On this screen we’ll generalize from thinking about the average rate of change of an object’s position with respect to time to more general rates of change, like temperature with respect to time, or vertical position (y) with respect to horizontal position (x).
In the preceding Topic, we saw that if we are given an object’s position at a particular time t, and the rate at which its position is changing at that moment, then we can determine its position a short time dt later:
\[ \text{position at time $(t+dt)$} = \text{position at time }t + \overbrace{\text{ small change in position}}^{\text{(rate at time $t$)} * \, dt} \]
For Hannah’s position s along a straight road, we had
\[s(\text{9:00:00} + dt) = s(\text{9:00:00}) + \overbrace{\text{ (Hannah’s rate at 9:00:00)}}^{20 \text{ m/s}} * dt, \]
while for the ball’s height y after Calvin tosses it upward
\[ y(0.50 \text{ s} + dt) = y(0.50 \text{ s}) + \overbrace{\text{ (balls’s rate at 0.50 s)}}^{5.00 \text{ m/s}} * dt \]
In the scenarios on the preceding screen, each object’s position changed as time smoothly flowed from one moment to the next. That is, our independent variable was time t, and we considered position as a function of time t. We indicate this functional dependence by writing
\[s = s(t) \quad \text{to indicate that position $s$ is a function of time $t$.}\]
We read that statement as “s is a function of t,” which means that if we tell you a particular time t (like 9:00:00), you can tell us Hannah’s position s at that instant. And as time smoothly slides from one moment to the next, Hannah’s position smoothly changes accordingly, as a function of time.
Parentheses in mathematics unfortunately have many meanings. One of course is multiplication: for example, $3(6) = 3*6 = 18.$A different use of parentheses is to indicate a function’s input. You’re used to seeing, for example, $f(x) = 5x^2 + 3$ to indicate that for any inputted value of x, you square it, multiply by 5, and then add 3. We could write instead $f(\Box) = 5\Box^2 + 3,$ which is the same function except that we’re using $\Box$ to represent the input value.
We emphasize this because we know that for most students, initially seeing
\[s = s(t) \quad \text{or} \quad f = f(t)\]
just seems weird. After a while you’ll come to read the statements as intended: “s is a function of t” and “f is a function of t.” (And not “s equals s times t” or “f equals f times t”.)
Similarly, we can write
\[y = y(t) \quad \text{to indicate that position $y$ is a function of time $t$.}\]
We read that statement as “y is a function of t.” This construction means that if we give you a particular time t (like 0.50 s), then you can tell us Calvin’s ball’s position y at that instant. And as time progresses from one moment to the next, the ball’s position continuously varies accordingly – again as a function of time in this formulation.
Indeed, we also considered the rate at which Hannah’s position was changing, and the rate at which the position of Calvin’s tossed ball was changing. In both cases the rates were time-based: at the moment of interest Hannah was traveling forward at 20 meters per second, and the ball was traveling upward at 5.00 meters per second. We thus say in these scenarios that we are considering the rate with respect to time.
We can of course consider functions that depend on different independent variables. For instance, in a typical standard math problem you’re given a function f whose independent input-variable is x, in which case we have
\[f = f(x), \quad \text{to indicate that the function $f$ is a function of $x$.}\]
In this formulation, the function’s output varies as the input variable x takes on different values.
Similarly, we can write
\[y = y(x), \quad \text{to indicate that $y$ is a function of $x$.}\]
Let’s consider one such physical situation, in which vertical height y is a function of horizontal position x.
Syd bikes steadily up the hill shown to the right. Let’s use the red dot on the center of his front tire as the indicator of his current position. We’ll measure Syd’s (the dot’s) horizontal and vertical position in meters, again abbreviated “m.”
We are interested in Syd’s vertical position y as a function of his horizontal position x: $y = y(x).$
Check Question:
In the preceding topic, we used ds to refer to a small change in Hannah’s position s, and dt to refer to a small change in time t. Now generalizing: in Calculus, we use $d\Box$ to refer to a small change in the quantity $\Box,$ whatever it is.
Time does not enter explicitly into the way we’re thinking about Syd’s ride; instead we’re focused on how much his vertical position changes as his horizontal position increases as he smoothly bikes along.
Hence when we think about the rate at which his vertical position changes, we don’t want to think in terms of how quickly his vertical position changes as time increases; instead, we must think about it in terms of how much his vertical position changes as his horizontal position increases.
We thus need to extend the way we conceive of “rate” so we can correlate small changes in his vertical position with the tiny changes in his horizontal position:
\[dy = \text{rate}*dx\]
Viewed differently, we can rearrange the preceding equation to see the rate-of-change in Syd’s scenario is the quantity $\dfrac{dy}{dx}:$
\[\text{rate } = \frac{dy}{dx}\]
The quantity $\dfrac{dy}{dx}$ is pronounced “dee-wye dee-ex,” rather than, for instance “dee-wye over dee-ex” or “dee-wye divided by dee-ex.” Instead, it’s just “dee-wye dee-ex.” And you can see that we’re now considering the rate with respect to (changes in) his horizontal position x: the $dx$ in the denominator reminds us that our independent variable is x, and we’re looking at what happens when we change x by a small amount.
As we saw earlier, for time-based rates we must specify the particular moment at which a given rate applies. For instance, when providing details about Hannah’s trip we stated that her rate of travel at 9:00:00 was $\left.\dfrac{ds}{dt}\right|_\text{at 9:00:00}=20 \, \tfrac{\text{meters}}{\text{second}}.$ At any other moment in time, her rate of travel could be (and most likely was) different. Similarly, Calvin’s tossed-ball was traveling upward at $t = 0.50$ s at the rate $\left.\dfrac{dy}{dt}\right|_\text{at 0.50 s} = 5.00\, \tfrac{\text{meters}}{\text{second}}$. A brief time later, at $t = 0.80$ s, it is traveling at a slower rate. The key point is that we need to know the relevant rate at the exact time of interest.
Considering now rates with respect to horizontal-position x, we must specify the position x at which the rate applies. We use the same vertical bar to do so, like $\left.\dfrac{dy}{dx}\right|_\text{at $x = 3.00$ m} = 6.00 \, \tfrac{\text{vertical m}}{\text{horizontal m}}$.
With this idea in mind, let’s continue gaining familiarity with the scenario of Syd’s ride.
As Syd passes the point on the path with $x = 3.00$ meters and $y = 9.00$ meters, due to the path’s shape he is climbing at the rate
\[\left.\dfrac{dy}{dx}\right|_\text{at x = 3.00 m} = 6.00 \, \tfrac{\text{vertical m}}{\text{horizontal m}}\]
Equivalently, we could state this rate in centimeters instead of meters:
\[\left.\dfrac{dy}{dx}\right|_\text{at x = 3.00 m} = 6.00 \, \tfrac{\text{vertical cm}}{\text{horizontal cm}}\]
This latter expression is a better way to think about the rate since we’re going to focus on small changes in horizontal position dx. So for instance, at this location for every 1.0 cm of horizontal-travel, his vertical-position changes by 6.0 cm.
Check Question
The trail slope gradually increases as Syd bikes along, and so it is steeper at $x = 5.00$ m than at $x = 3.00$ m. Specifically, at $x = 5.00$ m, the trail rises at the rate $\left.\dfrac{dy}{dx}\right|_\text{at x = 5.00 m} = 10 \, \tfrac{\text{vertical cm}}{\text{horizontal cm}},$ so
\[ \left.\dfrac{dy}{dx}\right|_\text{at x = 5.00 m} \gt \left.\dfrac{dy}{dx}\right|_\text{at x = 3.00 m} \]
Thinking about what dy and dx mean, why is it harder for Syd to bike along at $x = 5.00$ m than at $x = 3.00$ m? (Consider, for example, what happens when Syd bikes forward 1.0 cm in each location.)
Solution.
To think through why it’s harder for Syd to bike at $x = 5.00$ m than at $x = 3.00$ m, let’s imagine him moving forward horizontally the small distance dx = 1.0 cm in each location. The following animation illustrates; click “Begin” to start.
Now that you’re familiar with the various quantities associated with Syd’s ride, let’s return to the Calculus calculations we introduced in the preceding Topic.
Syd bikes along a path with a gradually increasing slope. Let’s again use “m” to signify meters.
As Syd passes the point with x = 3.00 m, he is at a vertical position of y(3.00 m) = 9.0 m. And the bit of path he is on rises at the rate of
\begin{align*} \left.\dfrac{dy}{dx}\right|_\text{at x = 3.00 m} &= 6.00 \, \tfrac{\text{vertical m}}{\text{horizontal m}} \\[8px]
&= 6.00 \, \tfrac{\text{vertical cm}}{\text{horizontal cm}}
\end{align*}
Do you agree with Ryan’s calculations? If not, is Syd’s actual change in vertical position after moving forward horizontally 2.00 m greater or less than 12.00 vertical m? Explain.
Solution.
(a) The calculation here mimics that of our time-based calculations of the previous screen, but now with dx in place of dt since our independent variable is horizontal position x instead of time t.
Specifically, Syd’s change in vertical position when he moves forward the small horizontal distance $dx = 0.01$ m is:
\begin{align*}
\text{change in vert. position} &= \text{(rate at horiz. position $x$)}*\text{change in horiz. position} \\[8px]
dy_\text{[at $x=$ 3.00 m]} &= \left(\left.\dfrac{dy}{dx}\right|_\text{at x = 3.00 m} \right)*dx \\[8px]
&= \left( 6.00 \, \tfrac{\text{vert. m}}{\text{horiz. m}}\right)*(0.01 \text{ horiz. m}) \\[8px]
&= 0.06 \text{ vert. m} \quad \cmark
\end{align*}
(b) This calculation should also feel familiar:
\begin{align*}
\text{final vert. position} &= \text{initial vert. position} + \text{small change $dy$} \\[8px]
\overbrace{y(x+dx)}^\text{vert. position $y$, at horiz. position $x+dx$} &= \overbrace{y(x)}^{\text{vert. position $y$, at horiz. position }x} + \overbrace{\text{ small change }dy}^{\text{(rate at horiz position $x$)} * \, dx} \\[8px]
y(3.01 \, \text{horiz. meters}) &= y(3.00 \, \text{horiz. meters}) + \left( 6.00 \, \tfrac{\text{vert meters}}{\text{horiz meters}}\right) \cdot (0.01 \, \text{horiz. meters}) \\[8px]
&= 9.0 \, \text{vert. meters} + 0.06 \, \text{vert. meters} \\[8px]
&= 9.06 \, \text{vert. meters} \quad \cmark
\end{align*}
(c) Ryan would be correct if the path Syd is on had a constant upward slope – but it does not. Instead, the path gets steeper and steeper as Syd bikes along, and so with every small bit he bikes forward horizontally, he gains vertical distance at a faster and faster rate. (And hence Ryan is overconfident and wrong, and still has some Calculus to learn!)
Later, when he passes $x = 5.00$ m and moves forward $dx = 0.01$ m, he gains height $dy_\text{[at $x=$ 5.00 m]} = 0.10$ m.
That is, Syd undergoes a larger vertical gain for the same $dx = 0.01$ m horizontal forward motion because the path is steeper and steeper as he goes.
Since on every bit of path beyond $x = 3.00$ m Syd gains height at a rate greater than $6.00 \, \tfrac{\text{vert. m}}{\text{horiz. m}},$ his total change in vertical position is greater than Ryan’s naive calculation. $\quad \cmark$
If the calculations in Syd’s scenario seem easy to you, great! You’ve made the switch to thinking about “rate” as it applies to an independent quantity other than time, like “rate with respect to x,” which we’ll need as we explore many other scenarios. And if considering such a rate doesn’t feel natural easy to you yet, please just keep going. With continued exploration rates of all sorts will come to feel more familiar (as is so often true when you’re learning something new).
And believe it or not, the scenarios about Syd above just set up our ability to quickly estimate $(3.01)^2$ to within 0.001% . . . as we’ll see on the next screen.
Syd’s tiny change in vertical position dy when he travels the short horizontal distance dx from $x=3.00$ m is then given by the formulation
\begin{align*}
{\rm\small small\,change\,in\,vert\,position} &= {\rm\small (rate\,at\,horiz\,position}\,\small { x=3.00}\rm\small m) * \rm\small{(small\,change\,in\,horiz\,position)} \\[8px]
dy &= \left( \left.\frac{dy}{dx}\right|_\text{at $x=3.00$ m} \right) \cdot dx
\end{align*}
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