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2 Trignometry problems

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1. A bicycle tire has a diameter of 20 inches and is revolving at a rate of 10 rpm. At t =0, a certain point is at height 0. What is the height of the point above the ground after 20 seconds?

2. The sun always illuminates half of the moon's surface, except during a lunar eclipse. The illuminated portion of the moon visible from Earth varies as it revolves around Earth resulting in the phases of the moon. The period from a full moon to a new moon and back to a full moon is called a synodic month and is 29 days, 12 hours, and 44.05 minutes long. Write a sine function that models the fraction of the moon's surface which is seen to be illuminated during a synodic month as a function of the number of days, d, after full moon. [Note: full moon equals illuminated.]

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Solution Summary

How to calculate height of a rotating wheel?

How to model phases of moon using a sine function?

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Dear Student,

The answers to these questions are given below.

Best of Luck

OTA 103881


Since the bicycle is rotating at 10 rpm, in 20 seconds it will finish 10 x (20/60) = 3.333 rotations.

Since the point is at height 0, after every complete rotation the height will still be 0.

Thus, after 3 complete rotations, the height will still be 0.

Now, we only have to calculate the height reached by the point in the last 0.333 rotation.

Note that because there is symmetry, it doesn't matter if the tire rotates clock or anti clockwise.

0.333 rotation corresponds to one third of a rotation and since full rotation is 360 degrees, this is

360/3 = 120 degrees.

The radius of the tire = 20 inches/2 = 10 ...

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