How many radians on the unit circle would the minute hand travel from 0° if it were to move 3π inches? What is the coordinate point associated with this radian measure?
@satellite73 @Loser66
I'm not sure if this is needed but here it is, 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.
@rational
Is the length of minute hand 4 in ?
Yes
The length travelled on "unit circle" equals the angle in radian measure. So simply find the angle intercepted in travelling 3pi inches on the clock
\[l=r\theta\] plugin \(l=3\pi\), \(r=4\) and solve \(\theta\)
is it just \[\frac{ 3\pi }{ 4 }\]
Yep thats the angle in radians also, thats the distance travelled on unit circle
that doesn't make sense could you show it in more steps
do you mean l = r(theta) 3pi = 4(theta) 3pi / 4 = theta ?
if so, looks good.
yes haha
just so you know, it is important to show the steps correctly. there is no way for others to predict what you think/assume
ok, what about the second part of the question for coordinate plane?
coordinate point
A point has two coordinates : x, y
you can find them using : \(x = r\cos(\theta)\) \(y=r\sin(\theta)\)
\(r\) = radius of clock = \(4\) \(y\) = angle from 0 degrees = \(\frac{3\pi}{4}\) plug them
just \[4\cos (\theta)\] ?
\(x = r\cos(\theta) = 4*\cos(\frac{3\pi}{4}) = ?\)
i get \[\sqrt[-2]{2}\] what does that simplify into?
you should be getting \(\large x = -2*\sqrt{2}\)
and \(\large y = 2*\sqrt{2}\) so the required coordinate point would be \[\large (-2\sqrt{2},~2\sqrt{2})\]
okay thank you, that's a weird coordinate,
np:)
btw, \(\sqrt{2}\) is just a number
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