Periodic Extrema and Rates of Change on a Ferris Wheel
A Ferris wheel's height above the ground, in meters, is modeled by h(t)=12+10 sin(πt/15), where t is in minutes and 0≤t≤60. A student differentiates to get h'(t)=(2π/3)cos(πt/15), sets cos(πt/15)=0, and solves πt/15=π/2 to get t=7.5. The student concludes that the wheel reaches its maximum height of 22 meters only at t=7.5 minutes. The student also computes the average rate of change from t=0 to t=15 as 0 and claims this must equal the instantaneous rate at t=15 because the function is smooth.
1. Find all times in 0≤t≤60 when the wheel reaches a maximum or minimum height. What did the student's solution omit?
Read the worked answer
Setting cos(πt/15)=0 gives πt/15=π/2+kπ, so t=7.5+15k. In the interval, k=0,1,2,3 give t=7.5, 22.5, 37.5, and 52.5. At t=7.5 and 37.5, sin=1 and h=22, so both are maxima. At t=22.5 and 52.5, sin=−1 and h=2, so both are minima. The endpoints give h(0)=12 and h(60)=12, neither extreme. The student found only the first maximum and did not use the periodicity of cosine, so the claim that the maximum occurs only at t=7.5 is false.
2. Evaluate the student's claim that the average rate of change from t=0 to t=15 equals the instantaneous rate at t=15.
Read the worked answer
Compute h(0)=12 and h(15)=12+10 sin(π)=12, so the average rate is (12−12)/15=0. The instantaneous rate is h'(15)=(2π/3)cos(π)=−2π/3≈−2.09 m/min. The values are not equal. Smoothness does not make an average rate equal a pointwise derivative; the Mean Value Theorem guarantees only that some intermediate time has the same instantaneous rate as the average. Here that occurs at t=7.5, where h'=0. The student confused an average over an interval with a derivative at an endpoint.
3. Suppose the wheel's rotation period is changed to 20 minutes while the radius and center height stay the same. Write the new height model and compare the maximum vertical speed with the original.
Read the worked answer
A 20-minute period gives h(t)=12+10 sin(2πt/20)=12+10 sin(πt/10). The derivative is h'(t)=10(π/10)cos(πt/10)=π cos(πt/10), so the maximum vertical speed is π≈3.14 m/min. The original derivative h'(t)=(2π/3)cos(πt/15) has maximum speed 2π/3≈2.09 m/min. The shorter period increases the maximum vertical speed by a factor of (π)/(2π/3)=3/2. This makes sense because the wheel completes the same vertical cycle in two-thirds the time.