Find the rate of change of the distance from the particle to the origin at this instant.

Question

A particle is moving along the curve y=3sqrt3x+7. As the particle passes through the point (3,12), its x-coordinate increases at a rate of 5 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

Answer 1

A function that gives the distance of a particle on the curve to the origin is given by:

D(x) = √[(x-0)²+((3√3x+7)²] = √[x²+9(3x+7)]=√(x²+27x+63)

Taking the derivative with respect to t:

dD/dt = 1/2 (x²+27x+63)^-1/2(2x+27) dx/dt

At this point substitute 3 for x and 5 for dx/dt to find out how fast the distance of the particle from the origin is changing at the instance when x = 3 (and y = 12).

Answer 2

dy/dt = 3/2(3x + 7)^-1/2·3dx/dt = 9/2(3x + 7)^-1/2dx/dt. At point (3, 12) dx/dt = 5 units/s; dy/dt = 9/2·(9 + 7)^-1/2·5 =

= 45/8 units/s = 5.625 units/s

Velocity of particle at this point v = (5² + 5.625²)^1/2 = 7.526 units/s