## an interminable conversation 10: more basic physics – integrals

Jacinta: So I watched episode 3 of the crash course physics videos, on integrals, and found it so overwhelming I had to immediately go and take a nap. I’m surely too old and thick for this stuff, but I must soldier on.

Canto: We can do battle together – so this episode is about integrals, the inverse of derivatives. I’m not sure we gleaned much from the episode about derivatives, but if we combine this with exercises from Brilliant and some other practical application-type videos and websites we might make some more progress before we die.

Jacinta: Okay so equations, at least some of them, can be plotted on graphs with x-y axes, and the integral of the equation is the area between the curve and the horizontal x axis. Dr Somara is going to teach us some shortcuts for calculating these integrals, which sounds ominous.

Canto: I didn’t really understand the stuff about derivatives, but I’ll keep going, hoping for a light-bulb moment. So integrals help us to understand how things move, she says, which in itself sounds weird. And then she mentions the displacement curve, which I’ve forgotten.

Jacinta: Looking elsewhere, I’ve found a simple video showing displacement-time graphs. Displacement, which is simply movement from one position to another, is shown on the y (vertical) axis, and time on the x axis. A graph showing a straight horizontal line would mean no displacement, therefore zero velocity. A graph showing a vertical straight line would mean displacement in zero time, which would indicate something impossible – infinite velocity? Anyway, a straight line between the horizontal and the vertical would indicate a fixed velocity – neither accelerating nor decelerating. A curve would indicate positive or negative shifts in velocity. I think. Sorry – the terms used are *constant *velocity and *variable *velocity. That’s much neater. Oh and there’s also negative velocity, but that’s a weird one.

Canto: Thanks, that’s useful. Need to point out though that ‘curve’ is just the term for representing the data on a graph – in that sense the curve could be a straight line, or whatever. So Dr Somara starts with a gravity problem. You want to know the distance between your bedroom window and the ground, in a multi-storey building. You have a ball, a stop-watch and a knowledge of gravity. The ball will fall at g, 9.81 m/sec². What is the distance? According to the Doctor, discussion so far about motion has involved three aspects, position, velocity and acceleration, and has focussed on velocity as the *derivative *of position, and acceleration as the derivative of velocity. The connection has to be reversed to work out the distance problem. So *velocity is the integral of acceleration. *

Jacinta: And of course position is the integral of velocity. And this is important – velocity is the area under the acceleration curve, and position the area under the velocity curve. Which might be difficult to calculate. Areas within polygons, without too many sides, is easy enough, sort of, but under complex curvy stuff, not so much. And when we talk about ‘under’ here, it’s the area, often, to the axis, which represents some sort of zero condition, I think. Anyway, one method of calculating this area is to treat it as a series of rectangles, growing more or less infinitely smaller. Imagine dividing a circle into squares to determine its area – a big one extending from four equidistant points on the circumference, which would account for most of the area, and then progressively smaller squares in the interstices. You get the idea?

Canto: Yes, with that method you’d get infinitely closer to the precise area… Anyway, Dr Somara shows us a curve which is apparently the graphic representation of a formula, x⁴- 3x² + 1, and shows us how to find the integral, or at least shows us how we can divide the curve and its connection to the x axis by dividing it into rectangles. But what’s a more practical way of doing it? Well I’ll follow her precisely here. ‘If you know that your velocity is equal to twice time (v = 2t), then you know this is the derivative of position. So to find the equation for position, you have to look for an equation whose derivative is 2t, as for example, x = t². So x = 2t is the integral of v = t²

Jacinta: Yeah I can barely follow that. But the good doctor assures us that integral calculation is a bit messy. But apparently we can use the ‘power rule’ which we used for derivatives, and reverse it, in some instances. To quote: ‘Basically, you add one to the exponent, then divide the variable by that number’. Here’s an example: with v = 2t, x = 2/2t¹+¹, so x = 2/2t², so x = t². x = t² is the integral of v = 2t. She shows another more complex example, but I can’t do the notation for it with my limited keyboard skills. It involves some division. Anyway, with these mathematical methods we can look at trigonometric derivatives and do them backwards, e.g. the integral of cos(x) is sin(x).

Canto: We need to look at a variety of explanations of all this to bed it down methinks. I can only say I know a little more than I did, and that’s progress. And next we get onto constants. So what’s a constant (c)? It’s a number, and can literally be any number, positive, negative, fractional, whatever. It can be a placeholder, as for gravity, g (here on Earth), or presumably the speed of light, or ye olde cosmological constant, which is apparently still alive and well. Anyway, the derivative of a constant is always 0. That’s because a derivative’s a rate of change, and constants don’t change, by definition.

Jacinta: The derivative of t² is 2t, which presumably works by the power rule. Add any number and the derivative will always be 2t. That’s to say, the derivative of t² +/- (any number) is 2t. So, ‘if you’re looking for the integral of an equation like x = 2t, you have *infinite *choices, all of which are equally correct’. It could be t² or perhaps t² -18 or t² + 0.456. But I’m not clear on what this has to do with constants.

Canto: We’re flying blind, but it’s not too dangerous. The idea seems to be that the integral of x = 2t is t² + c. And here, if not back there, is where it gets tricky. With a bit of practice, we might know what the graph of the integral would look like, but not so much where it will lie vis-a-vis the vertical axis. For that we need to know more about the constant, ‘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’. If it’s just t², it will intersect at zero, if it’s t² – 10, it’ll intersect at minus 10, etc. To avoid this infinity of integrals, the practice is to add c at the end of the integral, to stand for all the possible constants. So, saying that the integral of x = 2t is t² + c covers all the infinite options for c.

Jacinta: Well, Dr Somara next talks about the ‘initial value’, which you can apparently use to work out ‘where your integral is supposed to be on the y-axis’ without knowing the value for c – I think. For a graph of position, the initial value would be your starting position – where it intersects the vertical axis. This is the c value.

Canto: So returning to the bedroom window and the ball, the ball is dropped from the window sill at the same time the stopwatch starts. It hits the ground at 1.7 secs. So we know the time and the acceleration, 9.81m/sec². We need an equation for the ball’s position. We do this by finding its velocity, working out the integral of its acceleration. If you have a graph with the y-axis representing acceleration, which in this case is constant, and the x-axis representing time, the uniformly accelerating ball would be represented as a flat line, making the area under the ‘curve’ – between it and the x-axis – fairly easy to calculate. The area would be rectangular, and would be calculated by base x height. The base is t, the amount of time the ball took from release to hitting the ground, and the height is *a*, the acceleration, so it’s just a matter of *a* x *t*. The integral is *at *plus *c, *the constant. We need this constant, according to Dr Somara, because we can see that the velocity graph will be diagonal, a line ‘slanted in such a way that, every second, it rises by an amount equal to the acceleration’.

Jacinta: But, where to put it the line on the vertical axis? We’re looking for the *integral of the acceleration, *so we may use the power rule, which I still don’t get. So I’ll quote the doctor, for safety: ‘The acceleration, *a, *is a constant, but we could also say that it’s *a *x *t*º (*t *to the power zero)’. And anything ‘to the power zero’ is always 1. So, according to this mysterious power rule, the integral of acceleration (i.e the velocity) would be equal to the acceleration multiplied by the time – plus *c. *The *c* is added because we didn’t know where to place it on the y axis when time, on the x axis, is zero.

Canto: Yeah right. Let’s continue to quote the doctor – ‘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…’ the initial value, being the velocity at time zero. ‘So the integral of the acceleration *could *have just been *a* x *t, *or *at’. *Or *at *plus or minus whatever. The *c *in the integral represents these options. But if we can work out the velocity at time zero (v0) we won’t need *c. *So, according to our Doctor, ‘if we write our equation with that v0 in it, as a placeholder for the velocity when time equals zero, we end up with the full equation for velocity, *v* = *at* + *v0′. *This is the kinematic equation, the definition of velocity. So the equation tells us that the *final *velocity of out tennis ball is *at, *that’s to say 9.8m/sec x 1.7 secs, that’s to say, 16.7m/sec (down, towards the Earth’s centre of gravity).

Jacinta: All of which seems to complicate something not quite so complicated. Anyhow, the pain isn’t over yet – we need to link acceleration with position, and this requires further integration, apparently. So, according to the power rule, which we should have learned, the integral of *at *is half *at *squared, and to get the integral of v0 you multiply it by *t. *Get it?

Canto: No. Let me quote from a highlighted comment: ‘i cant imagine how the avg viewer with no prior knowledge of calculus would actually understand calculus just by watching this video’.

Jacinta: Hmmm. Maybe we’ll try Khan Academy next. Anyway, if you put these integrals together you’ll get something that looks like the kinematic equation, the displacement curve. So for our example, we can work out the height from which the ball was dropped, or the distance the ball has travelled, using the initial position and time (zero and zero) plus half *at *squared. *a* was 9.81 m/sec², and *t *was 1.7 secs. *t² *is 2.89. Multiplying them makes 28.35, which, halved, is 14.175. metres. At least I got the calculation right, but as to the why….

**References**

https://sites.google.com/a/vistausd.org/physicsgraphicalanalysis/displacement-position-vs-time-graph

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