Course:MATH102/Question Challenge/2001 December Q8

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Question

A ferris wheel 30 meters high turns counterclockwise at a uniform rate of 3 revolutions per minute. Find the rate of change in the altitude (h) of a passenger at the instant when they are located at point P. Use the fact that $\theta =\pi /3$ radians at that instant. Specify the units of your result.

Ferris wheel rotates counterclockwise.

Hints

Hint 1
Use the fact that 1 revolution is $2\pi$ radians to write down a statement about $\frac{d\theta }{dt}$

Hint 2
What is the radius of the ferris wheel? Can you express the coordinates of the point P in terms of that radius and a trigonometric function of the angle $\theta$ ?

Hint 3
Use the chain rule to relate $\frac{dh}{dt}$ to $\frac{d\theta }{dt}$.

Solutions

Solution

$\frac{d\theta }{dt}$ = 3 revolutions per minute $=3(2\pi )=6\pi$ radians per minute=constant The radius of the wheel is $r=15$ meters. We label the diagram so that we can relate the height to the angle. Then $y=r\sin \theta$ Heights labeled on the ferris wheel diagram. The height of the passenger above the ground is given by: $h=15+y=15+15\sin \theta$. Hence, $\frac{dh}{dt}=15\cos \theta \cdot \frac{d\theta }{dt}$ We substitute in the angles $\theta =\pi /3$ and $d\theta /dt=6\pi$ at that given instant and simplify: $\frac{dh}{dt}=15\cos (\pi /3)(6\pi )$ $\frac{dh}{dt}=15(0.5)(6\pi )$ $\frac{dh}{dt}=45\pi$ m/min.