Question

a particle moves on the circle $x^{2}+y^{2}=100$ in the $xy$-plane for time $tgeq0$. at the time when the particle is at the point $(8,6)$, the value of $\frac{dx}{dt}$ is 5. what is the value of $\frac{dy}{dt}$ at this time?
(a) $\frac{dy}{dt}=-\frac{20}{3}$
(b) $\frac{dy}{dt}=-\frac{4}{3}$
(c) $\frac{dy}{dt}=\frac{5}{3}$
(d) $\frac{dy}{dt}=7$

Explanation:

Step1: Differentiate circle equation

Differentiate $x^{2}+y^{2}=100$ with respect to $t$ using chain - rule. We get $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$.

Step2: Simplify the equation

Divide the entire equation $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$ by 2, resulting in $x\frac{dx}{dt}+y\frac{dy}{dt}=0$.

Step3: Substitute given values

We know that $x = 8$, $y = 6$, and $\frac{dx}{dt}=5$. Substitute these values into $x\frac{dx}{dt}+y\frac{dy}{dt}=0$. So, $8\times5+6\times\frac{dy}{dt}=0$.

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

First, calculate $8\times5 = 40$. The equation becomes $40+6\times\frac{dy}{dt}=0$. Then, subtract 40 from both sides: $6\times\frac{dy}{dt}=-40$. Finally, divide both sides by 6 to get $\frac{dy}{dt}=-\frac{40}{6}=-\frac{20}{3}$.

Answer:

A. $\frac{dy}{dt}=-\frac{20}{3}$