Math Is Fun Forum

Perhaps the chain rule will make more sense if you think about it this way. Imagine you put together a computer program to calculate derivatives. One way to do this would be to find the slope of your function at a given point. That is, for sufficiently small deltas:

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This can be made arbitrarily accurate at your point by shrinking delta so long as y(x) is well behaved near x. If it has infinities or other troublesome discontinuities within delta of x then this equality breaks down. When you're dealing with division instead of true derivatives cancelation makes more sense. Now say y and x both depend on t. Then,

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And you can make your substitutions:

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Does this seem reasonable? Makes sense to me anyway. This argument should work for the vast majority of functions you're likely to encounter in real world applications.

Yeah, the chain rule is a bit confusing. I'll try to give an explanation, but I really should be in bed right now, so it'll have to be quick!

As you said, a(t)  = v'(p(t))p'(t), but you have to be careful what you mean by the primes - they aren't all the same. The original expression that you had was:

a = dv/dp * dp/dt

As you said, this is correct. I know this will make the math majors wince, but you really can think of the dp's as cancelling on the right hand side. So the prime on v is a derivative with respect to p and the prime on the p is a derivative with respect to t. If you're confused about how to use this formula, here's an example:

Say
v(p) = 4p+7
p(t) = 9t^2

Then a(t) = (4)*(18t) = 72t

To check this, you could solve the problem without the chain rule by writing v(t) = 4p + 7 = 4(9t^2)+7 = 36t^2 + 7
So a(t) = dv/dt = 72t. You get the same answer either way.

That's a very strange diagram. The equalities you wrote down are definitely true for the case where all three vectors are in the same plane as you said, but just to make sure you understand, these forumulas are also true for any three vectors.

Instead of looking at the answer, try just writing everything in terms of t using the t = sinx substitution. Tell us what you get when you do that and if your answer comes out wrong, write down the steps you used. I'm guessing you'll figure it out for yourself if you go through this exercise.

That is definitely correct, and it should be straightforward from where Ganesh left it. Now, it should be pointed out, as mathsyperson said, that the solution you get for y=10 in this way is lower than for y=2.

But if you think about it, for x and y > 1, x < y implies that x^x^x^... is always less than y^y^y^... , so this implies that 10 < 2, which is nonsense. I'm still thinking about this, but it should mean that there is a cutoff value for y above  which there is no x solution.

So my final puzzle related to this is: find the largest value of y such that

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has a solution. Unfortunately, I think you'll need to know calculus to be able to figure this out, but the answer makes me wonder if there's something really deep going on here that I don't understand. I find this really interesting!

Based on what I've seen online it looks as though the steering matrix is used in specialized applications in analysis of electrical signals. I doubt you'll find anyone here who has any experience with such things - this looks like a question for an electrical engineer. Sorry.

Yes, but only bearly.

Espeon, I have a story you might find interesting:

My junior high math teacher once told us about a famous mathematician in elementary school (I can't remember who the mathematician was, I'll call her Janet). Janet's teacher was getting tired of teaching, so one day she assigned the class a quiz in which they had to add up all the numbers from 1 to 500. Janet didn't know it then (she hadn't learned much math yet), but she was being assigned to calculate the 500th triangular number.

Janet, of course, didn't know any triangle number formulas, and she thought adding up all those numbers would be really boring. So, while the other students started working, she instead did a bit of doodling. Without really thinking about it, she wrote out the problem twice on her page.

001 + 002 + 003 + ... + 500 = ?
001 + 002 + 003 + ... + 500 = ?

Great, if I add all those numbers up, I'll do twice as much work and get twice the answer that I'm looking for, she thought. Bored, she erased the second line and rewrote it backwards, like this:

001 + 002 + 003 + ... + 500 = ?
500 + 499 + 498 + ... + 001 = ?

That's strange, Janet thought, if I add those numbers vertically, they all add up to 501 (did you notice?). 1+500=501, 2+499=501, 3+498=501, etc. But, she also knew that if she added up all of the numbers, all 1000 of them, she'd get double the answer her teacher wanted. So she wrote a third line:

001 + 002 + 003 + ... + 500 = ?
500 + 499 + 498 + ... + 001 = ?
------------------------------
501 + 501 + 501 + ... + 501 = double ?

So if you add 501 to itself 500 times, you'll get twice the answer that you're looking for, Janet realized. But adding something to itself lots of time is just the same as multiplication. This meant that 500 multiplied by 501 was double the answer she wanted:

500 x 501 = 2 x ?

Suddenly, Janet knew how to get the answer. She realized this meant ? had to be half of 500 x 501. So Janet solved the problem:

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She wrote down the answer and turned her quiz in. It took the other students more than an hour to finish, but Janet was done in ten minutes, and she was the only person who got the right answer - everyone else made a mistake adding all those numbers together.

You might notice that this technique could work for any triangular number, not just 500. Let's say you're looking for the Nth triangle number. Then,

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This works for all triangle numbers. If you plug in N = 500, you'll get Janet's formula!

Blah ... Mathsyperson, you've found the answer to my followup puzzle before I even asked it. I'll bet your sense of direction is good - you must have really good bearings. And hear I thought bearly anyone would get this right. Oh well, I guess this wasn't such a bear of a problem after all. You guys probably can't bear any more of my puns now so I'll stop.

Congrats to both of you!

Howdy all, thanks for the warm welcome.

The wolves are entertaining at least, so I don't mind!