Question: A bug is crawling along the curve defined by the equation $x^2=y^3+3y$. When the bug is at the point $(2,1)$, it is moving in such a way that its vertical velocity, $\dfrac{dy}{dt}$, is equal to $6$. What is its horizontal velocity, $\dfrac{dx}{dt}$, at that time?
I know that at this point we can utilize some sort of optimization, i.e find a given point where the derivative is equal to 0, but after racking my brain at this problem for a couple hours now, I really have to conclude that I'm stuck. Any help would be greatly appreciated, and if you do give an answer, please justify it as well! Thank you!
Using implicit differentiation by $t$, we have $\large{2x\frac{dx}{dt}=3y^2\frac{dy}{dt}+3\frac{dy}{dt}}$. Now let's substitute $x, y, \frac{dy}{dt}$ with their values and find $\frac{dx}{dt}$: $$2\cdot 2\cdot \frac{dx}{dt}=3\cdot 1^2 \cdot 6 +3\cdot 6$$ $$4\frac{dx}{dt}=36$$ $$\frac{dx}{dt}=9$$