anonymous
  • anonymous
Will fan and medal Freddie is at chess practice waiting on his opponent's next move. He notices that the 4-inch-long minute hand is rotating around the clock and marking off time like degrees on a unit circle. Part 1: How many radians does the minute hand move from 3:35 to 3:55? (Hint: Find the number of degrees per minute first.) Part 2: How far does the tip of the minute hand travel during that time? Part 3: How many radians on the unit circle would the minute hand travel from 0° if it were to move 3π inches? Part 4: What is the coordinate point associated with this radian measure?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I just need to know if I'm right. To find the degree you would divide the degrees of a circle by the number of minutes in an hour which would be 360/60=6. So six per minute you would multiply by 5 and get 30° in 5 minutes. Now we need to find the radians by dividing 30 by 180 like this 30/180 which would give you 1/6 and the radian measure is π/6. I'm not exactly sure what to do now
anonymous
  • anonymous
@texaschic101 can you help?
anonymous
  • anonymous
@Michele_Laino can you help? Please.

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anonymous
  • anonymous
Can someone help?
Michele_Laino
  • Michele_Laino
the angular velocity of the 4-inch hand is 2*pi/60= pi/30 radians/min
anonymous
  • anonymous
I'm confused..
Michele_Laino
  • Michele_Laino
since the long minute hand travels one complete turn every 60 minutes
Michele_Laino
  • Michele_Laino
so we can write this condition: \[2\pi R = \omega R\Delta t\] where \Delta t = 60 minutes and R = 4 inches
Michele_Laino
  • Michele_Laino
\[2\pi R = \omega R60\] dividing by 60, we get: \[\omega = \frac{{2\pi }}{{60}} = \frac{\pi }{{30}}\]
Michele_Laino
  • Michele_Laino
so the answer for part 1, is: \[\alpha = \frac{\pi }{{30}} \times 20\]
Michele_Laino
  • Michele_Laino
since from 3.35 to 3.55 there are 20 minutes
Michele_Laino
  • Michele_Laino
so, after a simplification, we get: \[\alpha = \frac{{2\pi }}{3}\]
Michele_Laino
  • Michele_Laino
oops.. \[\alpha = \frac{{2\pi }}{3}\;radians\]
anonymous
  • anonymous
Sorry my computer froze. I think I am understanding a bit more
Michele_Laino
  • Michele_Laino
now, part 2 the requested distance L, is given by the subsequent computation: \[\Large L = R\alpha = R\frac{{2\pi }}{3}\] where R = 4 inches
anonymous
  • anonymous
The part I need help with is part 4 now
Michele_Laino
  • Michele_Laino
part 3) the requested angle is: \[\Large \beta = \frac{{3\pi }}{R}\]
Michele_Laino
  • Michele_Laino
and the requested coordinates are: \[\Large \left( {R,\frac{{3\pi }}{2}} \right)\] |dw:1433448393972:dw|
anonymous
  • anonymous
That is confusing^
Michele_Laino
  • Michele_Laino
the long-minute hand travels for a distance 3pi long, right?
anonymous
  • anonymous
Yes
Michele_Laino
  • Michele_Laino
so the final position is: |dw:1433449752947:dw|
Michele_Laino
  • Michele_Laino
at the final position correspond an angle of 3*pi/2 radians, so the polar coordinates are: (R, 3*pi/2) being R the length of the long minute hand
Michele_Laino
  • Michele_Laino
|dw:1433450239850:dw|

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