Let $u=s+t$ and $v=s^2+t^2$ how do I find $\left(\frac{\partial t}{\partial u}\right)_v$ and $\left(\frac{\partial t}{\partial v}\right)_u?$
The notation is confusing to me and the book didn't elaborate. I know the first notation is saying to take the partial of $t$ with respect to $u$ with $v$ held constant, but that will only make sense if $t$ was a function of those two independent variables. How do I solve this problem?
On
Is there any restriction on $t$? What I did was squaring both sides of the first equation to obtain $u^2=s^2+2st+t^2$. Combining with the 2nd equation gives $t=\frac{u^2-v}{2s}$. From the first equation, we could rewrite this as $t=\frac{u^2-v^2}{2(u-t)}$. To make $t$ the subject, we proceed by completing the square.
You would substitute $s$ from the first equation into the second, and then implicitly differentiate.
Since $u=s+t$, then $s=u-t$. So $$\begin{align}v &= (u-t)^2 + t^2 \\ v&= u^2 - 2ut + 2t^2 \tag1\end{align}$$
Now we take the partial derivative of $(1)$ with respect to $u$. But, remember that $t$ is a function of $u$ and $v$. Account for that. $$ \begin{align}\frac{\partial}{\partial{u}}v &= \frac{\partial}{\partial{u}}u^2 - \frac{\partial}{\partial{u}}(2ut) + \frac{\partial}{\partial{u}}(2t^2) \\ 0& = 2u - \left(2u\cdot\frac{\partial t}{\partial{u}} + 2t\right) + 4t \frac{\partial t}{\partial{u}} \tag2\end{align}$$
Look at that. Now all we have to do is solve for $\frac{\partial t}{\partial{u}}$ in $(2)$.
Similarly, we can find $\frac{\partial t}{\partial{v}}$.
Also as a note, the subscript $v$ in $\left(\frac{\partial t}{\partial u}\right)_v$ indicates that we take the partial derivative of $\frac{\partial t}{\partial u}$ with respect to $v$. After finding those, we will find that $$\left(\frac{\partial t}{\partial u}\right)_v = \left(\frac{\partial t}{\partial v}\right)_u.$$