Asked by Sam
Find the length of the curve defined by the parametric equations x = 2/3t,
y = 2ln((t/3)^2−1) from t =6 to t =7.
y = 2ln((t/3)^2−1) from t =6 to t =7.
Answers
Answered by
Steve
just crank it out:
s = ∫[6,7] √(x'^2+y'^2) dt
= ∫[6,7] √((2/3)^2+(4t/(t^2-9))^2) dt
= ∫[6,7] √((2/3)^2+(4t/(t^2-9))^2) dt
= ∫[6,7] √((t^2+9)^2/(9(t^2-9)^2)) dt
= 2/3 ∫[6,7] (t^2+9)/(t^2-9) dt
= 2/3 ∫[6,7] (t^2+9)/(t^2-9) dt
= 2/3 ∫[6,7] 1 + 3/(t-3) - 3/(t+3) dt
= 2/3 (t+3ln(t-3)-3ln(t+3))[6,7]
= 2/3 + 2ln(6/5)
s = ∫[6,7] √(x'^2+y'^2) dt
= ∫[6,7] √((2/3)^2+(4t/(t^2-9))^2) dt
= ∫[6,7] √((2/3)^2+(4t/(t^2-9))^2) dt
= ∫[6,7] √((t^2+9)^2/(9(t^2-9)^2)) dt
= 2/3 ∫[6,7] (t^2+9)/(t^2-9) dt
= 2/3 ∫[6,7] (t^2+9)/(t^2-9) dt
= 2/3 ∫[6,7] 1 + 3/(t-3) - 3/(t+3) dt
= 2/3 (t+3ln(t-3)-3ln(t+3))[6,7]
= 2/3 + 2ln(6/5)