Suppose you are downloading a large file to your computer. At $t=0$ you clicked “Save as” in your browser and started the download. Let $f(t)$ represent the size of the downloaded data. The number $f(t)$ is what you would see if, at time $t$, you clicked on the partially-downloaded file and checked how much space it takes on your disk. If you are downloading a $700$[Mb] file, then the download progress bar at time $t$ will correspond to the fraction $\frac{f(t)}{700 \text{[Mb]}}$.

The derivative function $f'(t)$ is a description of how the function $f(t)$ changes over time. In our example $f'(t)$ is the download speed. Indeed, if you are downloading at 100 [kb/s], then the function $f(t)$ must increase by 100[kb] each second. If you maintain this download speed the file size will grow uniformly: $f(0)=0$[kb], $f(1)=100$[kb], $f(2)=200$[kb], $\ldots$, f(100)=10[Mb]. The “estimated time remaining” is calculated by dividing the size of the part remaining to be downloaded by the current download speed: \[ \text{time remaining } = \frac{ 700 - f(t) }{ f'(t) } [s]. \] The bigger the derivative is, the faster the download will finish.

In many physics problems, you will be given the formula of some function $f(t)$ and asked questions about the slope of $f(t)$. If the slope is big for some value of $t$ this means that the “rise over run” of the function is big there. Rise over run works for lines since they have the same slope everywhere, but general functions have a different slope at different values of $t$.

OK, I am about to drop the “d” word on you soon, but first let me give some synonyms for the word “slope”. Instead of “slope of $f$”, we can say “rate of change of $f$”, and to be specific we should say the rate of change “with respect to $t$”. In other words when you do the “rise over run” you look at the change in $f$ divided by the change in $t$: \[ \text{slope}_f(t) = \frac{ \text{change in} \ f(t) }{ \text{change in}\ t } = \frac{ f(t+\Delta t) - f(t) }{ \Delta t }. \]

Now imagine that every time you had to talk about the “slope” of a function you had to write out $slope_f(t)$. Wouldn't this get annoying after a while? Let's try to come up with a better notation. Let's just write $f'$ instead of $\text{slope}_f$. How about that? But slope where? We need to explicitly show the value of $t$ for which we are talking about. Remember that only for lines is the slope the same everywhere. Curved functions have different slopes at different places in their graph. Our shorthand will be $f'(t) \equiv \text{slope}_f(t)$.

The mathematically precise way to say “the slope of $f$ at $t$” is “the derivative of $f$ at $t$”. So we have three synonyms now: slope, rate of change and derivative.

The entire course of Calculus I, half of your homework assignments and three quarters of your final exam will consist of questions where you are given some function $f(t)$ and you are supposed to calculate the derivative $f'(t)$. Calculating the derivative or finding the derivative or taking the derivative is the new skill that you will learn in Calculus. Another way of saying “taking the derivative of $f$” is to say you are differentiating $f$. This is why Calculus I is called “differential calculus”: because you take a lot of derivatives.

Why should we care about taking derivatives? It turns out that it is some pretty useful information to know the slope of a function. If $f(t)$ is the file size of some torrent file you are downloading, then $f'(t)$ is your download speed. If $x(t)$ is your position, then $x'(t)\equiv v(t)$ is your velocity. You pay your ISP for more derivative, and you get speeding tickets if you don't watch your derivative. If you understand derivatives you will understand all the concepts in physics much better too.

The derivative is very useful for finding the minima and maxima of the $f(t)$. A maximum of $f(t)$ occurs at $t=t_{max}$ if moving a little bit to the left or to the right makes $f$ decreases, i.e., we have: \[ f(t_{max} - 0.0001) \ \ \leq \ \ f(t_{max}) \ \ \geq \ \ f(t_{max} + 0.0001). \] This maximization skill, i.e., finding the maximum values of $f(t)$ is really useful in business. If $f(t)$ represents the financial gain of your company when it produces $t$ tons of widgets, and you want to maximize your gain, then you need to find the maximum value of $t$. Maybe producing too many widgets means you are flooding the market and the price drops so you make less money. Also if choose $t$ to be too small, then you are not maximizing the possibilities of the market.

Say that $f(t_{max})$ is a maximum of $f$. What does that tell you about the slope of $f$ at $t_{max}$? By definition this means that $f(t_{max} - 0.0001) \leq f(t_{max})$, and in fact we can subtract any small number from $t_{max}$ and the inequality would still hold. If it weren't the case, then $f(t_{max})$ wouldn't be a maximum. Going from $t=t_{max} - 0.0001$ to $t=t_{max}$ the function $f$ went up, i.e., it had positive slope. Then after $t_{max}$ we know that $f(t_{max}) \geq f(t_{max} + 0.0001)$ so the function went down, we can say that $f$ has negative slope after $t_{max}$. A maximum is therefore a place where the “sign” of the slope function changes from positive to negative. Exactly at the maximum, the slope of the function must be equal to zero: \[ t_{max} \text{ is a maximum of } f \quad \Rightarrow \quad f'(t_{max})= 0. \]

To see how finding the maximum can be a used in physics, we will now revisit the situation where the ball is thrown into the air with an initial velocity of $\vec{v}_i = 12\hat{y}$[m/s]. At what time $t$ will the ball reach its maximum height? What is the maximum height? The equations of for the ball in flight are as follows: \[ \begin{align*} y(t) &= \frac{1}{2}(-9.8)t^2 + 12 t + 0, \nl v(t) &= (-9.8)t + 12, \nl a(t) &= -9.8. \end{align*} \] Recall that the velocity $v(t)$ corresponds to the change in $y(t)$ at all times $v(t)=y'(t)$. Thus, if we are looking for the maximum value of $y(t)$ , we have to look for the time at which the velocity goes to zero $v(t)=0 = (-9.8)t + 12$. This gives us $t_{max}=1.22$[s] which happens at the top of the trajectory. Indeed at $t_{top}=1.22$[s] the ball has used up all its initial velocity, stops momentarily and then starts to fall back down. We can find $t_{top}$ from the fact that the ball momentarily stops at the top.

To find the maximum height, we plug the value $t_{max}$ which we found from $v(t)=0$ into the equation $y(t)$. The maximum height reached by the ball is $y(1.22)=\frac{1}{2}(-9.8)(1.22)^2 + 12 (1.22)= 7.3$[m].