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!
Best Answer
You can't assume $x$ increases at a constant rate so saying $x=at$ isn't guaranteed to work.
Instead, observe the constant speed means that
$$\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=2\sqrt{10}. \tag{*}$$
We also have by implicit differentiation,
$$\frac{dy}{dt}=(2x-1)\frac{dx}{dt}$$
so that at $x=2$ we can say $\displaystyle \frac{dx}{dt}=\frac{1}{3}\frac{dy}{dt}$. Plug this into $(*)$ and see if you can solve for $\displaystyle\frac{dy}{dt}$...
(You also need to see if $y\,'$ is positive or negative independently.)