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

When $x = 0$ and $y = 1$, from $z^{4}=x^{3}+y^{2}$, we get $z^{4}=0 + 1$, so $z = 1$ (since $z>0$).

Step3: Substitute the known values into the differentiated equation

Substitute $x = 0$, $y = 1$, $z = 1$, $\frac{dx}{dt}=2$, and $\frac{dy}{dt}=-1$ into $4z^{3}\frac{dz}{dt}=3x^{2}\frac{dx}{dt}+2y\frac{dy}{dt}$. We have $4(1)^{3}\frac{dz}{dt}=3(0)^{2}(2)+2(1)(-1)$.

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

$4\frac{dz}{dt}=0 - 2$, so $\frac{dz}{dt}=-\frac{1}{2}$.

Answer:

$-\frac{1}{2}$