Asked by anonymous
if given a certain value how would I find what angle does the unit tangent make with the x-axis?
Answers
Answered by
oobleck
that would be dy/dx
Answered by
anonymous
so my unit tangent vector T(w) = cos(w)i+sin(w)j and say w=-pi/3 how would i find what angle does the unit tangent vector T(w) make with the x-axis?
Answered by
oobleck
since you have parametric equations,
y = sinw
x = cosw
dy/dx = (dy/dw) / (dx/dw) = cosw/-sinw = -cotw
or note that the equation could have been written
x^2 + y^2 = 1
x + y y' = 0
y' = -x/y = -cosw/sinw = -cotw
y = sinw
x = cosw
dy/dx = (dy/dw) / (dx/dw) = cosw/-sinw = -cotw
or note that the equation could have been written
x^2 + y^2 = 1
x + y y' = 0
y' = -x/y = -cosw/sinw = -cotw
Answered by
anonymous
and then i would plug in my value for w to find the angle?
Answered by
oobleck
oops. sorry. I didn't notice that T was the tangent vector. I took it to be the original position vector. So you must have had
r(w) = sinw i - cosw j
so things are backwards. But you can see how to do it.
r(w) = sinw i - cosw j
so things are backwards. But you can see how to do it.
Answered by
anonymous
my r(w) = wi-ln(cosw)j so r'(w) = i + tanwj. so to find the angle it would be tan(w)
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