I’ve always had a great interest in math as a hobby, but was never very good at it, so I never completed any math course higher than Algebra II Trigonometry. My father has tried before to explain to me how Calculus works, but he moves too fast and doesn’t explain all of the terms he’s using like limit, sine, cosine, tangent, secant, etc. Yeah, I know I’m really way behind, but this is something I’d really like to learn. I’ve searched the Internet, but any lesson I can find assumes I know the basic functions and terminology. I’ll affirm right now that I am not in school right now and that this is not helping me with any sort of schoolwork. If it helps, here’s what I know already:
Calculus is the science of understanding and calculating change.
In Calculus, there are dependent variables and independent variables. As far as I remember, we’re interested in dependent variables. I belive a derivative is a change based on the dependent variable.
I can solve advanced algebraic problems including, but not limited to: Quadratic equations, complex mathematics, and two-dimensional graphing.
I have some training in intermediate geometry and can understand the measurements and principles therein.
Please don’t assume that I know any other definitions or principles. I don’t know much of logic or proofs. Explain this to me like I’ve never heard of Calculus before.
If it’s too much information for anyone to bother writing, I’ll just buy a book. I was just hoping to learn something here first.
Hold it. You’ve had Trigonometry, but sine, cosine, tangent, and secant are unfamiliar terms?
Of course, calculus is a book, but you already have some of the basics.
It’s just another way of finding out what we need to know, in an organized fashion–with short cuts.
The whole idea behind limits generally serves to justify those shortcuts. But, ultimately, you’re trying to find the area under the line y=x, between x=2 and x=4. You can do that geometrically, easily, but then the areas under y=x^2 or y=x^3 are harder. Calculus can give you those answers easily (even though they don’t really involve change)
Calculus is a bit much to explain from the ground up in a SD post. We can give you a brief summirization, but I don’t think you’d understand why Calc is so useful/important from such a brief overview. I would recommend checking this book out of your local library. You don’t need any math background to read it, the style of writting is amusing and it does a good job of giving one a good conceptual sense of what calculus is.
You need both. You use a function giving the relationship between the dependent and independent variables to determine the derivative. If you a function that gives the position as the distance from a point (dependent variable) given the time (independent variable), you can take the derivative; in this case, the derivative is a function that gives the velocity as a function of time.
This wouldn’t work as a formal definition—it’s possible to study all of Calculus from a theoretical standpoint without ever mentioning change—but it gives you a pretty good intuitive idea of what it’s about.
You’re talking about functions! When the value of one variable depends on another variable, we say the one variable is a function of the other variable. If you’ve taken algebra and trigonometry, you’ve studied functions, but in Calculus you really study the hell out of them; nary a day goes by when you’re not talking about functions. Given a function, Calculus enables you to determine things like [ul][li]Where does the graph go up? down? how steeply? (tells us about function increasing/decreasing)[/li][li]Any high points/low points on graph? If so, where? (tells us about maximum or minimum values & where they occur)[/li][li]Places where function is undefined, or holes or jumps in graph: what happens close to these places?[/li][li]What happens as we go off to the right or left (as x gets large or small)?[/li][li]The area enclosed by the graph of the function[/li][/ul]
The classic example, and one which motivated much of the development of Calculus, is motion. When you have a moving object, its position (height above the ground, or displacement from a starting point) is a function of time (how many seconds or whatever have elapsed).
The Calculus that you’d study if you took a Calculus class consists of “differential calculus” and “integral calculus.” Differential calculus is about something called “derivatives” of functions, which tell you how the dependent variable changes relative to the independent variable, and geometrically relate to the slope of the tangent line to a curve (the graph of the function). Integral calculus is about “integrals” of functions, which are in some sense the inverse of derivatives and which geometrically relate to the area underneath a curve.
Both of these concepts are defined in terms of “limits,” which are about what happens when the independent variable gets arbitrarily close to, but not equal to, some specific value. Historically, when Calculus was first developed, people thought about what happens when the variable changes by an infinitessimally small amount, but this didn’t really make total sense and led to problems over what an “infinitessimally small amount” could mean, so the concept of a limit was introduced to put things on a firmer theoretical footing.
Really, you aren’t really going to get it no matter what anyone here says unless you take a course or at least work through a textbook.
That said, the Calculus is basically two different things: differential calculus and integral calculus. My explanation will be as devoid of other technical terms as I can make it.
Differential calculus is basically all about slopes, just like the slope of a line from algebra. Given a function, draw its graph. Then take a point (x,y) on the graph. Draw a “tangent line” through that point – one which passes through the point “in the same direction as the graph” just like a tangent line to a circle touches the circle without crossing it. The slope of that line is the derivative of the original function at the point x.
This is connected with “rates of change” as follows: say two values are given by some functional relation, like the position of a car along the road is given as a function of time. If this function is linear (y = cx for some constant x), then as x increases an amount, y increases c times as much. Also, note that from the above definition c is the derivative of the function at every point. In this case the derivative is equal to the rate of change of y as compared to x. Now if the function is more complicated this is just generalized exactly. Most of setting up differential calculus is justifying the assertion that the derivative of a function at a point “is” the rate of change of the function at that point.
Integral calculus is all about areas. Yes, geometry class showed a bunch of ways to calculate area for various shapes, but it’s lost when it comes to curved edges (except for the circle, but I’ll get back to that in a moment). At a first pass, integral calculus takes a positive function and calculates the area between it, the x-axis, and a two lines parallel to the y-axis. It does this by slicing the region up into a bunch of thin strips by lines parallel to the y-axis so that the top edge of each is nearly constant, which means it’s pretty much just a rectangle. Then the area of each strip is calculated (as if it was a rectangle) and they’re all added up. The more strips there are the closer the calculation will be to the right answer. The big difficulty in integral calculus is taking this to the extreme: adding up an infinite number of infinitely-thin strips to get a real answer.
This actually is close to something that most people have seen, whether or not they remember. The standard justification to a geometry student that the area of a circle is \pi r[sup]2[/sup] is as follows: cut the circle into a bunch of equally-sized sectors and rearrange them into something that looks like a parallelogram. The thinner the sectors are cut the closer this gets to a rectangle with height equal to the radius of the circle. The circumference of the circle (always 2\pi r) is now the top and bottom sides of the rectangle, which must each have length \pi r. The area of the rectangle – and thus that of the original circle – is then \pi r[sup]2[/sup]. Here, notice the idea of slicing a shape up into infinitely many infinitely thin pieces and adding the areas up. This shows that really the ideas of integral calculus, if not the exact techniques, are the only way geometry has of handling curves.
The capstone of the calculus is the “fundamental theorem of calculus”, which relates the two haves rather elegantly.
Firstly: given a function, its integral involved picking two vertical lines. Letting the leftmost be fixed and the rightmost move the integral becomes a function of the position of the rightmost line. As such it has a derivative, which turns out to be the original function back again.
Secondly: given a function, its derivative depends on a point, and so is a function of that point. Graphing the function, fixing one vertical line, and placing another vertical line as above, and taking the integral gives back a new function depending on the position of the movable line. This is the original function back again plus a constant value.
Thus it’s fair to say that integration and differentiation are “opposites” of each other in some sense.
Ah, but it’s the specifics of figuring out just what this slope is that gives differential calculus its flavor.
Let’s say, Agent Foxtrot, that you want to find out just what the slope for a tangent line to a graph of a function is at a particular point, x. In other words, we would like to know in just which direction the graph is “going” at that point. There are a number of real-world reasons why we might wish to know such a thing, but we’ll stick with the pure math here. Let’s say the function in question is f(x) = x[sup]3[/sup]
Now, to figure out the equation for any line in two dimensions is quite simple: Take any two points, (x[sub]1[/sub], y[sub]1[/sub]) and (x[sub]2[/sub], y[sub]2[/sub]), and divide the change in vertical distance by the change in horizontal distance, or
Now, let’s apply that general concept to finding the slope between two points on the graph of our function, each of whose ordered pairs is (x, f(x)). We get
Taken a step further we realize that x[sub]2[/sub] is merely x[sub]1[/sub] plus some number we’ll call h. So we’ll just call x[sub]1[/sub]: x and x[sub]2[/sub]: x+h in the above equation.
(f(x+h) - f(x)) / (x+h - x)
or
(f(x+h) - f(x)) / h
So now we know how to find the slope of any two points on the graph of the function. So what? We want to know the slope at only one point.
Think of it this way: What if we found the slope between the point (x, f(x)) and the point (x+h, f(x+h)), for some positive number h. Then, we kept making h smaller and smaller, checking the slope along the way. The closer h got to zero, the closer our points would be on the graph, and the closer the slope between the two point would be to the tangent slope at x. We call this idea a limit:
lim[sub]h–>0[/sub] {(f(x+h) - f(x)) / h }
Which effectively means the result of the formula using the smallest positive value of h possible.
So our function is f(x) = x[sup]3[/sup], and, as an example, let’s take x = 2.
first we take a look at f(x+h) - f(x)) / h:
lim[sub]h–>0[/sub] {(f(x+h) - f(x)) / h } = lim[sub]h–>0[/sub]((2+h)[sup]3[/sup] - 2[sup]3[/sup]) / h
= lim[sub]h–>0[/sub](12h + 6h[sup]2[/sup] + h[sup]3[/sup]) / h
= lim[sub]h–>0[/sub]12 + 6h + h[sup]2[/sup]
Now, we find the value for this formula using the smallest value of h we can, preferably 0, which is feasible in this case.
= 12 + 6(0) + (0)[sup]2[/sup]
= 12
Which is the slope of the line representing f(x) = x[sup]3[/sup] at x = 2.
If we apply the same tricks to the original formula without subsituting a value for x, we get a general formula for the slope of f(x) = x[sup]3[/sup] for any x:
lim[sub]h–>0[/sub] {(f(x+h) - f(x)) / h } = lim[sub]h–>0[/sub]((x+h)[sup]3[/sup] - x[sup]3[/sup]) / h
slope of x[sup]3[/sup] at x = 3x[sup]2[/sup] for any x.
Makes sense: the graph of x[sup]3[/sup] rises sharply for negative values of x (starting leftwards and moving toward 0), then flattens out at the origin, and rises sharply again moving toward the right. In other words, the values for the slope are positive but falling for negative values of x, 0 when x is zero, and positive and rising for positive values of x. Sounds like a parabola to me.
Of course, this is an interesting result, because if we take the original exponent and make it a coefficient, then subtract one from the exponent, we get the same result. Which, if we care to check, holds true for all non-negative exponents (negative exponents too, for that matter). Which is much easier to remember than doing these mathematical gymnastics every time.
Many common equations have their own mnemonic formulae for figuring out the derivative function, and a good calculus text can list them for you. The text should also have a good section on limits, so that you can figure out how to handle situations such as h not cancelling out of a demoninator.
So the slope of the graph of a function can itself be a function. That’s differential calculus.
Integral calculus is doing the inverse. Given a function g(x), find f(x) such that:
lim[sub]h–>0[/sub] {(f(x+h) - f(x)) / h } = g(x)
There are, again, a number of real-world situations where such a result comes in mighty handy, and a good text should show you how to go about figuring it out.
I specifically avoided, as I said I would, the technicalities of how the Calculus achieves its goals. The OP asked to explain the Calculus, not to teach it.
Wan’t trying to dis your post, Mathochist, just trying to indulge my penchant for going a little bit into the math without losing the essence.
IMO, the barrier that a lot of people hit when approaching calculus is that the necessarily cumbersome notation often serves to mask the elegantly simple ideas contained therein.
OK Agent Foxtrot if you are still around maybe an example will help. Calculus is straightforward but is different than things you have done before so you must follow carefully and make sure you understand each step along the way.
Suppose the distance, x, of an object from some 0 starting point is equal to the sine of the angle t. t is time. Or:
x = sin(t)
and further suppose we want to know the velocity at time = 1 second.
We can approximate it by computing the distance at 1.1 sec and subtracting the distance at 0.9 sec then dividing the result by the elapsed time, 0.2 sec, between those two distances. I.e. distance traveled/elapsed time = velocity.
When we do that the resulting velocity to 7 decimal places is (we are assuming the distances and times are known exactly just to illustrate a method):
(x(1.1) - x(.9))/.2 = 0.5394023
This is a rather coarse difference in times and we can get a better approximation by doing the calculation over just 1/10th sec.
(x(1.05) - x(.95))/.1 = 0.540072
Still not good enough? OK do it over an elapsed time of 0.02 sec.
(x(1.01) - x(.99))/.02 = 0.5402933
Just to end the suspense, if we continue making the time interval shorter and shorter until it gets as close to 0 as we want, but doesn’t ever quite reach 0, the answer will come out to be 0.5403023 again to 7 decimal places.
Now let’s go back and see what we did. We computed the distance at some time, t. Then we added an increment of time and computed a new distance. Let’s call the increment of time Dt. So are two distances are:
x(t) and x(t + Dt)
we then subtracted x(t) from x(t + Dt) and divided the result by the time increment Dt.
Equation 1) (x(t + Dt) - x(t))/Dt.
Then we let the time increment, Dt, get shorter and shorter and get as close to zero as desired without ever quite reaching zero.
Now the limit of Equation 1) as Dt approaches zero is called the derivative of the function and it equals the rate of change of the function at any point on it. In this case we used a distance function of time and the rate of change of distance per unit of time is the velocity.
I got the exact answer by taking the derivative of sin(t) which equals cos(t) (take my word for it) and the cos(1) = .5403023
I am so confused. High school kids are expected to learn this stuff?
I read everyone’s post and wolf_meister’s tutorial, and I’m even more confused than before. The biggest issue is that, despite having it explained to me time and time again, I can’t quite grasp what it is I’m trying to solve. If I want to calculate the velocity in which an object falls over a given amount of time, can’t I just use simple arithmetic?
I read scotandrsn’s long post and I didn’t understand a word of it. (Oh, and Mathochist, he worked very hard to write a good long post to help me understand, so please give him a break.) I’m not sure what I’m trying to solve here plus the staggering equations are way too much for me to comprehend (where the hell did h come from?). I’m a pretty smart guy; I learn things fairly quickly and easily, but all this just seems way beyond my comprehension. Am I forever doomed to not understand this particular form of math?
Today I asked my good friend (who excelled in math in high school and went on to earn a Bachelor’s in Economics) exactly what sine/cosine/tangent/cotangent/secant/cosecant do, and even he couldn’t explain it. He said that those terms confused the hell out of him, too. I also went to the library today and checked out Malodorous’s book recommendation. I hope it helps. I really apologize to everyone who posted and tried to help, and I ask that you please not give up on me. Thank you.
The thing is, there isn’t really one problem that calculus does. For extremely simple problems, usually you can just use elementary arithmetic, like finding the speed of something given the time it took and distance travelled. But in that speed problem, suppose the speed isn’t constant, but that it’s accelerating. Now there’s another simple equation that you can use here (just add 0.5at^2), but that equation comes from calculus. For the most part, arithmetic can be used in the rare simple case, but for everything else calculus is necessary.
Calculus usually comes into play in physics (and was pretty much invented for solving physics problems). Say, for example, you want to figure out the total electric force acting on an area from a spherical ball of charge. The electric force is just a simple equation, F=q^2/r^2, but the problem here is that from the sphere of charge, the distance r isn’t constant. The “equator” of the sphere is closer than the “poles” of the sphere. You could calculate the charge from every little bit of latitude from the sphere and add them up, or you could use calculus and do the whole thing at once.
Another science calculus comes into play is statistics, in a more abstract form. The most typical distribution of probability is the Gaussian distribution, aka the “bell curve”, which you may have encountered if you took statistics in high school. If you’re familiar with terms like “mean” and “standard deviation”, then one standard deviation from the mean of a Gaussian comprises about 68% of the area of the graph (corresponding to a 68% chance probability). How did people come up with that? Calculus, of course. Things get more complicated when you want to figure out how likely a specific value is, given its distance from the mean.
From just a geometric sense, if you’ve ever wondered how people figured out stuff like the volume of a sphere is (4/3)pi*r^3, then that’s calculus at work. The functions sine and cosine aren’t exactly necessary to understand calculus itself, they just pop up from time to time as in any sort of geometrical math.
To learn it on your own is a pretty tough undertaking though, because you can do various problems and still have no idea what’s actually going on (like my freshman year of college).
You say you have had trigonometry but don’t know terms like sine, cosine, secant, etc. Go back and review your trigonometry.
Go back and re read my post. There is nothing in it but simple arithmetic. Take a look at the post. Draw pictures. Draw a sine wave. Draw a cosine wave. Do you see how the cos represents the slope of the sine? At time = 0 the sine wave has a value of zero but it is changing value at its maximum rate and 0 is the time when the cosine is a maximum amplitude.
Don’t worry about dependent and independent. And don’t get all worried about functions. A function is a rule of correspondence between two numbers. Take the function:
y = 2x.
If I tell you x is 5 you can then come back and tell me y is 10. If I tell you y is 9 you can come back and tell me x is 4.5. This is a single valued function, i.e. there is one and only one y for every x and vice versa. Some functions are multiple valued. For example:
y = x[sup]2[/sup]
if y is 9 then x can be either 3 or -3.
If you want to get anywhere in this as a hobby your have to take the time to go over and over and over until you understand each individual step in the process. The whole thing is pretty straightforward but it takes diligence and practice.
Beginning calculus is not something you should attempt to do on a message board.
You have to really, really understand the early concepts, and let it sink in. Calculus II is still the hardest math course I’ve taken, and that includes multivariable calculus, differential equations, linear algebra, number theory, proofs, and complex analysis.
So you shouldn’t feel bad if early calculus stuff is difficult, but if you really want to learn it on more than a superficial level you have to put in significant time and effort.
I’m just saying that it’s all the equations that being thrown around in lieu of actual ideas that are confusing you. You can’t see the forest for the trees.
As for whether you can use simple arithmetic or not, that’s also to be laid at the feet of using equations, since one must pick simple examples for a beginner. Yes, the simple examples can be done without calculus, but consider this one:
Calculate the area of the “curved triangle” bounded by the x-axis, the line x=1, and the parabola y=x[sup]2[/sup].
I’m not going to go into the full explanation of why the Calculus answer works the way it does because it would only confuse you further. This is just an example of a problem for which the new tools are pretty much necessary.
