Question

question. if $z^{3}=x^{3}+y^{2}$, $\frac{dx}{dt}=3$, $\frac{dy}{dt}=2$, 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 $z^{3}=x^{3}+y^{2}$ with respect to $t$

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

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

When $x = 4$ and $y = 0$, then $z^{3}=x^{3}+y^{2}=4^{3}+0^{2}=64$, so $z = 4$ (since $z>0$).

Step3: Substitute the known values into the differentiated equation

We know that $\frac{dx}{dt}=3$, $\frac{dy}{dt}=2$, $x = 4$, $y = 0$, and $z = 4$. Substituting these values into $3z^{2}\frac{dz}{dt}=3x^{2}\frac{dx}{dt}+2y\frac{dy}{dt}$, we get $3\times4^{2}\frac{dz}{dt}=3\times4^{2}\times3+2\times0\times2$.
Simplify the right - hand side: $3\times16\times\frac{dz}{dt}=3\times16\times3+0$.
$48\frac{dz}{dt}=144$.

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

Divide both sides of the equation $48\frac{dz}{dt}=144$ by 48: $\frac{dz}{dt}=\frac{144}{48}=3$.

Answer:

$3$