TheAspiringDoc's Math Thread

[TheAspiringDoc]

Ok, thought my math's posts were a bit out of place in the mainstream question threads so I thought I'd make my own
I'm trying to get an intuitive grasp of the derivatives and antiderivatives of trigonometric functions.. Any idea's how?
Also, Find the length of the largest square that will fit inside a regular hexagon of edge length 30cm?

[keltingmeith]

Ok, thought my math's posts were a bit out of place in the mainstream question threads so I thought I'd make my own
I'm trying to get an intuitive grasp of the derivatives and antiderivatives of trigonometric functions.. Any idea's how?

Try it by sight - sketch the graph of sin(x), and use the major points on it to see if you can figure out what its gradient is at each point, and then sketch it out.

Another thought - sin(x) is cyclic (or periodic), so its derivative should be cyclic, no?

[kinslayer]

I'm trying to get an intuitive grasp of the derivatives and antiderivatives of trigonometric functions.. Any idea's how?
Also, Find the length of the largest square that will fit inside a regular hexagon of edge length 30cm?

Picture the unit circle and suppose there is a particle travelling along it. Its y-position is sin(t) and its x-position is cos(t), where t is the particle's angular distance from the positive x-axis.

Now think about what happens to the particle's velocity as it travels around. In particular, when the particle starts at (1, 0), its x-position is hardly changing at all (actually a maximum), while its y-position is changing fast. As it reaches the top, pi/2 radians later, the x-position is now changing fast and the y-position is hardly moving at all.

If you continue around the circle you can see that maxima/minima of the sine and cosine functions should occur at each others' zeroes, so you shouldn't be surprised that they are each other's derivatives (up to multiplication by -1).

For the hexagon thing, see this page:

http://www.drking.org.uk/hexagons/misc/deriv3.html

[TheAspiringDoc]

Can someone please explain to me in simple terms how you would go about integrating the volume of a torus (doughnut)?
Also, am I right in thinking that standard deviation can only be applied to a normal distribution?
and one more thing, does the VCE at anytime use the n-1 approach for samples of the population when calculating standard deviation?
Thanks

[99.90 pls]

Holy shit you make me feel inadequate lol

[keltingmeith]

Can someone please explain to me in simple terms how you would go about integrating the volume of a torus (doughnut)?

Define its cross section as an ellipse/circle, centred by the distance of the difference by the major and minor lengths. Then, the volume of the torus is a simple volume of revolution around the x/y axis (depending on how you've centred it).

Also, am I right in thinking that standard deviation can only be applied to a normal distribution?

No - you are very, VERY, wrong. Standard deviation is only the second centred moment, and many different distributions have defined moments. (in fact, it's often harder to come up with an counter example).

In fact, the exponential distribution with parameter has standard deviation .

and one more thing, does the VCE at anytime use the n-1 approach for samples of the population when calculating standard deviation?

What do you mean by the "n-1" approach...?

[TheAspiringDoc]

Wikipedia:

A random sample drawn from some larger parent population (for example, they were 8 marks randomly chosen from a class of from a class of 20), then we would have divided by 7 (which is n−1) instead of 8 (which is n) in the denominator of the last formula, and then the quantity thus obtained would be called the sample standard deviation. Dividing by n−1 gives a better estimate of the population standard deviation than dividing by n. This is known as Bessel's correction.

[kinslayer]

Wikipedia:

That's talking about the sample variance, which is an estimator of the population variance. The correction is there to ensure that the estimator is unbiased. It is also called a "degrees of freedom" correction, because the use of another estimator (i.e. the sample mean) reduces the degrees of freedom by 1.

http://en.wikipedia.org/wiki/Degrees_of_freedom_%28statistics%29

And as Eulerfan101 pointed out, the standard deviation is a general concept, not limited to the Normal distribution. However, the Normal distribution has the characteristic that one of its parameters is its standard deviation/variance (the other being the mean) which is quite nice and not true of other common probability distributions.

The volume of the torus can be found in several ways, probably the easiest is by rotating a solid of revolution about the y-axis:

http://dean.serenevy.net/teaching/classes/Summer2007/M216/VolumeOfTorus.pdf

[EspoirTron]

Generally, dividing by n-1 is used to give an unbiased estimation of the population variance. If you're using a sample from some population N, your sample size can be, let's say n. This sample is generally random, but nevertheless is contained in N. Say the true mean (I'll call it U) is at some point. Depending on what sample you take, your sample mean (U') could sit either really close to U, or really far from U - i.e. U could be outside of the n. If we consider this fact, by dividing by n we don't truly consider the implications of the former, i.e. we underestimate the variance of the population. When dividing by n-1 we get a larger variance, and hence a better estimate from n to infer onto N.

I think that makes sense? I would imagine in VCE, yes you would divide my n-1. I do however doubt you'll be expected to provide rigorous meaning behind your choice of choosing n-1 over n.

[keltingmeith]

I would imagine in VCE, yes you would divide my n-1. I do however doubt you'll be expected to provide rigorous meaning behind your choice of choosing n-1 over n.

The proof of this actually isn't very difficult, and is definitely something I would expect a VCE student capable of (with a little guidance, of course). However, they don't go over what makes a good estimator (and hence, how to unbias a biased estimator), so I don't think it will pop up even in the new study design.

But yes - in further, methods AND specialist you should be diving by n-1 to find the sample standard deviation. There are some specific cases where you divide by n, but they will not appear in VCE (unless they suddenly change the study designs to include estimators - which would not be a bad move, IMHO).

[TheAspiringDoc]

Ok, cool, so I get the method used to find the volume of a torus [(2 Pi R)(Pi r^2) where R (the major radius) is the distance from the centre of the torus to the middle of the physical 'ring like part' and little r is the minor radius , (it's more or less just a cylinder curved into a ring/doughnut like shape) but I thought of a couple more things; is merely coincidence the 2 Pi r is the derivative of Pi r^2 and that Pi r^2 is the Antiderivative (integral) of 2 Pi r? Also, how do we find the volume of the hollow centre of the torus, i.e. The part you would cut out of the pastry to make the doughnut?
Thanks

[TheAspiringDoc]

Hey, just wondering, is sin(1)+sin(2)+sin(3)+...sin(358)+sin(359)+sin(360) equal to sin(0)+sin(45)+sin(90)+sin(135)+sin(180)+sin(225)+sin(270)+sin(315)+sin(360)? and are either of them equal to [tex]\int_{0}^{360} sin(x) dx\[int][tex]? It's just that my calculator gave different answers for each ones and also a different answer to what I'd expected.. Thanks
umm, LaTeX is confusing?

[keltingmeith]

is merely coincidence the 2 Pi r is the derivative of Pi r^2 and that Pi r^2 is the Antiderivative (integral) of 2 Pi r?

Nope - I have forgotten the logic behind it, though. Same happens for TSA and the volume of a sphere.

Also, how do we find the volume of the hollow centre of the torus, i.e. The part you would cut out of the pastry to make the doughnut?
Thanks

Welp, it depends on how you model the "hollow" part. Where does in start/end? One method could be to model a similar ellipsoid and take the volume of the torus from it.

Hey, just wondering, is sin(1)+sin(2)+sin(3)+...sin(358)+sin(359)+sin(360) equal to sin(0)+sin(45)+sin(90)+sin(135)+sin(180)+sin(225)+sin(270)+sin(315)+sin(360)? and are either of them equal to [tex]\int_{0}^{360} sin(x) dx\[int][tex]? It's just that my calculator gave different answers for each ones and also a different answer to what I'd expected.. Thanks
umm, LaTeX is confusing?

They're definitely not all equal. Why would they be equal?

[TheAspiringDoc]

Nope - I have forgotten the logic behind it, though. Same happens for TSA and the volume of a sphere.

Welp, it depends on how you model the "hollow" part. Where does in start/end? One method could be to model a similar ellipsoid and take the volume of the torus from it. They're definitely not all equal. Why would they be equal?

Inside of the torus - I am referring to the 'apple core' shaped part.
Well, why should they (the sine sums in my previous post) all be equal? isn't it like finding the mean of 5 + 10 + 15 or 5 + 7.5 + 10 + 12.5 + 15 or everything between 5 and 15 inclusive?

[keltingmeith]

Well, why should they (the sine sums in my previous post) all be equal? isn't it like finding the mean of 5 + 10 + 15 or 5 + 7.5 + 10 + 12.5 + 15 or everything between 5 and 15 inclusive?

No - because a mean has a normalisation factor. You've just listed sums without any factor.