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?
2019-05-01T05:45:00Z
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.
Indica2019-05-01T06:37:56Z
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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
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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§οΏ½οΏ½οΏ½β)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