ap calculus bc mcq practice test b\n28. in the xy - plane, a particle moves along the parabola y = x² - x…

Explanation:

Step1: Differentiate $y$ with respect to $t$

Using the chain - rule, if $y=x^{2}-x$, then $\frac{dy}{dt}=(2x - 1)\frac{dx}{dt}$.

Step2: Recall the speed formula

The speed $v$ of a particle moving in the $xy$ - plane is given by $v=\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}}$. We know that $v = 2\sqrt{10}$, so $v^{2}=(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}=40$.

Step3: Substitute $\frac{dy}{dt}$ into the speed - squared formula

Substitute $\frac{dy}{dt}=(2x - 1)\frac{dx}{dt}$ into $(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}=40$. When $x = 2$, $\frac{dy}{dt}=(2\times2 - 1)\frac{dx}{dt}=3\frac{dx}{dt}$. Then $(\frac{dx}{dt})^{2}+(3\frac{dx}{dt})^{2}=40$.

Step4: Solve for $\frac{dx}{dt}$

Expand the left - hand side: $(\frac{dx}{dt})^{2}+9(\frac{dx}{dt})^{2}=40$, so $10(\frac{dx}{dt})^{2}=40$. Then $(\frac{dx}{dt})^{2}=4$. Since $\frac{dx}{dt}>0$, $\frac{dx}{dt}=2$.

Step5: Solve for $\frac{dy}{dt}$

Substitute $\frac{dx}{dt}=2$ into $\frac{dy}{dt}=3\frac{dx}{dt}$. We get $\frac{dy}{dt}=6$.

Answer:

D. 6