question if z² = x³ + y², dx/dt = -2, dy/dt = -3, and z > 0, find dz/dt at (x,y) = (4,0). please give an…

question if z² = x³ + y², dx/dt = -2, dy/dt = -3, and z > 0, find dz/dt at (x,y) = (4,0). please give an exact answer.

question if z² = x³ + y², dx/dt = -2, dy/dt = -3, and z > 0, find dz/dt at (x,y) = (4,0). please give an exact answer.

Answer

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 $z$ when $(x,y)=(4,0)$

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

Step3: Substitute the known values into the differentiated equation

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

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

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

Answer:

$-6$