Find the arclength of the curve rt4sqrt2te4te4t 0t1Solutionr

Find the arclength of the curve r(t)=4sqrt(2t),e^4t,e^4t, 0t1

Solution

r(t)=<4sqrt(2t),e^4t,e^4t>

r \'(t)=<4/sqrt(2t),4e^4t,-4e^4t>

|r \'(t)|=sqrt[(4/sqrt(2t))²+(4e^4t)²+(-4e^-4t)²]

|r \'(t)|=sqrt[(8/t)+(16e^8t)+(16e^-8t)]

arclength =integral_{[0 to 1]} |r\'(t)| dt

=integral_{[0 to 1]} sqrt[(8/t)+(16e^8t)+(16e^-8t)] dt

=55.67 units

arclength =55.67 units