x=e^(-t)cost , y=e^(-t)sint , 0

The formula of arc length of a parametric equation on the interval alt=tlt=b is:
L = int_a^b sqrt((dx/dt)^2+(dy/dt)^2) dt
The given parametric equation is:
x = e^(-t)cost
y=e^(-t)sint
The derivative of x and y are with respect to t are:
dx/dt = e^(-t) * (cost)' + (e^(-t))'*cost
dx/dt = e^(-t)*(-sint) + e^(-t)*(-1)cost
dx/dt=-e^(-t)sint-e^(-t)cost
dy/dt = e^(-t)*(sint)' + (e^(-t))'*sint
dy/dt = e^(-t)cost + e^(-t)*(-1)sint
dy/dt=e^(-t)cost - e^(-t)sint
Plugging them to the formula, the integral needed to compute the arc length of the given parametric equation on the interval 0lt=tlt=pi/2 is:
L= int_0^(pi/2) sqrt( (-e^(-t)sint-e^(-t)cost)^2 + (e^(-t)cost - e^(-t)sint)^2) dt
The simplified form of the integral is:
L= int_0^(pi/2) sqrt( (-e^(-t)(sint + cost))^2+ (e^(-t)(cost-sint))^2)dt
L=int_0^(pi/2)sqrt( e^(-2t)(sint+cost)^2 + e^(-2t)(cost-sint)^2) dt
L=int_0^(pi/2) sqrt(e^(-2t)((sint+cost)^2 + (cost-sint)^2) )dt
L= int_0^(pi/2) e^(-t) sqrt((sint+cost)^2+(cost-sint)^2)dt
L=int_0^(pi/2) e^(-t) sqrt(sin^2t +2sintcost +cos^2t+cos^2t -2sintcost +sin^2t)dt
L= int_0^(pi/2) e^(-t) sqrt(2sin^2t + 2cos^2t)
L= int_0^(pi/2) e^(-t)sqrt(2(sin^2t+cos^2t))
L= int_0^(pi/2) e^(-t) sqrt(2*(1))dt
L= int_0^(pi/2) e^(-t) sqrt(2)dt
L= sqrt2 int_0^(pi/2) e^(-t)dt
L= -sqrt2 e^(-t) |_0^(pi/2)
L =-sqrt2 (e^(-pi/2) - e^0)
L=-sqrt2(e^(-pi/2)-1)
L=sqrt2 - sqrt2e^(-pi/2)
Therefore, the arc length of the curve is  sqrt2 - sqrt2e^(-pi/2) units.