Question

Observe only, please!

Answer 1

Evaluate the integral \[\int\limits \textbf A \times \textbf A'' dt\]

Answer 2

@Empty

Answer 3

This one is one of my favorites. Hahaha

Answer 4

Show me

Answer 5

Hint:\[ \textbf A' \times \textbf A' = 0\]

Answer 6

This is a fun problem, I actually just found this in a book and was stumped for a whole day last week solving it, so this is a nice coincidence.

Answer 7

So can I integrate A and then differentiate A'' at the same time

Answer 8

Hahaha I guess I came from this when considering a physics problem:

we can calculate torque as the derivative of angular momentum. Therefore the integral of torque is angular momentum.

So from this definition of angular momentum:

\[\bar L = \bar r \times \bar p\]

Take the derivative of \(\bar L\) to show that \[\frac{d \bar L}{dt} = \bar \tau\]
where
\[\bar \tau = \bar r \times \bar F\]

Answer 9

Using that differentiating trick you once posted, where you can differentiate in the integral, and yeah haha it's in my physics textbook, but we never did this chapter, it's one of the most difficult I read a bit of it and did not understand a thing, this is where I learnt vectors are transformations and stuff...haha.

Answer 10

That's ok, just calculate this derivative of this function, and we'll go from there:

\[\bar L = \bar r \times \bar p\]

L is the angular momentum, r is the position, p is the momentum.

Answer 11

\[\frac{ \vec dL }{ dt } = \frac{ \vec dr }{ dt } \times \frac{ \vec dp }{ dt }\] should I be doing this

Answer 12

\[\vec r \times \vec F = \vec r \times \vec p\]

Answer 13

You're on the right path, now you just need to separate it out like you would the product rule:

\[(fg)' = f'g+fg'\]

Similarly we have

\[(\bar a \times \bar b)' = \bar a ' \times \bar b + \bar a \times \bar b '\]

Also we can sorta just see this is true by remembering:

\[\bar a \times \bar b = | a| |b| \sin \theta \] If we interchange the order of a and b, then we will be changing the orientation of the angle between them to be in the opposite direction:

\[\bar b \times \bar a = | a| |b| \sin -\theta = -| a| |b| \sin \theta \]

so we can sorta see that we have discovered

\[\bar a \times \bar b = - \bar b \times \bar a \]

Which is the "right hand rule", a result of seeing that it is dependent on an odd function like this. Sorta weird but this is also the same reason interchanging rows of a matrix's determinant causes the sign to flip as well. The determinant is just like a more generalized cross product... Not to overwhelm you or anything! But you often see determinants used for calculating cross products normally, so this shouldn't be too much of a stretch and the linear algebra knowlege will come eventually. :)

Answer 14

I see what you're doing, and it looks like you're sorta skipping ahead so let's try to focus on just this single step: \[\frac{d}{dt} (\vec r \times \vec p) = \]

Apply the product rule here, you will get a sum of two terms.

Answer 15

\[\frac{ d }{ dt }(\vec r \times \vec p) = \vec r~' \times \vec p + \vec r \times \vec p ~ '\]

Answer 16

PERFECT!!!

I think the main problem is you know what the result should be so you're frustrated because what you're getting right now isn't that because there are some tricky steps. I remember us having this problem before, you're anticipating what it should be when you should just let yourself pretend you're dumb temporarily and don't know what's coming haha XD

Answer 17

Lol I'm actually not, pretty happy about this, I don't mind making mistakes in this kind of stuff, it's pretty hard and the only way I'll learn it is through the struggle :P

Answer 18

Ok so now let's resolve these two terms, what do we have, can we rewrite these at all?

Answer 19

I think we can use that good thinking! Let's look at the first term and plug this in:
\[\vec p = m \vec v\]

Answer 20

Oh ok lol I tried it out, but wasn't sure it was the right way, and yeah haha that sounds good

Answer 21

So what does multiplying a vector quantity by a scalar quantity do to it? Describe it visually, not mathematically.

Answer 22

Well if we multiply a scalar by a vector it gives us a scale factor of the vector without changing the direction

Answer 23

So