xt ln t t22 212 1

Solution

Let at time t, x(t)=(f(t),g(t),h(t)) [functional form of the 3 co-ordinates)
At time t+dt, x(t+dt)=(f(t+dt),g(t+dt),h(t+dt))
Now, f\'(t)=(f(t+dt)-f(t))/dt
or, f(t+dt)=f(t)+f\'(t)dt
Hence, x(t+dt)=(f(t)+f\'(t)dt,g(t)+g\'(t)dt+h(t)+h\'(t)dt)
So, length of the path between the 2 points x(t+dt) and x(t)
=[{f(t+dt)-f(t)}2+{g(t+dt)-g(t)}2+{h(t+dt)-h(t)}2]
=[{f\'(t)}2+{g\'(t)}2+{h\'(t)}2]dt [dt is outside the root]
=s(t)dt [Let us name it this way for easier referral]
Hence, distance b/w 2 points x(t1) and x(t2)
=Int(s(t)dt) from t1 to t2
In your question,
f(t)=lnt f\'(t)=1/t
g(t)=t^2/2 g\'(t)=t
h(t)=2^1/2 h\'(t)=0
So, s(t)=[(1/t)^2+t^2+0]
=[t^2+1/t^2]
Integrate between t=1 and t=4 to get the answer.