How Is the Tangent Function Applied in Calculating Curvature?

[fishingspree2]

Hello

I need help understanding this proof: http://mathworld.wolfram.com/Curvature.html"

I understand everything from step 1 to step 7. However I don't understand how the result of step 7 is applied in step 8 and in step 9.

I understand that
$$\tan \phi=\frac{y'}{x'}$$ (step 7)
and that
$$\frac{\mathrm{d} }{\mathrm{d} t}\tan \phi=\sec^{2}\phi\frac{\mathrm{d} \phi}{\mathrm{d} t}$$ (step 8)

But I don't understand how $$\tan \phi=\frac{y'}{x'}$$ is applied to $$\frac{\mathrm{d} }{\mathrm{d} t}\tan \phi=\sec^{2}\phi\frac{\mathrm{d} \phi}{\mathrm{d} t}$$ to get $$\frac{x'y''-y'x''}{x'^{2}}$$

Same for step 9
Can anyone help me?

 

[fishingspree2]

Ok, i now fully understand the proof.
I have another question though

The link defines $$k=\frac{d\phi}{ds}$$.

We also know that it can also be defined by the inverse of the radius of the osculating circle.

However the first proof does not talk about circles and radii at all. How can we relate the first proof that defines $$k=\frac{d\phi}{ds}$$. to $$k=\frac{1}{R}$$?

Thank you

 

[HallsofIvy]

You define the "osculating circle" as the circle whose center lies on the normal to the curve and whose radius is equal to 1/k, 1 over the curvature.