Ask a physicist anything. (7)

Tuddrussell said:

When the Hulk lifts a heavy object the ground under him usually breaks from the strain.

That's simple and logical, but when Jean Grey lifts a comparable object it's as if all the weight just disappears, she can even lift an object while flying!

Where does that weight go?

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Jean has different power and is able to lift up objects with her mind. But I would say in the air?

Why on Earth do I get fruit flies in my kitchen during the summer?

I swear these little buggers spontaneously generate.

They only live not even a month, I get harsh winters here in my State that probably radiates a good 700 miles everywhere from where I live.

So yeah, they all should be dead come summer time but somehow they find their way back every year and it makes me want to scream.

Tuddrussell said:

When the Hulk lifts a heavy object the ground under him usually breaks from the strain.

That's simple and logical, but when Jean Grey lifts a comparable object it's as if all the weight just disappears, she can even lift an object while flying!

Where does that weight go?

Click to expand...

Haha, well, if I recall, Jean Grey is a telekinetic. Presumably she doesn't actually exert a force from her body which lifts these objects, but instead uses her mind to cause them to be lifted through some other means.

Zippy the Wonderslug said:

Why on Earth do I get fruit flies in my kitchen during the summer?

I swear these little buggers spontaneously generate.

They only live not even a month, I get harsh winters here in my State that probably radiates a good 700 miles everywhere from where I live.

So yeah, they all should be dead come summer time but somehow they find their way back every year and it makes me want to scream.

Click to expand...

Well, they are small! It's rather easy for small holes in your house to go unnoticed....around the edges of screens on the windows, for example. Or for them to fly in while you're walking in.

I'm going to go back to highschool. Highschool for adults. I gonna read math.

So let's look at intergrals once again. Explain it to me. And remeber I got less then forty in IQ, be polite.

How do i deduce the rules given for the derivate of a function? The derivate is defined as lim h->0 f(x+h)-f(x)/h?

The rules are y=x^p, p whole number and y'=p*x^p-1

Is it possible for me to alter the rules so that y=zx^p, p whole number and y'= x^pz instead? what would the world look like if i did?

Chalnoth said:

Well, they are small! It's rather easy for small holes in your house to go unnoticed....around the edges of screens on the windows, for example. Or for them to fly in while you're walking in.

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That and more often than not their eggs are in everything you eat.

How are intergrals defined, i couldn't find the definition anywere in my book?!

Lillen said:

How are intergrals defined, i couldn't find the definition anywere in my book?!

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Simple -> intergrals defined

But I desire some interaction... that's why i didn't google it myself. Because the lack of feedback and diversity from other people.

Integrals are just shy outegrals.

Lillen said:

How do i deduce the rules given for the derivate of a function? The derivate is defined as lim h->0 f(x+h)-f(x)/h

The rules are y=x^p, p whole number and y'=p*x^p-1

Click to expand...

Well, it's a matter of taking the limit. And unfortunately there is no single way to do it. Different functions require different tricks. This is one of the easier limits to take. The first thing to do is to evaluate f(x+h) and f(x):

lim h->0 ((x+h)^p - x^p)/h

So to get any further, we have to take (x+h) to the power p. So if p=3, we have to evaluate (x+h)(x+h)(x+h). How do we do this?

Well, first we have to understand how the operations are done. Such multiplications are performed using the distributive property of multiplication over addition:

(x+h)(x+h)(x+h) = x(x+h)(x+h) + h(x+h)(x+h)
= xx(x+h) + xh(x+h) + hx(x+h) + hh(x+h)
= xxx + xxh + xhx + xhh + hxx + hxh + hhx + hhh

So, what we are doing here is taking every possible permutation: we're summing up every possible choice of x's and h's. But there's a simplification we can make. As h approaches zero, h^2 will approach zero much more quickly (e.g. h=0.1, h^2 = 0.01; h=0.01, h^2 = 0.0001). So we only need to take the terms that evaluate to x^3 or hx^2. Quick inspection of the above finds four terms: xxx, xxh, xhx, hxx, or x^3 + 3hx^2.

But what happens if we have any power of p? p not equal to 3? In this case we have a product of p (x+h)'s:

(x+h)(x+h) ... (x+h)

Obviously we only have one x^p term again. But how many hx^(p-1) terms do we have? Wel, to get a term hx^(p-1), we have to select h once and x the rest of the times. We can make p different choices here: one choice for each term in the multiplication. So the term hx^(p-1) will always have a factor of p in front of it. Plugging into the previous equation, we have:

df/dx = lim (h->0) (x^p + phx^(p-1) + ... - x^p)/h

Here the ... is a set of other terms that make up (x+h)^p. There's going to be a lot of other terms, but as we take the limit, they will all be smaller than phx^(p-1), so we don't care about them (unless the phx^(p-1) term cancels out, in which case we would have more work to do...).

So now this limit is pretty easy. First, we can immediately cancel the x^p terms, because x^p - x^p = 0. This leaves us:

df/dx = lim (h->0) (phx^(p-1) + ...)/h = px^(p-1)

Here I've simply canceled out the h on the top and bottom of the fraction, and in the limit as h->0, the remaining terms of the polynomial all go to zero, so we're just left with the answer, px^(p-1).

Zach Weiner has been studying up on some math and physics, and has been posting explanations of the concepts he's covering every step of the way:
The Weinerworks

That blog might be pretty useful to you.

Lillen said:

Is it possible for me to alter the rules so that y=zx^p, p whole number and y'= x^pz instead? what would the world look like if i did?

Click to expand...

It is conceivable that you could come up with a different operation, other than a derivative, that makes this change. But then the question would be: is that operation useful?

Lillen said:

But I desire some interaction... that's why i didn't google it myself. Because the lack of feedback and diversity from other people.

Click to expand...

so it's not the definition you want, but you just wish to talk about it?

Well, this is after all, "ask a physicist... *anything*!"

Lillen said:

Well, this is after all, "ask a physicist... *anything*!"

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... "related to physics" is implied. You wouldn't ask for legal advice in a topic named "Ask your doctor anything", would you?

I try to follow you... there are some things I was clueless on!

No I guess I wouldn't. Then again, i think i misinterpreted *anything*. The topics name is ambiguous.

Lillen said:

How are intergrals defined, i couldn't find the definition anywere in my book?!

Click to expand...

The idea of an integral is that you want to add up the values of some function over some range. For example, imagine, if you will, that your odometer is broken on your car, and you want to determine how far you have gone on a trip. How could you do this?

One way to do it would be to use the speedometer. The speedometer actually carries the exact same information as the odometer, but the odometer's value is the integral of the speedometer. How is this?

Well, consider, if you will, if you were driving in your car at 60 mph for one hour. Clearly you would travel 60 miles, correct? An integral is a way to figure out how far you have gone even if your speed is changing, but the concept is the same. Instead of just taking a constant speed over some time, the integral sums up the speed you travel at every instant of time. That is:

integral from a to b of (v(t) dt) = lim (dt->0) sum(n from 0 to N) of v(a + ndt) dt

Here, N is chosen such that a+Ndt = b, that is: N = (b-a)/dt. Note that this means that as we take the limit as dt goes to zero, N goes to infinity, so we are summing up an infinite number of terms.

Okay, with that out of the way, what is this? Examine the part inside the sum:

v(a+ndt) dt

This is the velocity of the car at a specific time (a+ndt), a time which, as n increases, goes over the whole trip from a to b. This velocity is multiplied by the amount of time we are traveling at that velocity, dt. In case our velocity is changing all the time, we shrink dt to zero by taking the limit.

In practice, the integral is much more difficult than the derivative to evaluate. But we are helped by noticing something: if we take the derivative of an integral, we get back our original function! Why is this?

Well, think about what it means to change the limit of an integral. Consider this integral:

f(x) = integral from a to x of (v(t) dt).

(this is the distance traveled by our car since time a)

Now, we can imagine a derivative of this distance:

lim (h->0) (f(x+h) - f(x))/h

So we have to figure out what the integral f(x+h) is. As it turns out, that's very easy! The integral f(x+h) is just like taking the original integral, then adding one teeny tiny sliver at the end: we're just adding one extra term to the infinite sum that defines the integral. And that one extra term is:

v(x+h) h

(In order to only add one extra term, we have to set dt = h)

So we have to evaluate:

lim (h->0) of v(x+h) h/h = lim (h->0) of v(x+h) = v(x).

So the derivative of an integral is just the function we were integrating in the first place! It turns out that the reverse is also true: the integral of a derivative is the original function as well. This fact that the integral and the derivative are inverses of one another means in order to evaluate an integral, we try to find a function whose derivative is the argument of the integral.

I should never have dropped out of highschool. My first choice was natural sciences, and after dropping out there I read estetics. But I couldn't keep up with the studies (A womans feet's dragging me down to sheol and so on)!

This is me playing the guitar: Film.wmv - YouTube It is a bad recording on my computer, were I am just improvising.

So my question is when I tune the guitar, why do I hear swings after taking two A's, (e fift fret, and a) and its untuned!? A is 440 hertz, and I hear the swings if the two A's are in incongruence?

Lillen said:

So my question is when I tune the guitar, why do I hear swings after taking two A's, (e fift fret, and a) and its untuned!? A is 440 hertz, and I hear the swings if the two A's are in incongruence?

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Because the two have different frequencies if the tuning is off. Imagine, for example, that one is tuned to 440Hz, and the other is tuned to 442Hz. When you add two sound waves like this, what happens is that they combine to make a sound at 441Hz (the average of the frequencies) which throbs at a rate of 1Hz (half the difference of the frequencies).

So when I strike three tunes, one tuned 220, second 260, third 300 it will create comined sound at 240 and 280? Will that 240 and 280 combine themself to 260 again?