The formula to obtain the arc length ss
s=int_(t_1)^(t_2) sqrt((dx/dt)^2+(dy/dt)^2)*dts=∫t2t1√(dxdt)2+(dydt)2⋅dt
From the given f(t)=(t-sqrt(t^2+2), t+t*e^(t-2))f(t)=(t−√t2+2,t+t⋅et−2)
We have x=t-sqrt(t^2+2)" "x=t−√t2+2 and y= t+t*e^(t-2)y=t+t⋅et−2
We need to obtain the derivatives (dx)/dtdxdt and (dy)/dtdydt
(dx)/dt=1-t/sqrt(t^2+2)" "dxdt=1−t√t2+2 and (dy)/dt=1+t*e^(t-2)+e^(t-2)dydt=1+t⋅et−2+et−2
We solve now the arclength ss
s=int_(t_1)^(t_2) sqrt((dx/dt)^2+(dy/dt)^2)*dts=∫t2t1√(dxdt)2+(dydt)2⋅dt
s=int_(-1)^(1) sqrt((1-t/sqrt(t^2+2))^2+(1+t*e^(t-2)+e^(t-2))^2)*dts=∫1−1√(1−t√t2+2)2+(1+t⋅et−2+et−2)2⋅dt
The integral is complicated and requires the use of calculator or Simpson's Rule formula
s=int_(-1)^(1) sqrt((1-t/sqrt(t^2+2))^2+(1+t*e^(t-2)+e^(t-2))^2)*dts=∫1−1√(1−t√t2+2)2+(1+t⋅et−2+et−2)2⋅dt
s=3.2352212144206" "s=3.2352212144206 units
God bless....I hope the explanation is useful.
