Question

A Ferris wheel 50ft in diameter makes one revolution every

Answer 1

Answer:

The required function is $y=-25\cos (\pi x)+30$.

The rider is 50ft above the ground at x=45.56.

Step-by-step explanation:

The general cosine function is

$y=A\cos (Bx+C)+D$             .... (1)

where, |A| altitude, $\frac{2\pi}{B}$ is period, C/B is phase shift and D is midline.

The center of the wheel is 30ft above the ground. So, D=30.

A Ferris wheel 50ft in diameter makes one revolution every 2 minutes.

$\frac{2\pi}{B}=2\Rightarrow B=\pi$

Minimum value = 30-25 = 5

It means the equation passes through the point (0,5) and phase shift is 0.

$5=A\cos (B(0)+(0))+30$

$5-30=A\cos (0)$

$-25=A$

Substitute A=-25, B=π, C=0 and D=30 in equation (1).

$y=-25\cos (\pi x+0)+30$

$y=-25\cos (\pi x)+30$

Therefore, the required function is $y=-25\cos (\pi x)+30$.

Substitute y=50 to find the value of x.

$50=-25\cos (\pi x)+30$

$50-30=-25\cos (\pi x)$

$20=-25\cos (\pi x)$

Divide both sides by -25.

$\dfrac{20}{-25}=\cos (\pi x)$

$-0.8=\cos (\pi x)$

$\cos ^{-1}(-0.8)=\pi x$

$143.13=\pi x$

Divide both sides by π.

$\frac{143.13}{\pi}= x$

$45.55969= x$

$x\approx 45.56$

Therefore, the rider is 50ft above the ground at x=45.56.