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I don't understand the derivation of the wave equation given below -

$$T \sin (\theta _1) - T \sin (\theta ) = T\tan (\theta _1 )-T\tan (\theta ) = T \left. \left(\frac{\partial f}{\partial z} \right|_{z + \Delta z} - \left. \frac{\partial f}{\partial z}\right| _z \right) = T \frac{\partial ^2 f}{\partial z^2} \Delta z$$

I understand that the small angle approximation was used, but I'm at a loss for figuring out we turned $\tan$ into a derivative, and then after made it become a second derivative.

The derivative of $\tan \theta$ is of course $\sec \theta$ which is equal to $\frac{1}{cos \theta}$, which was taken with respect to $\theta$, maybe there's a way to use the chain rule to find $\partial _z f$?

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As we can see from the previous answers, we have the following approximations:

$tan(\theta) = \frac{opposite}{adjacent} = \frac {rise}{run} = gradient = \frac {∂y}{∂x}$

and we can say, still through approximation:

slope(z+∆z)=slope(z)+slope_variation_tax*∆z, which mathematically means:

$$\frac{\partial f(z+∆z)}{\partial z} = \frac{\partial f(z)}{\partial z}+\frac{\partial }{\partial z}(\frac{\partial f(z)}{\partial z}) \partial z=\frac{\partial f(z)}{\partial z}+\frac{\partial ^2 f}{\partial z^2} \partial z$$

Then you have, after substitution: $$T\frac{\partial f(z+∆z)}{\partial z}-T\frac{\partial f(z)}{\partial z}=T\frac{\partial f(z)}{\partial z}+T\frac{\partial ^2 f}{\partial z^2} \partial z-T\frac{\partial f(z)}{\partial z}=T\frac{\partial ^2 f}{\partial z^2} \partial z$$ And that's the last member of your equation. Hope I helped ;)

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I just realized that the reason we're using $\tan \theta = \partial _z f$ is because $\tan \theta = \frac{df}{dz}$, which makes sense if we consider this a right triangle.

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$tan(\theta) = \frac{opposite}{adjacent} = \frac {rise}{run} = gradient$

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  • $\begingroup$ Sure, but that doesn't explain the second part, which has me confused. $\endgroup$
    – Astrum
    Commented Nov 9, 2013 at 20:50

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