A particle is moving along the curve y= 3sqrt3x+1. As the particle passes through the point (5,12), its x-coordinate increases at a rate of 3 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

Answer 1

Let P(x,y) be any point on the curve
then we could write P as (x,3√(3x+1))
and if D is the distance to the origin,
D^2 = x^2 + (3√(3x+1))^2
= x^2 + 27x + 9
2D(dD/dt) = 2x(dx/dt) + 27(dx/dt)

at (5,12) D^2 = 25+135+9 = 169
so D = 13

so dD/dt = (2(5)(3) + 27(3))/(2(13))
= 4.26

check my work please

Answer 2

your answer is good