Question

A particle is moving along the curve y= 5sqrt(5x+4) . As the particle passes through the point (1,15), its x-coordinate increases at a rate of 4 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

Answer 1

Let the distance of the particle from the origin be r.
r= sqrt(x^2+y^2)

From the equation of the curve y=5sqrt(5x+4)
we have y^2= 25(5x+4) = 125x+100

using y^2 in the definition of r:
r= sqrt(x^2 + 125x+100)
dr/dt = 0.5(2x dx/dt + 125 dx/dt)/sqrt(x^2 + 125x+100)

at x=1, dx/dt is given as 4
thus
dr/dt = 254/sqrt(226)

Answer 2

Thank you soooo much phi!