Math Is Fun Forum

This is a pretty complicated problem, so don't feel at all bad for struggling with it

First point: whenever you write down a force equation, make sure you understand what it is the forces are being applied on. Be sure to draw a free body diagram for this object before you start and then transfer the force arrows into your mathematical formulas.

I'm not going to draw a picture, but I'll list the forces that I think are acting on the sled:

-An upwards normal force from the incline
-A downwards normal force from the brick
-The tension force from the rope
-Gravitational force
-Friction from the incline
-Friction from the brick (the sled is not moving at the same speed as the brick, so this is an issue)

Once you have the arrows drawn, you can write down the math expressions for each force - breaking them up by components. Again, I really think it's a bad idea to plug in numbers when you do this. There are *many* advantages to working with symbols all the way up to the very end, and only plug them in for your final answer. It'll save you a lot of headaches when you end up with a wrong answer and you want to work backwards to find your mistake.

As I was saying, when you write down your equations you need to keep in mind what the forces are acting on. In this case they're all acting on the sled - not the sled+brick together. So when you use F=ma, the mass is only the mass of the sled. It looks as though you also have another mistake here - I think the -30 in your expression is wrong, but since you didn't leave it in symbolic form I can't tell how you got here. I think you did everything else fine though.

Here's what I get for an answer:
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Mu-sub-s is the coefficient of friction between the incline and the sled, and Mu-sub-b is between the sled and the brick here. Most of our expressions match up when you plug the numbers in, but I  apparently disagree with the official answer in that I have cos instead of sin for the mu_s term instead of sin. So let me take this opportunity to show you how cool leaving things as symbols is:  since I left this in symbolic notation I think I can easily prove I am right. Behold the power .

What if the incline weren't at 22 degrees but much steeper - almost 90 degrees instead? As you make the angle very close to 90 degrees, you should logically expect the normal forces (and as a result, the friction forces as well) to vanish - the objects no longer push against each other. Logically, all you should have left is gravity and the tension force, so there should be no mu's left.
Here's the test. Plug in theta = 90 degrees. My answer:

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This makes sense. Gravity is pulling down and the rope is pulling up. But that's it. Teacher's answer (I think I subbed the symbols back in correctly):

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And somehow friction from the incline is still resisting the motion. This does not make sense to me.

So I think this is a mistake in your teacher's answer, though it is possible I'm not thinking about this the right way (it's been a while since I did these kinds of assignments). But either way, I hope this can help your thinking about how to approach these problems, which is what is really more important.

I really need to learn more statistics myself, but I think this is how you should approach it:

Definition of the correlation coefficient:

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V is the covariance, and I'm using the sigmas in place of the lower case letters for the standard deviations here - it can get confusing otherwise. Now try to express everything in terms of X. For example, if you write out the definition of covariance you get Mean(XY) - Mean(X)*Mean(Y). You can plug in for Y in terms of X here. You should find that the covariance only depends on the standard deviation of X.

Then, also express the standard deviation of Y in terms of the standard deviation of X and if you plug everything in to your formula all the X dependence should cancel out - if not, you know you made a mistake. Also, remember that the standard deviation has to be positive, but the covariance can be negative. That's how you will get a negative answer for the case b < 0. Please ask again if that doesn't make sense!

Ok. Sounds like you're probably not sure how to start.  Just remember that a fourier transform is a way of changing what variable you're expressing your function in terms of. It is often used to switch variables from "time" to "frequency" (I imagine that's what you're using it for here). Speaking of time, I'm a bit short on it at the moment, so I'm afraid I'll have to hurry this, but the basic approach is to plug into this formula:

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Here the integral is from -infinity to +infinity, f is the function before transforming it, and F is the function after the transformation. It should make sense that the t will have been "integrated away" and you'll only be left with your "frequency", w. Does that make sense?

If you understand that, the only hard part would be to evaluate the integral, which is just calculus, however complicated it may be.

An avatar's a great idea. Let's see if this works ...

Hi pban,

Hate to say it, but those formulas aren't right. To take number 3 as an example, you have

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Simplify the left hand side:

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And this is true if and only if cos(x) = 0. This equation puts a constraint on x - namely, that it must be one of the values {...,-3pi/2,-pi/2,pi/2,3pi/2,...}, but this relationship is surely not true for all x.  You mentioned the euler formula, which allows you to write

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So if you wanted to you could write your three conditions in terms of this, but that does not make them any more or less true.

Wait .... now I see what's probably confusing you here. Your answers on the right side are all the "real parts" of the equations on the left. Real parts just mean this - any complex number can be written as F = a + ib, where a is the "real part" and b is the "imaginary part" because it is being multiplied by i. Sometimes you don't care what the imaginary part is so you can throw it away ... it depends on the context of the problem, but those equations as written aren't really true.

Ah, I see. What you want to solve is the integral:

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Where the integration is over S, the surface of the plane given by the equation you listed (in the first octant). Basically, since we don't know how to directly integrate over funny shapes, the trick is to project them onto a surface we do know how to integrate over - like the xy plane (xz or yz planes would work just as well). The formula you want to use here is this:

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Where D is the projection of the surface onto the xy-plane. So, you can just evaluate the right hand side in the same way you would solve other integrals you've seen in previous calculus classes. It's just a double integral over x and y, and anywhere you see a z you just plug in what that is in terms of x and y given by the plane equation (that's what the z(x,y) means). Just remember to limit D to the projection of the region of the plane you want to integrate over - that is, the part in the first octant.

You can find a number of websites that explain this in more detail if you google "surface integrals". The formula is pretty strange, so if you want to understand why it works, you can look here

http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/surface/surface.html#derivation

for a derivation. Also, it certainly won't be an issue on any school problems you're asked to solve, but the derivation for the surface integral formula relies on an assumption that you can approximate the surface at any given point by the tangent plane at that point. So I think this forumla would fail if you ever tried to integrate over a surface that had a sharp kink in it (meaning there is no tangent plane). Could possibly be an issue in certain real life applications.

is conventionally taken to be a^(b^c), not (a^b)^c. This convention was probably chosen because it makes it easier to simplify some expressions. For example, it allows you to write:

Incidentally, n!! is sometimes taken to mean n(n-2)(n-4)... , which is definitely not the same as (n!)! It's probably pretty rare to see it used this way though.

See http://mathworld.wolfram.com/DoubleFactorial.html

There's no real trick to it, it's just that 1/2 comes out of the formula for the sin(x) coefficient when you evaluate it.

Remember, the formula for the sin coefficients looks like this:

Of course, since f is 0 over [-pi,0], the integral is really only from 0 to pi. Plug everything in and you'll see that you get 1/2. The cos terms require more complicated integration, of course.

For small N, your points will be equivalent under translations/rotations to one or more of the basic crystal structures, see http://cst-www.nrl.navy.mil/lattice/

There are guaranteed to be equations you can find for the locations of the points of these structures if you poke around online.

Or, if you're trying to deal with large N and you want to use a computer, you could start out with your N points scattered at random around your cube. Then use some simple algorithm to move each point away from its nearest neighbors in small iterations until they're pretty uniformly distributed. I don't know what the best algorithm would be ... you'd probably have to play around with some ideas until you find one that converges.

This looks like a quite tedious problem. Sorry, I don't have time to go through and do it myself, but I'll try and give you some advice that might help if you're struggling to get started:

The first thing to do is draw a picture. From looking at the signs of the x^2 terms, you know that the first parabola opens up and the other two parabolas open down. If you draw some examples, you will see that two parabolas opening in oppositte directions can touch each other in 0, 1, or 2 places. From the way your problem is worded, I think it's saying that the up facing parabola touches each of the other ones in one place - otherwise there would be more than just points A and B. So draw your picture with The up facing parabola between the two down facing ones so that it just grazes them as it passes by (remember that the down facing parabolas probably don't start from the same height). Label the touching points A and B and draw a line between them. Now draw another line that is tangent to both of the down facing parabolas - there is only one way to do this. Your job is to show that these two lines are parallel to each other, meaning that they have the same slope.

So find the slopes, and show they are equal. To find the slope of the line tangent to the parabolas, first find the slope of a line tangent to each of the parabolas at some point along them. This is easy if you know calculus, but more complicated otherwise. Then you need to compare your results to find points at which the slope is the same for both parabolas - these are your tangent points, and take that slope as your answer.

Now you need to show that this slope is the same as the slope of the line joining A and B. First you have to find where points A and B are. To do this, simply set the equations for the intersecting parabolas equal to each other to find the point of intersection. Remember that the intersection only occurs in one place, so the stuff under the square root when you solve the quadratic equation must vanish. Once you have points A and B, find the slope of the line that joins them, and show it is equal to the slope of the tangent line - then you're done!

Tough problem!

I wouldn't be surprised if there's a trick that makes this problem much easier in this special case, I just don't know what it is.

Personally I think using the prime notation (it that what liebnitz notation is?) when there's more than one variable involved is just asking for trouble. It's confusing enough without ambiguous notation to get in the way.

Btw, anyone know how to write partial derivatives in latex? Lower case deltas don't quite cut it.