Derivative with parametric set of equations

Question

I am struggling for hours on a fairly simple problem. I have the following set of parametric equations:

$$i = I (q_1^2 + q_1 - q_2^2 - q_2)$$ $$v = V (2(q_1 - 1) + \log q_1 - 2(q_2 - 1) - \log q_2)$$

Now I would like to calculate $\frac{di}{dv}$. The canonical way would be to solve the second equation for $q$, plug it into the first and find the derivative. But the second equation is not easily invertible.

So I treat the $q$ as a function of $v$ and take the derivative of the first equation.

Then I take the derivative of the second w.r.t. $q_1$ and $q_2$ and obtain the expressions for $\frac{d v}{d q_1}$ and $\frac{d v}{d q_2}$. Taking the reciprocal value and inserting them into the derivative of the first equation results in:

$$\frac{d i}{d v} = \frac{I}{V}(q_1 + q_2)$$

However, from a paper (and from physical results) I know that the solution must be

$$\frac{d i}{d v} = 2 \frac{I}{V} \frac{q_1 q_2}{q_1 + q_2}$$

I went over it tens of times, tried with Mathematica but the result is always wrong.

Is this approach fundamentally wrong?

Answer 1

You’re running into trouble because for partial derivatives, which is what you’re really working with here, it’s not generally true that $\frac{\partial x}{\partial y}=\left. 1 \middle/ \right. {\frac{\partial y}{\partial x}}$.

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One way to attack this problem is to compute $di$ and $dv$, divide one by the other, and try to eliminate the $\mathrm dq$’s:$$ \mathrm di = I((2q_1+1)\;\mathrm dq_1 - (2q_2+1)\;\mathrm dq_2) \\ \mathrm dv = V((2+\left. 1 \middle/ \right. q_1)\;\mathrm dq_1 - (2+\left. 1 \middle/ \right. q_2)\;\mathrm dq_2) $$ It might make the algebraic manipulations easier by defining $$ \alpha(q_1) = q_1^2+q_1 \\ \beta(q_2) = q_2^2+q_2 $$ so that $$ \mathrm di = I(\mathrm d\alpha - \mathrm d\beta) \\ \mathrm dv = V\left(\frac 1{q_1}\mathrm d\alpha - \frac 1{q_2}\mathrm d\beta\right) \\ \frac{\mathrm di}{\mathrm dv} = \frac I V \frac{\mathrm d\alpha - \mathrm d\beta}{\frac 1{q_1}\mathrm d\alpha - \frac 1{q_2}\mathrm d\beta} $$ This seems promising. It might also be of use to note that $$\frac{q_1 q_2}{q_1+q_2}=\frac 1{\frac 1 {q_1}+\frac 1 {q_2}}$$

Answer 2

Total differentials are to be used as indicated by amd.

But makes no sense to me when the right hand side does not involve $ \frac {dq_2}{dq_1}. $

Is some other condition of charge variation missing?