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In the book of Dynamics by Horace Lamb, at page 103, is it given

that for a motion on a smooth curve, the equation of motion is given by $$mv \frac{dv }{ ds} = -mg \sin \theta \qquad \frac{ mv^2 }{r} = -mg \cos\theta + R, $$ where $\theta$ is the angle between the surface normal and vertical direction, and $R$ is the "pressure" exerted by the curve.

[...]

Then, we have $$\cos \theta = \frac{dx}{ds} \qquad \sin\theta = \frac{dy }{ds}, $$ where $s$ is length of the path taken over the surface, and $x,y$ are usual cartesian coordinates as $y$ is taken as upward.

However, I cannot understand how does the author derives the latter equations between $\theta$ and the derivatives of $x,y$ wrt $s$.

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  • $\begingroup$ This seems like a math question, but meta.stackexchange.com/a/10250 describes why the migration to Math SE failed and the post was closed as off-topic. $\endgroup$
    – user191954
    Nov 21, 2018 at 12:16

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Consider a small segment of the curve, and approximate this segment with a straight line $\Delta s$. Then Pythagoras' theorem gives you

$$ \Delta s^2 = \Delta x^2 + \Delta y^2 $$

According to basic trigonometry:

$$ \frac{\Delta x}{\Delta s} = \cos\, \alpha,\;\; \frac{\Delta y}{\Delta s} = \sin\, \alpha, $$

where $\alpha$ is the angle between $\Delta s$ and $\Delta x$. In the limit the same equations hold for $\dfrac{dx}{ds}$ and $\dfrac{dy}{ds}$.

I don't understand the definition of $\theta$ in the question, but either it can be identified with the angle $\alpha$, or more unlikely, the book is wrong.

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  • $\begingroup$ Generally the angle when looking at $\frac{\Delta x}{\Delta s}$ is different than when looking at $\frac{\text d x}{\text d s}$ $\endgroup$ Nov 17, 2018 at 19:30
  • $\begingroup$ Yes. But if $x=x(s)$ is a continuous function, the angle difference can be made arbitrarily small by choosing a sufficiently small $\Delta s$. $\endgroup$
    – Cuspy Code
    Nov 17, 2018 at 19:43
  • $\begingroup$ Typically $\Delta$ is used for "larger" values, and $\text d$ is used for infinitesimal values as we take the $\Delta$ value to $0$. So if you write $\frac{\Delta x}{\Delta s}=\cos\alpha$, then as you take the $\Delta$ values to $0$, the angle between $\Delta x$ and $\Delta s$ will change during this process. The angle will approach some value. $\endgroup$ Nov 17, 2018 at 19:56
  • $\begingroup$ That is exactly my point. $\endgroup$
    – Cuspy Code
    Nov 17, 2018 at 20:30
  • $\begingroup$ It's somewhat analagous to how in introductory calculus how the slope of the secant line approaches the slope of the tangent line. $\frac{\Delta y}{\Delta x}=m$ for the slope of the secant line. Then you take the $\Delta$ quantities $0$, and $m$ approaches a value generally different from the original $m$. The same thing is happening here. Just because the angle between $\Delta x$ and $\Delta s$ is $\alpha$ does not mean that it stays this way. Therefore, if you write $\frac{\Delta x}{\Delta s}=\cos\alpha$, you should note that $\alpha$ changes as you take the limit to get to $\frac{dx}{ds}$ $\endgroup$ Nov 17, 2018 at 20:32

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