of $c$ (i.e an open interval that contains $c$). Similarly, $f$ has a local minimum (or relative minimum) at $c$ if $f(c)\leq f(x)$ in some neighborhood of $c$.

Example.

Figure 1. The graph of $f(x)=3x^4-16x^3+18x^2$ on $[-1,4]$

 

Figure 1 shows the graph of $f(x)=3x^4-16x^3+18x^2$, $-1\leq x\leq 4$. It has a local maximum at $x=1$ and a local minimum at $x=3$. The local minimum $f(3)=-27$ is also an absolute minimum. $f$ has an absolute maximum $f(-1)=37$. This $f(-1)=37$ is not a local maximum by the way. The reason is that there is no local neighborhood around $x=-1$ as the domain is given by $[-1,4]$.

A natural question one may ask is whether a function always has an absolute maximum and an absolute minimum. You can easily find many examples that show that a function does not necessarily have an absolute maximum or an absolute minimum value. For instance, $y=x$ on $(-\infty,\infty)$ has neither an absolute maximum nor an absolute minimum. The function $y=x^2$ on $[0,1)$ has an absolute minimum 0 at $x=0$ but has no absolute maximum.

Theorem. [Max-Min Theorem, Fermat]
If $f$ is continuous on a closed interval $[a,b]$, then $f$ attains an absolute maximum and an absolute minimum on $[a,b]$.

The following theorem is also due to Fermat.

Theorem. If $f$ has a local maximum or a local minimum at $c$ and if $f'(c)$ exists, then $f'(c)=0$.

The converse of this theorem is not necessarily true i.e. $f'(c)=0$ does not necessarily mean that $f(c)$ is a local maximum or a local minimum. For example, consider $f(x)=x^3$. $f'(0)=0$ but $f(x)$ has neither a local maximum nor a local minimum at $x=0$ as shown in Figure 2.

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

The preceding theorem is important as an absolute maximum and an absolute minimum may be found among local maximum values, local minimum values and the evaluations of $f$ at the end points, $f(a)$ and $f(b)$. To find local maximum values and local minimum values, we first find points $c$ such that $f'(c)=0$. Such points are called critical points. The reason they are called critical points is that the graph of a function changes from increasing to decreasing or from decreasing to increasing at a critical point.

Definition. A critical point of a function $f(x)$ is a number $c$ in the domain of $f$ such that either $f'(c)=0$ or $f'(c)$ does not exist.

Recipe for Finding Absolute Maximum and Absolute Minimum

Let $f$ be a continuous function on a closed interval $[a,b]$.

Step 1. Find all critical points of $f$ in $(a,b)$.

Step 2. Evaluate $f$ at each critical point obtained in Step 1.

Step 3. Find $f(a)$ and $f(b)$.

Step 4. Compare all the values obtained in Steps 2 and 3. The largest value is the absolute maximum and the smallest value is the absolute minimum.

Example. Find the absolute maximum and the absolute minimum values of
$$f(x)=x^3-3x^2+1,\ -\frac{1}{2}\leq x\leq 4.$$

Solution.

Step 1. Find all critical points of $f$ in $\left(-\frac{1}{2},4\right)$.

$f'(x)=3x^2-6x$. Set $f'(x)=0$ i.e. $3x^2-6x=0$. $3x^2-6x$ is factored as $3x(x-2)$. So we find two critical points $0, 2$.

Step 2. Evaluate $f$ at each critical point obtained in Step 1.

$f(0)=1$ and $f(2)=-3$.

Step 3. Find $f\left(-\frac{1}{2}\right)$ and $f(4)$.

$f\left(-\frac{1}{2}\right)=\frac{1}{8}$ and $f(4)=17$.

Step 4. Compare all the values obtained in Steps 2 and 3.

The largest value is $f(4)=17$ so this is the absolute maximum value of $f$ on $\left[-\frac{1}{2},4\right]$. The smallest value is $f(2)=-3$ so this is the absolute minimum of $f$ on $\left[-\frac{1}{2},4\right]$.

Calculus 21: Linear Approximations and Differentials

 Linear Approximation

Figure 1. Linear Approximation

Let $y=f(x)$ be a differentiable function. The function $f(x)$ can be approximated by the tangent line to $y=f(x)$ at $a$ if $x$ is near $a$. Such an approximation is called a linear approximation.

If $x\approx a$ then $\Delta x=x-a\approx 0$, so we have
\begin{align*}
\frac{\Delta y}{\Delta x}&\approx \frac{dy}{dx}\\
&=f'(a).
\end{align*}
This means that
$$\frac{f(x)-f(a)}{x-a}\approx f'(a),$$
i.e.
$$
f(x)\approx f(a)+f'(a)(x-a). \tag{1}
$$
The equation (1) is called the linear approximation or tangent line approximation of $f$ at $a$. The linear function
$$
L(x):=f(a)+f'(a)(x-a)
$$
is called the linearization of $f$ at $a$. Notice that $L(x)$ is the equation of tangent line to $f$ at $a$.

Example. Find the linearlization of $f(x)=\sqrt{x+3}$ at $a=1$ and use it to approximate $\sqrt{3.98}$ and $\sqrt{4.05}$.

Solution. $f'(x)=\frac{1}{2\sqrt{x+3}}$, so
\begin{align*}
L(x)&=f(1)+f'(1)(x-1)\\
&=2+\frac{1}{4}(x-1)\\
&=\frac{x}{4}+\frac{7}{4}.
\end{align*}
When $x\approx 1$, we have the approximation
$$\sqrt{x+3}\approx \frac{x}{4}+\frac{7}{4}.$$

Figure 2. Linear approximation of $f(x)=\sqrt{x+3}$ at $a=1$

Setting $x+3=3.98$ we find $x=0.98$. Hence,
\begin{align*}
\sqrt{3.98}&\approx \frac{0.98}{4}+\frac{7}{4}\\
&=1.995.
\end{align*}
Setting $x+3=4.05$ we find $x=1.05$. Hence,
\begin{align*}
\sqrt{4.05}&\approx \frac{1.05}{4}+\frac{7}{4}\\
&=2.0125.
\end{align*}

Example. Use linear approximation to estimate $\sqrt{99.8}$.

Solution. In order to use linear approximation we need to choose $f(x)$, $x$ and $a$. First clearly from the given quantity we see that $f(x)=\sqrt{x}$ and thereby $x=99.8$. Since $f'(x)=\frac{1}{2\sqrt{x}}$, the linear approximation of $\sqrt{99.8}$ at $a$ is
$$\sqrt{99.8}\approx \sqrt{a}+\frac{1}{2\sqrt{a}}(99.8-a)$$
How do we choose a suitable $a$? There are two criteria you have to have in mind. One is $a$ has to be close to $x$ for the linear approximation to be useful. Second $a$ needs to be chosen so that $f(a)$ and $f'(a)$ can be calculated easily (meaning by hand without aid of a calculator). Why is this important? You have to understand that the use of linear approximation is not assuming any use of a calculator. (If you can use a calculator, what is the point of doing this approximation?) This is a method that was developed when there were no calculators available so people could calculate values like $\sqrt{99.8}$ by hand. Considering the two criteria, we find that $a=100$ is the one. Hence,
$$\sqrt{99.8}\approx \sqrt{100}+\frac{1}{2\sqrt{100}}(99.8-100)=10+\frac{1}{20}(-0.2)=9.99$$

Example. Use linear approximation to estimate $\cos 29^\circ$.

Solution. $f(x)=\cos x$ and $x=29^\circ=\frac{29\pi}{180}$ ($29^\circ$ is not a number but $\frac{29\pi}{180}$ is). Since $f'(x)=-\sin x$, the linear approximation of $\cos 29^\circ$ at $a$ is
$$\cos 29^\circ\approx \cos a-\sin a \left(\frac{29\pi}{180}-a\right)$$
The suitable $a$ is $=\frac{30\pi}{180}=\frac{\pi}{6}$ in the spirit of the two criteria we discussed in the example above. Therefore, we have
$$\cos 29^\circ\approx \cos\frac{\pi}{6}-\sin\frac{\pi}{6}\left(-\frac{\pi}{180}\right)=\frac{\sqrt{3}}{2}+\frac{\pi}{360}$$

Differentials

Figure 3. Differentials

As seen in Figure 3 above, when $\Delta x\approx 0$, $\Delta x=dx$ and $\Delta y\approx dy$. On the other hand, $\frac{dy}{dx}=f'(x)$. Hence, we obtain
$$
\Delta y\approx dy=f'(x)dx=f'(x)\Delta x. \tag{2}
$$

Example. The radius of a sphere was measured and found to be 21 cm with a possible error in measurement of at most 0.05 cm. What is the maximum error in using this value of the radius to compute the volume of the sphere?

Solution. Let $V$ denote the volume of a sphere of radius $r$. Then $V=\frac{4}{3}\pi r^3$. What we are trying to find is $\Delta V$ with $\Delta r\leq 0.05$ cm. As seen in (2), $\Delta V\approx dV$, so we find $dV$ instead because finding $dV$ is easier than finding the exact error $\Delta V$. Differentiating $V$ with respect to $r$, we obtain
\begin{align*}
\Delta V&\approx dV\\&=4\pi r^2 dr\\
&=4\pi r^2\Delta r\\
&\leq 4\pi\cdot(21)^2\cdot 0.05\\
&=277.
\end{align*}
So the maximum error in the calculated volume is about 277 $\mbox{cm}^3$.

Linear approximation and differentials may appear to be different entities but the two methods are indeed equivalent and they serve the same purpose. To illustrate this, let us take a look at the following example which will be answered by linear approximation and differentials.

Example. Approximate $\sqrt{81.1}$.

Solution by Linear Approximation. Let $f(x)=\sqrt{x}$ and choose $a=81$. The reason for this choice of $a$ is that one can easily calculate  without the aid of a calculator (which is the main point of using this method) and also $a=81$ is close to 81.1. Now we find the tangent line to $f(x)$ at $a=81$, or equivalently the linear approximation $L(x)$ at $a=81$. It is
$$L(x)=\frac{1}{2\cdot 9}(x-81)+9$$
Then
\begin{align*}
L(81.1)&=\frac{1}{18}(81.1-81)+9\\&=\frac{1}{180}+9\\
&=9.005555555555556
\end{align*}
approximates $\sqrt{81.1}$.

Solution by Differentials. Recall that $\Delta y=f(x+\Delta x)-f(x)$ is approximated by the differential $dy=f'(x)dx=f'(x)\Delta x$ for very small $\Delta x$. Now with $f(x)=\sqrt{x}$, $dy=\frac{1}{2\sqrt{x}}\Delta x$. From $\Delta y\approx dy$, we have
$$f(x+\Delta x)\approx f(x)+\frac{1}{2\sqrt{x}}\Delta x$$
If we set $f(x+\Delta x)=\sqrt{81.1}$, we can choose $x=81$ and $\Delta x=0.1$. Accordingly, we find
\begin{align*}
\sqrt{81.1}&\approx\sqrt{81}+\frac{1}{2\sqrt{81}}0.1\\
&=9+\frac{1}{180}=9.005555555555556
\end{align*}

Calculus 20: Related Rates

Related rates problems often involve (context-wise) real-life applications of the chain rule/implicit differentiation. Here are some of the examples that are commonly seen in calculus textbooks.

Example. Car A is traveling west at 50mi/h and car B is traveling north at 60mi/h. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when car A is 0.3 mi and car B is 0.4 mi from the intersection?

Solution.



Denote by $x$ and $y$ the distances from the intersection to car A and to car B, respectively. Then we have $\frac{dx}{dt}=-50$mi/h and $\frac{dy}{dt}=-60$mi/h. Let us denote $z$ the distance between $A$ and $B$. Then by Pythagorean law we have
$$z^2=x^2+y^2$$
Differentiating this with respect to $t$, we obtain
$$z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$$
and thus
\begin{align*}
\frac{dz}{dt}&=\frac{1}{z}\left[x\frac{dx}{dt}+y\frac{dy}{dt}\right]\\
&=\frac{1}{0.5}[0.3(-50)+0.4(-60)]=-78\mathrm{mi/h}
\end{align*}

Example. Air is being pumped into a spherical balloon so that its volume increases at a rate of $100\mathrm{cm}^3/\mathrm{s}$. How fast is the radius of the balloon increasing when the diameter is 50 cm?

Solution. Let $V$ and $r$ denote the volume and the radius of the spherical balloon. Then $V=\frac{4}{3}\pi r^3$. Differentiating this with respect to $t$, we obtain
$$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$$
So,
\begin{align*}
\frac{dr}{dt}&=\frac{1}{4\pi r^2}\frac{dV}{dt}\\
&=\frac{1}{4\pi(25)^2}100\\
&=\frac{1}{25\pi}\mathrm{cm/s}
\end{align*}

Example. Gravel is being dumped from a conveyor belt at a rate of $30 \mathrm{ft}^3/\mathrm{min}$  and its coarseness is such that it forms a pile in the shape of a cone  whose base diameter and height are the same. How fast is the height of  the pile increasing when the pile is 10 ft high?

Solution. The cross section of the gravel pile is shown in the figure below.

The amount of gravel dumped is the same as the volume of the cone. Let us denote the volume by $V$, its base radius by $r$, and its height by $h$. Then $V=\frac{1}{3}\pi r^2h$. Since $h=2r$, $V$ can be written as
$$V=\frac{1}{12}\pi h^3$$
Differentiating this with respect to $t$, we obtain
$$\frac{dV}{dt}=\frac{1}{4}\pi h^2\frac{dh}{dt}$$
So, we have
\begin{align*}
\frac{dh}{dt}&=\frac{4}{\pi h^2}\frac{dV}{dt}\\
&=\frac{4}{\pi(10)^2}(30)=\frac{1.2}{\pi}\mathrm{ft/min}\approx 0.38\mathrm{ft/min}
\end{align*}

Example. A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft/s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall?

Solution.

Let us denote by $x$ and $y$ the distance from the wall to the bottom of the ladder and the distance from the top of the ladder to the floor, respectively. By Pythagorean law, we have $x^2+y^2=100$. Differentiating this with respect to $t$, we obtain
$$x\frac{dx}{dt}+y\frac{dy}{dt}=0$$
Hence, we have
\begin{align*}
\frac{dy}{dt}&=-\frac{x}{y}\frac{dx}{dt}\\
&=-\frac{6}{8}(1)=-\frac{3}{4}\mathrm{ft/s}
\end{align*}

Example. A water tank has the shape of an inverted circular cone with base radius  2m and heigh 4 m. If water is being pumped into the tank at a rate of $2 \mathrm{m}^3/\mathrm{min}$, find the rate at which the water level is rising when the water is 3 m deep.

Solution. The cross section of the water tank is shown in the figure below.

The amount of water $V$ when the water level is $h$ and the surface radius is $r$ is $V=\frac{1}{3}\pi r^2h$. From the above figure we have the following ratio holds $$\frac{2}{4}=\frac{r}{h}$$ i.e. $r=\frac{h}{2}$. SO $V$ can be written as
$$V=\frac{1}{12}\pi h^3$$
Differentiating this with respect to $t$, we obtain
$$\frac{dV}{dt}=\frac{1}{4}\pi h^2\frac{dh}{dt}$$
Hence,
\begin{align*}
\frac{dh}{dt}&=\frac{4}{\pi h^2}\frac{dV}{dt}\\
&=\frac{4}{\pi(3)^2}(2)\\
&=\frac{8}{9\pi}\mathrm{m/min}\approx 0.28\mathrm{m/min}
\end{align*}

Example. A street light is at the top of a 15 feet tall pole. A 6 feet tall woman walks away from the pole with a speed of 5 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 45 feet from the base of the pole?

Solution. Let $x$ be the distance from the light pole to the woman and $y$ be the distance from the light pole to the tip of her shadow as shown in the figure below.

By similar triangles, we have $\frac{15}{y}=\frac{6}{y-x}$. Solving this equation for $y$, we obtain $y=\frac{5}{3}x$. Differentiating this with respect to $t$, we find how fast the tip of her shadow is moving:
$$\frac{dy}{dt}=\frac{5}{3}\frac{dx}{dt}=\frac{25}{3}\mathrm{ft/s}$$
As seen regardless of how far the woman is from the pole, the speed of the tip is constant.

Example. The fish population, $N$, in a small pond depends on the amount of algae, $a$ (measured in pounds), in it. The equation modeling the fish population is given by $N=(3a^2-20a+26)^4$. If the amount of algae is increasing at a rate of 2 lb/week, at what rate is the fish population changing when the pond contains 5 lb of algae?

Solution. By the chain rule, we obtain
\begin{align*}
\frac{dN}{dt}&=4(3a^2-20a+26)^3\left(6a\frac{da}{dt}-20\frac{da}{dt}\right)\\
&=4(3a^2-20a+26)^3(6a-20)\frac{da}{dt}
\end{align*}
$\frac{da}{dt}=2$lb/week, so when $a=5$lb, $\frac{dN}{dt}$ is
$$\frac{dN}{dt}=4(3(5)^2-20(5)+26)^3(6(5)-20)(2)=-80\ \mathrm{lb/week}$$
What this means is that the fish population is decreasing by 80 lb/week at the instant when the pond contains 5 lb of algae.

Example. The retail price per gallon of gasoline is increasing at 0.02 dollars per week. The demand equation is given by $$10p-\sqrt{356-x^2}=0$$ where $p$ is the price per gallon (in dollars), when $x$ million gallons are demanded. At what rate is the revenue changing when 10 million gallons are demanded?

Solution. The total revenue $R$ is given by the equation
$$R=xp$$
Differentiating this equation with respect to $t$, we obtain
$$\frac{dR}{dt}=\frac{dx}{dt}p+x\frac{dp}{dt}$$
The only quantity we don't have to calculate $\frac{dR}{dt}$ is $\frac{dx}{dt}$. To find it, let us differentiate the demand function with respect to $x$. By the chain rule, we obtain
$$10\frac{dp}{dt}+\frac{x}{\sqrt{351-x^2}}\frac{dx}{dt}=0$$
When $x=10$ million gallons, with $\frac{dp}{dt}=0.02$, we find from this equation that
$$\frac{dx}{dt}=-0.02\sqrt{256}=-0.02\cdot 16= -0.32\ \mbox{million gallons/week}$$
When $x=10$, from the demand function, we find $p$ as
$$p=\frac{\sqrt{256}}{10}=\frac{16}{10}=1.6$$
Therefore, the rate of change of revenue when 10 million gallons of gasoline is demanded is
$$\frac{dR}{dt}=-0.32(1.6)+10(0.02)=-0.312$$
What this means is that the revenue is decreasing by about 0.31 million dollars per week when the price increases 0.02 dollars per week (consequently the demand decreases by 0.32 million gallons per week as we saw earlier).

Example. A plane flying with a constant speed of 29 km/min passes over a ground radar station at an altitude of 13 km and climbs at an angle of 20 degrees. At what rate is the distance from the plane to the radar station increasing 3 minutes later?

Solution. First, take a look at the following picture

The Law of Cosine

The law of cosine says that the sides $a$, $b$, $c$ and the angle $\theta$ are related by
$$a^2=b^2+c^2-2bc\cos\theta$$

The question above can be pictorially represented as the following sketch

Using the law of cosine with $b=13$ km, we have
$$a^2=13^2+c^2-2\cdot 13\cdot c\cdot\cos\left(\frac{11}{18}\pi\right)$$
Here, the angle $\theta$ is given by $\theta=90^\circ+20^\circ=110^\circ=\frac{11}{18}\pi$ (Note: for this question, you can use degree instead of radian but in that case make sure that your calculator is set to use degree angle measurement.) Since we are to find $\frac{da}{dt}$, differentiate the above equation with respect to $t$:
$$2a\frac{da}{dt}=2c\frac{dc}{dt}-2\cdot 13\frac{dc}{dt}\cos\left(\frac{11}{18}\pi\right)$$
Since the airplane is flying along the side $c$ at the constant speed 29 km/min, it would have traveled $c=29\cdot 3=87$ km in three minutes. Thus,
$$a=\sqrt{13^2+87^2-2\cdot 13\cdot 87\cos\left(\frac{11}{18}\pi\right)}=92.2586$$
and hence $\frac{da}{dt}$ in three minutes is
\begin{align*}
\frac{da}{dt}&=\frac{c}{a}\frac{dc}{dt}-\frac{13}{a}\frac{dc}{dt}\cos\left(\frac{11}{18}\pi\right)\\
&=\frac{1}{a}\frac{dc}{dt}\left(c-13\cos\left(\frac{11}{18}\pi\right)\right)\\
&=\frac{1}{92.2586}29\left(87-13\cos\left(\frac{11}{18}\pi\right)\right)\\
&=28.7447\ \mathrm{km/min}
\end{align*}

Calculus 19: Derivatives of Logarithmic and Exponential Functions

In this note, we study derivatives of logarithmic and exponential functions.

Derivatives of Logarithmic Functions

First recall that
$$\lim_{t\to 0}(1+t)^{\frac{1}{t}}=e \tag{1}$$
\begin{align*}
\frac{d}{dx}\ln x&=\lim_{h\to 0}\frac{\ln(x+h)-\ln x}{h}\\
&=\lim_{h\to 0}\frac{1}{h}\ln\left(\frac{x+h}{x}\right)\\
&=\frac{1}{x}\lim_{h\to 0}\ln\left(1+\frac{h}{x}\right)^{\frac{x}{h}}\\
&=\frac{1}{x}\lim_{t\to 0}\ln(1+t)^{\frac{1}{t}}\\
&=\frac{1}{x}\end{align*}
with $t=\frac{h}{x}$.
$$\frac{d}{dx}\ln x=\frac{1}{x} \tag{2}$$
Using the change of base formula $\log_ax=\frac{\ln x}{\ln a}$, we obtain
$$\frac{d}{dx}\log_ax=\frac{1}{x\ln a}$$

Derivatives of Exponential Functions

We can find the derivative of the natural exponential function $y=e^x$ using the relationship $x=\ln y$ and implicit differentiation. Differentiating $x=\ln y$ with respect to $x$ we obtain $1=\frac{1}{y}\frac{dy}{dx}$ i.e. $\frac{dy}{dx}=y=e^x$. Hence
$$\frac{d}{dx}e^x=e^x$$
Note that $a^x=e^{x\ln a}$. So by the chain rule we find
$$\frac{d}{dx}a^x=\frac{d}{dx}e^{x\ln a}=e^{x\ln a}\ln a=a^x\ln a$$
Hence
$$\frac{d}{dx}a^x=a^x\ln a$$

The Power Rule (General Form)

Let us consider $x^n$ for any $x>0$ and any real number $n$. As we have seen above $x^n=e^{n\ln x}$ so by the chain rule
$$\frac{d}{dx}x^n=\frac{d}{dx}e^{n\ln x}=e^{n\ln x}\frac{n}{x}=nx^{n-1}$$
This completes the proof of the general power rule.

Logarithmic Differentiation

The derivatives of functions involving products, quotients, and powers may be found more easily (quickly) by taking the natural logarithm of such functions before differentiating. This allows us to break a complicated function into simpler pieces using properties of the natural logarithm. This whole process, which is called logarithmic differentiation, makes differentiation much easier and quicker.

Example. Use logarithmic differentiation to find the derivative of $y=\frac{x\sqrt{x^2+1}}{(x+1)^{\frac{2}{3}}}$.

Solution.
\begin{align*}\ln y&=\ln \frac{x\sqrt{x^2+1}}{(x+1)^{\frac{2}{3}}}\\
&=\ln x+\frac{1}{2}\ln(x^2+1)-\frac{2}{3}\ln(x+1)
\end{align*}
Differentiating with respect to $x$,
$$\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}+\frac{x}{x^2+1}-\frac{2}{3(x+1)}$$
Therefore,
$$\frac{dy}{dx}=\left[\frac{1}{x}+\frac{x}{x^2+1}-\frac{2}{3(x+1)}\right]\frac{x\sqrt{x^2+1}}{(x+1)^{\frac{2}{3}}}$$

Example. Let $y=x^x$, $x>0$. Find $\frac{dy}{dx}$.

Solution 1. $y=x^x=e^{x\ln x}$ and by the chain rule we obtain
$$\frac{dy}{dx}=x^x(1+\ln x)$$

Solution 2. Use logarithmic differentiation. $\ln y=x\ln x$ and differentiating this with respect to $x$, we have
$$\frac{1}{y}\frac{dy}{dx}=1+\ln x$$
Hence, $$\frac{dy}{dx}=x^x(1+\ln x)$$

Alternative Approach

In the earlier approach we started out with $e^x$ and regarded $\ln x$ as its inverse function. It can also be done the other way around, namely we first define $\ln x$ and regard $e^x$ as its inverse function. The natural logarithmic function $\ln x$ can be defined by
$$\ln x=\int_1^x\frac{1}{t}dt,\ x>0 \tag{3}$$
The number $x$ that satisfies the equation $\ln x=1$ is denoted by $e$. All properties of natural logarithm can be derived from definition (3). Also from definition (3), we obtain (2) by the Fundamental Theorem of Calculus. Using (2) one can show the limit $$\lim_{x\to 0}(1+x)^{\frac{1}{x}}=e$$

Proof. Let $f(x)=\ln x$. Then $f'(x)=\frac{1}{x}$ and so $f'(1)=1$. On the other hand,
\begin{align*}
f'(1)&=\lim_{x\to 0}\frac{f(1+x)-f(1)}{x}\\
&=\lim_{x\to 0}\frac{\ln(1+x)}{x}\\
&=\lim_{x\to 0}\ln(1+x)^{\frac{1}{x}}\\
&=\ln[\lim_{x\to 0}(1+x)^{\frac{1}{x}}]
\end{align*} Therefore,
$$\lim_{x\to 0}(1+x)^{\frac{1}{x}}=e$$

Remark. By substituting $y=\frac{1}{x}$,
$$e=\lim_{y\to\infty}\left(1+\frac{1}{y}\right)^y$$

Remark. An alternative definition of $e$ is as an infinite series
$$e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots$$

SDE: Itô's Formula

 Let us consider the 1-dimensional case ($n=1$) of the stochastic differential equation (4) in here
$$dX=b(X)dt+dW \tag{1}$$
with $X(0)=0$.
Let $u: \mathbb{R}\longrightarrow\mathbb{R}$ be a smooth function and $Y(t)=u(X(t))$ ($t\geq 0$). What we learned in calculus (the chain rule) would dictate us that $dY$ is
$$dY=u'dX=u'bdt+u'dW,$$
where $'=\frac{d}{dx}$. It may come to you as a surprise to hear this but this is not correct in case of stochastic processes. First, by Taylor series expansion, we obtain
\begin{align*}
\Delta Y&=u(X+\Delta X)-u(X)\\
&=u(X)+u'(X)\Delta X+\frac{u''(X)}{2!}(\Delta X)^2+\cdots -u(X)\\
&=u'(X)\Delta X+\frac{u''(X)}{2!}(\Delta X)^2+\cdots
\end{align*}
and thus we have
\begin{align*}
dY&=u'dX+\frac{1}{2}u^{\prime\prime}(dX)^2+\cdots\\
&=u'(bdt+dW)+\frac{1}{2}u^{\prime\prime}(bdt+dW)^2+\cdots
\end{align*}
Now, we introduce the following striking formula
$$(dW)^2=dt \tag{2}$$
The proof of (2) is beyond the scope of this note and so it won't be discussed here. However it can be found, for example, in [1]. Using (2) $dY$ can be written as
$$dY=\left(u'b+\frac{1}{2}u^{\prime\prime}\right)dt+u'dW+\cdots$$
The terms beyond $u'dW$ are of order $(dt)^{\frac{3}{2}}$ and higher. Neglecting these terms, we have
$$dY=\left(u'b+\frac{1}{2}u^{\prime\prime}\right)dt+u'dW \tag{3}$$
(3) is the stochastic differential equation satisfied by $Y(t)$ and it is called the Itô's Formula named after a Japanese mathematician Kiyosi Itô.

Example. Let us consider the stochastic differential equation
$$dY=YdW,\ Y(0)=1 \tag{4}$$
Comparing (3) and (4), we obtain
\begin{align*}
u'b+\frac{1}{2}u^{\prime\prime}&=0 \tag{5}\\u'&=u \tag{6}
\end{align*}
The equation (6) along with the initial condition $Y(0)=1$ results in $u(X(t))=e^{X(t)}$. Using this $u$ with equation (5) we get $b=-\frac{1}{2}$ and so the equation (1) becomes
$$dX=-\frac{1}{2}dt+dW$$
in which case $X(t)=-\frac{1}{2}t+W(t)$. Hence, we find $Y(t)$ as
$$Y(t)=e^{-\frac{1}{2}t+W(t)}$$

Example. Let $P(t)$ denote the price of a stock at time $t\geq 0$. A standard model assumes that the relative change of price $\frac{dP}{P}$ evolves according to the stochastic differential equation
$$\frac{dP}{P}=\mu dt+\sigma dW \tag{7}$$
where $\mu>0$ and $\sigma$ are constants called the drift and the volatility of the stock, respectively. Again using Itô's formula similarly to what we did in the preceding example, we find the price function $P(t)$ which is the solution of
$$dP=\mu Pdt+\sigma PdW,\ P(0)=p_0$$
as
$$P(t)=p_0\exp\left[\left(\mu-\frac{1}{2}\sigma^2\right)\right]t+\sigma W(t).$$

Example. In this example, we solve the stochastic population growth model
\begin{align*}
\frac{dN}{dt}&=(r(t)+\xi(t))N(t) \tag{8}\\
&=r(t)N(t)+\xi(t)N(t)
\end{align*}
with $N(0)=N_0$. Here, $r(t)$ is a known function (this $r(t)$ may be considered as the relative growth rate at time $t$ in the deterministic population growth model) and $\xi(t)=\frac{dW}{dt}$ is a white noise. Let $N(t)=u(X(t))$. Then
$$dN=r(t)udt+udW \tag{9}$$
Comparing (3) and (9), we obtain
\begin{align*}
u'b+\frac{1}{2}u''&=r(t)u\\
u'&=u
\end{align*}
$u(t)=N_0e^{X(t)}$ and hence we find $b=r(t)-\frac{1}{2}$. From the equation (1), we have
$$dX=\left(r(t)-\frac{1}{2}\right)dt+dW$$
whose solution $X(t)$ is given by
$$X(t)=\int_0^t r(s)ds-\frac{t}{2}+W(t)$$
Therefore,
$$N(t)=N_0\exp\left[\int_0^t r(s)ds-\frac{t}{2}+W(t)\right]$$

References:

1. Bernt Øksendal, Stochastic Differential Equations, An Introduction with Applications, 5th Edition, Springer, 2000