An approximation is used to derive $\Delta r≈r\Delta \phi$. From the Taylor series expansion of sine function about 0. Show that the approximation holds for small angle $\Delta \phi$.
This question is related to centripetal motion. But I haven't learnt Taylor series before so I have no idea how to solve this question.
For me, as I know that sin $x ≈ x - \frac{x^3}{3!} + \frac{x^5}{5!}+\cdots$
For small angle, $\sin(\phi)$ can be considered as $\phi$. But how can I show the approximation holds for small angle $\Delta \phi$?
Should I just use $\lim x$ approaches to 0 $\phi ≈ x - \frac{x^3}{3!} + \frac{x^5}{5!}+\cdots$?
BC=2r sin θ/2
=2r (θ/2−θ^3/(3!*8)+θ^5/(5!32)....)
=2r (θ/2) [cause x is very very small]
=rθ
if you haven't learnt taylor series than this explanation will be better and simpler:- When θ becomes very small we can easily say that BC becomes small. When ∠BAC becomes very very small we can treat BC as a line segment.This step can be done because arc(BC) is so small that it can now be considered to be a line segment.
NOW we can apply the approximation Δr≈rΔø as you said and it would make sense