Given $\mathit{z}\mathbf{=}{\mathit{r}}^{\mathbf{2}}\mathbf{+}{\mathit{s}}^{\mathbf{2}}\mathbf{+}\mathit{r}\mathit{s}\mathit{t}$,${\mathit{r}}^{\mathbf{4}}\mathbf{+}{\mathit{s}}^{\mathbf{4}}\mathbf{+}{\mathit{t}}^{\mathbf{4}}\mathbf{=}\mathbf{2}{\mathit{r}}^{\mathbf{2}}{\mathit{s}}^{\mathbf{2}}{\mathit{t}}^{\mathbf{2}}\mathbf{+}\mathbf{10}$, find ${\left(\partial z/\partial r\right)}_{\mathbf{t}}$when$\mathit{r}\mathbf{=}\mathbf{2}$ ,$\mathit{s}\mathbf{=}\mathit{t}\mathbf{=}\mathbf{1}$ .

The given equations$z=r^2+s^2+rst\left(1\right)$ , $r^4+s^4+t^4=2r^2{s}^2{t}^2+10\left(2\right)$.

Differentiate the equations as:

$$\begin{gathered} z=2rdr+2sds+rsdt+trds+tsdr\left(3\right) \\ 4r^{3}dr+4s^{3}ds+4t^{3}dt=4r^2{s}^{2}tdt+4t^2{r}^{2}sds+4t^2{s}^{2}rdr\left(4\right) \end{gathered}$$

Since, $t$is a constant, so$dt=0$ .

Substitute $dt=0$in the equation (3) as:

$$\begin{gathered} dz=2rdr+2sds+rsdt+trds+tsdr \\ d{z}_t=2rd{r}_t+2sd{s}_t+rs\left(0\right)+trd{s}_t+tsd{r}_t \\ d{z}_t=2rd{r}_t+2sd{s}_t+trd{s}_t+tsd{r}_t \\ d{z}_t=\left(2r+ts\right)d{r}_t+\left(2s+tr\right)d{s}_t\left(5\right) \end{gathered}$$

Here, the subscript$t$ indicates that $t$is a constant.

Substitute$dt=0$ in the equation (4) as:

$$\begin{gathered} 4r^{3}dr+4s^{3}ds+4t^{3}dt=4r^2{s}^{2}tdt+4t^2{r}^{2}sds+4t^2{s}^{2}rdr \\ 4r^{3}d{r}_t+4s^{3}d{s}_t+4t^3\left(0\right)=4r^2{s}^{2}t\left(0\right)+4t^2{r}^{2}sd{s}_t+4t^2{s}^{2}rd{r}_t \\ 4r^{3}d{r}_t+4s^{3}d{s}_t=4t^2{r}^{2}sd{s}_t+4t^2{s}^{2}rd{r}_t \\ \left(4r^3-4t^2{s}^{2}r\right)d{r}_t=\left(4t^2{s}^{2}s-4s^3\right)d{s}_t \end{gathered}$$

Solving further as:

$$\begin{gathered} \left(4r^3-4t^2{s}^{2}r\right)d{r}_t=\left(4t^2{s}^{2}s-4s^3\right)d{s}_t \\ \frac{\left(4r^3-4t^2{s}^{2}r\right)}{\left(4t^2{s}^{2}s-4s^3\right)}d{r}_t=d{s}_t \end{gathered}$$

Substitute the value of $ds$in the equation (5) as:

$$\begin{gathered} d{z}_t=\left(2r+ts\right)d{r}_t+\left(2s+tr\right)d{s}_t \\ d{z}_t=\left(2r+ts\right)d{r}_t+\frac{\left(2r+ts\right)\left(4r^3-4t^2{s}^{2}r\right)}{\left(4t^2{r}^{2}s-4s^3\right)}d{r}_t \\ d{z}_t=\left[\left(2r+ts\right)+\frac{\left(2r+ts\right)\left(4r^3-4t^2{s}^{2}r\right)}{\left(4t^2{r}^{2}s-4s^3\right)}\right]d{r}_t \end{gathered}$$

Now divide by $dr$, then:

${\left(\frac{\partial z}{\partial r}\right)}_t=\left(2r+ts\right)+\frac{\left(2r+ts\right)\left(4r^3-4t^2{s}^{2}r\right)}{\left(4t^2{r}^{2}s-4s^3\right)}\left(6\right)$

Substitute the values of $r$, $s$, and$t$ in (6) equation as:

$$\begin{gathered} {\left(\frac{\partial z}{\partial r}\right)}_t=\left(2r+ts\right)+\frac{\left(2r+ts\right)\left(4r^3-4t^2{s}^{2}r\right)}{\left(4t^2{r}^{2}s-4s^3\right)} \\ {\left(\frac{\partial z}{\partial r}\right)}_t=\left(2\left(2\right)+\left(1\right)\left(1\right)\right)+\frac{\left(2\left(1\right)+\left(1\right)\left(2\right)\right)\left(4{\left(2\right)}^3-4{\left(1\right)}^2{\left(1\right)}^2\left(2\right)\right)}{\left(4{\left(1\right)}^2{\left(2\right)}^2\left(1\right)-4{\left(1\right)}^3\right)} \\ {\left(\frac{\partial z}{\partial r}\right)}_t=\left(4+1\right)+\frac{\left(2+2\right)\left(4\left(8\right)-4\left(2\right)\right)}{\left(4\left(4\right)-4\right)} \\ {\left(\frac{\partial z}{\partial r}\right)}_t=5+\frac{\left(4\right)\left(32-8\right)}{\left(16-4\right)} \end{gathered}$$

Solving further as:

$$\begin{gathered} {\left(\frac{\partial z}{\partial r}\right)}_t=5+\frac{\left(4\right)\left(24\right)}{\left(12\right)} \\ {\left(\frac{\partial z}{\partial r}\right)}_t=5+8 \\ {\left(\frac{\partial z}{\partial r}\right)}_t=13 \end{gathered}$$

Therefore, the answer is ${\left(\frac{\partial z}{\partial r}\right)}_t=13$.