Question

PLEASE HELP! WILL MARK BRAINLIEST!

Answer 1

Answer:

B

Step-by-step explanation:

Answer 2

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.

$X(t)=(x,y)$

In which $X(t)$ are the coordinates of the seat (the x - coordinate and the y - coordinate) that depend from time.

$''x''$ and $''y''$ are functions that depend from the variable ''t''.

For this exercise :

$X(t)=[-25sin(\frac{\pi}{30}t);-25cos(\frac{\pi}{30}t)+30]$

In order to find the maximum height a rider will experience we will study the behaviour of the y - component from the function $X(t)$.

The function to study is $y(t)=-25cos(\frac{\pi}{30}t)+30$

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.

⇒ $y(t)=-25cos(\frac{\pi}{30}t)+30$  ⇒

$y'(t)=\frac{5}{6}\pi sin(\frac{\pi}{30}t)$

Now we equalize $y'(t)$ to 0 ⇒

$y'(t)=0$ ⇒ $\frac{5}{6}\pi sin(\frac{\pi}{30}t)=0$

In this case it is easier to look for the values of ''t'' that verify :

$sin(\frac{\pi}{30}t)=0$

Now we need to find the values of ''t''. We know that :

$sin(0)=0\\\\sin(\pi)=0\\sin(-\pi)=0$

Therefore we can write the following equivalent equations :

$\frac{\pi}{30}t=0$ (I)

$\frac{\pi}{30}t=\pi$ (II)

$\frac{\pi}{30}t=-\pi$ (III)

From (I) we obtain $t_{1}=0$

From (II) we obtain $t_{2}=30$

And finally from (III) we obtain $t_{3}=-30$

We found the three critical points of $y(t)$. To see if they are either maximum or minimum we will use the second derivative test. Let's calculate the second derivate of $y(t)$ :

$y'(t)=\frac{5}{6}\pi sin(\frac{\pi}{30}t)$ ⇒

$y''(t)=\frac{\pi ^{2}}{36}cos(\frac{\pi }{30}t)$

Now given that we have an arbitrary critical point ''$t_{n}$'' ⇒

If $y''(t_{n})>0$  then we will have a minimun at $t_{n}$

If $y''(t_{n})$ then we will have a maximum at $t_{n}$

Using the second derivative test with $t_{1},t_{2}$ and $t_{3}$ ⇒

$y''(t_{1})=y''(0)=\frac{\pi ^{2}}{36} >0$ ⇒ We have a minimum for $t_{1}=0$

$y''(t_{2})=y''(30)=\frac{-\pi^{2}}{36}$ ⇒ We have a maximum for $t_{2}=30$

$y''(t_{3})=y''(-30)=\frac{-\pi^{2}}{36}$ ⇒ We have a maximum for $t_{3}=-30$

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 $y(t)$ the critical points $t_{2}$ and $t_{3}$ ⇒

$y(t_{2})=y(30)=55$

$y(t_{3})=y(-30)=55$

We found out that the maximum height a rider will experience is 55 feet.