anonymous
  • anonymous
question...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • 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??
UsukiDoll
  • UsukiDoll
oh wow what subject is this?
anonymous
  • anonymous
vector calculus

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UsukiDoll
  • UsukiDoll
really? I wasn't given that type of problem when I took the course. maybe I got lucky :S
anonymous
  • anonymous
Maybe u can help?Take a look anyway pls, I'm only preparing for college, so there's no hurry :)
UsukiDoll
  • UsukiDoll
besides there could be dot product and cross product O_O!
anonymous
  • 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
anonymous
  • anonymous
I was thinking the same but then the 3rd term makes no sense
anonymous
  • anonymous
@dan815
anonymous
  • anonymous
it does.. look closely its box product and two vectors are same
anonymous
  • 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
anonymous
  • anonymous
yep..
IrishBoy123
  • IrishBoy123
second last term you are cross same thing - theta =0 last term you are dotting something with its cross product => 0
anonymous
  • 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

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