Question

question. if (z^{2}=x^{3}+y^{2}), (\frac{dx}{dt}=-2), (\frac{dy}{dt}=-3), and (z > 0), find (\frac{dz}{dt}) at ((x,y)=(4,0)). please give an exact answer. provide your answer below: (\frac{dz}{dt}=square)

Explanation:

Step1: Differentiate $z^{2}=x^{3}+y^{2}$ with respect to $t$

Using the chain - rule, we have $2z\frac{dz}{dt}=3x^{2}\frac{dx}{dt}+2y\frac{dy}{dt}$.

Step2: Find the value of $z$ at $(x,y)=(4,0)$

When $x = 4$ and $y = 0$, $z^{2}=x^{3}+y^{2}=4^{3}+0^{2}=64$. Since $z>0$, $z = 8$.

Step3: Substitute the given values into the differentiated equation

We know that $\frac{dx}{dt}=-2$, $\frac{dy}{dt}=-3$, $x = 4$, $y = 0$ and $z = 8$. Substitute these into $2z\frac{dz}{dt}=3x^{2}\frac{dx}{dt}+2y\frac{dy}{dt}$.
$2\times8\frac{dz}{dt}=3\times4^{2}\times(-2)+2\times0\times(-3)$.

Step4: Solve for $\frac{dz}{dt}$

First, simplify the right - hand side: $3\times4^{2}\times(-2)+2\times0\times(-3)=3\times16\times(-2)+0=-96$.
Then, from $16\frac{dz}{dt}=-96$, we get $\frac{dz}{dt}=\frac{-96}{16}=-6$.

Answer:

$-6$