## anonymous one year ago 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?

1. 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

2. anonymous

@texaschic101 can you help?

3. anonymous

4. anonymous

Can someone help?

5. Michele_Laino

the angular velocity of the 4-inch hand is 2*pi/60= pi/30 radians/min

6. anonymous

I'm confused..

7. Michele_Laino

since the long minute hand travels one complete turn every 60 minutes

8. Michele_Laino

so we can write this condition: $2\pi R = \omega R\Delta t$ where \Delta t = 60 minutes and R = 4 inches

9. Michele_Laino

$2\pi R = \omega R60$ dividing by 60, we get: $\omega = \frac{{2\pi }}{{60}} = \frac{\pi }{{30}}$

10. Michele_Laino

so the answer for part 1, is: $\alpha = \frac{\pi }{{30}} \times 20$

11. Michele_Laino

since from 3.35 to 3.55 there are 20 minutes

12. Michele_Laino

so, after a simplification, we get: $\alpha = \frac{{2\pi }}{3}$

13. Michele_Laino

oops.. $\alpha = \frac{{2\pi }}{3}\;radians$

14. anonymous

Sorry my computer froze. I think I am understanding a bit more

15. 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

16. anonymous

The part I need help with is part 4 now

17. Michele_Laino

part 3) the requested angle is: $\Large \beta = \frac{{3\pi }}{R}$

18. Michele_Laino

and the requested coordinates are: $\Large \left( {R,\frac{{3\pi }}{2}} \right)$ |dw:1433448393972:dw|

19. anonymous

That is confusing^

20. Michele_Laino

the long-minute hand travels for a distance 3pi long, right?

21. anonymous

Yes

22. Michele_Laino

so the final position is: |dw:1433449752947:dw|

23. 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

24. Michele_Laino

|dw:1433450239850:dw|