Asked by Kaitlyn
Graph the curve. Find the exact length of the curve.
x = 2et cos(t), y = 2et sin(t), 0 ≤ t ≤ π
x = 2et cos(t), y = 2et sin(t), 0 ≤ t ≤ π
Answers
Answered by
Steve
Assuming you mean
x = 2e^t cos(t)
y = 2e^2 sin(t)
you know that
x=cos(t)
y=sin(t)
is a circle, So, this is just a spiral, of radius 2e^t for angle t.
http://www.wolframalpha.com/input/?i=x+%3D+2e^t+cos%28t%29%2C+y+%3D+2e^t+sin%28t%29+for+0%3C%3Dt%3C%3Dpi
Now, the arc length is
∫[0,π] √((dx/dt)^2 + (dy/dt)^2) dt
= ∫[0,π] √((2e^t(cost-sint))^2 + (2e^t(cost+sint))^2) dt
= ∫[0,π] √(8e^(2t)) dt
= ∫[0,π] 2√2 e^t dt
= 2√2 (e^π - 1)
x = 2e^t cos(t)
y = 2e^2 sin(t)
you know that
x=cos(t)
y=sin(t)
is a circle, So, this is just a spiral, of radius 2e^t for angle t.
http://www.wolframalpha.com/input/?i=x+%3D+2e^t+cos%28t%29%2C+y+%3D+2e^t+sin%28t%29+for+0%3C%3Dt%3C%3Dpi
Now, the arc length is
∫[0,π] √((dx/dt)^2 + (dy/dt)^2) dt
= ∫[0,π] √((2e^t(cost-sint))^2 + (2e^t(cost+sint))^2) dt
= ∫[0,π] √(8e^(2t)) dt
= ∫[0,π] 2√2 e^t dt
= 2√2 (e^π - 1)
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