Finding $\frac{dy}{dt}$ given a curve and the speed of the particle Today I was doing some practice problems for the AP Calculus BC exam and came across a question I was unable to solve.

In the xy-plane, a particle moves along the parabola $y=x^2-x$ with a constant speed of $2\sqrt{10}$ units per second. If $\frac{dx}{dt}>0$, what is the value of $\frac{dy}{dt}$ when the particle is at the point $(2,2)$?

I first tried to write the parabola as a parametric equation such that $x(t)=at$ and $y(t)=(at)^2-at$ and then find a value for $a$ such that $\displaystyle\int_0^1\sqrt{(x'(t))^2+(y'(t))^2}dt=2\sqrt{10}$. However, since it was a multiple choice question we were probably not supposed to spend more than 3min on the question so I though that my approach was probably incorrect. The only information that I know for sure is that since $\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}\rightarrow\frac{dy}{dt}=(2x-1)\frac{dx}{dt}$ and we are evaluating at $x=2$ and so $\frac{dy}{dt}=3\frac{dx}{dt}$. Other than that I am not sure how to proceed and so any help would be greatly appreciated!

As you noticed, $dy=(2x-1)dx$. The speed is constant, so that $\dot{x}^2+\dot{y}^2=40$. So you get the system $$\begin{cases} \dot{y}=(2x-1)\dot{x} \\\ \dot{x}^2+\dot{y}^2=40 \end{cases}.$$ By substitution, $\left[1+(2x-1)^2 \right]\dot{x}^2 = 40$, whence, at $x=2$, $\dot{x}^2=4$. Since $\dot{x}>0$, you find $\dot{x}=2$, and then $\dot{y}=(2x-1)\dot{x}=3 \cdot 2 = 6$. I hope I did not make mistakes.
