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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[\frac{d}{dt}(\vec V.\frac{d \vec V}{dt} \times \frac{d^2 \vec V}{dt^2})\]\[\vec V . \frac{d \vec V}{dt} \times \frac{d^3 \vec V}{dt^3}+\vec V.\frac{d^2 \vec V}{dt^2} \times \frac{d^2 \vec V}{dt^2}+\frac{d \vec V}{dt}.\frac{d \vec V}{dt} \times \frac{d^2 \vec V}{dt^2}\] 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??
oh wow what subject is this?
vector calculus

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really? I wasn't given that type of problem when I took the course. maybe I got lucky :S
Maybe u can help?Take a look anyway pls, I'm only preparing for college, so there's no hurry :)
besides there could be dot product and cross product O_O!
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
I was thinking the same but then the 3rd term makes no sense
it does.. look closely its box product and two vectors are same
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
yep..
second last term you are cross same thing - theta =0 last term you are dotting something with its cross product => 0
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

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