A particle moves along the curve y = ln(x/2) where x > 0

Question

A particle moves along the curve y = ln(x/2) where x > 0 As the particle passes through the point (2e, 1) its x-coordinate increases at a rate of sqrt(4e ^ 2 + 1) * cm / s How fast is the distance from the particle to the origin changing at this instant?

Answer 1

Dist = √(x² + y²)

D(x) = √(x² + ln²(x/2))

dD/dt = (2x + 1/x·2ln(x/2) / (2√(x² + ln²(x/2)))·dx/dt where dx/dt = √(4e² - 1). Plug in x = 2e and evaluate.