Let's make a homework help thread!

Err... both of those are partially-right.

Some things to note:

• Diatomic hydrogen has no net charge.
• In an aqueous solution, none of the component ions are physically bonded together.
• Don't forget to balance the equation.

edited 28th Feb '12 10:51:50 PM by Noaqiyeum

Ok, calling on the hive mind!

This question has come up in my calculus class, and the professor says he is unsure of the answer and that if anyone works it out they should tell him. The question is:

Take a curve rotated around the X-axis. Is its surface area the same as that of a conic frustum whose two ends are circles identical to the ends of the original solid, and whose side length is equal to the arc length of the original curve?

This can be thought of as "stretching" the solid of rotation so that the sides become straight and it becomes an easily-calculated section of a cone.

As a real world example, if you take a soda can and crush in the middle of it, its surface area will not change, but it will go from being a cylindrical solid to a solid of revolution of some curve. Or will it? We're trying to either prove that it works, or show that it doesn't work and explain why.

Umm.

My instinctive response is to say it should work, but I've just thought of a good reason why it wouldn't.

Consider the curve y = f(x), and rotate it about the x-axis. Every x-value now corresponds to a circle of radius y, and therefore circumference 2πy, and the surface area is the integral sum of all the circumferences of the circles between whatever two x-values you've chosen.

Now consider the frustum of the same arc length as y=f(x), as you described. The two end circles may have the same circumference, but most of the time that won't be the case for the circles in between them, so the integral sum of the frustum circumferences will differ from that of the rotated curve.

EDIT - Better explanation.

Take the curve y = f(x) = sin(x), with endpoints x = 0 and x = π. It should be obvious that this is a well-behaved, non-pathological shape, with both a positive, finite volume and surface area. But y = f(0) = 0, and y = f(π) = 0, so the circles on the ends will have radius 0 and circumference 2π(0) and the frustum will be a line segment with no surface to speak of. This contradicts our initial assumption, ergo, R.A.A., our assumption is false and this method does not generally hold.

edited 7th Mar '12 2:44:08 PM by Noaqiyeum

The problem with your first argument is that the bounds of integration won't be the same. If the original curve was mostly convex, a frustum with the same arc length will be longer than the original solid, and vice versa. This is moving us in the right direction: If the original circles were mostly larger than the new ones there will be more of them in the new solid.

The problem when both ends are single points is a good point though.

edited 7th Mar '12 3:02:33 PM by EdwardsGrizzly

I think saying that it doesn't work, and bringing up the situation where both ends are a single point is sufficient. After all, to disprove a method such as that, only a single counter-example is needed.

The thing is, when your counterexample is something so neat like that there's a possibility that only the zero case breaks the rule.

My guess would be no. My function is the following over interval [0,3]:

for x = [0,1], y = 1
for x = (1,2), y = 0
for x = [2,3], y = 1

Revolving this around the x axis yields two cylinders of length and radius 1. Total surface area is 4π.

The arc length along this curve is 5 — three horizontal segments, two vertical. Your conic frustum would be a cylinder of length 5, radius 1, and surface area 10π.

EDIT: wait, forgot the ends. Derp.

The piecewise would have 4π for the cylindrical surface, and another 4π for the "caps", for a total of 8π. The frustum would have 10π for the cylindrical surface and another 2π for the caps, for a total of 12π.

If the piecewise, un-smooth nature of this function creeps you out (fair enough, there's a discontinuity I handwaved over), just approximate the general shape with half a cycle of sin¹⁰⁰(x) or something that will be 1 at both ends of the relevant interval, near-zero in the middle, and almost vertical bits connecting them. The original surface area would basically be both sides of both caps, or 4π, and the frustum would be a cylinder of length π+2 and radius 1, which gives you something like 2π + 2π(π+2) surface area.

Another counterexample would be, say, 1/x³ over a very long interval. The surface area should converge, but the arc length doesn't — so your conic frustum would get arbitrarily long and have arbitrarily large surface area.

Conceptually, imagine you have some generic wibbly-wobbly function. Assume that its surface area and that of its corresponding frustum are already equal due to whatever prior luck. Now splice a horizontal line segment into it somewhere. You're changing the arc length (and thus, the frustum's surface area) by the same amount no matter where you put it, but depending on the height you might be adding a whole lot of area to the original surface or very little.

edited 7th Mar '12 7:02:01 PM by Pykrete

Um, here's another thing.

We have a hamster freefalling in air. The hamster's terminal velocity is 32 ft/sec.

The answer key says that the differential for this is y''+(32/32)y' = -32.

-32 is the acceleration. I'm wondering about the fraction. Is the 32 on top or the 32 on the bottom the hamster's terminal velocity? I've been trying to look this up and I haven't been able to figure it out.

It's the 32 on the bottom which is velocity; the 32 on the top I think is supposed to be 'g', the gravity constant.

The way I figured it out is that y" has to have units of ft/s^2 (it's an acceleration), as does the acceleration on the other side. Since y' is velocity (speed) measurement, it would be in terms of ft/s. In order units to balance out (when ever you add two quantities the units must be the same) the 32/32 term must have units of 1/s.

If the 32 on top is gravity and the 32 (ft/s^2) on the bottom is velocity (ft/s), when you multiply it out you get:

ft/s^2 * (ft/s)^-1 —> Flip for division
= ft/s^2 * (s/ft)
= (ft*s)/(ft*s^2) —> Combine top and bottom
= 1/s —> cancel like terms

Not sure how to get it into the y''+(32/32)y' = -32 form in the first place though. What was the original question?

EDIT: Okay, so I can get there mathematically/physics-ally, but I have no idea why one would want to write it in such a form. Unless it's for an exercise, I guess.

edited 8th Mar '12 6:17:48 PM by zzzdragon

Yes, I managed to figure this out by changing one of the 32s. It is a math exercise so there's a lot of extra work involved.

edit: nvm I'm just stupid

edited 8th Mar '12 7:33:35 PM by ohsointocats

I can never stop having problems with Chemistry can I? The problem goes as:

If I'm correct this is a phase and temperature change, so the equation used is q = (L x m) + (mc(delta)t)

I'm basically having overall problems just trying to solve it, so I'll just say what I think I know.

• L=6.02 kJ/mol
• C= 2.03 J/g C and 4.184 J/g C for some reason

...that's pitifully little, I guess I just need help understanding it some more and a lot of helping solving it. As always, large thank you in advance.

Well, the ice warms up to 0 degrees, then melts, then after melting, it warms up to 15 degrees.

So, we're dealing with H 2 O which has molar mass 18 g / mol.

18 g of ice -> 2.03 J/g*C * 10 C = 365.4 J
1 mol ice -> 6,020 J/mol = 6,020 J to melt the ice
18 g water -> 4.184 J/g*C * 15 C = 1,129.68

And in total, we have a total energy absorption of 7,555.08 J.

edited 8th Mar '12 9:47:45 PM by DarkConfidant

I'm sorry, but I honestly couldn't follow/understand any of that in the slightest.

Basically, you should have two cases of mcΔt. One to bring the ice from -10 to 0, and another to bring the liquid water from 0 to 15.

Plus the heat-of-fusion term for the transition between them.

The bounds of integration won't be the same, but moving them based only on arc length won't change them enough to keep the same surface area. The zero-radius ends are an extreme example of this. If you use the same function f(x) = sin(x) and move the end circles arbitrarily close to x = 0 and π, the surface area of the frustum should get arbitrarily close to zero, while the surface area of the rotated function shouldn't change much at all.

It's probably equivalent to a linearisation approximation or something.

Also, I trust Pykrete's math. :P

edited 10th Mar '12 9:28:03 PM by Noaqiyeum

Dude, there's a gamble if I've ever seen one. There's a reason I picked examples where I could cheese my way around actual math

I'm just starting limits now and there's one problem that's tripping me up.

lim (cube root of) [(1 + 8^2)/(x^2 + 4)]
x→∞

How do I solve for this? Should I divide the numerator and denominator by x^2? If so, what do I do after that?

edited 15th Mar '12 7:28:29 AM by NatTheWriter

Unless I'm misreading it somewhere, that should just be 0: You have only constants in the numerator, and infinity in the denominator, to constant/infinity is 0, and cube root of 0 is 0.

That's what I thought. The book claims it's 2, though. :/ I guess the text is wrong.

EDIT: Eep. I forgot to put x after 8. Sorry, it's supposed to be 8x^2.

edited 15th Mar '12 8:54:46 AM by NatTheWriter

With limits, a fraction like you have there will, as x -> infinity, tend towards the ratio of the highest power if both the top and the bottom have the same largest exponent on the variable, regardless of what else is in the fraction. So:

lim x->inf (5x^2+2)/(2x^2+7) = lim x->inf (5x^2)/(2x^2) = 5/2

Same principle applies to the problem you gave earlier.

Who here is good enough with chemistry to help a writer/historian who doesn't get this newfangled atom business?

Okay, with an x sitting there, yeah it's 2. All lower-order terms become negligible as x -> infinity, the fraction goes to 8, and cube root that.

I command this thread to LIVE because I want to start posting more OFTEN again and to do that I need good grades in SPANISH and to get those I need help checking some SENTENCES.

@Exe: depends what you're doing. I can do some atomic things but not others - what is it you're on with?