Asked by Desperate Student
Find the arc length of the curve
π₯(π‘)=cosπ‘+π‘sinπ‘, 0β€π‘β€π/2
π¦(π‘) = sin π‘ β π‘ cos π‘ ^2
π₯(π‘)=cosπ‘+π‘sinπ‘, 0β€π‘β€π/2
π¦(π‘) = sin π‘ β π‘ cos π‘ ^2
Answers
Answered by
Steve
(ds)^2 = (dx/dt)^2 + (dy/dt)^2
= (-sint + sint + tcost)^2 + (cost - cost + tsint)^2
= t^2cos^2(t) + t^2sin^2(t)
= t^2
so,
ds = dt
that's pretty easy to integrate, right?
Hmmm. I missed that ^2 hanging out there. Is
y(t) = sint - t(cost)^2
or
y(t) = sint - tcos(t^2)?
In either case, adjust the expression for ds above.
= (-sint + sint + tcost)^2 + (cost - cost + tsint)^2
= t^2cos^2(t) + t^2sin^2(t)
= t^2
so,
ds = dt
that's pretty easy to integrate, right?
Hmmm. I missed that ^2 hanging out there. Is
y(t) = sint - t(cost)^2
or
y(t) = sint - tcos(t^2)?
In either case, adjust the expression for ds above.
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