Helping Ada Lovelace with her Homework: Classroom Exercises from a Victorian Calculus Course – Solution to Problem 3

Since $\tan \theta=b/a$, Lovelace assumed that $\sin⁡\theta=b$ and $\cos⁡\theta=a$ in her attempt to solve Problem 3. However, $\sin⁡\theta$ and $\cos⁡\theta$ are actually equal to $b/r$ and $a/r$, respectively; so the expression Lovelace derived should have been
$$\begin{array}{lcl} (a+bk)^{m+nk}&=& r^{m+nk} (\cos⁡\theta+k \sin⁡\theta)^{m+nk}\\ &=&r^{m+nk}  (e^{k\theta})^{m+nk}\\ &=&r^{m+nk}\cdot e^{k(m\theta)} \cdot e^{-n\theta}\\ &=&r^m\cdot e^{-n\theta} \cdot r^{nk} \cdot e^{k(m\theta)}\\ &=&e^{m \log ⁡r-n\theta}\cdot e^{k(n \log ⁡r+m\theta)}\\ &=&e^A (\cos ⁡B + k \sin ⁡B ).\end{array}$$

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