Related Rates, Help!!!!!?

Question

a particle is moving along the curve y= sqrt(x). As the particle passes through the point (4,2) , its x-coordinate increases at a rate of 3 cm/s. How fast is the distance from the particle to the origin at this instant?

Answer 1

The distance that the particle is from the origin is:
√(x^2 + y^2)
But y = √x ===> y = x^2
d/dt √(x^2 + y^2) = d/dt √(2x^2) = (√2)dx/dt = (√2)3cm/s