3.9 #42 Related Rates Ferris Wheel Angular Velocity and Right Triangle Trig

The script discusses a physics problem involving a Ferris wheel that completes one revolution every two minutes. The problem is to determine the rate at which a person is rising when they are 16 feet above the ground, given the radius of the wheel is 10 meters. The script begins by establishing the rotational velocity of the Ferris wheel, which is converted into angular velocity in radians per minute. The angular velocity \( \frac{d\theta}{dt} \) is calculated to be \( \pi \) radians per minute. The script then sets up a right triangle to relate the vertical position \( y \) of the person to the angle \( \theta \) at the center of the wheel, aiming to find \( \frac{dy}{dt} \) when \( h \) (the height above the ground) is 16 meters.

Continuing from the setup in the previous paragraph, the script derives an equation relating \( y \) and \( \theta \) using the sine function, resulting in \( y = 10 \sin(\theta) \). The derivative of \( y \) with respect to time \( t \), \( \frac{dy}{dt} \), is then calculated to be \( 10 \cos(\theta) \frac{d\theta}{dt} \). Knowing \( \frac{d\theta}{dt} \) as \( \pi \) radians per minute, the script focuses on finding \( \cos(\theta) \) when \( y \) is 6 meters (which corresponds to \( h \) being 16 meters). Using the properties of a right triangle and the Pythagorean theorem, \( \cos(\theta) \) is determined to be \( \frac{8}{10} \) or 0.8. Substituting the known values, \( \frac{dy}{dt} \) at that moment is calculated to be \( 8\pi \) meters per minute, indicating the person's rate of ascent.