`x=e^(-t)cost , y=e^(-t)sint , 0<=t<=pi/2` Find the arc length of the curve on the given interval.

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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.

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