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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