Question

an equation relating the variables $x$ and $y$, the values of $x$ and $y$, and the value of $\frac{dx}{dt}$ at a particular instant of $t$ are given. find the value of $\frac{dy}{dt}$. note: both $x$ and $y$ are functions of $t$. $x^{2}y=-36;x = 3,y=-4,\frac{dx}{dt}=-5;\frac{dy}{dt}=?$ -$\frac{40}{3}$ $\frac{40}{3}$ no correct answer choice is given. -$\frac{36}{25}$ -$\frac{36}{25}$

Explanation:

Step1: Differentiate both sides with respect to $t$

Using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = x^{2}$ and $v = y$. The derivative of $x^{2}y$ with respect to $t$ is $2x\frac{dx}{dt}y + x^{2}\frac{dy}{dt}$, and the derivative of $- 36$ with respect to $t$ is $0$. So we have $2x\frac{dx}{dt}y+x^{2}\frac{dy}{dt}=0$.

Step2: Substitute the given values

We are given that $x = 3$, $y=-4$, and $\frac{dx}{dt}=-5$. Substitute these values into the equation $2x\frac{dx}{dt}y+x^{2}\frac{dy}{dt}=0$.
\[

$$\begin{align*} 2\times3\times(-5)\times(-4)+3^{2}\times\frac{dy}{dt}&=0\\ 120 + 9\times\frac{dy}{dt}&=0 \end{align*}$$

\]

Step3: Solve for $\frac{dy}{dt}$

First, subtract $120$ from both sides of the equation: $9\times\frac{dy}{dt}=-120$. Then divide both sides by $9$ to get $\frac{dy}{dt}=-\frac{120}{9}=-\frac{40}{3}$.

Answer:

$-\frac{40}{3}$