Question

1. (exam 2 review #8) suppose 4x² + 9y² = 36, where x and y are functions of t. a. 2.5 pts if $\frac{dy}{dt}=\frac{1}{3}$, find $\frac{dx}{dt}$ when x = 2 and y = $\frac{2}{3}sqrt{5}$; b. 2.5 pts if $\frac{dx}{dt}=3$, find $\frac{dy}{dt}$ when x = -2 and y = $\frac{2}{3}sqrt{5}$.

Explanation:

Step1: Differentiate the given equation

Differentiate $4x^{2}+9y^{2}=36$ with respect to $t$ using the chain - rule.
The derivative of $4x^{2}$ with respect to $t$ is $8x\frac{dx}{dt}$, and the derivative of $9y^{2}$ with respect to $t$ is $18y\frac{dy}{dt}$, and the derivative of the constant 36 with respect to $t$ is 0. So we get $8x\frac{dx}{dt}+18y\frac{dy}{dt}=0$.

Step2: Solve for $\frac{dx}{dt}$ in part A

We are given $\frac{dy}{dt}=\frac{1}{3}$, $x = 2$ and $y=\frac{2}{3}\sqrt{5}$.
Substitute these values into $8x\frac{dx}{dt}+18y\frac{dy}{dt}=0$.
$8\times2\frac{dx}{dt}+18\times\frac{2}{3}\sqrt{5}\times\frac{1}{3}=0$.
$16\frac{dx}{dt}+4\sqrt{5}=0$.
$16\frac{dx}{dt}=- 4\sqrt{5}$.
$\frac{dx}{dt}=-\frac{\sqrt{5}}{4}$.

Step3: Solve for $\frac{dy}{dt}$ in part B

We are given $\frac{dx}{dt}=3$, $x=-2$ and $y = \frac{2}{3}\sqrt{5}$.
Substitute into $8x\frac{dx}{dt}+18y\frac{dy}{dt}=0$.
$8\times(-2)\times3+18\times\frac{2}{3}\sqrt{5}\times\frac{dy}{dt}=0$.
$-48 + 12\sqrt{5}\frac{dy}{dt}=0$.
$12\sqrt{5}\frac{dy}{dt}=48$.
$\frac{dy}{dt}=\frac{4}{\sqrt{5}}=\frac{4\sqrt{5}}{5}$.

Answer:

A. $\frac{dx}{dt}=-\frac{\sqrt{5}}{4}$
B. $\frac{dy}{dt}=\frac{4\sqrt{5}}{5}$