Find an arc length parameterization of the curve that has the same orientation as the given curve and for which the reference point corresponds to t=0. Use an arc length s as a parameter. r(t) = 3(e^t) cos (t)i + 3(e^t)sin(t)j; 0<=t<=(3.14/2)

The derivatives are $$x'(t) = \frac{d}{dt}(3e^tcost) = 3e^tcost - 3e^tsint = 3\sqrt{2}cos(t+\frac{\pi}{4})$$ $$y'(t) = \frac{d}{dt}(3e^tsint) = 3e^tsint + 3e^tcost = 3\sqrt{2}sin(t+\frac{\pi}{4})$$ So norm of the derivative is $$||r'(t)|| = \sqrt{x'(t)^2 + y'(t)^2} = 3\sqrt{2}e^t$$ The arc length parametrization is $$s(t) = \int_0^t ||r'(\tau)|| d\tau$$ $$x(t) = \int_0^t 3\sqrt{2}e^{\tau}d\tau$$$$= (3\sqrt{2}e^{\tau}]_0^t = 3\sqrt{2}(e^t-1)$$ So, the arc length parametrization is $$s(t) = 3\sqrt{2}(e^t-1)$$