Question

a point is moving along the graph of $x^{3}y^{2}=12348$. when the point is at $(7,6)$, its $x$-coordinate is changing at the rate of $-3$ units per minute. how fast is the $y$-coordinate changing at that moment? the $y$-coordinate is changing by $square$ (type an integer or a fraction. simplify your answer.)

Explanation:

Step1: Differentiate the equation implicitly

Differentiate $x^{3}y^{2}=12348$ with respect to time $t$ using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$. Here $u = x^{3}$ and $v = y^{2}$.
The derivative of $x^{3}$ with respect to $t$ is $3x^{2}\frac{dx}{dt}$, and the derivative of $y^{2}$ with respect to $t$ is $2y\frac{dy}{dt}$. So, $3x^{2}y^{2}\frac{dx}{dt}+2x^{3}y\frac{dy}{dt}=0$.

Step2: Substitute the given values

We are given that $x = 7$, $y = 6$, and $\frac{dx}{dt}=-3$. Substitute these values into the differentiated equation:
$3\times(7)^{2}\times(6)^{2}\times(-3)+2\times(7)^{3}\times6\times\frac{dy}{dt}=0$.
First, calculate $3\times49\times36\times(-3)=3\times49\times(- 108)=-15876$.
And $2\times343\times6\times\frac{dy}{dt}=4116\frac{dy}{dt}$.
The equation becomes $-15876 + 4116\frac{dy}{dt}=0$.

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

Add $15876$ to both sides of the equation: $4116\frac{dy}{dt}=15876$.
Then, divide both sides by $4116$: $\frac{dy}{dt}=\frac{15876}{4116}=\frac{1323}{343}=\frac{189}{49}=\frac{27}{7}$.

Answer:

$\frac{27}{7}$