How does author derives $\cos \theta = \frac{dx}{ds} \qquad \sin\theta = \frac{dy }{ds},$? [closed]

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 $$s$$.

• 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.
– user191954
Nov 21, 2018 at 12:16

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.

• Generally the angle when looking at $\frac{\Delta x}{\Delta s}$ is different than when looking at $\frac{\text d x}{\text d s}$ Nov 17, 2018 at 19:30
• Yes. But if $x=x(s)$ is a continuous function, the angle difference can be made arbitrarily small by choosing a sufficiently small $\Delta s$. Nov 17, 2018 at 19:43
• 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. Nov 17, 2018 at 19:56
• That is exactly my point. Nov 17, 2018 at 20:30
• 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 $\frac{dx}{ds}$ Nov 17, 2018 at 20:32