Placeholder Image

字幕表 動画を再生する

  • If you watched our last episode -- and really, if you haven't, you should.

  • ...you now know all about derivatives, and how to use them, to describe the way an equation is changing.

  • Which means that now we can talk about the other main part of calculus -- basically,

  • the inverse of derivatives, called integrals.

  • Integrals are useful because they also tell you a lot about an equation: if you plotted

  • an equation on a graph, the integral is equal to the area between the curve and the horizontal axis.

  • Finding an integral is a little less straightforward than finding a derivative, but, as with derivatives,

  • there are shortcuts we can use to make things a little easier.

  • We'll even be able to use integrals to talk about how things move -- specifically, the

  • equation we've been calling the displacement curve, and why it looks the way it does.

  • So let's get started.

  • [Theme Music]

  • Say you want to know how high your bedroom window is above the ground outside below.

  • But you don't have anything to measure it with --

  • just a ball, the stopwatch app on your phone, and your impressive knowledge of physics.

  • The force of gravity is what makes the ball fall,

  • so you know that its acceleration is small g 9.81 ms^2, downward.

  • But you're trying to find its change in position -- how FAR it falls.

  • We've spent a lot of time talking about the connection between the qualities of motion:

  • position, velocity, and acceleration.

  • But so far, we've been describing that connection in a particular direction: velocity is the derivative of position

  • -- that is, a measure of its change -- and acceleration is the derivative of velocity.

  • To figure out how far the ball falls, you need to use the reverse of that connection.

  • Expressed mathematically, that means that velocity is the integral of acceleration,

  • and position is the integral of velocity.

  • In other words: if you draw these equation on a graph, velocity is equal to the area

  • under the acceleration curve, and position is equal to the area under the velocity curve.

  • Finding that area is the tricky part.

  • There are simple ways to find the area of pretty much any shape, as long as that shape

  • is made of nothing but straight lines and corners.

  • And when you think about it, a curve is just a set of infinitely tiny lines.

  • So the area under a curve can be divided into a set of infinitely tiny rectangles.

  • Integrals tell you what happens when you divide the area under a curve into those infinitely

  • tiny rectangles, take the area of each of them, and add up those areas.

  • So, how do you find that integral?

  • Well, you start by using the fact that integrals are basically the OPPOSITE of derivatives.

  • If you know that your velocity is equal to twice time, for example, then you know that's the derivative of the position.

  • So, to find the equation for your position, you just have to look for an equation whose DERIVATIVE is 2t … like x = t^2, for example.

  • It's kind of a roundabout way of doing things, compared to the neat equation we were able to use to find derivatives.

  • But there is no tidy equation that we can use to calculate any integral we want.

  • But! As with derivatives, there ARE shortcuts for finding certain, useful ones.

  • For instance, you can take the power rule that we used for derivatives, and run it backward.

  • Basically: you add one to the exponent, then divide the variable by that number.

  • So the integral of 2t -- which is written like this -- becomes t^2.

  • In the same way, the integral of 42t^5 is 7t^6.

  • You can take the trigonometric derivatives that we talked about, and do those backward, too.

  • The integral of cos(x) is sin(x), and so on. And the integral of e^x is just e^x.

  • But there's one complication that we haven't talked about yet -- maybe you've already spotted it: constants.

  • A constant is basically just a number. It can LITERALLY be a number -- like 2, or a half, or negative 4.

  • Or it can be a placeholder for a number, like the small g, we've been using to represent the acceleration caused by gravity.

  • And constants pose a problem when it comes to integrals because: the derivative of a constant is just 0.

  • A derivative is a rate of change, after all, so a constant, which by definition DOESN'T

  • change, will always have a derivative of zero.

  • Which means that lots of different equations - an infinite number, in fact - can all have the same derivative.

  • Like, the derivative of t^2 is 2t. But you can add ANY number -- or a letter representing

  • a number -- to it, and the derivative will STILL be 2t. So the derivative of t^2 + 1

  • is also 2t. And the same is true for t^2 - 7.

  • Which means: If you're looking for the INTERGRAL of an equation like x = 2t, you have INFINITE

  • CHOICES, all of which are equally correct.

  • t^2 would work, but so would t^2 + 1 … or t^2 - 7 … or t^2 + 0.256.

  • In these cases, we might know what the SHAPE of the integral should look like on a graph

  • -- like, whether it's a straight line, or how it curves -- but we don't know where to put it along the vertical axis.

  • So we need to know what the constant is, in order to know where to start drawing its integral.

  • Whatever the constant is equal to, that's where the curve will intersect with the vertical axis.

  • So t^2 would intersect at 0, but t^2 - 7 would intersect at -7. You get the idea.

  • Mathematicians had to figure out how to get around the problem of having infinite integrals

  • to choose from, so they came up with a way to represent ALL of them: just add a C at the end of the integral.

  • That C stands for all of the constants that we know we COULD put there.

  • So if we say that the integral of 2t is t^2+ C,

  • then we're including t^2 + 1 and t^2 - 7 and all those other infinite options --

  • every equation whose derivative is 2t.

  • But sometimes you don't need the C at all, because you CAN figure out where your integral

  • is supposed to be on the y axis. Like if you have what's known as the initial value.

  • In the case of a position graph, for instance, the initial value would be where you started out,

  • so you'd know to draw the rest of the graph's shape from there.

  • If you started at the 2 meter mark, say, and moved one meter every second, you'd put

  • the graph here. But if you started at the 4 meter mark, you'd shift it up a little.

  • Basically, it gives you the point where your integral intersects the vertical axis -- which is the value of C.

  • Let's try it, and at the same time, we might as well figure out the height of your bedroom window.

  • You're standing in your room, holding a tennis ball out the window with your arm resting on the sill.

  • Now you drop the ball and start your stopwatch app at the same time.

  • Turns out that the ball takes 1.7 seconds to hit the ground.

  • Like we said earlier, we know the ball's acceleration -- 9.81 ms^2 -- and we know the time involved.

  • Somehow, we have to get from there to the equation for the ball's position.

  • So, first, let's find its velocity -- the middle step -- by taking the integral of its acceleration.

  • Take a look at this graph of the ball's acceleration over time.

  • It's just a flat line, which means that it should be pretty easy to find the area between it and the horizontal axis.

  • It's a rectangle! And the area of a rectangle is just its base times its height.

  • In this case, the base is t, the amount of time the ball took to fall.

  • And the height is a, the acceleration.

  • So, the area between the acceleration graph and the horizontal axis is just (a) times (t).

  • And the integral is (a) times (t), plus that constant we add -- C.

  • For now, we need the C, because we know the general shape of the velocity graph:

  • It's a diagonal line slanted in such a way that every second, it rises by an amount equal to the acceleration.

  • But still we don't know where to PUT that line on the vertical axis. Not yet, anyway.

  • Now, we could have figured out the integral of acceleration just as easily by using the power rule:

  • The acceleration, a, is a constant, but we could also say that it's (a) x (t^0) --

  • because anything raised to the power 0 is just 1.

  • So, according to the power rule, the integral of acceleration -- which is the velocity -- would

  • be equal to the acceleration, multiplied by time, plus C.

  • That's the same answer we got earlier!

  • Now, here's where the initial value comes in. The velocity graph tells you what the

  • velocity is for each moment in time. But we had to add the C, because we didn't know

  • where to place it on the vertical axis -- when time equals zero.

  • So, the integral of the acceleration COULD have just been (acceleration) x (time), or

  • (a)(t). But it could also have been (at) + 4. Or (at) - 6.

  • So we put a C in the integral instead, to represent all those options.

  • But we can get rid of that C if we can figure out the velocity, when time equals zero --

  • what we've been calling v(0).

  • And if we write our equation with that v(0) in it, as a placeholder for the velocity when

  • time equals zero, we end up with the the full equation for velocity.

  • That should look familiar, because it's one of our kinematic equations -- the definition of acceleration!

  • Neat how everything works out like that.

  • This equation tells us that the final velocity of our falling tennis ball, when it hit the

  • ground, was 16.7 ms downward.

  • But we aren't done yet. We're looking to link acceleration and POSITION,

  • so we'll need to go one step further by integrating again.

  • There are a couple of different ways we could do it, but let's just use the power rule again.

  • The integral of (a * t) is (half)(a t)(squared),

  • and the integral of v_0 is just v_0 * t.

  • Put 'em together, and you end up with this, which is starting to look a whole lot like

  • ANOTHER kinematic equation -- the one we've been calling the displacement curve.

  • Now, what about that C?

  • Well, just like before with the starting velocity, the starting position will tell us where to

  • stick this equation on the vertical axis. So we'll just make C equal to the starting

  • position, which we'll write as x_0.

  • And that's our integral -- the displacement curve equation.

  • Which means that now, we have everything we need to figure out, how high your window is.

  • The starting velocity is zero, because you just dropped the ball without throwing it.

  • The acceleration is 9.81 ms^2. And it took 1.7 seconds to land.

  • And now you know everything there is to know about calculus. (long pause) No you don't.

  • As you can probably imagine, we've barely scratched the surface here -- there's a

  • reason it normally takes two semesters of university, just to cover the basics.

  • And, you know some people spend their whole lives studying this stuff.

  • But we've at least established enough background, that when those things do come up in this

  • course, we'll be able to use what we've covered here, to talk about them.

  • Today, you learned that integrals are the area between an equation on a graph and the horizontal axis.

  • You also learned a few shortcuts to help find them, and how to find C using an initial value.

  • Crash Course Philosophy is produced in association with PBS Digital Studios. You can head over

  • to their channel to check out amazing shows like Shanks FX, Gross Science, and PBS Game Show.

  • This episode of Crash Course was filmed in the Doctor Cheryl C. Kinney Crash Course Studio

  • with the help of these amazing people and our equally amazing Graphics Team is Thought Cafe.

If you watched our last episode -- and really, if you haven't, you should.

字幕と単語

ワンタップで英和辞典検索 単語をクリックすると、意味が表示されます

B1 中級

積分。クラッシュコース物理学 #3 (Integrals: Crash Course Physics #3)

  • 16 0
    Idoz に公開 2021 年 01 月 14 日
動画の中の単語