Precalc Question! Please Help!?

Question

As a fan turns counterclockwise, a bug sits 1.8 feet from the center of rotation on one of its blades. The bug is at the 3 o'clock position on the fan when it begins to turn.

If the fan makes less than one full rotation, determine the distance the bug has traveled along its arc from the 3 o'clock position when its final position is...

a) ... 1 ft above the horizontal diameter of the fan for the second time.

b) ... 0.7 ft below the horizontal diameter of the fan for the first time.

c) ... 1.3 ft below the horizontal diameter of the fan for the second time.

Answer 1

Alissa,

 

So you’re bugged by this problem? Let’s then get to it methodically:

    Let r = radial distance of bug from center of rotation of fan blade (1.8 ft).

         C = circumference distance of circle at bug’s radius (1 full rotation).

         S = length of the arc along the circumference that bug travels on a partial rotation.

         A = angle (in degrees) that fan rotates to calculate S.

               (One full rotation = 360 degrees, with zero at 3 o’clock position = horizontal line).

         h = distance above or below horizontal line in cases below where bug stops.

 

Then, C = 2Πr = 2(3.14159)(1.8 ft) = 11.31 ft.

         S = (A/360)C = 11.31 (A/C) , note that A/C = fraction of one full rotation.

 Sin (A) = h/r = h/1.8

 

Case (a): h = 1.0 ft (2nd time above horizontal line so in Quadrant 2)

        Sin (A) = 1.0/1.8 = 0.5555 implies A = 180 – 33.749 = 146.25 degrees

                S = 11.31 (146.25/360) = 4.60 ft.

 

Case (b): h = - 0.7 ft (1st time below horizontal line so in Quadrant 3)

      Sin (A) = -0.7/1.8 = -0.38889 implies A = 180 + 22.8854 = 202.885 degrees

              S = 11.31 (202.885/360) = 6.37 ft.

 

Case (c): h = - 1.3 ft (2nd time below horizontal line so in Quadrant 4)

      Sin (A) = -1.3/1.8 = -0.7222 implies A = 360 – 46.238 = 313.761 degrees

              S = 11.31 (313.761/360) = 9.86 ft.

 

Now you're bug free :)