Question

A particle is moving along the curve y=4sqrt(3x+3). As the particle passes through the point (2,12), 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 of this instant.

Answer 1

r^2 = x^2 +y^2
2r dr/dt = 2x dx/dt +2y dy/dt
\[\frac{ dr }{dt } = \frac{ x \frac{ dx }{dt }+ 4\sqrt{3x+3}*\frac{ 3 }{ 8 }\frac{ 1 }{ \sqrt{3x+3} }\frac{ dx }{dt }}{\sqrt{x^2+y^2} }\]
\[\frac{ dr }{dt } = \frac{ x \frac{ dx }{dt }+ \frac{ 12 }{ 8 }\frac{ dx }{dt }}{\sqrt{x^2+y^2} }\]