Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Finding the Principle Unit Vector?
Okay so I find T which is (dr/dt)/|dr/dt|
This is the unit tangent vector, and we plug in the t value.
Then we find dT/dt/|dT/dt|, but this usually results in 3 quotient rules. Is there a way to do this that doesn't involve the quotient rule?
The problem I keep running into is I simply do not have enough room on a line to do this for my x coordinate, y coordinate and z coordinate, and I do not write big. I usually have a square root in the denominator, which creates a nasty chain rule inside the quotient rule, and I feel like there is a simpler way to do this that I missed.
1 Answer
- IndicaLv 72 years agoFavorite Answer
π» = π§β /|π§β| where β means d/dt
The alternative definition of π΅ = (dπ»β/dt) / |dπ»β/dt| is (π§βΓ(π§ββΓπ§β)/ (|π§β||π§ββΓ π§β|)
So just work out π§β, π§ββ, π§ββΓπ§β, π§βΓ(π§ββΓπ§β) and then normalize it and you are there
β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.
Itβs derivation is β¦
By quotient rule π»β = ( |π§β|π§ββ β |π§β|βπ§β ) / |π§β|Β² β¦ (i)
To find |π§β|β differentiate |π§β|Β² = π§ββ’π§β to get 2|π§β||π§β|β = 2π§ββ’π§ββ giving |π§β|β = (π§ββ’π§ββ) / |π§β|
Sub this in (i) for π»β = ( |π§β|π§ββ β (π§ββ’π§ββ)π§β/|π§β| ) / |π§β|Β² = ( |π§β|Β²π§ββ β (π§ββ’π§ββ)π§β ) / |π§β|Β³
This can be written π»β = ( (π§ββ’π§β)π§ββ β (π§ββ’π§ββ)π§β ) / |π§β|Β³ = (π§βΓ(π§ββΓπ§β)) / |π§β|Β³ using vector triple product
β΄ π΅ = (π§βΓ(π§ββΓπ§β)) / |π§βΓ(π§ββΓπ§β)| = (π§βΓ(π§ββΓπ§β)) / (|π§β||π§ββΓπ§β|) since π§β and π§ββΓ π§β are perp
