If $\mathit{z}\mathbf{=}\mathit{x}\mathit{y}$and$\{\begin{array}{l}2x^3+2y^3=3t^2\\ 3x^2+3y^2=6t\end{array}$ , find $\mathit{d}\mathit{z}\mathbf{/}\mathit{d}\mathit{t}$ .

The given equations are $z=xy$and $\left\{\begin{array}{l}2x^3+2y^3=3t^2\\ 3x^2+3y^2=6t\end{array}\right.$.

Finding $\frac{dz}{dt}$as:

$$\begin{gathered} \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}\left(1\right) \\ 2x^3+2y^3=3t^2 \end{gathered}$$

Now calculate the value as:

$$\begin{gathered} 6x^2\frac{dx}{dt}+6y^2\frac{dy}{dt}=6t \\ {x}^2\frac{dx}{dt}+y^2\frac{dy}{dt}=t\left(2\right) \\ 3x^2+3y^2=6t \\ x\frac{dx}{dt}+y\frac{dy}{dt}=1\left(3\right) \end{gathered}$$

Now solving the equation (2) and (3) to get an expression for$\frac{dx}{dt}$ and $\frac{dy}{dt}$as:

$$\begin{gathered} x^2\frac{dx}{dt}+y^2\frac{dy}{dt}=t\left(1\right) \\ x\frac{dx}{dt}+y\frac{dy}{dt}=1\left(2\right) \\ {x}^2\frac{dx}{dt}+xy\frac{dy}{dt}=x \end{gathered}$$

Subtracting equation (2) from equation (1) as:

$$\begin{gathered} y^2\frac{dx}{dt}-xy\frac{dy}{dt}=t-x \\ \frac{dy}{dt}=\frac{t-x}{y^2-xy} \end{gathered}$$

Substituting back in equation (1) for $\frac{dy}{dt}$as:

$$\begin{gathered} x^2\frac{dx}{dt}+y^2\left(\frac{t-x}{y^2-xy}\right)=t \\ \frac{dx}{dt}=\frac{t}{x^2}-\frac{y\left(t-x\right)}{x^2\left(y-x\right)} \end{gathered}$$

We will use the results in the two boxed equations to substitute back in the $\frac{dz}{dt}$equation as:

$$\begin{gathered} z=xy \\ \frac{\partial z}{\partial x}=\frac{\partial \left(xy\right)}{\partial x}=y \\ \frac{\partial z}{\partial y}=\frac{\partial \left(xy\right)}{\partial y}=x \\ \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt} \end{gathered}$$

Solving further as:

$$\begin{gathered} \frac{dz}{dt}=y\left(\frac{t}{x^2}-\frac{y\left(t-x\right)}{x^2\left(y-x\right)}\right)+x\left(\frac{t-x}{y^2-xy}\right) \\ \frac{dz}{dt}=y\left(\frac{ty-tx-yt+xy}{x^2\left(y-x\right)}\right)+\frac{xt-x^2}{y\left(y-x\right)} \\ \frac{dz}{dt}=\frac{y^2-ty}{x\left(y-x\right)}+\frac{xt-x^2}{y\left(y-x\right)} \\ \frac{dz}{dt}=\frac{y^3-6t{y}^2+x^{2}t-x^3}{xy\left(y-x\right)} \end{gathered}$$

Simplifying further as:

$$\begin{gathered} \frac{dz}{dt}=\frac{\left(y^3-x^3\right)-t\left(y^2-x^2\right)}{xy\left(y-x\right)} \\ \frac{dz}{dt}=\frac{\left(y-x\right)\left(y^2+xy+x^2\right)-t\left(y-x\right)\left(y+x\right)}{xy\left(y-x\right)} \\ \frac{dz}{dt}=\frac{y^2+xy+x^2-t\left(y+x\right)}{xy} \\ \frac{dz}{dt}=1+\frac{x^2+y^2-t\left(x+y\right)}{xy} \end{gathered}$$

So, it can also be written as:

$$\begin{gathered} xy=z \\ {x}^2+y^2=2t \\ \frac{dz}{dt}=1+\frac{t\left(2-x-y\right)}{z},z\ne 0 \end{gathered}$$

Therefore, the value of $\frac{dz}{dt}$is $1+\frac{t\left(2-x-y\right)}{z},z\ne 0$.