question\nif $z^{3}=x^{3}+y^{2}$, $\\frac{dx}{dt}=-2$, $\\frac{dy}{dt}=3$, and $z > 0$, find…

question\nif $z^{3}=x^{3}+y^{2}$, $\\frac{dx}{dt}=-2$, $\\frac{dy}{dt}=3$, and $z > 0$, find $\\frac{dz}{dt}$ at $(x,y)=(4,0)$.\nplease give an exact answer.

question\nif $z^{3}=x^{3}+y^{2}$, $\\frac{dx}{dt}=-2$, $\\frac{dy}{dt}=3$, and $z > 0$, find $\\frac{dz}{dt}$ at $(x,y)=(4,0)$.\nplease give an exact answer.

Answer

Explanation:

Step1: Differentiate both sides with respect to $t$

Using the chain - rule, if $z^{3}=x^{3}+y^{2}$, then $\frac{d}{dt}(z^{3})=\frac{d}{dt}(x^{3}+y^{2})$. By the chain - rule, $\frac{d}{dt}(z^{3}) = 3z^{2}\frac{dz}{dt}$, $\frac{d}{dt}(x^{3})=3x^{2}\frac{dx}{dt}$, and $\frac{d}{dt}(y^{2}) = 2y\frac{dy}{dt}$. So, $3z^{2}\frac{dz}{dt}=3x^{2}\frac{dx}{dt}+2y\frac{dy}{dt}$.

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

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

Step3: Substitute the given values into the differentiated equation

We know that $x = 4$, $y = 0$, $z = 4$, $\frac{dx}{dt}=-2$, and $\frac{dy}{dt}=3$. Substitute these values into $3z^{2}\frac{dz}{dt}=3x^{2}\frac{dx}{dt}+2y\frac{dy}{dt}$: $3\times4^{2}\frac{dz}{dt}=3\times4^{2}\times(-2)+2\times0\times3$. $48\frac{dz}{dt}=3\times16\times(-2)+0$. $48\frac{dz}{dt}=-96$.

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

Divide both sides of the equation $48\frac{dz}{dt}=-96$ by 48: $\frac{dz}{dt}=\frac{-96}{48}=-2$.

Answer:

$-2$