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give vector v = 1/2
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um, the i and j seem out of place there; those tend to be reserved for a more equation looking vector notation
ok try this then...u=1/2(sqrt(3)i-j)
thats better :)
so, what we have is: \[\vec v=<\frac{\sqrt{3}}{2},-\frac{1}{2}>\] which looks to me like points on the unit circle
sqrt((3+1)/4) = 1
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ok.... so is it a unit vector as is?
as is, its a unit vector; since its magnitude = 1
thanks quick question if you could. When we say directional derivative, that means DuF right.
http://mathworld.wolfram.com/DirectionalDerivative.html in a sense I would say yes
What i mean is, when we say find the dirctional derrivative its reffering to that.
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