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Contents Index Dark Mode Prev Up Next \(\DeclareMathOperator{\erf}{erf}
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Chapter 6 Calculus and Trigonometry
Recall that in
Problem 5.10.7 we did not explain how we knew that the centripetal force was
\begin{equation}
mx\omega^2\text{.}\tag{6.1}
\end{equation}
We said that we would be able to show that this is the correct formula for the centripetal force once we had extended the scope of our differentiation rules. Extending our differentiation rules to include the trigonometric rules, which we will do next, will enable us to explain where
formula (6.1) came from.
According to Newton,
Force = mass \(\times\) acceleration , so to verify the centripetal force formula the task is to show that centripetal acceleration is given by
\(x\omega^2\text{.}\) We will finally do this in
Problem 6.2.15 but first, we’ll need to be able to differentiate both the sine and the cosine functions. In fact, since acceleration is the second derivative of position, we will need the second derivatives of both. Once the derivatives of the sine and cosine are known, computing the derivatives of the other four trigonometric functions is straightforward.
But before we begin we will remind you of some useful facts from Trigonometry.
