Respuesta :
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
[tex]I(t)=-3\cos (\frac{2\pi}{365}(t+10))+12[/tex]
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.
[tex]I(355)=-3\cos (\frac{2\pi}{365}(355+10))+12[/tex]
[tex]I(355)=-3\cos (2\pi)+12[/tex]
[tex]I(355)=-3(1)+12[/tex]
[tex]I(355)=9[/tex]
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.
[tex]I'(t)=-3(-\sin (\frac{2\pi}{365}(t+10)))(\frac{2\pi}{365})[/tex]
[tex]I'(t)=\frac{6\pi}{365}\sin (\frac{2\pi}{365}(t+10))[/tex]
Substitute t=265 in the above function.
[tex]I'(t)=\frac{6\pi}{365}\sin (\frac{2\pi}{365}(265+10))[/tex]
[tex]I'(265)\approx −0.05163[/tex]
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.
[tex]I''(t)=\frac{6\pi}{365}\cos (\frac{2\pi}{365}(t+10))(\frac{2\pi}{365})[/tex]
[tex]I''(t)=\frac{12\pi^2}{(365)^2}\cos (\frac{2\pi}{365}(t+10))[/tex]
Equate [tex]I''(t)=0[/tex] to find critical points.
[tex]\frac{12\pi^2}{(365)^2}\cos (\frac{2\pi}{365}(t+10))=0[/tex]
[tex]\cos (\frac{2\pi}{365}(t+10))=0[/tex]
[tex]\cos (\frac{2\pi}{365}(t+10))=\cos (\frac{\pi}{2})[/tex]
On comparing both sides we get
[tex]\frac{2\pi}{365}(t+10)=\frac{\pi}{2}[/tex]
[tex]t+10=\frac{\pi}{2}\times \frac{365}{2\pi}[/tex]
[tex]t+10=\frac{365}{4}[/tex]
[tex]t=\frac{365}{4}-10[/tex]
[tex]t=81.25[/tex]
Check the value of the function I'''(t) at t=81.25.
Since [tex]I'''(81.25)=-0.0000153<0[/tex], therefore the value of l′(t) is largest at 81.25.
we got that 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, 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 and value of l′(t) is largest at 81.25.
What is trigonometric function ?
Function which involves trigonometric ratios are trigonometric function .
Here given that
[tex]I(t)=-3 \cos \left(\frac{2 \pi}{365}(t+10)\right)+12[/tex]
where l is given in hours and t is the day of the year.
Putting t=355
[tex]I(355)=-3 \cos \left(\frac{2 \pi}{365}(355+10)\right)+12 \\[/tex]
[tex]I(355)=-3 \cos (2 \pi)+12 \\&I(355)=9[/tex]
Hence we can say that length of the day in Boulder is 9 hours at 355 day of the year.
Now Differentiating the given function with respect to t.
[tex]I^{\prime}(t)=-3\left(-\sin \left(\frac{2 \pi}{365}(t+10)\right)\right)\left(\frac{2 \pi}{365}\right)[/tex]
[tex]I^{\prime}(t)=\frac{6 \pi}{365} \sin \left(\frac{2 \pi}{365}(t+10)\right)[/tex]
Putting t=265
[tex]I^{\prime}(t)=\frac{6 \pi}{365} \sin \left(\frac{2 \pi}{365}(265+10)\right) \\\\&I^{\prime}(265) =-00.05163[/tex]
Hence we can say that length of the day in Boulder decreased by 0.05163 hours at 265 day of the year.
Now Differentiating the I'(t) function with respect to t
[tex]&I^{\prime \prime}(t)=\frac{6 \pi}{365} \cos \left(\frac{2 \pi}{365}(t+10)\right)\left(\frac{2 \pi}{365}\right) \\\\\\\I^{\prime \prime}(t)=\frac{12 \pi^{2}}{(365)^{2}} \cos \left(\frac{2 \pi}{365}(t+10)\right)[/tex]
Now simplify I''(t)=0 for critical points of I'(t)
[tex]\frac{12 \pi^{2}}{(365)^{2}} \cos \left(\frac{2 \pi}{365}(t+10)\right)=0 \\\\\\cos \left(\frac{2 \pi}{365}(t+10)\right)=0\\\\ \\\cos \left(\frac{2 \pi}{355}(t+10)\right)=\cos \left(\frac{\pi}{2}\right)[/tex]
[tex]\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[/tex]
Now finding value of I''(t) at t=81.25
we will get negative value of l'''(81.25)
Hence I'(t) is maximum at t=81.25
we got that 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, 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 and value of l′(t) is largest at 81.25.
To learn more about trigonometric function visit :https://brainly.com/question/26827051
