Note: I posted this firstly in the math forum, but I realised it fits here better!(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

I'm having a difficult time trying to re-derive an already-driven result I saw in a book. I put my calculations and result, and the book's result here. I hope someone could help...

Let H be a function of a, b, tan(θ) as,

[tex] H^2 = 2 \{ \frac{a^2 - b^2 tan^2(\theta)}{tan^2(\theta) - 1} \} [/tex]

and,

[tex] sin(2\theta) = \frac{2c^2}{a^2 + b^2} [/tex]

Now, a, b, c, and tan(θ) depend on some a parameter m!

Now, I want to express [tex] \frac{\partial H^2}{\partial m} [/tex] explicitly as a partial derivatives of a, b, and c with respect m. [tex] \frac{\partial \ a^2}{\partial \ m}, \frac{\partial \ b^2}{\partial \ m}, \frac{\partial \ c^2}{\partial \ m} [/tex].

The book I'm reading says the result is,

[tex] \frac{\partial H^2}{\partial m} = \frac{2}{tan^2(\theta) - 1} \{ \frac{\partial \ a^2}{\partial \ m} - tan^2{\theta} \ \frac{\partial \ b^2}{\partial \ m} - \frac{tan(\theta)}{cos(2\theta)}[1 + \frac{H^2}{a^2 + b^2}][2 \frac{\partial \ c^2}{\partial \ m} - sin(2 \theta) (\frac{\partial \ a^2}{\partial \ m} + \frac{\partial \ b^2}{\partial \ m} ) ] \} [/tex]

3. The attempt at a solution

But I get a slightly different result. It's either I'm doing something wrong, or there's a trick which I'm unaware of. I would appreciate it if anyone would like to check my calculations...

I have:

[tex] \frac{\partial H^2}{\partial m} = \frac{2}{tan^2(\theta) - 1} [ \frac{\partial \ a^2}{\partial \ m} - tan^2(\theta) \frac{\partial \ b^2}{\partial \ m} - b^2 \frac{\partial \ tan^2(\theta)}{\partial \ m} - \frac{H^2}{2} \frac{\partial \ (tan^2(\theta) -)}{\partial \ m} ] [/tex]

Where, I used the definition of H above.

Now, for the tan^{2}(θ) I use:

[tex]\frac{\partial \ tan^2(\theta)}{\partial \ m} = \frac{\partial \ (tan^2(\theta) - 1)}{\partial \ m} = \frac{\partial \ tan^2(\theta)}{\partial \ tan(\theta)} \ \frac{\partial \ tan(\theta)}{\partial \ sin(2\theta)} \frac{\partial \ sin(2\theta)}{\partial \ m} [/tex]

And for calculating [tex] \frac{\partial \ tan(\theta)}{\partial \ sin(2\theta)} [/tex], I express tan in terms of sin2θ as follows (maybe my mistake is here?)

[tex] tanθ = sin(2\theta)[1 + tan^2(\theta)]/2 [/tex]

Then I put the value of sin(2θ) in terms a,b, and c. (given at the beginning) and I get,

[tex] \frac{\partial H^2}{\partial m} = \frac{2}{tan^2(\theta) - 1} \{ \frac{\partial \ a^2}{\partial \ m} - tan^2{\theta} \ \frac{\partial \ b^2}{\partial \ m} - \frac{tan(\theta)}{cos(2\theta)}(1-tan^2(\theta))[\frac{b^2}{(a^2 + b^2)} + \frac{H^2}{2 (a^2 + b^2)}][2 \frac{\partial \ c^2}{\partial \ m} - sin(2 \theta) (\frac{\partial \ a^2}{\partial \ m} + \frac{\partial \ b^2}{\partial \ m} ) ] \} [/tex]

So the difference between my result and the book's result (see above) is that I have the factor,

[tex] (1-tan^2(\theta))[\frac{b^2}{(a^2 + b^2)} + \frac{H^2}{2 (a^2 + b^2)}] [/tex]

Whereas, in the book it's,

[tex] [1 + \frac{H^2}{a^2 + b^2}] [/tex]

Can anyone spot my mistake?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Trying to re-derive a formula!

Can you offer guidance or do you also need help?

**Physics Forums | Science Articles, Homework Help, Discussion**