Here's a question for you to try out.

[dcc]

Im still not convinced that saying that

What about , does that mean still holds true?

[Mao]

Im still not convinced that saying that

What about , does that mean still holds true?

that is a valid point (as for all vectors)

but when neither are zero vectors, that would be true.

[dcc]

Im still not convinced that saying that

What about , does that mean still holds true?

that is a valid point (as for all vectors)

but when neither are zero vectors, that would be true.

So the question perhaps should of been framed:

"Prove that if the speed of a particle is constant and the particle is accelerating, then the acceleration of the particle is perpendicular to the velocity."

[Mao]

"Prove that if the speed of a moving particle is constant and the particle has a non-zero acceleration, then the direction of acceleration of the particle is perpendicular to the velocity."

[Lycan]

Eh, the particle was stated to be moving. I also in a 'curve' which was meant to imply non-zero acceleration but then lines are also considered curves aren't they. Anywho, the dot product rule is correct. Although I had:

x^2 + y^2 = c^2 and differentiated both sides in terms of T for a more elegant solution.

[Mao]

Eh, the particle was stated to be moving. I also in a 'curve' which was meant to imply non-zero acceleration but then lines are also considered curves aren't they. Anywho, the dot product rule is correct. Although I had:

x^2 + y^2 = c^2 and differentiated both sides in terms of T for a more elegant solution.

but that only works for the particular case of (you can similarly show for higher dimensions, etc)

[Lycan]

It was just a general idea, of course you can simply extend it to n dimensions.

x^2 + y^2 + z^2 ... = c^2

[Mao]

mind posting up your proof?

[Lycan]

Sure.

Let's just pretend these are all vectors and look nice and stuff.

Let v(t) = x(t)i + y(t)j + z(t)k + ... n dimensions

Speed = ([x(t)]^2 + [y(t)]^2 + [z(t)]^2)^(1/2) + ...= k, where k is a constant

Therefore [x(t)]^2 + [y(t)]^2 + [z(t)]^2 + ... = k^2

d/dt([x(t)]^2 + [y(t)]^2 + [z(t)]^2 + ... ) = d/dt(k^2) = 0

Using chain rule:

2x(t)x'(t) + 2y(t)y'(t) + 2z(t)z'(t) + ... = 0

x(t)x'(t) + y(t)y'(t) + z(t)z'(t) + ... = 0

a(t)= d/dt(v(t)) = x'(t)i + y'(t)j + z'(t)k + ....

a.v = x(t)x'(t) + y(t)y'(t) + z(t)z'(t) + ...= 0

Hence they are perpendicular.

[Mao]

lolol, your solution is subjected to the same critique dcc gave to mine xD

[dcc]

lolol, your solution is subjected to the same critique dcc gave to mine xD

Not if he is answering the revised question, which implies that neither or are zero


p.s. I might add, that if you are trying to prove that two vectors are perpendicular using the dot product, then it is simply not enough to state:

as the last line of your proof

rather, you MUST state:

hence

Otherwise you have not proved as to whether the two vectors are perpendicular

[Lycan]

I realise that. I was just rushing as I was typing it up before heading off to school.