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Problem statement:
Find the arc length of the curve defined by x = √t and y = 3t -1 on the interval 0 < t < 1attempted solution:
dx/dt = 1/2t-1/2 , dy/dt = 3 and dx = dt/ 2√t
dy/dx = 6√t
length = ∫01 √(1 + (6√t)2) .dt/ 2√t
= ∫01 √1 + 36t) dt/2√t
now I'm stuck with a product that is very difficult to solve using integration by parts or by trig substitution. Please shed some light.
Find the arc length of the curve defined by x = √t and y = 3t -1 on the interval 0 < t < 1attempted solution:
dx/dt = 1/2t-1/2 , dy/dt = 3 and dx = dt/ 2√t
dy/dx = 6√t
length = ∫01 √(1 + (6√t)2) .dt/ 2√t
= ∫01 √1 + 36t) dt/2√t
now I'm stuck with a product that is very difficult to solve using integration by parts or by trig substitution. Please shed some light.
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