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

For the first part, you want to compare $\frac{\mathrm dx}{\mathrm dt}$ and $\frac{\mathrm dy}{\mathrm dt}$, not $x$ and $y$. Remember, the problem is referred to as "related rates" for a reason. Using the Chain Rule, you get
$$\frac{\mathrm dy}{\mathrm dt} = \frac{\mathrm dy}{\mathrm dx} \cdot \frac{\mathrm dx}{\mathrm dt} \tag{1}$$
Hence, $\frac{\mathrm dx}{\mathrm dt} = \frac{\mathrm dy}{\mathrm dt}$ when $\frac{\mathrm dy}{\mathrm dx} = 1$. You have $y = \sqrt{x}$, so where does this occur?
For the second part, you have distance too. Whenever you have distance, you're dealing with $r^2 = (\Delta x)^2 + (\Delta y)^2$, by the Pythagorean Theorem. Since you start from the origin, this just becomes $r^2 = x^2+y^2$. Differentiating implicitly with respect to time gives
$$\frac{\mathrm d}{\mathrm dt}r^2 = \frac{\mathrm d}{\mathrm dt} \left(x^2+y^2\right)$$
$$2r\frac{\mathrm dr}{\mathrm dt} = 2x\frac{\mathrm dx}{\mathrm dt} + 2y\frac{\mathrm dy}{\mathrm dt}$$
$$r\frac{\mathrm dr}{\mathrm dt} = x\frac{\mathrm dx}{\mathrm dt} + y\frac{\mathrm dy}{\mathrm dt}$$
Well, you're given $x = 4$, so you can find $y$ by $y = \sqrt{x}$, meaning you can find $r$ as well. You're also given $\frac{\mathrm dx}{\mathrm dt}$. This just leaves $\frac{\mathrm dy}{\mathrm dt}$, which you can find by using $(1)$. Finally, plug in the values and find $\frac{\mathrm dr}{\mathrm dt}$.
