Related Rates (point moving along a curve) Consider a point moving along the curve $$f(x) = \sqrt x$$.
a). Find the position of the point on the curve where both coordinates of the point are changing at the same rate.
b). If $\dfrac{dx}{dt}$ is $2 \text{ m/sec}$ at the point $(4,f(4))$, how fast is the point moving away from the origin?

My attempt:
Find point where dy/dt = dx/dt,
Given $y = \sqrt x$ ==>  $dy/dt = 1/2\sqrt x$* dx/dt
and this is where I'm still lost

a) No.  You're not looking for the point where the x and y have the same value, you're looking for the point where the values are changing at the same rate.  Whenever you see "change" in calculus, that is a free clue that you should be thinking about the derivative.  So you are looking for the point where $\frac{dy}{dx}=1$.
b) For this, we are being asked about how quickly a different quantity is changing, so we will need a new function to take the derivative of.  The distance from the origin  to $(x,(f(x))$ is given by $$g(x)=\sqrt{x^2+(f(x))^2}=\sqrt{x^2+(\sqrt x)^2}=\sqrt{x^2+x}$$  The problem is asking you to calculate $\frac {dg}{dt}$ at $x=4$ given that $\frac{dx}{dt}=2$.  That looks like a lot to work through, but remember that the Chain Rule tells us that $\frac{dg}{dt}=\frac{dg}{dx}\cdot\frac{dx}{dt}$, so you've got just enough clues to work it out.
