Parametric equation Length of Curve?

Length Of Curve Equation

ds = ?(4/9 + 16t^2/(t^2-9)^2) dt = (4/9)(t^2+9)/(t^2-9) for this reason the arc length of the curve (no longer of the equation!!!!!) is the quintessential of this expression from 4 to 5. This equals 4/9 cases a million + [3(ln((t-3)/(t+3))] the position the 2d time period is from 4 to 5, giving us the curve length of (4/9)(a million + 3 ln(7/4)).

Since x' = 19(-sin t + (sin t + t cos t)) = 19t cos t

and y' = 19(cos t - (cos t - t sin t)) = 19t sin t, we have √((x')^2 + (y')^2) = 19t.

Hence, the arc length equals

∫(t = 0 to π/2) 19t dt = 19π²/8.

I hope this helps!

Let t = theta for simplified typing.

dx/dt = 19*(-sin(t) + t*cos(t) + sin(t)) = 19t*cos(t)

dy/dt = 19*(cos(t) + t*sin(t) - cos(t)) = 19t*sin(t)

Length = ∫√(dx/dt)² + (dy/dt)² dt from 0 to pi/2

= ∫19t dt from 0 to pi/2

= 19/2 * (pi²/4) = 19pi²/8