1) The hour hand on a clock is 4 inches long, the minutehand 6 inches long, and

Question

1) The hour hand on a clock is 4 inches long, the minutehand 6 inches long, and the second hand 7 inches long. Assumethat t = 0 corresponds to 12:00PM.

a) Determine the total distance travelled by the 3 handsbetween noon and 2:45 PM.

Explanation / Answer

The hour hand moves 2 radians every 12 hours It makes an angle of (/6)t with the vertical (where t ismeasured in hours) The minute hand moves 2 radians every 1 hours It makes an angle of (2)t with the vertical (where t ismeasured in hours) The second hand moves 2 radians every 1/60 hours It makes an angle of (120)t with the vertical (where t ismeasured in hours) (a) between 12:00 and 2:45, t = 2.75hrs The hour hand moves (/6)(.75)r = (/6)(2.75)4 =11/6 The minute hand moves (2)(.75)r = (2)(2.75)6 =33 The second hand moves (120)(.75)r = (120)(2.75)7 =1980 Total distance traveled by the tips is 12089/6 inches (approx527 feet 6 inches). For the positions of the tips of the hands (assuming an origin atthe center), we have in polar form: (all hands start vertical (corresponds to /2) and then runbackwards (clockwise)) hour hand ----> (r,) = (4,/2 - (/6)t) minute hand ----> (r,) = (6,/2 - 2t) second hand ----> (r,) = (7,/2 - 120t) Turning these into parametric equations for x and y, we have hour hand ----> x = 4cos(/2 - (/6)t); y =4sin(/2 - (/6)t) minute hand ----> x = 6cos(/2 - 2t); y =6sin(/2 - 2t) second hand ----> x = 7cos(/2 - 120t); y =7sin(/2 - 120t) We can simplify these to: hour hand ----> x = -4sin((/6)t); y =4cos((/6)t) minute hand ----> x = -6sin(2t); y = 6cos(2t) second hand ----> x = -7sin(120t); y =7cos(120t)

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