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As I finished calc 1. I can use the product rule and chain rule and resolve integrals. But I feel like its too mechanical for my taste. I know the procedure and I execute on paper without really understanding or experiencing the "ahaa moment".

For example, when I was learning geometry in elementary school, the "ahaa moment" for me was when I had to move furniture in my room and needed to find areas of stuff. I'm trying to find the equivalent application of the derivative and integral as I learn calculus. Could someone demystify this?

Thank you in advance.

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There aren't many elementary applications of calculus (unlike geometry). On the other hand, calculus is the foundation of optimization problems. This is probably the main kind of problem you'll be concerned with if you're an economist or an engineer. – Deathkamp Drone Jun 26 '14 at 16:56
Learn about Mean Value Theorem, and/or optimization, and it'll become much more interesting :) – Sawarnik Jun 26 '14 at 20:30

10 Answers 10

A slope is a rate of change. If a car moves at 30 miles per hour, and you graph its distance traveled as a function of time, the slope of that graph is 30 miles per hour.

Isaac Newton was the first person ever to show that the laws of nature that govern objects we see every day here on Earth are the same as the laws of nature that govern things in the heavens: planets, moons, comets (including the planet Earth as a whole).

In order to do that he had to think about the rates at which they move as a function of their locations.

Rates of change also occur in the following way: Newton asked himself whether the gravitational pull of the earth on the moon or other orbiting object is the same as if the whole mass of the earth were concentrated at the center. (The bottom line, after much effort, turns out to be "yes".) So how do you find the sum of the infinitely many infinitely small quantities involved? The answer is that you ask how fast that sum changes as progressively more of the earth is taken into account. This is not a rate of physical motion as time passes, but is something else.

Unfortunately far too much of the conventional mathematics curriculum is designed to prepare students for later courses, in such a way that you find out the motivations only if you take certain later courses that most of the students will never take. It leaves students not understanding the whole thing. Honest education does not do that. A few professors here and there are working on calculus courses that are in that respect honest. At least two professors at Macalester College have been working on that, and at various other places that I can't name right now.

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Could you break down how I would graph the distance traveled as a function of time? Does that mean, time is on the x axis and on the y axis there is distance? So are you saying the slope is the speed at which I was traveling at time x? – TazMan Jun 26 '14 at 3:20
WOAHHH. That's sort of magical. I don't know how that's possible though. I mean we have the time and we have the distance, how can we get the rate at which we were traveling at some historical time x?? That does not seem possible. – TazMan Jun 26 '14 at 3:22
You don't just have the total time and the total distance; rather you have the distance traveled so far at every moment in some time interval. – Michael Hardy Jun 26 '14 at 3:31
I am not trying to sound critical of you personally, but I think it is a sad commentary on what "Calc 1" is like in your school that you managed to complete that course without learning that velocity is the derivative of position. I'm sure it's not your fault, but come on -- that is the most fundamental idea in Calculus; it literally is what motivates the entire subject. – mweiss Jun 26 '14 at 18:29
I studied business. They never taught physics. Basically they taught how to find derivatives. with respect to one variable, then it went on to partial derivatives, then rieman sum and integrals. And some garbage at the end on differential equations that I did not understand much. Actually this was wayy to mechanical of a course. I wish more time was spent explaining the Why's rather than the Hows. – TazMan Jun 27 '14 at 21:41

The derivative might just be the most important thing in mathematics. In physics, if the function that gives the motion of a particle with respect to time is defined as $x(t)$, then the derivative of this function $$x'(t)=v(t)$$ gives the velocity of the particle at time $t$. Also, the derivative of the function $v(t)$, or $$v'(t)=a(t)$$ gives the acceleration of the particle. Even Newton's famous law $F=ma$ can be rewritten as the derivative of the momentum with respect to time, or $$\Sigma{F}=\frac{dp}{dt}.$$ Also, the power exerted by a force is equal to the derivative of the work with respect to time $$P=\frac{dW}{dt}.$$ In fluid mechanics, the definition of pressure is given as the derivative of the force with respect to the area that the force is acting upon, or $$p=\frac{dF}{dA}.$$ In physics the list goes on and on, there would be no physics with calculus. The derivative can also be applied inside of mathematics, given the gradient of any curve, which is useful, and the second derivative giving information about concavity and points of inflection.

Basically, it is really important, and if you go into the sciences or engineering as well as studying mathematics, you will notice the derivative, and all other aspects of calculus.

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ehhh, I kinda know or read the stuff you mentioned.. Its super important in the sciences. But I need an intuitive understanding in real layman terms. What you explained about physics and acceleration/pressure is still wayyy to abstract for me. What about an everyday example that I can relate to? – TazMan Jun 26 '14 at 3:11
Fair enough. Basically, calculus is used when studying quantities that are changing. The rate at which something changes, which is the derivative, is very important in the sciences. It is an abstract concept at first, but the more you study it the more you will understand it, and be amazed by its power. – M. Knight Jun 26 '14 at 3:13
Your answer starts by claiming that "The derivative might just be the most important thing in mathematics" but what you actually explain is why the derivative might just be the most important thing in physics. – David Richerby Jun 26 '14 at 9:24
What is pedantic about that? Mathematics and physics are kind of different things. – demonkoryu Jun 26 '14 at 12:36
@MitchKnight I don't think I was being pedantic. Mathematics is a subject in its own right, not just a tool for physicists. The technique that is most important for physicists may or may not be the most important part of mathematics as a whole (if such a thing even exists). – David Richerby Jun 26 '14 at 13:05

There are two other ways to think of the derivative:

  1. as the "instantaneous rate of change" of the function -- that is, how much it is changing at one particular moment were it viewed as a function of time. Namely, the rate of change of a quantity over an interval is $$\mathrm{rate\ of\ change} = \frac{\mathrm{change\ in\ quantity}}{\mathrm{interval\ over\ which\ change\ occurs}}$$ or, in symbols, $$R = \frac{\Delta f}{\Delta x}$$. Then the derivative is the limiting value of this rate when $\Delta x \rightarrow 0$, that is, in the "ideal" case where the interval is a single instant of time. We denote this by $f'(x)$ or $\frac{df}{dx}$, the latter calling to mind the ratio above.

  2. as a "best linear approximation" of the function. That is, the derivative tells you what the linear function is which is "closest" to the function at a given point in that every other linear function will have a worse error for points near the point of differentiation. This is what the tangent line represents -- note that it too is really a function. If you've done a little linear algebra, you might have heard of a "linear transformation", which on the real numbers is a function of the form $g(x) = ax$. You can think of this as finding the linear transformation (the $a$-value, more precisely), translated to be relative to the point at which you differentiate, which best approximates the deforming action of the function on a small piece of the real number line about the point at which you take the derivative. Since the linear transformation shown can be thought of geometrically as a "stretch" or "scaling" (that's what it means when you multiply a real number by another real number -- you scale one number by another), you can also think of the derivative as telling you how much the function "stretches" or "scales" a small piece of the real number line about the point of differentiation. This is actually a good concept to keep in mind for when you get to more advanced forms of math, where generalizations of the derivative to more exotic kinds of spaces than the Real Numbers will appear, and it is this sense in which it is used.

Usually in physics, one is most often interested in interpretation 1), whereas in more advanced forms of math, one may be more interested in interpretation 2).

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Thank you. So when I take the derivative of a function, why am I reducing it? whats the reason for that? For example the derivative of, 3x^5 + 4x is 15x^4 +4. What does this derived 15x^4 + 4 give us? It gives us the a linear line, not a curve? why? – TazMan Jul 1 '14 at 17:47
@TazMan: The derivative gives you the linear approximation at the point by giving the number which, when it acts via multiplication, approximates the behavior of the function near that point. In particular, the translated function $f(x + x_0) - f(x_0)$, which is just $f$ recentered so the point of interest $(x_0, f(x_0))$ is at $(0, 0)$, behaves like the linear map $f'(x_0) x$ for small $x$ (which, in untranslated coordinates, means $x$ close to $x_0$). The linear approximation's variation from point to point need not be linear. – mike4ty4 Jul 2 '14 at 1:19
And of course, the graph of that linear approximation is just the tangent line, and that's the connection between the two. The tangent line is the best linear approximation and conversely. – mike4ty4 Jul 2 '14 at 1:22
So for your specific example, $15x_0^4 + 4$ is the value $f'(x_0)$ which multiplies $x$ to produce the linear function $(15x_0^4 + 4)x$ which approximates the behavior of $f$ close to the point $x_0$ (when appropriately translated). There is such a linear map associated with every point $x_0$, giving how the function behaves close to that point. – mike4ty4 Jul 2 '14 at 1:26
In translated coordinates, you can write $T_{x_0}(x) = (15x_0^4 + 4)(x - x_0) + (3x_0^5 + 4x_0) = f'(x_0)(x - x_0) + f(x_0)$, which is centered about $x_0$ and is the equation for the tangent line. – mike4ty4 Jul 2 '14 at 1:31

Taking your example of the furniture in the room we can think of the integral as the area under the curve. We divide the area up into small blocks and fit them in there like the furniture. Consider the integral $$ \int_0^1 x^2 dx = {1 \over 3} $$ we can think of this as the area beneath this graph. integral of 1/3 x squared evaluated from 0 to 1 If we were to grind up the unit square so we could fit small pieces of it into this area we could fit $1\over3$ of a piece.

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Imagine you would like to recreate some smooth curve you have gotten (perhaps a sine curve ), but you are only allowed to use small straight line components. This is actually a very common task; I would imagine computers do this all the time. The shorter line segments you use, the 'closer' you are to actually recreate the actual curve. You can see this by imagining your recreation of the sine curve with small straight lines will get less 'pointy' the shorter you make the straight line segments. Imagine you would use infinitely smaller and smaller straight line bits to rebuild this curve, then your recreation would be indistinguishable from the curve you were given.

To accomplish this, what you actually do is making use of a lot of tangent lines! (Remember that the tangent is always a straight line.) So you are actually using the derivative for this. So in this sense the derivative actually recreates the curve you are given. (And it's also a bit cool that you can build something as 'non-straight' as a sine curve with straight lines.)

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The differential is the slope or the rate of change.

On a roller-coaster:

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so is it safe to say that if I get the derivative of the derivative of the derivative of velocity I can reach to jerk? The equation keeps getting smaller, how does the graph change? – TazMan Jul 1 '14 at 18:24
@TazMan - Remember that derivative of sin is cos and vice-versa so not everything diminishes to zero like polynomials do. – OldCurmudgeon Jul 1 '14 at 19:36
What is this diminishing. Whats the big deal behind it? Are all derivatives the slope of a tangent line? The tangent line must be straight right? So the equation of the line can never be more complex than y = mx + b.. – TazMan Jul 2 '14 at 1:37
@TazMan - In the case of polynomials yes all derivatives diminish inevitably because d/dx x^n = x^(n-1) (more or less) but that is a feature of polynomials. And yes, the derivative is the slope of the tangent which is the rate of change but the world is not all polynomials. There are curves that do not decay to nothing as you differentiate - there are also some that behave in very unusual ways but the core of my answer is the insight that the differential is the rate of change which is exactly the equation for the slope at any point. – OldCurmudgeon Jul 2 '14 at 8:10

(in addition to OldCurmudgeon)

If a roller-coaster is too wild for you, consider a merry go round with constant rotation around $(0, 0)$.

You can notice by differentiating x(t) and y(t) twice that acceleration is always towards the rotation-axis and if you differentiate two more times, you'll see that the jounce-vector is a positive constant multiple of the position vector.

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First, don't be fooled by how derivatives are introduced in looking at slopes. This is an artifact of how functions are introduced by graphing them. A function is just a black box that takes in an input number and spits out an output number. I cannot tell you how many people I've run into who think the derivative is the slope of a tangent line or that the integral is the area under a curve. These are applications of derivatives and integrals, not fundamental definitions.

From a purely mathematical perspective, consider a Taylor series expansion: in a neighborhood around a value $a$, the behavior of a function $f(x)$ could be well approximated by looking at the derivative:

$$f(x) \approx f(a) + (x-a) f'(a) + O([x-a]^2)$$

Such an expansion is at the heart of, for example, the small angle approximation: you can approximate the sine of an angle by the value of the angle itself, which simplifies problems in many applications, including optics.

The integral gives us a way to generalize weighted sums on a discrete number of points or objects to a continuously weighted sum on a continuous interval. Consider, for instance, the notion of assigning probabilities to events and then performing a weighted sum to get the expected value. The integral allows us to assign a probability density to an interval and to integrate that density, weighted by the value function, in an analogous way to the discrete case.

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THis is really what I'm looking for. Could you explain why taking a derivative that results in "reducing" a function (as I understand it) results in a better approximation? is that what derivative is all about? getting an approximation? – TazMan Jul 2 '14 at 1:31
It's about looking at how sensitive the function is to changes in the input. That's why you can use the derivative to generate an estimate through Taylor series: if the change in the input is quite small, then everything else becomes insignificant compared to the derivative. - Finally, do note that how sensitive a function is to changes in its input may vary with different inputs. This emphasizes that the derivative itself can often be thought of as a function in its own right. – Muphrid Jul 2 '14 at 1:37

I think you'll enjoy this example. :)

Remember how to calculate the mean of a list of $n$ numbers?
Of course you do, it's almost an insult to ask: you just add them all up, and divide by $n$.

But why is this the "right" thing to do?! You probably don't know. Well, here's why.

Say you have this list of numbers: $1$, $2$, $7$, $10$

Let's say we want to find the "center" of all of these numbers. How could you go about doing this?
Well, obviously, anything outside the range $1$ to $10$ isn't the "center" by any reasonable definition.

But, then, what's the center? Well, let's call $\mu$ the "center". What properties should $\mu$ satisfy?
Ideally, it should be as close to all the numbers as possible, as a whole.

Let's see what happens if we take the distance of each number with $\mu$, square it, and add it up:

$$d_2(\mu) = |\mu - 1|^2 + |\mu - 2|^2 + |\mu - 7|^2 + |\mu - 10|^2$$

Since the absolute values don't affect the squaring operation, we can remove them:

$$d_2(\mu) = (\mu - 1)^2 + (\mu - 2)^2 + (\mu - 7)^2 + (\mu - 10)^2$$

What happens if we find the $\mu$ that minimizes $d_2$? Well, we find the root of its derivative $d_2'$:

$$ \begin{align*} d_2'(\mu) = 0 &= 2(\mu - 1) + 2(\mu - 2) + 2(\mu - 7) + 2(\mu - 10) \\ 0 &= \mu - 1 + \mu - 2 + \mu - 7 + \mu - 10 \\ 4\mu &= 1 + 2 + 7 + 10 \\ \implies\ \ \ \ \ \ \mu &= (1 + 2 + 7 + 10) \div 4 \end{align*} $$

Hey look, the mean $\mu$ is the minimizer of the total squared deviation from our list!

Now you may wonder: why did we use squared deviation? Why not just absolute deviation?

Well, let's try that instead:

$$ \begin{align*} d_1(m) &= |m - 1| + |m - 2| + |m - 7| + |m - 10| \end{align*} $$

Now take the derivative (thanks to @Ian's comment below for the suggestion on the compact notation). Let $\mathrm{sgn}(x)$ be the signum function (equivalently, $\mathrm{sgn}(x) = |x| / x$):

$$ \begin{align*} d_1'(m) &= \mathrm{sgn}(m - 1) + \mathrm{sgn}(m - 2) + \mathrm{sgn}(m - 7) + \mathrm{sgn}(m - 10) = 0 \end{align*} $$

Remember that $\mathrm{sgn}$ can be either $-1$, $0$, or $1$. Does does the $m$ that satisfies this sound familiar?
It should: any $m$ that satisfies this equation is is a median of this list!

(You may have been taught that the mean of $2$ and $7$ is "the median". In fact, that's arbitrary and there is no reason to choose $5$ as the median. In reality, if you had to choose a single number, it should be $\approx 4.85$ (I think that's 34/7), and I'll leave it as an exercise for you to figure out why.)

We see that the mean and median are very intimately related -- they are both measures of dispersion. And using the derivative, we can prove why their formulas are correct! (And if we didn't know how to calculate the mean, the derivative would tell us.)
Bonus points: if you replace the exponent with $0$ instead of $1$ or $2$, you will get back the mode as well!

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You didn't differentiate your absolute values in the $d'_1$ line. Doing so actually makes it clearer that you get a median: you get $\text{sign}(m-1) + \text{sign}(m-2) + \text{sign}(m-7) + \text{sign}(m-10)$. Any root of this has two of these numbers positive and two negative, i.e. is between 2 and 7. – Ian Jun 26 '14 at 16:47
@Ian: Ooh, yeah I totally missed that, sorry. I also totally forgot I could use $\mathrm{sgn}$ for the notation, that's a handy notation. Thanks, I'll update that. – Mehrdad Jun 26 '14 at 18:36
@Downvoter: Care to comment? – Mehrdad Jul 4 '14 at 1:48

An application, which if applied world wide on all roads, would make it impossible to go over the speed limit and go unnoticed.

If you place sensors with timers that correctly identify passing cars throughout every stretch of road spaced 1km apart and the worldwide speed limit is now 50 km/h = 50 km/3600 seconds, a car going at top legal speed (50km/h) would take 1/50th of an hour = 3600seconds/50 = 72 seconds to cross adjacent sensors. If a car takes less than 72 seconds to cross adjacent sensors, then the car MUST have gone over the speed limit at some point between the sensors. This conclusion comes from the fact that we have a mapping between time and distance, which allows us to make accurate (in my example not exact) deductions about the collected information.

If it still isn't clear enough, I will demonstrate how exactly the program coordinating the sensors would go:

var max_speed=50
Sensor A:Car[x] Passes at Time Stamp 1:15:15PM
Sensor B:Car[x] Passes at Time Stamp 1:16:10PM
//average* speed=difference in distance/difference in time
var Car[x].speed=[1/Date(1:16:10)-Date(1:15:15)][3600]=65.45/h
65km/h>50km/h = EMIT SPEEEDING TICKET FOR Car[x]

*average speed: you might actually go at various instantaneous speeds throughout continuous length of road (even if you put it on cruise control).

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