[Calculus] arc length problem

scfan77 · 2 months

Let me point out that it's not

g(s) = s, (s/2)2/3

Rather, it's

g(s) = (s, (s/2)2/3 )

The parenthesis are very important because they are what tells you it's a parametric equation. Again, g(s) is a parametric function. In other words, it's a vector, but each component is itself a function (instead of a constant). For the wikipedia article, you'd be interested in the 2D and 3D examples which should look familiar from your lecture. Also look at the Usefulness and Conversion sections.

Since I'm not familiar with how your class is taught/what you know. Feel free to ask more questions.

cheatatmath · 2 months

Thank you! It makes sense now in an informal sense, but thank you for the little extra rigor. Really helps nail it down.

cheatatmath · 2 months

Ooooh, okay. At first I didn't get it, but the point is, that out of all the different ((j+1)/n), j/n) there has to be one of them inside that interval. Is there a rigorous way to show that there's one inside? (I have a feeling that picado answers this.)

Thank you!

scfan77 · 2 months

This is a good read.

Basically, length(y) = Integral from 0 to 14sqrt(7) of |g'(s)| ds, where g(s) is the parameterization of the curve. In this case, I believe (been a while since I've done these problems), g(s) = (s, (s/2)^2/3 ). Therefore, g'(s) = (1, 1/3***(2/x)^1/3 ). Thus |g'(s)| = Sqrt(1² + [1/3(2/x)^1/3 ]² ). Integrate that with your bounds, and you're done.

cheatatmath · 3 months

You can show that E' partitions into two disjoint sets, E^o and the set of limits points.

Do you mean Closure(E) partitions? If that's the case, is any non-interior point of E a limit point?

Also, I'm playing with the definitions and applying them to the problems, but all I've gotten is Intersection of (G_a)^C = E^C U (Intersection H_b), where G_a is the collection of all open sets in E, and H_b is the collection of all closed sets containing E. How can I simplify this further?

Thanks

josh1360 · 3 months

I'm not quite sure, but this is what I have. When the trough is at theta = 0, V = (1/2) Pi L r² . So my initial guess is that V(theta) = (1/2) Pi L r² cos(theta). Also, when the trough is at 90 degrees, V should be 0, yes? Well the V(theta) I gave above also satisfies that.

However, I feel this is imprecise. It might satisfy the obvious conditions, but I'm not sure about what happens at different theta. Perhaps this will be a starting point.

Edit: In your problem, L = 1m, r = 100mm, and 0 <= theta <= Pi/2.

cheatatmath · 3 months

I started to do (d), but got stuck; hoping you could help :).

Take a p in E^{o^C} . <=> Take a p not in E^o . Thus B_r(p) is not a subset of E for any r. Take a y in B_r(p). Thus y can either be in E or not be in E. If y is not in E, then y is in E^C . If y is in E and y != p, then B_r(p) Intersect (E{p}) is not empty. Thus y is a limit point of E. (i.e. y is in E').

I'm trying to find some connect between y being in E' to y being in E^C' . Then B_r(p) is a subset of Closure(E^C ). Then I'd need to show Closure(E^C ) is a subset of E^{o^C} , but this is what I got so far. Thanks!

(B_r(p) - A ball of radius r > 0 centered at point p).

cheatatmath · 3 months

Sweet! I understand all of it except the last line. So G1 is a cover which contains 1, right? And G2 is the cover which contains 1/2. And GN contains 1/N. So any n>N are covered in GN, and any n<N are covered in G1...GN-1? Is that how this is supposed to work?

cheatatmath · 3 months

We have the first part of the Heine-Borel theorem (Compact -> Closed & Bounded), but I think we'll be covering it soon, so I can use it in the homework. Thanks again! I get excited when there's actually problems I can understand :D

cheatatmath · 3 months

Okay cool. I thought that was the case. Now I'm trying to see how exactly we get limit points. Let me see if I got this right. We 'fix' n and let k go from 2 to infinity. Then that approaches 1/n (thus 1/n is a limit point). Then we look at another n and do the same thing. So it approaches that 1/n. Is this right?

cheatatmath · 3 months

So in your example, the first and last parts make the set closed, right? Now the middle part makes the limit points, yes?

cheatatmath · 3 months

But what is z an element of? Is z in E, int(E)? Maybe I should ask why d(z,y) < r/2. So now assuming I understand why d(z,y) < r/2, I get why d(z,x) < r. But how does that tell us that the ball of radius r/2 at x is a subset of E^o ?

cheatatmath · 3 months

For (a), what is z?

cheatatmath · 3 months

I don't think it's open by definition. For it to be open, it needs to be shown that for every point in interior(E), there exists a neighborhood at that point with some radius such that the neighborhood is in interior(E). What we do know by definition is that there does exist some neighborhood at every point with some radius such that the neighborhood is in E, but we need interior(E).

Or am I just completely mistaken?

cheatatmath · 3 months

I sent this to picado as well, but for your benefit: This one is "Let X be a metric space. A point p is an interior point of E (a subset of X) if there is a neighborhood N of p such that N is a subset of E." A neighborhood (I'm assuming this can have multiple definitions as well) is defined as "A neighborhood of p is a set N_r (p) consisting of all q such that d(p,q) < r, for some r > 0."

I got the feeling that these are all simple and based on definitions, which is how I started. I just couldn't see a way to rephrase the definitions in a way that would prove these.

cheatatmath · 3 months

I didn't realize there were multiple definitions for interior point. This one is "Let X be a metric space. A point p is an interior point of E (a subset of X) if there is a neighborhood N of p such that N is a subset of E." A neighborhood (I'm assuming this can have multiple definitions as well) is defined as "A neighborhood of p is a set N_r (p) consisting of all q such that d(p,q) < r, for some r > 0."

cheatatmath · 3 months

We never did mathematical induction in class (nor have I heard of it before). This is what the homework says about it:

Theorem (mathematical induction). Let A(n) (n = 1, 2, ...) be a sequence of assertions. Suppose that (a) and (b) or (a) and (c) hold. Here

(a) The assertion A(1) is valid

(b) If the assertion A(n) is valid, A(n+1) is valid.

(c) If assertions A(k) are valid for 1 <= k <= n, then A(n + 1) is valid.

Then A(n) are valid for n = 1, 2, 3, ...

What the hell is this talking about?

Also, thanks for trying to explain this. A lot is going over my head (and thus I'm doing lots of wikipedia-ing).

casos92 · 5 months

This is my tentative guess for the first problem. I took Calc 3 last year, so my memory may be a tad off. But my answers are on your answer sheet, so let's try this:

We know that we're going to switch the two integrals, the problem is, the bounds are off now. So at our upper bound, if we know the x value, what's y going to be? Just use the equation y=4x-x² for the upper bound and y=2x for the lower bound. Thus our y upper bound is 4 and lower bound is 0. Now you need x as a function of y for the inner integral. Just invert the y bounds. y = 4x-x² -> x = 2 + sqrt(4-y) and y = 2x -> x = y/2. Tada! The function inside the integral stays the same because you're integrating the exact same function, you're just doing it in a different order.

Hope that helped!

Edit: For the second you're just setting up the integral. So on a test you'd graph by hand, but I have Wolfram, so let's see what our situation looks like. The x bounds are super easy, 0<=x<=2. What about y? Well a problem, we're going to have to setup two integrals because the y function changes at x = 1. So we have integral from 0 to 1 of x² (this is the value between y = 0 and y = x²⁾ dx and the second integral is the integral from 1 to 2 of 2-x dx.

Okay, this isn't exactly right because your professor wants a double integral. The thing is, after you set up the double integral, it simplifies into the single integrals I made. Can you see what the double integral should be? (I don't want to retype all this :P)

stevenwentworth · 6 months

So if you stick something in one eye it comes out the other, right?

cheatatmath

TROPHY CASE

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