## anonymous one year ago question...

1. anonymous

$\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??

2. UsukiDoll

oh wow what subject is this?

3. anonymous

vector calculus

4. UsukiDoll

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

5. anonymous

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

6. UsukiDoll

besides there could be dot product and cross product O_O!

7. anonymous

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

8. anonymous

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

9. anonymous

@dan815

10. anonymous

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

11. anonymous

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

12. anonymous

yep..

13. IrishBoy123

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

14. anonymous

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