Question

In Example \(1.3.4\) we arrived at a model for the length of daylight (in hours) in Philadelphia on the \({t^{th}}\) day of the year:

\(L\left( t \right) = 12 + 2.8\sin \left( {\frac{{2\pi }}{{365}}\left( {t - 80} \right)} \right)\)

Use this model to compare how the number of hours of daylight is increasing in Philadelphia on March 21\(\left( {t = 80} \right)\) and May 21\(\left( {t = 141} \right)\).

Short answer

At \(t = 80\), the rate of increase of daylight is \(0.0482\) hours per day and at \(t = 141\), the rate of increase of daylight is \(0.02398\) hours per day.

Step-by-step solution

The chain rule

If \(g\) is differentiable at \(x\) and \(f\) is differentiable at \(g\left( x \right)\), then the composite function \(F = f \circ g\) defined by \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) is differentiable at \(x\) and \(F'\) is given by the product \(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\). In Leibniz notation, if \(y = f\left( u \right)\) and \(u = g\left( x \right)\) are both differentiable functions, then \(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}\).

The rate of increase of daylight

Differentiate \(L\left( t \right)\) with respect to \(t\) as follows:

\(\begin{aligned}L'\left( t \right) &= 0 + 2.8\cos \left( {\frac{{2\pi }}{{365}}\left( {t - 80} \right)} \right)\frac{d}{{dt}}\left( {\frac{{2\pi }}{{365}}\left( {t - 80} \right)} \right)\\ &= 2.8\cos \left( {\frac{{2\pi }}{{365}}\left( {t - 80} \right)} \right)\left( {\frac{{2\pi }}{{365}}} \right)\\ &= \frac{{5.6\pi }}{{365}}\cos \left( {\frac{{2\pi }}{{365}}\left( {t - 80} \right)} \right)\end{aligned}\)

At \(t = 80\), the rate of increase of the daylight can be obtained as follows:

\(\begin{aligned}L'\left( {80} \right) &= \frac{{5.6\pi }}{{365}}\cos \left( {\frac{{2\pi }}{{365}}\left( {80 - 80} \right)} \right)\\ &= \frac{{5.6\pi }}{{365}}\cos \left( 0 \right)\\ &= \frac{{5.6\pi }}{{365}}\left( 1 \right)\\ &\approx 0.0482\end{aligned}\)

Hence, the number of hours of daylight is increasing in Philadelphia on March 21 is \(0.0482\) hours per day.

At \(t = 141\), the rate of increase of the daylight can be obtained as follows:

\(\begin{aligned}L'\left( {141} \right) &= \frac{{5.6\pi }}{{365}}\cos \left( {\frac{{2\pi }}{{365}}\left( {141 - 80} \right)} \right)\\ &= \frac{{5.6\pi }}{{365}}\cos \left( {61} \right)\\ &= \frac{{5.6\pi }}{{365}}\left( {0.484} \right)\\ &\approx 0.02398\end{aligned}\)

Hence, the number of hours of daylight is increasing in Philadelphia on May 21 is \(0.02398\) hours per day.

The increase on May 21 is approximately half to that of March 21.