A particle is moving along the curve y=5 sqrt (2x+6). As the particle passes through the point (5,20 , 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

y = 5√(2x+6)
dy/dt = 5/√(2x+6) dx/dt

d^2 = x^2+y^2
d dd/dt = x dx/dt + y dy/dt
at (5,20), d=5√17, dy/dt = 5/4
5√17 dd/dt = 5*5 + 20*5/4
dd/dt = (25+25)/(5√17) = 10/√17