Say you have a curve y = f(x). I am looking at a tangent line at point A (line A) and a tangent line at point B (line B). Now let’s also define angle A as the angle between line A and, say, the x axis, and likewise angle B as the angle between line B and the x axis. If the only information you have is the absolute difference in the slopes of lines A and B, can you calculate the difference between angles A and B?
For some reason my first answer was yes, but after thinking through some examples I’m not so sure. For example, say the difference in slopes between lines A and B is 1. Wouldn’t the difference between angles A and B be quite different if:
- slope of line A is 3 and slope of line B is 2
- vs slope of line A is 2 and slope of line B is 1?
Am I right that you need more information to calculate the difference between angles A and B? Would this answer be different if the difference between slopes of lines A and B was expressed as a percentage instead of an absolute difference?
Let $\theta\in\left[0,\dfrac\pi2\right]$ be the angle between two lines with gradients $a$ and $b$ such that $ab\ne-1.$
Then \begin{align}\theta&=\Bigg|\arctan\bigg(\tan\Big(\arctan (a) - \arctan (b)\Big)\bigg)\Bigg|\\&= \arctan\left|\frac{a−b}{1+ab}\right|,\end{align} since $$\arctan (a) - \arctan (b)= \begin{cases}\arctan\left(\dfrac{a−b}{1+ab}\right)+\pi\:\operatorname{sgn}(a) &\text{if }ab<-1; \\ \arctan\left(\dfrac{a−b}{1+ab}\right) &\text{if }ab>-1.\end{cases}$$
Hence, to determine the angle between two non-vertical, non-perpendicular lines, their gradient difference is insufficient (for example, the angles between $y=2x$ and $y=x,$ and between $y=2x$ and $y=3x,$ are $18^\circ$ and $8^\circ,$ respectively), but their absolute gradient difference together with their gradient product is sufficient.
P.S. $$-\pi<\arctan(a) - \arctan(b)<\pi.$$