A particle is moving along the curve y=sqrt(x)

Question

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

 

=_____cm/s

Answer 1

If P is a point on the curve y = √x, then P = (x, √x).

 

D = distance from P to (0,0) = √ [(x-0)² + (√x-0)²] = √(x²+x)

 

Given:  dx/dt = 3, when the particle passes through (4,2).

 

dD/dt = (1/2)(x²+x)^-½[2x(dx/dt) + dx/dt]

 

At (4,2), dD/dt = (1/2)(18)^-½[8(3) + 3]

 

                       = (1/2)(1/(3√2))(27) = 9/(2√2)

 

                                                       = 9√2/4