In this section, we just want to remind ourselves of what this object—the derivative—is, how we should hold it in our minds as we move through the course, and then practice being flexible with this interpretation.
The equation of a line with slope \(m\) that passes the point \((0,b)\) (this is another way of saying that the \(y\)-intercept of the line is \(b\)) is:
This tangent line is very similar to the actual curve of the function \(f(x)\) near \(x=2\text{.}\) Another way of saying this is that while the slope of \(f(x)\) is not always the value you found for \(f'(2)\text{,}\) it is close to that for \(x\)-values near \(2\text{.}\)
We know that slope is \(\dfrac{\Delta y}{\Delta x}\) and we’re using the fact that \(\dfrac{dy}{dx} \approx \dfrac{\Delta y}{\Delta x}\) for small values of \(\Delta x\text{.}\)
Activity2.2.2.Interpreting the Derivative as a Rate of Change.
This is going to feel somewhat redundant, since we hopefully know that a slope is really just a rate of change. But we’ll be able to explore this a bit more and see how we can use a derivative to tell us information about some specific context.
Let’s say that we want to model the speed of a car as it races along a strip of the road. By the time we start measuring it (we’ll call this time 0), the position of the car (along the straight strip of road) is:
where \(t\) is time measured in seconds and \(s(t)\) is the position measured in feet. Let’s say that this function is only relevant on the domain \(0\leq t\leq 15\text{.}\) That is, we only model the position of the car for a 15-second window as it speeds past us.
Calculate \(s'(t)\text{,}\) the derivative of \(s(t)\text{,}\) using Definition 2.1.2 The Derivative Function. What information does this tell us about our vehicle?
Find \(v'(0)\text{.}\) Why does this make sense when we think about the difference between the average velocity on the time interval and the value of \(v(0)\) that we calculated?
What does it mean when we notice that \(v'(t)\) is constant? Explain this by interpreting it in terms of both the velocity of the vehicle as well as the position.
Look back at the definition of Derivative at a Point. The end of it is interesting: “provided that the limit exists.” We need to keep in mind that this is a limit, and so a derivative exists or fails to exist whenever that limit exists or fails to exist.
Find the coordinates of another point at some other \(x\)-value on both the graph of \(j(x)\) and \(j'(x)\text{.}\) Plot the point on each graph, and explain what the output of \(j'\) tells us at this point.
Use your graph of \(j'(x)\) to find the \(x\)-intercept of \(j'(x)\text{.}\) Locate the point on \(j(x)\) with this same \(x\)-value. How do we know, visually, that this point is the \(x\)-intercept of \(j'(x)\text{?}\)
Use your graph of \(j'(x)\) to find where \(j'(x)\) is positive. Pick two \(x\)-values where \(j'(x)\gt 0\text{.}\) What do you expect this to look like on the graph of \(j(x)\text{?}\) Find the matching points (at the same \(x\)-values) on the graph of \(j(x)\) to confirm.
Use your graph of \(j'(x)\) to find where \(j'(x)\) is negative. Pick two \(x\)-values where \(j'(x)\lt 0\text{.}\) What do you expect this to look like on the graph of \(j(x)\text{?}\) Find the matching points (at the same \(x\)-values) on the graph of \(j(x)\) to confirm.
Let’s wrap this up with one final pair of points. Let’s think about the point \((-3,5)\) on the graph of \(j(x)\) and the point \((-3,-6)\) on the graph of \(j'(x)\text{.}\) First, explain what the value of \(-6\) (the output of \(j'\) at \(x=-3\)) means about the point \((-3,5)\) on \(j(x)\text{.}\) Finally, why can we not use the value \(5\) (the output of \(j\) at \(x=-3\)) means about the point \((-3,-6)\) on \(j'(x)\text{?}\)
So far we’ve been using the “prime” notation to represent derivatives: the derivative of \(f(x)\) is \(f'(x)\text{.}\) We will continue to use this notation, but we’ll introduce a bunch of other ways of writing notation to represent the derivative. Each new notation will emphasize some aspect of the derivative that will serve to be useful, even though they all represent essentially the same thing.
We can find slopes on any curve, not just functions. This sometimes is also used as a way to simplify the notation, especially when we want to manipulate equations involving \(y'\text{.}\)
The derivative is a slope. It is \(\frac{\Delta y}{\Delta x}\) as \(\Delta x \to 0\text{,}\) and we use \(dx\) and \(dy\) (called differentials) to represent \(\Delta x\) and \(\Delta y\) as the limits as \(\Delta x \to 0\text{.}\) This notation also is useful to tell us what the rate of change is: what is changing (in this case \(y\)) and what is it changing based on (in this case \(x\))?
The derivative is an action that we do to some function. We can call it an operator although we won’t formally define that term in this text. We’ll look at this idea more in Section 3.1. We can specify what we are expecting the input variable to be, based on the differential \(dx\) in the denominator.
For each of the following functions, use the definition of the derivative at a point to find \(f'(a)\) and then find the equation of the line tangent to the graph of \(f(x)\) at \(x=a\text{.}\)
We start watching a car driving down a straight road towards a stop sign. We can tell that their position function is \(s(t) = -7.5t^2 + 81t\) with position measured in feet from their location when we began observing them and time measured in seconds.
The monetary value of an ancient mathematical artifact (the pencil that Isaac Barrow used to prove the Fundamental Theorem of Calculus in 1667) follows the function \(m(t) = \sqrt{1600t}+10\) where \(t\) is the number of years since 1667 and \(m(t)\) is the monetary value of the pencil in US Dollars.
