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[Zrk2]

Yeah. With these types of questions you always just set up SA and Volume equations and take the derivative of the non-defined one, and set it equal to zero and solve. Easy marks on your test.

[Lectorog]

take the derivative

on a quiz in Precalculus

He may have difficulties with that.

[rutsber]

take the derivative

on a quiz in Precalculus

He may have difficulties with that.

I could probably do it, but it would more than likely be counted wrong because we just started limits on Friday. I never thought of doing it like that though, it could be useful, especially next semester when I take calculus.

[da_nang]

It's not just the derivative that tells you where the extrema is.  Fermat's theorem.

[chewie]

A math thread on bay12 starting at the same time as my studies in geography, where we also have to do math about which I've no idea?

I love you. Expect many questions in the future.

Edit: Except a friend of mine got hold of the official solutions.

[rutsber]

A friend of mine has a math problem. She needs to figure out how deep a state would be in corn, based on a certain number of bushels. She was told to take the volume of the bushels over the area of the state, or
lwh/lw

The area of the state is in square miles but the volume of the bushel is in inches cubed. Does she need to convert the square miles into square inches, and if so what exactly would you multiply by?

[Karlito]

Or convert the volume into cubic miles, yeah. Otherwise you'd end up with an answer that had units in³/mi², which isn't terribly useful. 1 mile is 63360 inches, so a mi² is (63360 in)²=4014489600 in².

Fortunately, WolframAlpha is already extremely useful for this kind of calculation.

[rutsber]

Or convert the volume into cubic miles, yeah. Otherwise you'd end up with an answer that had units in³/mi², which isn't terribly useful. 1 mile is 63360 inches, so a mi² is (63360 in)²=4014489600 in².

Fortunately, WolframAlpha is already extremely useful for this kind of calculation.

Ah thanks. I got the answer by plugging the area and total volume into WolframAlpha, but she needed to show her work. I didn't know you could just plug that last bit into WolframAlpha, so thank you for that.

[palsch]

A fun little problem.

There are solutions in the comments, so I'll just copy it here in hard to read plaintext.

Prove that;

(1/2)(3/4)(5/6)...(99/100) < 1/10

Using only middle school mathematics and in only a couple of lines.

Then prove the same for < 1/11 and < 1/12

[Another]

A fun little problem.

There are solutions in the comments, so I'll just copy it here in hard to read plaintext.

Prove that;

(1/2)(3/4)(5/6)...(99/100) < 1/10

Using only middle school mathematics and in only a couple of lines.

Then prove the same for < 1/11 and < 1/12

I remember doing this in 8th grade for <1/10. A trivial modification of removing first few terms in the additional construct and a few resulting inequalities will give <20/220, <7/80 (slightly easier result) and <40/480.

[Zrk2]

I have a screenshot here from Maple TA (a software that I have to use for calculus quizzes). Can anyone tell me what's going on here?

[Karlito]

Acceleration is the second derivative of position (with respect to time), not the first.

[Zrk2]

Dammit.

I keep doing dumb shit like this, and I really need to cut it out.

[FearfulJesuit]

Question about transfinite ordinals.

I am aware that you can order the ordinal numbers like so:

0, 1, 2, 3...ω, ω+1, ω+2...ω·2, ω·2+1...ω·3...ω·4...ω²... (followed eventually by ϵ₀ and its ilk of course)

In this sequence, any given number is the order of everything before it. So, 0 is preceded by nothing; order of 0. {0} has one member, so the order of that set is 1. All the natural numbers have no final member, so their order is ω. Follow up the natural numbers with a one-member set {x} at the end and you get ω+1. And so on and so forth.

Now, multiplication and addition with ordinals is not quite like with natural numbers. Specifically, ω+1 is its own little ordinal number, being the order of the set {0, 1, 2...x}, where x is just something hanging out at the end. However, 1+ω is...equal to ω, because {x, 0, 1, 2...} just gets renumbered to {0, 1, 2, 3...}; no final member, therefore ordinal number ω.

Now, though, let's look at multiplication. ω·2 is the order of something like this: {0, 1, 2, 3...a, b, c...}. But 2·ω is equal to ω. This isn't making any sense to me because if you take the natural numbers and then put a set of order ω before it, like this... {0, 1, 2...} -> {a, b, c...0, 1, 2...}

wouldn't this just reorder to a set of ω·2?

What am I doing wrong here?

[Another]

For certain formal definition of "equal" 2*ω=ω. If that definition does not fully coincide with intuitive definition - that is a problem with intuition.