A particle moves along the curve y=lnx so that its abscissa is increasing at a rate of 2 units per second. At what rate is the particle moving away from the origin as it passes through the point (1,e)?

Answer 1

y = lnx
The distance z from the origin is thus

z^2 = x^2 + (lnx)^2
2z dz/dt = 2x dx/dt + 2lnx (1/x) dx/dt
z dz/dt = (x+lnx/x) dx/dt
at (1,e), z = √(1+e^2), so

dz/dt = (1+1/1)(2)/√(1+e^2) = 4/√(1+e^2)