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  1. MHB Craftsman
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    #1
    Please be patient as I struggle with latex here .....
    Part 1 of the problem says to start with:
    $ \frac{\partial\bar{r}}{\partial{q}_{1}} ={h}_{1} \hat{q}_{1} $ and then to find an expression for $ {h}_{1} $ that agrees with $ {g}_{ij}=\sum_{l} \frac{\partial{x}_{l}}{\partial{q}_{i}}\frac{\partial{x}_{l}}{\partial{q}_{j}} $

    My attempt is:
    $ \hat{q}_{1} = \frac{1}{h}_{1} \frac{\partial\bar{r}}{\partial{q}_{1}}$
    but $ \hat{q}_{1}.\hat{q}_{1}=1$
    Then $ \left({h}_{1}\right)^{\!{2}} =\left(\frac{\partial\bar{r}}{\partial{q}_{1}}\right)^{\!{2}}$
    Now $ \left(\frac{\partial\bar{r}}{\partial{q}_{1}}\right)= \left(\frac{\partial{x}}{\partial{q}_{1}}\right)+\left(\frac{\partial{y}}{\partial{q}_{1}}\right)+\left(\frac{\partial{z}}{\partial{q}_{1}}\right) $
    so $ \left(\frac{\partial\bar{r}}{\partial{q}_{1}}\right)^{\!{2}}= \left(\frac{\partial{x}}{\partial{q}_{1}}\right)^{\!{2}}+\left(\frac{\partial{y}}{\partial{q}_{1}}\right)^{\!{2}}+\left(\frac{\partial{z}}{\partial{q}_{1}}\right)^{\!{2}} $
    Now $ {g}_{ii} $ is defined = $ {h}_{i}^{\!{2}}$ and comparing with $ {g}_{ij}=\sum_{l} \frac{\partial{x}_{l}}{\partial{q}_{i}}\frac{\partial{x}_{l}}{\partial{q}_{j}} $ above...
    Therefore $ {h}_{1}= \sqrt{ \left(\frac{\partial{x}}{\partial{q}_{1}}\right)^{\!{2}}+\left(\frac{\partial{y}}{\partial{q}_{1}}\right)^{\!{2}}+\left(\frac{\partial{z}}{\partial{q}_{1}}\right)^{\!{2}}} $ QED - but have I done anything illegal here?
    -----
    Part 2 is a derivation:
    Again starting with $ \hat{q}_{1} = \frac{1}{h}_{1} \frac{\partial\bar{r}}{\partial{q}_{1}}$
    Then $ \frac{\partial\hat{q}_{1}}{\partial{q}_{2}}=\frac{1}{{h}_{1}}\frac{\partial{}^{2}\bar{r}}{\partial{q}_{1}\partial{q}_{2}} = \frac{1}{{h}_{1}}\frac{\partial}{\partial{q}_{1}}\left(\frac{\partial\bar{r}}{\partial{q}_{2}}\right)^{\!{}} $ (${h}_{1}$ constant w.r.t. 2)
    but $ \frac{\partial\bar{r}}{\partial{q}_{2}}={h}_{2} \hat{q}_{2} $
    so $ \frac{\partial\hat{q}_{1}}{\partial{q}_{2}}=\frac{1}{{h}_{1}} \frac{\partial\left({h}_{2}\hat{q}_{2}\right)}{\partial{q}_{1}}= \hat{q}_{2}\frac{1}{{h}_{1}}\frac{\partial{h}_{2}}{\partial{q}_{1}}$
    Which is QED - but again I have this uncertain feeling so would appreciate confirmation there I have done nothing untoward
    -----------------
    Part 3 has me so far, I would appreciate a hint ....
    Derive $ \frac{\partial\hat{q}_{1}}{\partial{q}_{1}}= -\sum_{j\ne{1}}^{} \hat{q}_{2}\frac{1}{{h}_{2}}\frac{\partial{h}_{1}}{\partial{q}_{2}}$
    I have tried a few things without success ... probably there is a trick I haven't encountered before?

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  3. MHB Craftsman
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    #2 Thread Author
    In addition: I have assumed perhaps incautiously, that people would recognize the subject here - it is around generalised curvilinear coordinates, with the expression for $ {g}_{ij}$ representing the mixed coordinates, but ${h}_{i}$ excludes the mixed coordinates, IE it is for an orthogonal basis. I feel fairly confident of my working for the first 2 parts, but would like confirmation that I haven't done anything that isn't justifiable. For the 3rd part, I have spent ages trying different things - I thought I was onto it when I differentiated the starting eqn w.r.t. BOTH i and j, but still could not get a term in $ \frac{\partial{\bar{q}}_{i}}{\partial{q}_{i}} $. So I'd also really appreciate a tip as to what approach might product that term, thanks.

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