Section 7.2 Velocity
Early in this text, we motivated the definition of the derivative by studying the tangent problem. But, that is not the only historical context in which the derivative arises. For example, suppose we have a function, say \(p\text{,}\) that is giving us how far away we are from some location at a given time, say \(t\text{.}\) For sake of argument, suppose that at 1 o’clock we have \(p(1)=60\) where the \(60\) here is miles. In other words, at 1 o’clock we are 50 miles from where we started. Now, suppose we also know that \(p(2)=110\text{.}\) In other words, at 2 o’clock we are 120 miles from home. Under these circumstances, if someone asked, "How fast are you going?", what would you say? Well, that probably pops to mind quickly. Namely, 60 mph (i.e. miles per hour). How’d we get that? Really think about it. We found the distance travelled by looking at the 110 and the 50 to realize we’d gone 60 miles (i.e. we subtracted the \(p(1)=50\) from the \(p(2)=110\)) and that it had taken from 1 o’clock to 2 o’clock which is 1 hour (i.e. 2 o’clock - 1 o’clock).
\begin{equation*}
\frac{p(2)-p(1)}{2 - 1} = \frac{110 - 50}{1} = 60
\end{equation*}
Ok... but does that mean we were doing 60mph right at 1 o’clock? Hmmm... not necessarily. After all, we might have been moving a bit slower at 1 o’clock and then sped up a bit before 2 o’clock. So, this is more like an average. So, what if we knew also that \(p(1:30)=80\text{?}\) Well, that gives
\begin{equation*}
\frac{p(1:30)-p(1)}{1:30 - 1} = \frac{80 - 50}{1/2} = 60
\end{equation*}
So now are we convinced that we were doing 60mph at 1 o’clock? Still no. But, since only a half hour passed after 1 o’clock, this average feels more like it might represent how fast we were going right at 1 o’clock. So, let’s take even a shorter time span after 1 o’clock. Suppose we check how far we’ve gone after just 6 minutes (i.e. 1/10 of an hour). What if that gave \(p(1+1/10)=56\text{.}\) Now we have
\begin{equation*}
\frac{p(1+1/10)-p(1)}{1+1/10 - 1} = \frac{56 - 50}{1/10} = 60
\end{equation*}
Now that’s only been 6 minutes. So, there wasn’t a lot of time for the speed to change. So, the speed at 1 o’clock may have been pretty close to 60. But, it was *still* 6 minutes. So, suppose that \(h\) is a very teeeensy piece of an hour. Even less than the 6 minutes. Maybe \(h\) is like 1/1000000 of an hour. Now, suppose it turns out that
\begin{equation*}
\frac{p(1+h)-p(1)}{1+h - 1} = \frac{p(1+h)-p(1)}{h} = 60
\end{equation*}
Now what do we figure the speed was at 1 o’clock? Well, a millionth of an hour is a tiny fraction of a second so there wasn’t much time for the speed to change so it is pretty reasonable to expect that our speed at 1 o’clock was, in fact, 60mph. Notice that what we’ve been doing is computing
\begin{equation*}
\frac{p(1+h)-p(1)}{h}
\end{equation*}
for smaller and smaller values of \(h\text{.}\) In other words, it is apparent that the speed right at 1 o’clock is
\begin{equation*}
\displaystyle \lim_{h\to 0} \frac{p(1+h)-p(1)}{h}
\end{equation*}
Of course, this is just the derivative of \(p\) at \(t=1\text{.}\) This is generally called *instantaneous velocity* (which just means velocity at a specific time). For any function that gives how far away we are from a given position (i.e. a *displacement* function), we see that intantaneous velocity at any given time is \(p'(t)\text{.}\) So, our knowledge of differentiation now gives us the skills to compute velocities when we have a displacement function.
