A group of scientists, a group of engineers,
and a group of mathematicians were all given the same applied calculus problem. It
sounds like the beginning of a joke, but it was actually part of a research study.
The Scientists and Engineers solved the problem quickly But the mathematicians struggled. Why?
Because mathematicians think about derivatives differently than scientists or engineers
and the way that mathematicians tend to think about derivatives isn't very
useful for many real world problems. Why does this matter? Because if you
have had a calculus class, you were probably taught to think about derivatives like a
mathematician, not like a scientist or engineer. And if you can't think about derivatives
like a scientist or engineer, then it will be harder for you to recognize and use
derivatives in real-world situations. So let's see what is so powerful
about the the way that scientists and engineers think about derivatives.
Near the end of the video, we will illustrate the power of this kind of thinking by
analyzing the fairness of an event in the world strongman competition, and how the derivative
gives competitors a strategy to improve their performance.
intro If you ask a student that recently took
a calculus class, what is a derivative? They will most likely say, the derivative is the
slope of a tangent line to a function or curve. This is the meaning that is pushed the
most in calculus, and it is one of the things that most students remember from
calculus class. But there are different ways to think about the meaning of a derivative.
Here are five meanings: 1) ratio of small changes, 2) the solution to one of these limits 3) the rate
of change of one quantity with respect to another, or 4) the new function after applying derivaitve
rules 5) Slope of a tangent line. Ideally, students leave calculus being able to think
about derivatives as all of these, since each of these could be helpful for different situations.
In a previous video, we did a little math to find out how fast my potato gun shoots potatoes, We
didn't need calculus to find its muzzle velocity, we just used some algebra and the meaning of
speed. Let's use the potato gun example to talk about each of these 5 meanings of derivative.
If you haven't had differential calculus or it has been a while, you may want to watch our
potato gun video first called "Math meets Mayhem" We introduce the idea of derivatives
in that video which will be helpful next. In that video we shot my potato gun straight
up in the air and timed the potato to see how long it was in the air. We then used this
physics equation and values from our timed shot to solve for our unknowns. We ended
up with this formula that gives the height of the potato for any time t from t=0 to t=10.
We wanted to know the speed of the potato when it left the barrel. We used meaning #1, the ratio
of small changes to calculate that speed because that is how chronographs work. chronographs
are used to measure velocities of bullets, and they do it by timing the bullet
as it passes through the two sensors, then takes a ratio of the distance between
the sensors and the difference in time. We did a similar thing, but used our mathematical
model to get values for the change in distance and change in time. We picked a small time difference,
and looked to see how far the potato traveled in that time interval using our model. It traveled
about 4.826 meters in the first 0.1 seconds, so our estimate of the muzzle velocity was
[4.26/.1=] 48.26 meters/second. [just say the value I'll put the math on screen]
We wanted a more accurate estimate so we took an even smaller interval, and as we
keep taking smaller and smaller intervals, it appears we get closer to the value of 48.75
meters per second as our time interval goes to zero. This lead us to meaning #2, finding the
limit of this expression as t goes to zero. If we start with a ratio of small changes,
then we can build the limit definitions, since the limits are ratios of small
changes with the denominator going to zero. So in essense the limit was doing
what we did in the last video as we calculated better approximations of the muzzle
velocity with smaller and smaller intervals. Now that we have a value of 48.75, we
can interpret it as the muzzle velocity, but we can also use meaning #3 and
interpret it as a converting tool that converts a small change in time, around
t=0, to a change in distance or height. The change in time from 0 to .01 seconds
can be converted to change in distance with (48.75m/sec)*(.01-0) seconds. 0.4875 meters.
The units can help us with the interpretation. If you have had calculus, you probably are very
good at meaning #4 and remember derivative rules that would allow us to calculate the derivative
at t=0 very quickly. We can use the power rule, plug in t=0, and get 48.75. The derivative rules
conception is good for calculations, but not so great for recognizing or interpreting derivatives.
In this example we have touched on all but one of the conceptions for derivatives,
#5, slope of a tangent line. It doesn't seem connected to our question at all.
The slope of the tangent line is a graphical interpretation of the derivative and the question
about how fast the potato is going doesn't seem to be connected to a graph. So it would feel
wierd to try and use it for this problem. And that is actually true of a lot of real-world
derivative situations. Real-world questions rarely connect on the surface to the slope interpretation
of the derivative, so if students only think about derivatives as slopes of tangent lines, then
they are not prepared to recognize or interpret derivatives in many real-world situations.
Thinking of a derivative as the slope of a tangent line wouldn't be so bad if we
emphasized the units for the slope. But that is rarely done in math class.
If students used their knowledge of calculus and algebra to find the slope of the
tangent line of our height function at t=0, they would get y=48.75*x+**. but when do
we ever talk about the units of that 48.75? And if students or teachers
write it in terms of x and y, it is even harder to see why 48.75 makes sense in
this situation, since the units are meters/second, but it isn't clear from here that y is a
height in meters and x is time in seconds. It would be better written as (delta y) meters=
48.75 m/sec * t secs+ ** meters, with all of the units for each value, so we can recognize that
the slope is actually a measure of velocity. So which of these conceptions about
derivatives are helpful in real-world problem solving? Well, let's see how the
engineers, scientists, and mathematicians tried to solve a real-world problem.
The problem the three groups of professors were asked to solve was to find a certain
derivative in a system of weights, strings, and springs. Adding or taking away the weights
hanging off the table would move the position of the strings, and the two strings had rulers
and an indicator to show the string position. The question the groups were to answer was to find
the derivaitive of x (the marker position of the string on the left) with respect to the force (the
weights hanging off the edge on the left side). Both the Scientists and engineers went about
collecting some data to see how a small change in force impacted the change in the marker on
the string. They did this by adding or taking away a small amount of weight, and calculating
the change in position of the string marker. For example, if adding 10 grams of
weight moved the x marker down by 2 mm, then they could estimate the derivative
as -2/10=-.2 mm/gram. Both of those groups were able to write down an estimate.
What did the mathematicians do? They spent a lot of time trying to write down
a formula for the positions and weights, so then they could apply the derivative
rules and calculate the derivative that way. However, they never could write down a
formula, and even though later they did gather some data to estimate the slope of
the function they couldn't complete, they never did find any estimate for the derivative.
This study illustrates the power of thinking about derivatives as a ratio of small changes
in real-world situations. But each of the 5 conceptions of a derivative has advantages.
For example, the limit definition [2] is used to develop differentiation rules
[number 4], and differentiation rules make it easier to calculate derivatives.
Thinking about derivatives as slopes of a tangent line can help in reasoning about derivatives
in graphical contexts or to justify techniques like why setting the derivative equal to zero
can help to find maximum or minimum values. But it is the reasoning about the ratio of
small changes that is used to develop the thinking of all the others. That makes it the
heart of derivation. It is also the conception that is usually most helpful in interpreting the
meaning of a derivative in a real-world context. I want to be clear, we are not saying that the
way mathematicians think about derivatives is wrong or bad, because each of the conceptions of a
derivative is important. Every mathematician has, and can, think about derivatives
as a ratio of small changes. What we are saying is that for many, it just
isn't their go-to, natural conception. Here's a really interesting example. I enjoy
watching the strongman competitions. But one of the things I wonder about is if a certain body
type has an advantage in a particular event. If you are tall, it might be easier to push
over these poles, because getting your hands higher improves leverage. But if you have
a short torso and long arms deadlifts might be easier as you don't have to reach as far.
In this event, called the pillars of hercules, competitors are timed to see how long they can
hold the heavy pillars before they let go. This event seemed suspect to me, because it seems to
favor athletes with short arms for their height, because the farther out the pillars
lean, the harder they are to hold up. Taking some measurements from
an image, I developed this model [ f(d)=3100(sin(arctan(d/132))) ]to tell me
the force the athlete feels based on how far out the pillars are leaning past vertical (measured
at the height of a typical athlete's arm height. so at 60 cm, or about 2 feet, out, the force
the athlete is working against can be found by plugging 60 into that function. We get an
output of 1283 newtons, or about 290 lbs. Although they try to adjust the chains so the
pillars lean out about the same, it is far from exact. So what happens if someone has longer arms
and it is leaning out more than another athlete? Well, we can use the derivative to help us
make sense of the situation. The derivative of our function at d=60 gives us a
value of 17.7. What does that mean? Well, let's first figure out the units for that
value to help us make sense of it. our input was a distance in cm, our output is a force in newtons.
Derivatives are ratios of small changes with the change in output divided by the change in input,
so in this case, we have a change in newtons divided by a change in cm. So Newtons-per-cm.
The value of the derivative is positive, so our value of 17.7 newton/cm means that
for every increase of 1 cm (on both sides), the force the athlete feels increases by 17.7
newtons. 17.7 newtons/cm is about 10 lbs per inch. So if an athlete has longer arms for their
height and it puts the pillars out 5 cm or 2 inches further out on each side, then they have
about 35.4 newtons or 20 pounds more force to fight against, Those are not trivial numbers.
A competitor can use this finding to help them by trying to keep the pillars pulled in, either
by bending their arms, or bending their knees, or both, to reduce the effective weight they feel.
Notice that in our pillar of hercules example, the interpretation of the derivative as a ratio
of small changes was very helpful in making sense of the meaning of our derivative value, 17.7.
Sure, we could have used the limit definitions or derivative rules to help us find the value,
but when it comes to interpreting and using the derivative, it was the ratio of small changes
that comes to the rescue more often than not.
