2.10: Instantaneous Velocity

What can we say about velocity if the position graph is a curve?

For example, if \(p(t) = 15t - t^2\), we’d get a graph that looks like this.

What is the velocity at \(t = 5\)? Seriously, what is it? I want to know.

Well, a first stab might be that it is the slope, since that is what we said before. But generally slope is only applied to lines, not curves like this. How could we find slope for a curve?

Well, recall that slope is just how steeply something is increasing. It turns out, we can take a line that is increasing with the same steepness as the curve at a given point, and measure the slope of the line.

Here, the red line is the same steepness as the curve at \(t = 5\). This is called a tangent line. Note that often the tangent line only touches the curve once. Since it is a line, we can measure the slope, and this should represent the velocity at \(t = 5\). But since it touches one time, we don’t have two points to compute the slope. This may seem like a minor problem, but to find the exact slope takes one of the major insights in calculus: we need to develop a process to get closer and closer, and then use a limit to find the exact value. This might seem like a lot of work but it is worth it, as it demonstrates the power of calculus.

Let’s zoom in a bit on the graph:

Okay, now we can start to see some things. First, notice we know the \(y\)-value or function value at the point \(t = 5\). Why? Since the function \(p(t) = 15t - t^2\) is giving the position, we can plug in \(t = 5\). We see \(p(5) = 15(5) - (5)^2 = 50\).

But this doesn’t give us the slope. To find a slope, you need two points — then you can use the rise over run formula. But we only have one point. So instead let’s look at a line that does have two intersection points, but is not quite the line we want.

Let’s say this new green line hits the curve at \(t = 6\). A line like this green one that hits in two locations is called a secant line. What is the \(y\)-value at \(t = 6\)? Well, it’s \(p(6) = 15(6) - (6)^2 = 54\).

Now we can find the slope of the green line. It’s \({\color{green} \frac{54 - 50}{6-5} = \frac{4}{1} = 4}\), since that is what rise over run tells us. But again, it isn’t quite the line we want. Instead, we could choose a blue line that’s even closer.

Now the slope of the blue line is \({\color{blue} \frac{52.25 - 50}{5.5-5} = \frac{2.25}{0.5} = 4.5}\). Still not there. But we can get closer and closer and closer. Instead of repeating this calculation every time, let us use variables. So instead of the blue line crossing at \(t = 5.5\), let’s just say it crosses at \(t = 5 + h\), or \(h\) to the right of where the red line crosses.

What is the blue question mark? Well, we just plug \(t = 5 + h\) in to \(p(t) = 15t - t^2\). We see

\[\begin{align*} p(5 + h) & = 15(5 + h) - (5 + h)^2 \\ & = 75 + 15h - (25 + 10h + h^2) \\ & =75 + 15h - 25 - 10h - h^2 \\ & = 50 + 5h - h^2 \end{align*}\]

There you go.

Now the slope is rise over run, and we have

\[\begin{align*} {\color{blue} \text{slope} } & = {\color{blue} \frac{50 + 5h - h^2 - 50}{5 + h - 5} }\\ & = {\color{blue}\frac{5h - h^2}{h}} \end{align*}\]

Okay, what we are really interested in is when \(h\) gets really, really small. That’s when the blue line slope will equal the red line slope, and the red line slope is what we want. But we can’t plug in \(h = 0\), since that would give a division by zero explosion. So instead, we can use a limit!

\[\begin{align*} {\color{red}\text{slope}} & = \lim_{h \to 0} {\color{blue} \frac{5h - h^2}{h}} \\ & = \lim_{h \to 0} {\color{blue} \frac{h(5 - h)}{h}} \\ & = \lim_{h \to 0} {\color{blue} (5 - h)} \\ & = 5 - (0) \\ & = 5 \end{align*}\]

And there you have it. The slope of the red line is \({\color{red} 5}\)!

Now, suppose instead of \(t = 5\), we were interested in the instantaneous velocity at \(t = x\). And suppose instead of \(15t^2 - t^2\), we had some other function describing position, which we call \(f(t)\). What would the picture look like then? Well, it would look very similar to the last picture.

And this is the picture that gives us the definition of the derivative. What is the blue slope? It’s

\({\color{blue} \text{Blue Slope} = \frac{f(x + h) - f(x)}{(x + h) - x} = \frac{f(x + h) - f(x)}{h}} \)

So how do we find the red slope? We just take the limit as the blue line approaches the red — that is, see what happens as \(h\) goes to zero.

\({\color{red} \text{Definition of Derivative:}} \)
\({\color{red} f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}}\)

Here is the notation for the derivative: \(f'(x)\), \(\frac{df}{dx}\), or \(\frac{d}{dx} f(x)\).