Ferris Wheel Trig Problems

A ferris wheel spins at a constant rate, so a rider's height above the ground rises and falls in a smooth, repeating pattern. That kind of repeating up-and-down motion is exactly what sine and cosine functions describe, which is why ferris wheels are one of the most common real-world settings for periodic trig problems. The goal in every ferris wheel problem is the same: build a function \(h(t)\) that gives the rider's height above the ground at any time \(t\), then use that function to answer questions about height, time, or period.

These problems rely on the same graph shapes covered in the sine graph lesson and the transformations covered in transformation of trigonometric functions, so it helps to be comfortable with amplitude, period, and vertical shift before working through these.

Every ferris wheel height function has the same general form:

\( h(t) = A\sin(B(t - C)) + D \)   or   \( h(t) = A\cos(B(t - C)) + D \)

Each letter comes directly from a piece of information about the wheel:

• Amplitude, \(A\): equal to the radius of the wheel, since that is how far the rider moves above and below the center.
• Midline, \(D\): the height of the wheel's center above the ground.
• Period: the time for one full revolution. Since period \( = \dfrac{2\pi}{B}\), you solve for \(B\) using \( B = \dfrac{2\pi}{\)period\(} \).
• Horizontal shift, \(C\): shifts the graph left or right, used when the rider does not start at the very bottom or very top.

Choosing sine or cosine, and whether it is positive or negative, depends only on the rider's starting position:

• Starting at the bottom: use \( -\cos \) (height starts low, at the minimum).
• Starting at the top: use \( +\cos \) (height starts high, at the maximum).
• Starting level with the center, moving up: use \( +\sin \).

Suppose a ferris wheel has a radius of 15 meters, its center is 17 meters above the ground, it takes 40 seconds to complete one revolution, and a rider boards at the very bottom of the wheel.

1. Amplitude: \( A = 15 \) (the radius).
2. Midline: \( D = 17 \) (the center height).
3. Period: one revolution takes 40 seconds, so \( B = \dfrac{2\pi}{40} = \dfrac{\pi}{20} \).
4. Starting point: the rider starts at the bottom, so use \( -\cos \) with no horizontal shift.

Putting the pieces together gives:

\( h(t) = -15\cos\left(\dfrac{\pi}{20}t\right) + 17 \)

Check it at \(t = 0\): \( h(0) = -15\cos(0) + 17 = -15 + 17 = 2 \). That matches the lowest point of the wheel, since the bottom sits 2 meters above the ground (center height minus radius). The graph of this function over one and a half revolutions looks like this:

Using the same wheel, find the rider's height 5 seconds after boarding.

\( h(5) = -15\cos\left(\dfrac{\pi}{20}(5)\right) + 17 = -15\cos\left(\dfrac{\pi}{4}\right) + 17 \)

Since \( \cos\left(\dfrac{\pi}{4}\right) \approx 0.707 \):

\( h(5) \approx -15(0.707) + 17 \approx -10.6 + 17 = 6.4 \) meters

At 10 seconds, a quarter of the way through the revolution, the rider is level with the center: \( h(10) = -15\cos\left(\dfrac{\pi}{2}\right) + 17 = 0 + 17 = 17 \) meters. At 20 seconds, halfway around, the rider is at the top: \( h(20) = -15\cos(\pi) + 17 = 15 + 17 = 32 \) meters.

Now work in the other direction: find every time in the first revolution when the rider is at a height of 24.5 meters.

\( 24.5 = -15\cos\left(\dfrac{\pi}{20}t\right) + 17 \)

\( 7.5 = -15\cos\left(\dfrac{\pi}{20}t\right) \)

\( \cos\left(\dfrac{\pi}{20}t\right) = -0.5 \)

Cosine equals \(-0.5\) at \( \dfrac{2\pi}{3} \) and \( \dfrac{4\pi}{3} \) within one revolution, so:

\( \dfrac{\pi}{20}t = \dfrac{2\pi}{3} \ \Rightarrow\ t = \dfrac{40}{3} \approx 13.3 \) seconds (rider is rising)

\( \dfrac{\pi}{20}t = \dfrac{4\pi}{3} \ \Rightarrow\ t = \dfrac{80}{3} \approx 26.7 \) seconds (rider is descending)

Two answers make sense because a rider passes through every height twice per revolution: once on the way up, once on the way down.

Not every ferris wheel problem starts at the bottom, and some use sine instead of cosine. A few things to watch for:

• If the rider starts at the top, switch to \( +\cos \) instead of \( -\cos \).
• If the problem gives a starting angle or a boarding platform partway up the wheel, you may need a horizontal shift \(C\), just like in graph transformations of trig functions.
• If diameter is given instead of radius, divide by 2 before using it as the amplitude.
• The same modeling approach appears in other periodic contexts, such as tides and water depth trig problems, where depth rises and falls instead of height.

• Always identify amplitude, midline, and period first, before worrying about sine versus cosine.
• Sketch or picture the wheel and mark where the rider starts; this makes the sign and function choice obvious.
• When solving for time, remember trig equations usually have more than one solution within a period.
• Check your answer by plugging it back into \(h(t)\) to confirm it gives a sensible height between the minimum and maximum.

Related lessons

• Trigonometry Help