Respuesta :
Answer:
The required function is [tex]h(T)=30\sin (\frac{\pi T}{15})+25[/tex].
Step-by-step explanation:
The general sine function is
[tex]y=A\sin (Bx+C)+D[/tex] .... (1)
Where, A is amplitude, [tex]\frac{2\pi}{B}[/tex] is period, C is phase shift and D is midline.
It is given that the maximum height above the ground is 55 feet and the minimum height above the ground is 5 feet.
The amplitude of the function is
[tex]A=\frac{Maximum+Minimum}{2}=\frac{55+5}{2}=30[/tex]
The Midline of the function is
[tex]D=\frac{Maximum-Minimum}{2}=\frac{55-5}{2}=25[/tex]
A ferris wheel rotates around in 30 seconds. So, the period of the function is 30.
[tex]\frac{2\pi}{B}=30\Rightarrow B=\frac{2\pi}{30}=\frac{\pi}{15}[/tex]
[tex]2\pi=30B[/tex]
Substitute A=30, [tex]B=\frac{\pi}{15}[/tex], C=0 and D=25 in equation (1), to find the required function.
[tex]y=30\sin (\frac{\pi}{15}x+0)+25[/tex]
The required variable is T. Replace the variable x by T. So the height function is
[tex]h(T)=30\sin (\frac{\pi}{15}T+0)+25[/tex]
Therefore the required function is [tex]h(T)=30\sin (\frac{\pi T}{15})+25[/tex].
Following are the calculation to the function:
Given:
[tex]\to \theta= 30 \ second\\\\\to \text{maximum height}= 55 \ feet\\\\\to \text{minumum height}= 5 \ feet[/tex]
To find:
[tex]\delta t=?[/tex]
Solution:
period [tex]\sin \theta = 2n\\\\[/tex]
Time taking [tex]= 30 \ second\\\\[/tex]
period of our function[tex]= \sin \frac{\pi}{15} \ t\\\\[/tex]
maximum function [tex]= 55 \ feet\\\\[/tex]
minimum function[tex]= 5 \ feet\\\\[/tex]
[tex]\to 25 \sin \frac{\pi}{15}t +30\\\\[/tex]
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