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 time are given. find the value of $\frac{dy}{dt}$. $x^{2}y=-36; x = 3, y=-4, \frac{dx}{dt}=-5; \frac{dy}{dt}=?$

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 $2x\frac{dx}{dt}y+x^{2}\frac{dy}{dt}=0$.
\[2\times3\times(- 5)\times(-4)+3^{2}\frac{dy}{dt}=0\]
\[120 + 9\frac{dy}{dt}=0\]

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

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

Answer:

$-\frac{40}{3}$