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 given. find the value of $\frac{dy}{dt}$. $x^{2}y=-36, x = 3, y=-4, \frac{dx}{dt}=-5; \frac{dy}{dt}=?$
$-\frac{36}{25}$
$-\frac{40}{3}$
$\frac{40}{3}$
$-\frac{36}{25}$

Explanation:

Step1: Differentiate the equation $x^{2}y=- 36$ with respect to $t$ using the product - rule.

The product - rule states that if $u$ and $v$ are functions of $t$, then $\frac{d(uv)}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}$. Let $u = x^{2}$ and $v = y$. Then $\frac{d(x^{2}y)}{dt}=x^{2}\frac{dy}{dt}+2xy\frac{dx}{dt}$. Since the derivative of a constant ($-36$) with respect to $t$ is $0$, we have $x^{2}\frac{dy}{dt}+2xy\frac{dx}{dt}=0$.

Step2: Substitute the given values $x = 3$, $y=-4$, and $\frac{dx}{dt}=-5$ into the differentiated equation.

Substituting into $x^{2}\frac{dy}{dt}+2xy\frac{dx}{dt}=0$, we get $(3)^{2}\frac{dy}{dt}+2(3)(-4)(-5)=0$.
This simplifies to $9\frac{dy}{dt}+120 = 0$.

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 $9$ to get $\frac{dy}{dt}=-\frac{120}{9}=-\frac{40}{3}$.

Answer:

$-\frac{40}{3}$