6.1.5 Continued

So let's do an example or two illustrating the function analysis procedure.

Example: analyze the function
$$f(x)=\frac{x^{3}-4x}{x^{2}+3}$$
on the interval $[-10,10]$, including finding the global extrema.

1. The domain is all real numbers, normally, but we've restricted it to $[-10,10]$. So that's our domain.

2. There are no vertical asymptotes, because the function is continuous (in particular, the denominator is never zero). There are no horizontal asymptotes, because the degree of the numerator is strictly greater than the degree of the denominator. However, there are slant asymptotes. If we perform polynomial division, we get that
$$f(x)=x-\frac{7x}{x^{2}+3}.$$
Hence, the fraction part goes to zero as $x\to\infty$, and we're left with $x$. Therefore, the slant asymptote for both infinities is $y=x$.

3. The first derivative is
$$f'(x)=\frac{(x^{2}+3)(3x^{2}-4)-(x^{3}-4x)(2x)}{(x^{2}+3)^{2}}=\frac{3x^{4}+5x^{2}-12-2x^{4}+8x^{2}}{(x^{2}+3)^{2}}=\frac{x^{4}+13x^{2}-12}{(x^{2}+3)^{2}}.$$
Setting this equal to zero is tantamount to setting $x^{4}+13x^{2}-12=0$, as the denominator is never zero. To solve this, we let $z=x^{2}$ and use the quadratic formula: $z^{2}+13z-12=0$ implies that
$$z=\frac{-13\pm\sqrt{169-4(-12)}}{2}=\frac{-13\pm\sqrt{217}}{2}.$$
Now, $217=7\times 31$, so we can't simplify the square root any more. Note that $\sqrt{217}\approx 14.73$, so we have one positive $z$ root, and one negative $z$ root. The negative $z$ root yields complex numbers for $x$, so we discard those, and focus on
$$z=\frac{\sqrt{217}-13}{2} \quad \implies \quad x=\pm\sqrt{\frac{\sqrt{217}-13}{2}}.$$
I agree it's not pretty, but in the real world, answers are hardly ever pretty. Be thankful you have an explicit answer like this at all! So, dividing up our domain according to these numbers, we have the following partition of $[-10,10]:$
$$\left[-10,-\sqrt{\frac{\sqrt{217}-13}{2}}\:\right),\quad \left(-\sqrt{\frac{\sqrt{217}-13}{2}},\sqrt{\frac{\sqrt{217}-13}{2}}\:\right),\quad \left(\sqrt{\frac{\sqrt{217}-13}{2}},10 \right].$$

4. We pick the three values $-5,0,5$ as the points in each non-overlapping region. I get that $f'(-5)=67/56 >0$, so the function is increasing there. Then $f'(0)=-4/3<0$, so the function is decreasing there. Finally, $f'(5)=67/56>0$, so we're increasing again.

5. The second derivative is
$$f''(x)=\frac{((x^{2}+3)^{2})(4x^{3}+26x)-2(x^{4}+13x^{2}-12)(x^{2}+3)(2x)}{((x^{2}+3)^{2})^{2}} =\frac{(x^{2}+3)(4x^{3}+26x)-4x(x^{4}+13x^{2}-12)}{(x^{2}+3)^{3}}$$
$$=\frac{4x^{5}+26x^{3}+12x^{3}+78x-4x^{5}-52x^{3}+48x}{(x^{2}+3)^{3}}=\frac{-14x^{3}+126x}{(x^{2}+3)^{3}}=-\frac{14x(x^{2}-9)}{(x^{2}+3)^{3}}.$$
The zeros of this function occur at $x=0,\pm 3$. Evaluating the second derivative at the critical points yields that the smaller critical point is a local maximum (the second derivative is negative there), and the larger critical point is a local minimum (the second derivative is positive there). The function is concave up from $-10$ to $-3$, concave down from $-3$ to $0$, concave up from $0$ to $3$, and concave down from $3$ to $10$. Evaluate the second derivative yourself in each of these regions to confirm these statements.

6. You can verify that the third derivative is
$$f'''(x)=\frac{42(x^{4}-18x^{2}+9)}{(x^{2}+3)^{4}}.$$
This is nonzero at $0$ and $\pm 3$. Hence, all three zeros of the second derivative are inflection points.

7. Plotting up the function, you get this picture.

One feature of this function of which I was not aware in advance are the inflection points at $\pm 3$. However, they are necessary in order to approach the slant asymptotes properly. Now, we were tasked with finding the global extrema. We have the critical points and the endpoints. If you evaluate the original function at all those points, you'll find that $-10$ is where the global min occurs, and $10$ is where the global max is.

Summary: we followed the procedure in the previous post, and we have successfully found all local and global extrema, points of inflection, regions of increase or decrease, and regions of concavity of the function. We found one slant asymptote. There is nothing further of interest about this function that we don't already know (for the purposes of the vast majority of applications).

Example: analyze the function
$$f(x)=\frac{e^{x}}{x^{2}-4}$$
on the entire real line.

1. The domain is $\mathbb{R}\setminus\{-2,2\}$. That is, the domain is all real numbers that are not $\pm 2$.

2. There are three asymptotes. There is a horizontal asymptote as $x\to-\infty$, because the numerator is getting small, and the denominator is getting big. It goes to zero very quickly as $x\to-\infty$. In addition, we have two vertical asymptotes at $x=\pm 2$, which are fairly straight-forward to find (just factor the denominator and set it equal to zero to find the vertical asymptotes; since the numerator is never zero, all such points are vertical asymptotes).

3. The first derivative is
$$f'(x)=\frac{(x^{2}-4)(e^{x})-(e^{x})(2x)}{(x^{2}-4)^{2}}=\frac{e^{x}(x^{2}-2x-4)}{(x^{2}-4)^{2}}.$$
Setting it equal to zero, we find that
$$x=\frac{2\pm\sqrt{4-4(-4)}}{2}=\frac{2\pm 2\sqrt{5}}{2}=1\pm\sqrt{5}.$$
So, the critical points are the vertical asymptotes at $\pm 2$, and the points $1\pm\sqrt{5}$. If we divide the real line up according to all these points, we find that we have
$$(-\infty,-2),\;(-2,1-\sqrt{5}),\;(1-\sqrt{5},2),\;(2,1+\sqrt{5}),\;(1+\sqrt{5},\infty).$$

4. We find, by plugging in numbers, that the function $f$ is increasing on $(-\infty,-2),\;(-2,1-\sqrt{5}),$ and $(1+\sqrt{5},\infty)$. It's decreasing on the other two intervals: $(1-\sqrt{5},2)$ and $(2,1+\sqrt{5})$.

5. We find that the second derivative is
$$f''(x)=\frac{e^{x}(x^{4}-4x^{3}-2x^{2}+16x+24)}{(x^{2}-4)^{3}}.$$
Great. How do we deal with finding the roots of a quartic? There is technically a formula for the roots - it's quite complicated. Let's see if we can be a little clever with this one, though. I claim there are no real roots of this quartic (meaning $g(x)=x^{4}-4x^{3}-2x^{2}+16x+24$). Proving this is tricky. I propose dividing up the real line into three regions: $(-\infty,0],\;(0,3],\;(3,\infty)$. I will outline the proof for you, and you can fill in the gaps. Setting $g(x)=0$ is tantamount to setting $x^{4}+16x+24=4x^{3}+2x^{2}$. Why did I choose that way of splitting it up? Because now everything has positive signs. Call the LHS "the quartic" and the RHS "the cubic". For the first region, $(-\infty,0]$, minimize the quartic. It turns out that the quartic is always greater than $4$, which is always greater than the cubic on that interval, hence there are no intersections. For the region $(3,\infty)$, introduce a shift function, $z=x-3$, or $x=z+3$. If you plug this substitution into both the quartic and the cubic, you will find that all the coefficients of the quartic are greater than all the corresponding coefficients of the cubic, and hence the quartic is always greater than the cubic for $z>0$, which corresponds to $x>3$.

The hardest region is $(0,3]$. I will show you two methods for showing the result in this region. One, you do a tangent line to the quartic at $x=0$. The quartic is concave up for all $x>0$, so it lies above its tangent line. You can show that the cubic is less than the tangent line up to $x=2$. Then piece together another straight line from where the tangent line leaves off at $x=2$, and make the second line go through the point $(3,140)$. You can show that the quartic is greater than both these lines on the desired interval, and that the two lines as pieced together are greater than the cubic on the desired interval.

Another method for the interval $(0,3]$ is to make the LHS look like $(x-1)^{4}$, and separate out the terms that do not show up in the actual quartic. That is, you have the following:
$$(x-1)^{4}=x^{4}-4x^{3}+6x^{2}-4x+1.$$
If you compare this to "the quartic", you find that you need
$$(x-1)^{4}=\underbrace{(x^{4}+16x+24)}_{\text{the quartic}}-\underbrace{(4x^{3}-6x^{2}+20x+23)}_{\text{helper function}}=x^{4}-4x^{3}+6x^{2}-4x+1.$$
So, we want to show that
$$x^{4}+16x+24>\underbrace{4x^{3}-6x^{2}+20x+23}_{\text{helper function}}>4x^{3}+2x^{2}$$
on $(0,3]$. The left-hand inequality is obvious, because $(x-1)^{4}\ge 0$. For the right-hand inequality, you get the comparison of a quadratic to zero, which is straight-forward in the region $(0,3]$.

So there are no roots of the second derivative. This makes the rest of step 5 and all of 6 unnecessary.

7. Here's the plot:

You can see the local max and min that we found, along with the asymptotes, regions of increase and decrease, and regions of upward and downward concavity.

6.2 Related Rates

Some students struggle with related rates, but they're not actually that bad. The idea of related rates is that one quantity is changing, and you want to find out how fast another, related quantity is changing at the same time. The general procedure is as follows:

1. Find the relevant equation that relates the two quantities of interest.

2. Differentiate the relevant equation with respect to time, treating both quantities of interest as functions of time, and using the chain rule and implicit differentiation as necessary.

3. Solve for the target variable; if necessary, plug in values as needed to get numbers.

Great. We've got a procedure. Let's see it in action.

No calculus book or text is complete without this problem, but I'm going to go a bit further with it than is typical. It'll need a little physics, but bear with me.

A 5 m ladder is leaning against a vertical wall. The foot of the ladder is being pulled away from the wall horizontally at 1 m/s. When the foot of the ladder is 3 m away from the wall, how fast is the tip of the ladder (the end that's resting on the wall) moving?

1. We need a formula for relating the position of the foot of the ladder, which we'll call $x$, to the position of the other tip of the ladder that's resting on the wall, which we'll call $y$. The keywords "vertically" and "horizontally" tell us that we have a right triangle, and when we have a right triangle, the Pythagorean Theorem applies. In fact, it is true that $x^{2}+y^{2}=5^{2}=25$. So that's it, then: $x^{2}+y^{2}=25$.

2. Before differentiating, we write each variable as a function of time, so that we actually have $x^{2}(t)+y^{2}(t)=25$. We use the chain rule here and implicit differentiation to discover that
$$2x(t)\,\dot{x}(t)+2y(t)\,\dot{y}(t)=0,\quad\text{or}\quad x(t)\,\dot{x}(t)+y(t)\,\dot{y}(t)=0.$$
Now the question is asking how fast the tip of the ladder resting against the wall is moving. That is $\dot{y}$. Solving for that quantity (and dropping the $t$-dependence from our notation), we have that
$$\dot{y}=-\frac{x\dot{x}}{y}.$$

3. All we need to do now is plug in values to get that
$$\dot{y}=-\frac{(3)(1)}{(4)}=\frac{3}{4}\,\text{m/s}.$$
One question you might have: how did I get the $4$ in the denominator? Well, don't forget that the Pythagorean Theorem still holds, and $y^{2}=25-x^{2}$. If you plug in $x=3$, you get that $y=4$ (you can throw out the negative solution $y=-4$ on physical grounds).

One further note: what happens to this ladder when the tip of the ladder approaches the ground? That is, what happens to $\dot{y}$ when $y\to 0$?
Well, we have that $\dot{x}=1$, as always. It's also true that $x\to 5$, the length of the ladder. Thus,
$$\lim_{y\to 0}\dot{y}=-\lim_{y\to 0}\frac{x\dot{x}}{y}=-\infty.$$
Real ladders don't achieve an infinite negative velocity when they slide down a wall like this, do they? Evidently, something in our model is incorrect. It turns out that what is incorrect is the assumption that the tip of the ladder resting on the wall rests on the wall the whole way down. It has to separate from the wall.​ The big question is, at what position does the ladder have to separate from the wall? Kudos to anyone who PM's me the answer to this question. I might even add your solution to this post!

6.2.2 Balloon Problem

A spherical balloon's volume is increasing constantly (due to being filled with helium) at a rate of 10 cc/min (that's cubic centimeters per minute). When the balloon has a volume of 10 cc's, how fast is its radius changing?

1. The relevant formula here is that
$$V=\frac{4\pi}{3}\,r^{3}.$$
Here $V$ is volume in cc's, and $r$ is the radius in centimeters. Proving this formula, incidentally, is something we will do in the Integral Calculus Tutorial!

2. Differentiating yields
$$\dot{V}=4\pi r^{2}\dot{r}.$$

3. Solving for the target variable yields
$$\dot{r}=\frac{\dot{V}}{4\pi r^{2}},$$
but we can eliminate the $r$ on the RHS by plugging in our original volume formula. Why would we want to do that? Because we are not given an $r$-value in the original question. True, we can figure it out by solving the volume equation for $r$, but it's a little more elegant to do that once algebraically, and get your final target variable in terms of given quantities. That way, you don't have to have a multi-stage computation. In addition, if the problem asks for several different values, you can simply plug in. Finally, it's a little easier to check if you've done your work correctly. Bottom line: don't plug in values until the end, preferably in one step!

We have that
$$\frac{3V}{4\pi}=r^{3}\quad\implies\quad r=\left(\frac{3V}{4\pi}\right)^{1/3}.$$
Plugging this into our formula for $\dot{r}$ yields
$$\dot{r}=\frac{\dot{V}}{4\pi \left(\frac{3V}{4\pi}\right)^{2/3}}=\frac{\dot{V}}{(4\pi)^{1/3}(3V)^{2/3}}.$$
Now we get to plug in our values, to obtain
$$\dot{r}=\frac{(10)}{(4\pi)^{1/3}(30)^{2/3}}\approx 0.4455\;\text{cm/min}.$$

So there's your related rates procedure in action.

6.3 Position, Velocity, and Acceleration

This application is historically one of the most important, as a good deal of calculus was invented to solve these problems.
Kinematics is merely the description of motion, not an attempt to explain why motion happens the way it does. The latter is called dynamics. We'll touch on that a bit in this tutorial, but we'll do more with it in the Integral Calculus Tutorial, since we don't have the tools necessary completely to solve dynamics problems. The entire goal of classical mechanics is to find the position of a particle given the forces on it, an initial position, and an initial velocity. Once you know the position of a particle, you can find out any other quantity of interest (momentum, velocity, energy, etc.) whatever.

6.3.1 One-Dimensional Motion

Let's start with one-dimensional motion: motion on a line. We'll call the particle's position $x$, relative to some origin where $x=0$. If the particle is moving relative to that origin, then it has a velocity, which we'll call $v$. What is the relationship between position and velocity? Well, velocity is the rate of change of position, and is hence its derivative with respect to time. That is, $v=\dot{x}$. Note that this is a definition, not a theorem or something you prove. We are defining velocity this way. We also define the acceleration $a$ to be the rate of change of velocity. That is, $a=\dot{v}$. If you plug in what $v$ is, you find that acceleration is then the second derivative of position, so that $a=\ddot{x}$. Finally, we define the jerk $j$ to be the rate of change of acceleration, so that $j=\dot{a}=\ddot{v}=\dddot{x}$.

The speed of a particle we define to be the magnitude of the velocity. So the speed has no direction information in it, although the velocity does. So we have $s=|v|=|\dot{x}|$. The velocity can tell us which direction a particle is heading as well as its speed, but the speed alone just tells us how fast it's moving.

There are other quantities of interest. One is the kinetic energy, defined by $T:=\frac{1}{2}\,ms^{2}$. The $:=$ there means "defined to be". Another way of saying that would be $T\equiv\frac{1}{2}\,ms^{2}$. Here the $m$ is the mass of the particle having speed $s$ and kinetic energy $T$.

If the one-dimensional motion is parallel to a uniform gravitational field, as is approximately the case on the surface of the earth, then you can define the potential energy to be $U:=mgy$. Note that, due to symmetries, the origin of the $y$-axis might change the absolute value of the potential energy; but since it is only a difference in potential energy that is physically meaningful, the choice of origin doesn't matter.

Let's do some examples.

6.3.1.1 Velocity Example

Suppose the position of a particle is $x=4t^{2}-3t+2$. What is the particle's velocity as a function of time? Well, we differentiate to obtain
$$v=\dot{x}=8t-3.$$

6.3.1.2 Jerk Example

Suppose a particle's velocity is $v=e^{-t}\cos(t)$. What is the particle's jerk? We must differentiate twice to obtain the jerk. We get
\begin{align*}
a&=\dot{v}=-e^{-t}\cos(t)-e^{-t}\sin(t)=-e^{-t}(\cos(t)+\sin(t))\\
j&=\dot{a}=\ddot{v}=e^{-t}(\cos(t)+\sin(t))-e^{-t}(-\sin(t)+\cos(t))=2e^{-t}\sin(t).
\end{align*}

6.3.1.3 Kinetic Energy Example

Suppose a particle's position is $y=\sin(t)$, and its mass is $m$. What is its kinetic energy? We differentiate once to obtain
$$v=\dot{y}=\cos(t).$$
The speed is the magnitude of the velocity, so we get that $s=|\cos(t)|$. Finally, we formulate the kinetic energy as
$$T=\frac{1}{2}\,m|\cos(t)|^{2}.$$
However, since the squaring eliminates the need for the absolute value, we can just say that
$$T=\frac{1}{2}\,m\,\cos^{2}(t).$$

6.3.2 Two-Dimensional Motion

In higher dimensions than one, you must use vectors to describe motion. What is a vector? Think of it as an arrow: it has a magnitude (or length), and a direction in which it's pointing. The stereotypical (and historically important) example of two-dimensional motion is projectile motion, such as a cannonball. Such motion has a horizontal component and a vertical component. So it makes sense that you'd need two independent numbers to record both pieces of information. The idea of a vector is that it keeps track of those kinds of independent variables in a convenient form. Synonyms for "vector" include list, ordered list, one-dimensional matrix, and perhaps a few others. You can write them in different ways. One notation is $(x,y)$, but I dislike that notation since it looks like interval notation. Typically, the context would make it clear whether you mean a vector or an interval. But why not do better? Another common notation is $\langle x,y\rangle$. This is better, but it still has a problem with being identical to the inner product notation in linear algebra and quantum mechanics. The best, most unambiguous notation for vectors of which I am aware is the following:
$$\mathbf{r}=\begin{bmatrix}x\\y\end{bmatrix}.$$
I will attempt to use this notation the most. Now I've just defined a vector has having a magnitude and a direction. What is the magnitude and direction for $\mathbf{r}$? Well, you just use the Pythagorean Theorem for the magnitude. That is, the magnitude of what I just wrote down is $|\mathbf{r}|=\sqrt{x^{2}+y^{2}}=\sqrt{\mathbf{r} \cdot \mathbf{r}}$. The direction you could think of as the angle this vector makes with the positive $x$-axis. And that we can get by writing $\text{atan2}(y,x)$, where the $\text{atan2}$ function is defined as follows:
$$\text{atan2}(y,x)=\begin{cases} \tan^{-1}(y/x)\qquad &x>0\\ \tan^{-1}(y/x)+\pi\qquad &y\ge 0,\; x<0\\ \tan^{-1}(y/x)-\pi\qquad &y<0,\; x<0\\ +\frac{\pi}{2}\qquad &y>0,\; x=0\\ -\frac{\pi}{2}\qquad &y<0,\; x=0\\ \text{undefined}\qquad &y=x=0. \end{cases}$$
Basically, the $\text{atan2}$ function is a more sophisticated version of the $\tan^{-1}$ function in that it gives you quadrant information in all four quadrants. That is, the range of the $\text{atan2}$ function is $(-\pi,\pi]$, instead of the normal $\tan^{-1}$ function, whose range is only $(-\pi/2,\pi/2)$.

The velocity, acceleration, jerk, speed, and kinetic and potential energies are all defined in an analogous fashion. The tricky part, though, is that differentiation is defined component-wise. That is, if you have the position vector
$$\mathbf{r}(t)= \begin{bmatrix}x(t)\\y(t) \end{bmatrix},$$
then
$$\mathbf{v}(t)=\dot{ \mathbf{r}}(t)= \begin{bmatrix} \dot{x}(t)\\ \dot{y}(t) \end{bmatrix}.$$
Also, the speed is defined as a dot product: $s=|\mathbf{v}|=\sqrt{\mathbf{v}\cdot\mathbf{v}}= \sqrt{\dot{x}^{2}+\dot{y}^{2}}$. The kinetic energy is defined as before. If the gravitational field is acting only in the $y$-direction (as is typical), then the potential energy definition does not change: it is independent of $x$.

Some examples:

6.3.2.1 Circular Motion Example

Given a particle's position vector as
$$\mathbf{r}(t)=\begin{bmatrix} \cos(t)\\ \sin(t) \end{bmatrix},$$
what is the particle's acceleration, kinetic, and potential energies, given that the particle's mass is $m$?

We differentiate twice to get
$$\mathbf{v}(t)=\dot{\mathbf{r}}(t)= \begin{bmatrix} -\sin(t)\\ \cos(t) \end{bmatrix}, \quad \text{and}$$
$$\mathbf{a}(t)=\dot{\mathbf{v}}(t)= \begin{bmatrix} -\cos(t)\\ -\sin(t) \end{bmatrix}.$$
The kinetic energy is
$$T=\frac{1}{2}\,m\,s^{2}=\frac{1}{2}\,m\,(\mathbf{v} \cdot \mathbf{v})=\frac{1}{2}\,m\,(\sin^{2}(t)+\cos^{2}(t))=\frac{m}{2}.$$
The potential energy is just
$$U=mgy=mg \sin(t).$$

6.3.2.1 Spiraling Inwards Example

Do the same as the first example, but with position vector
$$\mathbf{r}(t)=e^{-t} \begin{bmatrix} \cos(t)\\ \sin(t) \end{bmatrix}.$$

Differentiating yields
$$\mathbf{v}(t)=\dot{\mathbf{r}}(t)= \begin{bmatrix} -e^{-t} \cos(t) - e^{-t} \sin(t)\\ -e^{-t} \sin(t)+e^{-t} \cos(t) \end{bmatrix} = e^{-t} \begin{bmatrix} -\cos(t) - \sin(t)\\ -\sin(t)+ \cos(t) \end{bmatrix}, \quad \text{and}$$
$$\mathbf{a}(t)= \dot{ \mathbf{v} }(t)= \begin{bmatrix} e^{-t}( \cos(t)+ \sin(t))+e^{-t}( \sin(t)- \cos(t))\\ e^{-t}( \sin(t) - \cos(t)) + e^{-t} ( -\cos(t) - \sin(t))\end{bmatrix}=2\,e^{-t} \begin{bmatrix} \sin(t) \\ -\cos(t) \end{bmatrix}.$$
The kinetic energy is
$$T=\frac{m}{2}\,e^{-2t} \left[ (-\cos(t)-\sin(t))^{2}+(-\sin(t)+\cos(t))^{2} \right]$$
$$= \frac{m}{2}\,e^{-2t} \left[ \cos^{2}(t)+2 \sin(t) \cos(t) + \sin^{2}(t) + \sin^{2}(t) - 2 \sin(t) \cos(t) + \cos^{2}(t) \right] = m e^{-2t}.$$
The potential energy is
$$U=m g e^{-t} \sin(t).$$

6.3.2.3 Projectile Motion Example

A projectile of mass $m$ has a position given by
$$\mathbf{r}(t)= \begin{bmatrix} v_{0x}t\\ v_{0y}t - (g/2)t^{2} \end{bmatrix},$$
where $v_{0x}$ is the initial $x$ component of the velocity, and $v_{0y}$ is the initial $y$ component of the velocity. $g$ is the acceleration due to gravity, assumed to be constant. Find the vector sum of the forces on the particle.

Newton's Second Law states that the vector sum of the forces on the particle (vector sums are just component-wise sums) is equal to the mass of the particle times its acceleration. We know the mass. The acceleration we can find by differentiating to obtain:
$$\mathbf{v}(t)= \begin{bmatrix} v_{0x}\\ v_{0y} - gt \end{bmatrix},\; \text{and}$$
$$\mathbf{a}(t) = \begin{bmatrix} 0\\ - g \end{bmatrix}.$$
Hence, since $\sum \mathbf{F} = m\,\mathbf{a}$, where the $\sum$ indicates "vector sum", we have that
$$\sum \mathbf{F}=m \begin{bmatrix} 0\\ - g \end{bmatrix}.$$
That is, the only force on the particle is gravity acting in the negative $y$ direction, as expected.

6.4 Miscellaneous Real-World Examples

In the course of my work, I have come across one or two real-world examples of applications of differentiation. Here's one of them.

6.4.1 The Mathematics of a Leak Check

Suppose you have a gas, like hydrogen, contained in some system with a fixed volume, and you want to know if the system leaks. You can measure the temperature and pressure of the gas. How would you do it?

Well, the ideal gas law says that $PV=nRT$, where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles of gas, $R$ is the ideal gas constant, and $T$ is the temperature. The volume is constant, so if there is a leak, $n$ would be changing. However, suppose you do this leak check overnight? The temperature would change, and thus the pressure would almost certainly change. How could you be sure that the number of moles was changing? That is, how could you be sure that it was $n$ that was changing? First step: rearrange the ideal gas law equation:
$$\frac{P}{T}=\frac{nR}{V}.$$
Now the only thing on the RHS that might be changing is $n$. If there is no leak, the RHS should be constant. So, you measure the quantity
$$\frac{d(P/T)}{dt}=\frac{R}{V}\,\frac{dn}{dt}$$
to see if you have a leak.

Ah, but does it matter what units you measure in? Suppose you measured pressure in psig instead of psia? Or suppose you measured temperature in degrees Celsius instead of Kelvins? Would it matter? The answer is yes. The ideal gas law is only valid for absolute scales. Why is that? Suppose I measured pressure in psig $(P_{g})$ instead of psia $(P_{a})$. Well, we know that $P_{a}=P_{g}+P_{0}$, where $P_{0}$ is the atmospheric pressure. I claim that
$$\frac{d(P_{g}/T)}{dt}\not=\frac{R}{V}\,\frac{dn}{dt}.$$
Proof:
$$\frac{d(P_{g}/T)}{dt}=\frac{d}{dt} \left[ \frac{P_{a}-P_{0}}{T} \right]=\frac{T( \dot{P}_{a}-\dot{P}_{0} )-(P_{a}-P_{0}) \dot{T}}{T^{2}}.$$
We will assume the barometric pressure doesn't change much relative to the system pressure, so that $\dot{P}_{0}=0$. Then we get
$$\frac{d(P_{g}/T)}{dt} \;= \frac{T \dot{P}_{a}- \dot{T} P_{a}+ \dot{T}P_{0}}{T^{2}}$$
$$= \frac{T \dot{P}_{a}- \dot{T} P_{a}}{T^{2}}\;+ \frac{\dot{T}P_{0}}{T^{2}}$$
$$=\frac{d}{dt} \left[ \frac{P_{a}}{T} \right]\;+ \frac{\dot{T}P_{0}}{T^{2}}$$
$$\not= \frac{R}{V}\,\frac{dn}{dt}.$$

Kudos to the reader if you can show that using degrees Celsius doesn't work, either. So the moral of the story is that you must use absolute pressure and absolute temperature, just like the equation is meant to be used!

6.4.2 Temperature Dependence of a Wheatstone Bridge

A Wheatstone bridge is a resistor configuration often used in pressure transducers, force transducers, and other applications. It looks like a diamond, with voltage applied at two opposite corners. You measure the voltage between the two other corners. Here is a circuit diagram for the Wheatstone bridge:

Now the transfer function is defined as the ratio of the output voltage to the input voltage. For a Wheatstone bridge, using mesh current analysis, you can show that the transfer function is
$$H=\frac{V_{0}}{V_{1}}=\frac{R_{3}}{R_{2}+R_{3}}-\frac{R_{4}}{R_{1}+R_{4}}.$$
Now each of these resistor values depends on temperature in a linear (actually, affine) fashion thus:
\begin{align*}
R_{1}(T)&=\epsilon_{1}(T-T_{0})+R_{01}\\
R_{2}(T)&=\epsilon_{2}(T-T_{0})+R_{02}\\
R_{3}(T)&=\epsilon_{3}(T-T_{0})+R_{03}\\
R_{4}(T)&=\epsilon_{4}(T-T_{0})+R_{04}.
\end{align*}
The question is, how does the transfer function depend on temperature? You can show that the transfer function as a function of temperature is
$$H(T)=\frac{R_{03}+(T-T_{0})\epsilon_{3}}{R_{02}+R_{03}+(T-T_{0})(\epsilon_{2}+\epsilon_{3})} \quad -\frac{R_{04}+(T-T_{0})\epsilon_{4}}{R_{01}+R_{04}+(T-T_{0})(\epsilon_{1}+\epsilon_{4})}\quad .$$
Finally, you just take the derivative of this expression with respect to $T$ to obtain the rate of change of the transfer function (which governs how the pressure transducer or force transducer behaves) with respect to temperature:
$$\frac{dH}{dT}= \frac{R_{02}\epsilon_{3}-R_{03}\epsilon_{2}}{(R_{02}+R_{03}+(T-T_{0})(\epsilon_{2}+\epsilon_{3}))^{2}} \quad - \frac{R_{04}\epsilon_{1}-R_{01}\epsilon_{4}}{(R_{01}+R_{04}+(T-T_{0})(\epsilon_{1}+\epsilon_{4}))^{2}}\quad .$$

Knowing this equation allows you to find out how sensitive the measurement is relative to temperature changes. Does a small temperature change cause a big change in the measurement value?

This marks the end of the differential calculus tutorial. I hope it's been useful to you, even as it has been fun to write!

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