Answer:
The maximum height a rider will experience is 55 feet.
Step-by-step explanation:
Let's start writing the function that defines the path of a seat on the new Ferris wheel. This function will depend of the variable ''t'' which is time.
[tex]X(t)=(x,y)[/tex]
In which [tex]X(t)[/tex] are the coordinates of the seat (the x - coordinate and the y - coordinate) that depend from time.
[tex]''x''[/tex] and [tex]''y''[/tex] are functions that depend from the variable ''t''.
For this exercise :
[tex]X(t)=[-25sin(\frac{\pi}{30}t);-25cos(\frac{\pi}{30}t)+30][/tex]
In order to find the maximum height a rider will experience we will study the behaviour of the y - component from the function [tex]X(t)[/tex].
The function to study is [tex]y(t)=-25cos(\frac{\pi}{30}t)+30[/tex]
To find its maximum, we will derivate this function and equalize it to 0. Doing this, we will find the ''critical points'' from the function.
⇒ [tex]y(t)=-25cos(\frac{\pi}{30}t)+30[/tex] ⇒
[tex]y'(t)=\frac{5}{6}\pi sin(\frac{\pi}{30}t)[/tex]
Now we equalize [tex]y'(t)[/tex] to 0 ⇒
[tex]y'(t)=0[/tex] ⇒ [tex]\frac{5}{6}\pi sin(\frac{\pi}{30}t)=0[/tex]
In this case it is easier to look for the values of ''t'' that verify :
[tex]sin(\frac{\pi}{30}t)=0[/tex]
Now we need to find the values of ''t''. We know that :
[tex]sin(0)=0\\\\sin(\pi)=0\\sin(-\pi)=0[/tex]
Therefore we can write the following equivalent equations :
[tex]\frac{\pi}{30}t=0[/tex] (I)
[tex]\frac{\pi}{30}t=\pi[/tex] (II)
[tex]\frac{\pi}{30}t=-\pi[/tex] (III)
From (I) we obtain [tex]t_{1}=0[/tex]
From (II) we obtain [tex]t_{2}=30[/tex]
And finally from (III) we obtain [tex]t_{3}=-30[/tex]
We found the three critical points of [tex]y(t)[/tex]. To see if they are either maximum or minimum we will use the second derivative test. Let's calculate the second derivate of [tex]y(t)[/tex] :
[tex]y'(t)=\frac{5}{6}\pi sin(\frac{\pi}{30}t)[/tex] ⇒
[tex]y''(t)=\frac{\pi ^{2}}{36}cos(\frac{\pi }{30}t)[/tex]
Now given that we have an arbitrary critical point ''[tex]t_{n}[/tex]'' ⇒
If [tex]y''(t_{n})>0[/tex] then we will have a minimun at [tex]t_{n}[/tex]
If [tex]y''(t_{n})<0[/tex] then we will have a maximum at [tex]t_{n}[/tex]
Using the second derivative test with [tex]t_{1},t_{2}[/tex] and [tex]t_{3}[/tex] ⇒
[tex]y''(t_{1})=y''(0)=\frac{\pi ^{2}}{36} >0[/tex] ⇒ We have a minimum for [tex]t_{1}=0[/tex]
[tex]y''(t_{2})=y''(30)=\frac{-\pi^{2}}{36}<0[/tex] ⇒ We have a maximum for [tex]t_{2}=30[/tex]
[tex]y''(t_{3})=y''(-30)=\frac{-\pi^{2}}{36}<0[/tex] ⇒ We have a maximum for [tex]t_{3}=-30[/tex]
The last step for this exercise will be to find the values of the maximums.
We can do this by replacing in the equation of [tex]y(t)[/tex] the critical points [tex]t_{2}[/tex] and [tex]t_{3}[/tex] ⇒
[tex]y(t_{2})=y(30)=55[/tex]
[tex]y(t_{3})=y(-30)=55[/tex]
We found out that the maximum height a rider will experience is 55 feet.
