Question

If${\mathit{P}}^{\mathbf{3}}\mathbf{+}\mathit{s}\mathit{q}\mathbf{=}\mathit{t}$and${\mathit{q}}^{\mathbf{3}}\mathbf{+}\mathit{t}\mathit{p}\mathbf{=}\mathit{s}$, find${\left(\frac{\partial p}{\partial s}\right)}_{\mathbf{t}}\mathbf{,}{\left(\frac{\partial p}{\partial s}\right)}_{\mathbf{q}}$at$\left(p,q,r,s\right)\mathbf{=}\left(-1,2,3,4\right)$.

Short answer

The value of${\left(\frac{\partial p}{\partial s}\right)}_t$ is$-\frac{9}{7}$ .

The value of${\left(\frac{\partial p}{\partial s}\right)}_q$ is$\frac{3}{2}$ .

Step-by-step solution

Given information.

Given$p^3+sq=t,q^3+tp=s$,

Definition of partial differentiation.

Partial differentiation is defined as the process, in which find the partial derivative of a function.

In Partial differentiation, the function has more than one variable and finds the partial derivative of a function with respect to one variable and keeping the other variable constant.

Take the partial differentiation of each functions.

Now find the partial differentiation of function,$p^3+sq=t$ with respect to .

$3p^2\frac{\partial p}{\partial s}+s\frac{\partial p}{\partial s}+q=\frac{\partial t}{\partial s}.........\left(1\right)$

Now takepartial differentiation of function,$q^3+tp=s$ with respect to .

$3q^2\frac{\partial p}{\partial s}+p\frac{\partial p}{\partial s}+t\frac{\partial p}{\partial s}.........\left(2\right)$

Now from equation (1), findrole="math" localid="1660820981399" ${\left(\frac{\partial p}{\partial s}\right)}_t$ by keeping is constant.

$$\begin{gathered} 3p^2{\left(\frac{\partial p}{\partial s}\right)}_t+s\left(\frac{\partial p}{\partial s}\right)+q=0 \\ {\left(\frac{\partial p}{\partial s}\right)}_t=-\frac{\left(s\frac{\partial p}{\partial s}+q\right)}{3p^2}...........\left(3\right) \end{gathered}$$

Now from equation (2), findrole="math" localid="1660821116982" ${\left(\frac{\partial p}{\partial s}\right)}_t$ by keeping t is constant.

$$\begin{gathered} 3q^2\frac{\partial q}{\partial s}+0+t{\left(\frac{\partial p}{\partial s}\right)}_t=1 \\ \frac{\partial q}{\partial s}=\frac{1-t{\left(\frac{\partial q}{\partial s}\right)}_t}{3q^2}......\left(4\right) \end{gathered}$$

Now substitute from equation (4) to equation (3).

$$\begin{gathered} {\left(\frac{\partial q}{\partial s}\right)}_t=-\frac{1}{3p^2}\left(s\left(\frac{1-t{\left(\frac{\partial q}{\partial s}\right)}_t}{3q^2}\right)+q\right) \\ =-\frac{1}{3p^2}.\left[\frac{s-st{\left(\frac{\partial q}{\partial s}\right)}_t}{3q^2}+q\right] \end{gathered}$$

Solve (∂p∂s)t .

Now solve the expression further.

$$\begin{gathered} {\left(\frac{\partial p}{\partial s}\right)}_t=-\frac{s}{9p^2{q}^2}+\frac{st{\left(\frac{\partial p}{\partial s}\right)}_t}{9p^2{q}^2}-\frac{q}{3p^2} \\ {\left(\frac{\partial p}{\partial s}\right)}_t\left(1-\frac{st}{9p^2{q}^2}\right)=-\left(\frac{s}{9p^2{q}^2}+\frac{q}{3p^2}\right) \\ {\left(\frac{\partial p}{\partial s}\right)}_t=\frac{-\left(\frac{s}{9p^2{q}^2}+\frac{q}{3p^2}\right)}{\left(1-\frac{st}{9p^2{q}^2}\right)}............\left(5\right) \end{gathered}$$

Now substitute the value of from$p^3+sq=t$.

${\left(\frac{\partial p}{\partial s}\right)}_t=\frac{-\left(\frac{s}{9p^2{q}^2}+\frac{q}{3p^2}\right)}{\left(1-\frac{st}{9p^2{q}^2}\right)}............\left(6\right)$

Substitute$\left(p,q,s,t\right)=\left(-1,2,3,4\right)$ in equation (6) and solve the expression.

$$\begin{gathered} {\left(\frac{\partial p}{\partial s}\right)}_t=\frac{-\left({\displaystyle \frac{3}{9\times 1\times 4}}+{\displaystyle \frac{2}{3\times 1}}\right)}{\left(1-{\displaystyle \frac{3\times 5}{9\times 1\times 4}}\right)} \\ {\left(\frac{\partial p}{\partial s}\right)}_t=-\frac{9}{7} \end{gathered}$$

Hence the value of${\left(\frac{\partial p}{\partial s}\right)}_t$ is$-\frac{9}{7}$ .

Solve (∂p∂s)t .

From equation (1), find${\left(\frac{\partial p}{\partial s}\right)}_q$ by keeping is constant.

$$\begin{gathered} 3p^2{\left(\frac{\partial p}{\partial s}\right)}_q+0+q=\frac{\partial t}{\partial s} \\ {\left(\frac{\partial p}{\partial s}\right)}_q=\frac{\frac{\partial t}{\partial s}-q}{3p^2}............\left(7\right) \end{gathered}$$

Now from equation (2), find${\left(\frac{\partial p}{\partial s}\right)}_q$ by keeping is constant.

$3q^2.0+p\frac{\partial t}{\partial s}+t{\left(\frac{\partial p}{\partial s}\right)}_q=1..........\left(8\right)$

Find the value of from equation (8). substitute in equation (7).

$$\begin{gathered} 3q^2.0+p\frac{\partial t}{\partial s}+t{\left(\frac{\partial p}{\partial s}\right)}_q=1..........\left(8\right) \\ \frac{\partial t}{\partial s}=\frac{1}{p}\left[1-t{\left(\frac{\partial p}{\partial s}\right)}_q\right] \end{gathered}$$

Now substitute$\frac{\partial t}{\partial s}$ in equation (7).

$$\begin{gathered} {\left(\frac{\partial p}{\partial s}\right)}_q=\frac{{\displaystyle \frac{\partial t}{\partial s}}-q}{3p^2} \\ =\frac{1}{3p^2}.\frac{1}{p}\left[1-t{\left(\frac{\partial p}{\partial s}\right)}_q\right]-\frac{q}{3p^2} \\ {\left(\frac{\partial p}{\partial s}\right)}_q\left(1+\frac{t}{3p^3}\right)=\frac{1}{3p^3}-\frac{q}{3p^2} \\ {\left(\frac{\partial p}{\partial s}\right)}_q=\frac{\frac{1}{3p^3}-\frac{q}{3p^2}}{\left(1+\frac{t}{3p^3}\right)} \end{gathered}$$

Substitute$\left(p,q,s,t\right)=\left(-1,2,3,4\right)$ in equation (6) and solve the expression.

$$\begin{gathered} {\left(\frac{\partial p}{\partial s}\right)}_q=\frac{{\displaystyle \frac{1}{3\left(-1\right)}}-{\displaystyle \frac{2}{3\left(1\right)}}}{\left(1+{\displaystyle \frac{5}{3\left(-1\right)}}\right)} \\ {\left(\frac{\partial p}{\partial s}\right)}_q=\frac{3}{2} \end{gathered}$$

Hence the value of${\left(\frac{\partial p}{\partial s}\right)}_q$ is$\frac{3}{2}$ .