73. ferris wheel a ferris wheel 50 ft in diameter makes one revolution every 40 sec. if the center of the…

Explanation:

Step1: Model the Ferris wheel motion

The height ( h(t) ) of a rider on a Ferris wheel can be modeled using a sinusoidal function. The general form for the height as a function of time ( t ) (after reaching the low point) is ( h(t)=A\sin(\omega t)+k ), where ( A ) is the amplitude, ( \omega ) is the angular frequency, and ( k ) is the vertical shift (center height).

• The diameter of the Ferris wheel is 50 ft, so the radius ( A = \frac{50}{2}=25 ) ft.
• The center of the wheel is 30 ft above the ground, so ( k = 30 ) ft.
• The period ( T = 40 ) sec. The angular frequency ( \omega=\frac{2\pi}{T}=\frac{2\pi}{40}=\frac{\pi}{20} ) rad/sec.

Since we start at the low point, the sine function is shifted. At ( t = 0 ), ( h(0)=30 - 25 = 5 ) ft (low point). The correct model considering the start at low point is ( h(t)=25\sin\left(\frac{\pi}{20}t-\frac{\pi}{2}\right)+30 ) (because the sine function starts at 0, to start at -1 (low point), we shift it by ( -\frac{\pi}{2} )) or we can also use a cosine function with a phase shift. Alternatively, we can use ( h(t)=25\cos\left(\frac{\pi}{20}t+\pi\right)+30 ), but it's easier to adjust the sine function. Let's use the sine function with a phase shift such that at ( t = 0 ), ( h(0) = 5 ).

We want to find ( t ) when ( h(t)=50 ) ft.

So, set up the equation: [ 50=25\sin\left(\frac{\pi}{20}t-\frac{\pi}{2}\right)+30 ]

Step2: Solve for ( t )

First, subtract 30 from both sides: [ 50 - 30=25\sin\left(\frac{\pi}{20}t-\frac{\pi}{2}\right) ] [ 20 = 25\sin\left(\frac{\pi}{20}t-\frac{\pi}{2}\right) ] Divide both sides by 25: [ \sin\left(\frac{\pi}{20}t-\frac{\pi}{2}\right)=\frac{20}{25}=\frac{4}{5} ]

Recall that ( \sin\left(x-\frac{\pi}{2}\right)=-\cos(x) ), so: [ -\cos\left(\frac{\pi}{20}t\right)=\frac{4}{5} ] [ \cos\left(\frac{\pi}{20}t\right)=-\frac{4}{5} ]

We know that ( \cos(\theta)=-\frac{4}{5} ), so ( \theta=\arccos\left(-\frac{4}{5}\right) ). Also, ( \cos(\theta)=\cos(2\pi - \theta) ), but we are looking for the time when the rider is 50 ft above the ground. Let's solve ( \frac{\pi}{20}t=\arccos\left(-\frac{4}{5}\right) )

First, calculate ( \arccos\left(-\frac{4}{5}\right) ). We know that ( \cos(\alpha)=-\frac{4}{5} ), so ( \alpha=\pi-\arccos\left(\frac{4}{5}\right) ). Let's compute ( \arccos\left(\frac{4}{5}\right)\approx0.6435 ) radians, so ( \alpha=\pi - 0.6435\approx2.4981 ) radians.

Now, solve for ( t ): [ t=\frac{2.4981\times20}{\pi}\approx\frac{49.962}{\pi}\approx16. \text{Wait, let's check the equation again.}

Wait, the correct model: when at the low point (( t = 0 )), the height is ( 30 - 25=5 ) ft. The height function can also be written as ( h(t)=25\cos\left(\frac{\pi}{20}t+\pi\right)+30 ), because ( \cos(\pi)= - 1 ), so at ( t = 0 ), ( h(0)=25\times(-1)+30 = 5 ) ft (correct). Let's use this cosine model.

Set ( h(t)=50 ): [ 50=25\cos\left(\frac{\pi}{20}t+\pi\right)+30 ] Subtract 30: [ 20 = 25\cos\left(\frac{\pi}{20}t+\pi\right) ] Divide by 25: [ \cos\left(\frac{\pi}{20}t+\pi\right)=\frac{4}{5} ] Using the identity ( \cos(A + \pi)=-\cos(A) ), so: [ -\cos\left(\frac{\pi}{20}t\right)=\frac{4}{5} ] [ \cos\left(\frac{\pi}{20}t\right)=-\frac{4}{5} ] Which is the same as before. Now, the general solution for ( \cos(x)=-\frac{4}{5} ) is ( x = 2n\pi\pm\arccos\left(-\frac{4}{5}\right) ), ( n\in\mathbb{Z} ). We are looking for the first positive ( t ) (after reaching the low point) when ( h(t)=50 ).

We know that ( \arccos\left(-\frac{4}{5}\right)=\pi-\arccos\left(\frac{4}{5}\right)\approx3.1416 - 0.6435 = 2.4981 ) radians.

So, ( \frac{\pi}{20}t=2.4981 ) (taking the principal solution in ( [0, 2\pi] ) for the first time when the height is 50, since after the low point, the rider moves up, reaches the center at ( t = 10 ) sec (quarter period), top at ( t = 20 ) sec (half period, height ( 30 + 25 = 55 ) ft), wait, wait, the top height is ( 30 + 25 = 55 ) ft? Wait, the problem says "how long after reaching the low point is a rider 50 ft above the ground". Wait, the center is 30 ft, radius 25 ft, so top height is ( 30 + 25 = 55 ) ft, so 50 ft is below the top.

Wait, maybe my model is wrong. Let's re - model the height. Let's take the low point as ( t = 0 ), so the height function can be written as ( h(t)=30 - 25\cos\left(\frac{\pi}{20}t\right) ). Because at ( t = 0 ), ( \cos(0)=1 ), so ( h(0)=30 - 25 = 5 ) ft (low point). At ( t = 20 ) sec (half period), ( \cos(\pi)=-1 ), so ( h(20)=30 + 25 = 55 ) ft (top point). At ( t = 10 ) sec (quarter period), ( \cos\left(\frac{\pi}{2}\right)=0 ), so ( h(10)=30 ) ft (center height). This model makes more sense. Let's use this: ( h(t)=30 - 25\cos\left(\frac{\pi}{20}t\right) )

Now, set ( h(t)=50 ): [ 50=30 - 25\cos\left(\frac{\pi}{20}t\right) ] Subtract 30 from both sides: [ 20=-25\cos\left(\frac{\pi}{20}t\right) ] Divide both sides by - 25: [ \cos\left(\frac{\pi}{20}t\right)=-\frac{20}{25}=-\frac{4}{5} ] Now, we need to solve for ( t ): [ \frac{\pi}{20}t=\arccos\left(-\frac{4}{5}\right) ] We know that ( \arccos\left(-\frac{4}{5}\right)=\pi-\arccos\left(\frac{4}{5}\right) ). Let's calculate ( \arccos\left(\frac{4}{5}\right)\approx0.6435 ) radians, so ( \arccos\left(-\frac{4}{5}\right)\approx3.1416 - 0.6435 = 2.4981 ) radians.

Then, [ t=\frac{2.4981\times20}{\pi}\approx\frac{49.962}{3.1416}\approx16 \text{ sec? Wait, no, let's check with the correct model.}

Wait, ( h(t)=30 - 25\cos\left(\frac{\pi}{20}t\right) ). Let's plug ( t = \frac{40}{3}\approx13.33 ) sec? Wait, let's solve the equation ( 30 - 25\cos\left(\frac{\pi}{20}t\right)=50 ) [ -25\cos\left(\frac{\pi}{20}t\right)=20 ] [ \cos\left(\frac{\pi}{20}t\right)=-\frac{4}{5} ] The general solution for ( \cos(x)=-\frac{4}{5} ) is ( x = 2n\pi\pm\arccos\left(-\frac{4}{5}\right) ), ( n\in\mathbb{Z} ). We want the smallest positive ( t ), so we take ( x=\arccos\left(-\frac{4}{5}\right) ) (since ( \cos(x) ) is negative in the second quadrant, and we are moving from the low point ( ( t = 0 ), ( x = 0 ), ( \cos(0)=1 )) up, so as ( t ) increases, ( \frac{\pi}{20}t ) increases, and ( \cos\left(\frac{\pi}{20}t\right) ) decreases from 1 to - 1. We want to find when ( \cos\left(\frac{\pi}{20}t\right)=-\frac{4}{5} )

We know that ( \cos(\theta)=-\frac{4}{5} ), so ( \theta=\arccos\left(-\frac{4}{5}\right)\approx2.498 ) radians. Then, [ t=\frac{2.498\times20}{\pi}\approx\frac{49.96}{3.1416}\approx16 \text{ sec? Wait, but the top is at } t = 20 \text{ sec with height } 55 \text{ ft. So 50 ft is between } t = 10 \text{ sec (center, 30 ft)} \text{ and } t = 20 \text{ sec (top, 55 ft)? Wait, no, from low point (t = 0, 5 ft) to center (t = 10, 30 ft) to top (t = 20, 55 ft). So 50 ft is between t = 10 and t = 20.

Wait, let's recalculate ( \arccos\left(-\frac{4}{5}\right) ). Using a calculator, ( \arccos\left(-\frac{4}{5}\right)\approx2.498 ) radians. Then ( t=\frac{2.498\times20}{\pi}\approx16 ) sec? Wait, ( \frac{\pi}{20}t = 2.498 ), so ( t=\frac{2.498\times20}{\pi}\approx16 ). But let's check the height at t = 16:

( h(16)=30 - 25\cos\left(\frac{\pi}{20}\times16\right)=30 - 25\cos\left(\frac{4\pi}{5}\right) )

( \cos\left(\frac{4\pi}{5}\right)=\cos(144^{\circ})=-\cos(36^{\circ})\approx - 0.8090 )

So ( h(16)=30 - 25\times(-0.8090)=30 + 20.225 = 50.225\approx50 ) ft. So that's correct.

Step3: Verify the solution

We used the height function ( h(t)=30 - 25\cos\left(\frac{\pi}{20}t\right) ), which is valid because:

• At ( t = 0 ), ( h(0)=30 - 25\cos(0)=30 - 25 = 5 ) ft (low point).
• The period ( T=\frac{2\pi}{\omega}=\frac{2\pi}{\frac{\pi}{20}} = 40 ) sec, which matches the given period.

When we solved ( 30 - 25\cos\left(\frac{\pi}{20}t\right)=50 ), we got ( \cos\left(\frac{\pi}{20}t\right)=-\frac{4}{5} ), and then found ( t\approx\frac{\arccos\left(-\frac{4}{5}\right)\times20}{\pi}\approx16 ) sec (more precisely, let's calculate ( \arccos\left(-\frac{4}{5}\right) ) in radians:

( \arccos\left(-\frac{4}{5}\right)=\pi-\arccos\left(\frac{4}{5}\right) )

( \arccos\left(\frac{4}{5}\right)\approx0.6435 ) radians, so ( \arccos\left(-\frac{4}{5}\right)\approx3.1416 - 0.6435 = 2.4981 ) radians.

Then ( t=\frac{2.4981\times20}{\pi}\approx\frac{49.962}{3.1416}\approx16. ) (actually, ( \frac{49.962}{3.1416}\approx16 ) when rounded to a reasonable decimal place. But let's do it more accurately.

( \frac{2.4981\times20}{3.1415926536}\approx\frac{49.962}{3.1415926536}\approx16. ) (exactly, let's see, ( \frac{4}{5}) of the way from 10 to 20? Wait, no, the angle ( \frac{\pi}{20}t ) when ( t = 16 ) is ( \frac{4\pi}{5}=144^{\circ} ), which is in the second quadrant, and ( \cos(144^{\circ})=-\cos(36^{\circ})\approx - 0.8 ), and ( - 25\times(-0.8)=20 ), so ( 30 + 20 = 50 ), which is correct.

Answer:

(\approx 16.7) seconds (more precisely, (\frac{20}{\pi}\arccos\left(-\frac{4}{5}\right)\approx\frac{20}{\pi}(2.4981)\approx16.7) seconds) or approximately 17 seconds. But from the calculation above, when we did the exact substitution, we saw that ( t\approx16 ) seconds (with a small error due to rounding). The exact value is ( t=\frac{20}{\pi}\arccos\left(-\frac{4}{5}\right) \approx 16.7 ) seconds.