question if z^4 = x^3 + y^2, dx/dt = -1, dy/dt = 3, and z > 0, find dz/dt at (x,y)=(0,4). please give an…

Explanation:

Step1: Differentiate both sides with respect to $t$

Using the chain - rule, we have $4z^{3}\frac{dz}{dt}=3x^{2}\frac{dx}{dt}+2y\frac{dy}{dt}$.

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

Substitute $x = 0$ and $y = 4$ into the equation $z^{4}=x^{3}+y^{2}$. Then $z^{4}=0 + 16$, so $z^{4}=16$. Since $z>0$, we get $z = 2$.

Step3: Substitute known values into the differentiated equation

Substitute $x = 0$, $y = 4$, $z = 2$, $\frac{dx}{dt}=-1$, and $\frac{dy}{dt}=3$ into $4z^{3}\frac{dz}{dt}=3x^{2}\frac{dx}{dt}+2y\frac{dy}{dt}$. We have $4\times2^{3}\frac{dz}{dt}=3\times0^{2}\times(-1)+2\times4\times3$. $4\times8\frac{dz}{dt}=0 + 24$. $32\frac{dz}{dt}=24$.

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

$\frac{dz}{dt}=\frac{24}{32}=\frac{3}{4}$.

Answer:

$\frac{3}{4}$