a particle moving calculus question

Question

A particle is moving along the curve y=5(sqrt 5x-1) . As the particle passes through the point (3,20), , its x coordinate increases at a rate of 2 units per second. Find the rate of change of the distance from the particle to the origin at that instant.

Answer 1

The curve must have equation

y = 5 √(5 x+ 1) otherwise the particle does not pass through the point (3, 20)

We are given that d x/ d t = 2 units per second and we are ask to find d s/ d t

where s is the distance or the moving particle from the origin.

But S² = x² + y² that is s² = x² + 125 x + 25 then

(d / dt) (s² ) = 2 x d x / d t + 125 d x/ d t

2 s ds/dt = 2 x d x / d t + 125 d x/ d t . Also when the particle passes through the point ( 3, 20 )

then s = √(409) hence

2 √(409) ds/dt = 2·3 ·2 + 250

√(409) ds/dt = 131

ds/dt = 131 / √(409) units per second when the particle is getting away from the vertex of the parabola.

When the particle approaches the vertex the d x / d t = -2 hence ds/dt = -131 / √(409