# Thread: Help with Unit Circle Problem

1. ## Help with Unit Circle Problem

"Suppose t is the length of the arc on the unit circle whose initial point is (1,0), and terminal point is P (x,y). If t=3.5, sketch on the unit circle below the approximate location of P(x,y)"

Okay, I understand this part of the question, I know the point is in quadrant 3, etc.

Then the question asks "find the x-coordinate and y-coordinate of P accurate to two decimal places"

I have the answers, but I am dumbfounded as to how they were acquired.

(.-94, .-35) are the coordinates.

Any help would be much appreciated.

Thank you

2. Originally Posted by NYKnicks

"Suppose t is the length of the arc on the unit circle whose initial point is (1,0), and terminal point is P (x,y). If t=3.5, sketch on the unit circle below the approximate location of P(x,y)"

Okay, I understand this part of the question, I know the point is in quadrant 3, etc.

Then the question asks "find the x-coordinate and y-coordinate of P accurate to two decimal places"

I have the answers, but I am dumbfounded as to how they were acquired.

(.-94, .-35) are the coordinates.

Any help would be much appreciated.

Thank you
Hi NYKnicks,

I suppose you found where the point lies by determining what part of the circumference 3.5 was.

$\frac{3.5}{2 \pi}$

Now, that's how much of 360 degrees the point has traveled from (1, 0).

$\frac{3.5}{2 \pi}\left(360\right) \approx 200.54^{\circ}$, which we'll call angle $\theta$.

The point $P(x, y) = P(\cos \theta, \sin \theta)$

Now find cos 200.54 and sin 200.54 and you have your coordinates.

3. You are given the arc length, and you know that the radius is r = 1. Plug this into the arc-length formula to find the subtended angle $\theta$.

Once you have the angle, use the fact that the x-value is the cosine on the unit circle, and the y-value is the sine.

4. Wow, thanks for the quick replies guys!

Understand now

5. It t = 3.5, then moving anti-clockwise with radius = r = 1, that is more that half of the circle of radius 1, half arc = pi = 3.1415 . . .

So the point is in the 3rd quardant . . . .
P(x,y) = P(cos 200.54, sin 200.54)