Course:MATH102/Question Challenge/2001 December Q8

	Questions? Click here to add or read comments for this problem
Please rate how easy you found this problem: {{#w4grb_rate:}} Hard                                              Easy

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.