Question

The length of the day in Boulder (Latitude 40 N) can be modeled approximately by l(t) = −3 cos ( 2π 365 (t + 10)) + 12 where l is given in hours and t is the day of the year.
(a) Evaluate l(355); fully interpret the result in the context of this problem, including units.
(b) Evaluate l ′ (265); fully interpret the result in the context of this problem, including units.
(c) Calculate when l ′ (t) is largest. Explain.

Answer 1

Answer:

(a) I(355)=9

(b) I'(265)=−0.05163

(c) The l′(t) is largest at 81.25.

Step-by-step explanation:

The given function is

$I(t)=-3\cos (\frac{2\pi}{365}(t+10))+12$

where I(t) the length of the day in Boulder and l is given in hours and t is the day of the year.

(a)

Substitute t=355 in given function.

$I(355)=-3\cos (\frac{2\pi}{365}(355+10))+12$

$I(355)=-3\cos (2\pi)+12$

$I(355)=-3(1)+12$

$I(355)=9$

Therefore, the value of I(355) is 9. It means the length of the day in Boulder is 9 hours at 355 day of the year.

(b)

Differentiate the given function with respect to t.

$I'(t)=-3(-\sin (\frac{2\pi}{365}(t+10)))(\frac{2\pi}{365})$

$I'(t)=\frac{6\pi}{365}\sin (\frac{2\pi}{365}(t+10))$

Substitute t=265 in the above function.

$I'(t)=\frac{6\pi}{365}\sin (\frac{2\pi}{365}(265+10))$

$I'(265)\approx −0.05163$

Therefore, the value of I'(265) is −0.05163. It means the length of the day in Boulder decreased by 0.05163 hours at 265 day of the year.

(c)

Differentiate the I'(t) function with respect to t.

$I''(t)=\frac{6\pi}{365}\cos (\frac{2\pi}{365}(t+10))(\frac{2\pi}{365})$

$I''(t)=\frac{12\pi^2}{(365)^2}\cos (\frac{2\pi}{365}(t+10))$

Equate $I''(t)=0$ to find critical points.

$\frac{12\pi^2}{(365)^2}\cos (\frac{2\pi}{365}(t+10))=0$

$\cos (\frac{2\pi}{365}(t+10))=0$

$\cos (\frac{2\pi}{365}(t+10))=\cos (\frac{\pi}{2})$

On comparing both sides we get

$\frac{2\pi}{365}(t+10)=\frac{\pi}{2}$

$t+10=\frac{\pi}{2}\times \frac{365}{2\pi}$

$t+10=\frac{365}{4}$

$t=\frac{365}{4}-10$

$t=81.25$

Check the value of the function I'''(t) at t=81.25.

Since $I'''(81.25)=-0.0000153$, therefore the value of l′(t) is largest at 81.25.