Chapter 3.4, Problem 72E

To determine

To find: The number of hours of daylight increases in Philadelphia on March 21 and May 21.

Answer to Problem 72E

The number of hours of daylight increases in Philadelphia on March 21 and May 21 is approximately 0.0482 and 0.02398.

Explanation of Solution

Given:

The function $L\left(t\right)=12+2.8\sin \left[\frac{2\pi }{365}\left(t-80\right)\right]$.

Derivative rule:

(1) Constant Multiple Rule: $\frac{d}{dx}\left[cf\left(x\right)\right]=c\frac{d}{dx}\left[f\left(x\right)\right]$

(2) Sum Rule: $\frac{d}{dx}\left(f+g\right)=\frac{d}{dx}\left(f\right)+\frac{d}{dx}\left(g\right)$

(3) Quotient Rule: $\frac{d}{dx}\left(\frac{f}{g}\right)=\frac{g\frac{d}{dx}\left(f\right)-f\frac{d}{dx}\left(g\right)}{{\left(g\right)}^2}$

Result used: 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 $F^{\prime }$ is given by the product

$F^{\prime }\left(x\right)=f^{\prime }\left(g\left(x\right)\right)\cdot g^{\prime }\left(x\right)$ (1)

Calculation:

Obtain the derivative of $L\left(t\right)$.

$\begin{array}{c}{L}^{\prime }\left(t\right)=\frac{d}{dx}\left(L\left(t\right)\right)\\ =\frac{d}{dx}\left(12+2.8\sin \left[\frac{2\pi }{365}\left(t-80\right)\right]\right)\end{array}$

Apply the sum rule (1) and the constant multiple rule (2),

$\begin{array}{c}{L}^{\prime }\left(t\right)=\frac{d}{dx}\left(12+2.8\sin \left(\frac{2\pi }{365}\left(t-80\right)\right)\right)\\ =\frac{d}{dt}\left(12\right)+\frac{d}{dt}\left(2.8\sin \left(\frac{2\pi }{365}\left(t-80\right)\right)\right)\\ =0+2.8\frac{d}{dt}\left(\sin \left(\frac{2\pi }{365}\left(t-80\right)\right)\right)\end{array}$

$L^{\prime }\left(t\right)=2.8\frac{d}{dt}\left(\sin \left(\frac{2\pi }{365}\left(t-80\right)\right)\right)$

Apply the chain rule as shown in equation (1),

Let $g\left(t\right)=\frac{2\pi }{365}\left(t-80\right)$ and $f\left(u\right)=2.8\sin u\text{ where }u=g\left(t\right)$.

$L^{\prime }\left(t\right)=f^{\prime }\left(g\left(t\right)\right)\cdot g^{\prime }\left(t\right)$ (3)

The derivative of $f^{\prime }\left(g\left(t\right)\right)$ is computed as follows,

$\begin{array}{c}{f}^{\prime }\left(g\left(t\right)\right)=f^{\prime }\left(u\right)\\ =\frac{d}{du}\left(f\left(u\right)\right)\\ =\frac{d}{du}\left(2.8\sin u\right)\\ =2.8\cos u\end{array}$

Substitute $u=\frac{2\pi }{365}\left(t-80\right)$ in the above equation,

$f^{\prime }\left(g\left(t\right)\right)=2.8\cos \left(\frac{2\pi }{365}\left(t-80\right)\right)$

Thus, the derivative is $f^{\prime }\left(g\left(t\right)\right)=2.8\cos \left(\frac{2\pi }{365}\left(t-80\right)\right)$.

The derivative of $g\left(t\right)$ is computed as follows,

$\begin{array}{c}{g}^{\prime }\left(t\right)=\frac{d}{dt}\left(\frac{2\pi }{365}\left(t-80\right)\right)\\ =\frac{2\pi }{365}\frac{d}{dt}\left(t-80\right)\\ =\frac{2\pi }{365}\left(1-0\right)\\ =\frac{2\pi }{365}\end{array}$

Thus, the derivative is $g^{\prime }\left(t\right)=\frac{2\pi }{365}$.

Substitute $2.8\cos \left(\frac{2\pi }{365}\left(t-80\right)\right)$ for $f^{\prime }\left(g\left(t\right)\right)$ and $\frac{2\pi }{365}$ for $g^{\prime }\left(t\right)$ in equation (3),

$L^{\prime }\left(t\right)=2.8\cos \left(\frac{2\pi }{365}\left(t-80\right)\right)\left(\frac{2\pi }{365}\right)$

Therefore, the derivative is $L^{\prime }\left(t\right)=2.8\cos \left(\frac{2\pi }{365}\left(t-80\right)\right)\left(\frac{2\pi }{365}\right)$.

Given that, the rate of the length of daylight is increasing in Philadelphia on March 21 and May 21. That is, $L^{\prime }\left(t\right)>0$.

Obtain the value of t for which $L^{\prime }\left(t\right)>0$.

$\begin{array}{l}2.8\cos \left(\frac{2\pi }{365}\left(t-80\right)\right)\left(\frac{2\pi }{365}\right)>0\\ \cos \left(\frac{2\pi }{365}\left(t-80\right)\right)>0\end{array}$

Note that, $\cos x>0\text{ when 0 < }x\text{ < }\frac{\pi }{2}$ and it can be concluded that $0<\frac{2\pi }{365}\left(t-80\right)<\frac{\pi }{2}$.

$\begin{array}{l}0<t-80<\frac{\pi }{2}\cdot \frac{365}{2\pi }\\ 0<t-80<91.25\\ 80<t<91.25+80\\ 80<t<171.25\end{array}$

Here, t is from 21st March to 20 th June.

On March 21, the value of $t=80$.

Substitute $t=80$ in $L^{\prime }\left(t\right)$,

$\begin{array}{c}{L}^{\prime }\left(80\right)=2.8\cos \left(\frac{2\pi }{365}\left(80-80\right)\right)\left(\frac{2\pi }{365}\right)\\ =2.8\cos \left(0\right)\left(0.0172\right)\\ =2.8\left(1\right)\left(0.0172\right)\left(Q\cos \left(0\right)=1\right)\\ =0.0482\end{array}$

Thus, the number of hours of daylight is increasing in Philadelphia on March 21 is approximately 0.0482.

On May 21, the value of t is $t=141\left(80+30+21\right)$.

Substitute $t=141$ in above equation,

$\begin{array}{c}{L}^{\prime }\left(141\right)=2.8\cos \left(\frac{2\pi }{365}\left(141-80\right)\right)\left(\frac{2\pi }{365}\right)\\ =2.8\cos \left(\frac{2\pi }{365}\left(61\right)\right)\left(\frac{2\pi }{365}\right)\\ =2.8\cos \left(\frac{122\pi }{365}\right)\left(0.0172\right)\\ =2.8\left(0.4975\right)\left(0.0172\right)\end{array}$

$\approx 0.02396$

Thus, the number of hours of daylight increases in Philadelphia May 21 is approximately 0.02398.

Note: $L^{\prime }\left(141\right)$ is approximately two times the $L^{\prime }\left(80\right)$.