Determining the Length of a Parametric Curve (Parametric Form)
Key Questions
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Answer:
6sqrt36√3 .Explanation:
The answer is
6sqrt36√3 .The arclength of a parametric curve can be found using the formula:
L=int_(t_i)^(t_f)sqrt(((dx)/(dt))^2+((dy)/(dt))^2)dtL=∫tfti√(dxdt)2+(dydt)2dt . Sincexx andyy are perpendicular, it's not difficult to see why this computes the arclength.It isn't very different from the arclength of a regular function:
L=int_a^b sqrt(1+((dy)/(dx))^2)dxL=∫ba√1+(dydx)2dx . If you need the derivation of the parametric formula, please ask it as a separate question.We find the 2 derivatives:
(dx)/(dt)=3-3t^2dxdt=3−3t2
(dy)/(dt)=6tdydt=6t And we substitute these into the integral:
L=int_0^(sqrt3)sqrt((3-3t^2)^2+(6t)^2)dtL=∫√30√(3−3t2)2+(6t)2dt And solve:
=int_0^(sqrt3)sqrt(9-18t^2+9t^4+36t^2)dt=∫√30√9−18t2+9t4+36t2dt
=int_0^(sqrt3)sqrt(9+18t^2+9t^4)dt=∫√30√9+18t2+9t4dt
=int_0^(sqrt3)sqrt((3+3t^2)^2)dt=∫√30√(3+3t2)2dt
=int_0^(sqrt3)(3+3t^2)dt=∫√30(3+3t2)dt
=3t+t^3|_0^(sqrt3)=3t+t3∣∣√30
=3sqrt3+3sqrt3=3√3+3√3
=6sqrt3=6√3 Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.
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By taking the derivative with respect to
tt ,{(x'(t)=6t),(y'(t)=6t^2):} Let us now find the length
L of the curve.L=int_0^1 sqrt{[x'(t)]^2+[y'(t)]^2}dt =int_0^1 sqrt{6^2t^2+6^2t^4} dt by pulling
6t out of the square-root,=int_0^1 6t sqrt{1+t^2} dt by rewriting a bit further,
=3int_0^1 2t(1+t^2)^{1/2}dt by General Power Rule,
=3[2/3(1+t^2)^{3/2}]_0^1=2(2^{3/2}-1) I hope that this was helpful.
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Please see how I solved this problem, it has a full explanation.
How do you find the length of the curve
x=3t-t^3 ,y=3t^2 , where0<=t<=sqrt(3) ?