Question

question...

Answer 1

\[\frac{d}{dt}(\vec V.\frac{d \vec V}{dt} \times \frac{d^2 \vec V}{dt^2})\]\[\vec V . \frac{d \vec V}{dt} \times \frac{d^3 \vec V}{dt^3}+\vec V.\frac{d^2 \vec V}{dt^2} \times \frac{d^2 \vec V}{dt^2}+\frac{d \vec V}{dt}.\frac{d \vec V}{dt} \times \frac{d^2 \vec V}{dt^2}\]
This becomes\[\vec V . \frac{d \vec V}{dt} \times \frac{d^3 \vec V}{dt^3}+0+0=\vec V . \frac{d \vec V}{dt} \times \frac{d^3 \vec V}{dt^3}\]
Why are the last 2 terms zero??

Answer 2

oh wow what subject is this?

Answer 3

vector calculus

Answer 4

really? I wasn't given that type of problem when I took the course. maybe I got lucky :S

Answer 5

Maybe u can help?Take a look anyway pls, I'm only preparing for college, so there's no hurry :)

Answer 6

besides there could be dot product and cross product O_O!

Answer 7

i do have an idea about the second one.. both the terms are same in cross product so the angle between them is 0 ... so the result will be zero

Answer 8

I was thinking the same but then the 3rd term makes no sense

Answer 9

@dan815

Answer 10

it does.. look closely its box product
and two vectors are same

Answer 11

Oh so it's like one those methods where u re-arrange triple product and it remains the same??In that way u can arrange the same vectors to come into the cross product so that it becomes 0

Answer 12

yep..

Answer 13

second last term you are cross same thing - theta =0
last term you are dotting something with its cross product => 0

Answer 14

I thought that stuff was not important at all, I just didn't want to learn the order you have to know when re-arranging
but now I've got it down....
also dot product of a vector with it's cross product with another vector is 0 ??
let's see,
cross product of A and B will be perpendicular to both A and B
so the dot product of this new vector with A will be....theta will be 90 cos90=0...
I understand it all now