A particle moves along the top of the
parabola $y^2 = 2x$ from left to right at a constant speed of 5 units
per second. Find the velocity of the particle as it moves through
the point $(2, 2)$.
So I isolate $y$, giving me $y=\sqrt{2x}$. I then find the derivative of $y$, which is $1/\sqrt{2x}$. And $\sqrt{2x}=y$ so the derivative of also equal to $1/y$. At $(2,2)$ the derivative is 0.5. Not sure where to go from there though. Any help is appreciated.
Best Answer
$(2y) dy/dx=2$; $dy/dx =1/y$.
Slope at $(2,2)$: $dy/dx =1/2= \tan \alpha$.
$\cos \alpha =2/√5$, $\sin \alpha =1/√5.$ (Pythagoras).
$v_x= v \cos \alpha$; $v_y = v \sin \alpha$, where $v = |\vec v|$ .
$\vec v = (v_x,v_y) $.