Find coordinates of point P(t)

Hi!

If $\displaystyle P(t)$ has coordinates $\displaystyle \left(-\frac{\sqrt{5}}{3},\frac23\right)$, find the coordinates of the point.

a. $\displaystyle P(t + \pi) \rightarrow \left(\frac{\sqrt{5}}{3},-\frac23\right)$

Quadrant 2: moving by π goes half a circle which ends up in Quadrant 4.

b. $\displaystyle P(-t) \rightarrow \left( \frac{\sqrt{5}}{3},-\frac23 \right)$

I just switch the signs.

c. $\displaystyle P(t - \pi) \rightarrow \left( \frac{\sqrt{5}}{3},-\frac23 \right)$

Moving clockwise from Quadrant 2, ending up in Quadrant 4.

d. $\displaystyle P(-t - \pi) \rightarrow \left(-\frac{\sqrt{5}}{3},-\frac23\right)$

From b., moving clockwise from Quadrant 4, ending up in Quadrant 2.

Re: Find coordinates of point P(t)

Parts b and d are incorrect. Draw a diagram, or think of the trigonometric identities (sine and cosine).

Re: Find coordinates of point P(t)

Also remember that the relationship between polar coordinates ( rcos theta, r sin theta ) for a point with Cartesian coordinates ( x , y ) is given by

x = r cos theta and y = r sin theta.

This may help you in understanding the fundamental principal.

Re: Find coordinates of point P(t)

My first attempt was with a diagram.

So, with sine and cosine I got the same: cosine of a negative is positive and sine of a positive is positive.

b. $\displaystyle \left(\frac{\sqrt{5}}{3},\frac23\right)$

d. $\displaystyle \left( \frac{\sqrt{5}}{3},-\frac23 \right)$

Re: Find coordinates of point P(t)

Quote:

Originally Posted by

**Unreal** My first attempt was with a diagram.

So, with sine and cosine I got the same: cosine of a negative is positive and sine of a positive is positive.

b. $\displaystyle \left(\frac{\sqrt{5}}{3},\frac23\right)$

d. $\displaystyle \left( \frac{\sqrt{5}}{3},-\frac23 \right)$

b) You want to use:

$\displaystyle \cos(-t)=\cos(t)$

$\displaystyle \sin(-t)=-\sin(t)$

This shows you the x-coordinate is the same, while the y-coordinate is negated.

On a diagram, you would see the original is a quadrant II angle, the the negative is then a quadrant III angle.

d) Here you can use:

$\displaystyle \cos(-t-\pi)=-\cos(t)$

$\displaystyle \sin(-t-\pi)=\sin(t)$

On a diagram, move half a revolution in the negative direction from part b) to wind up in quadrant I.

Re: Find coordinates of point P(t)

OK, I have got that explanation.

b. $\displaystyle \left(-\frac{\sqrt{5}}{3}, -\frac23\right)$

d.$\displaystyle \left(\frac{\sqrt{5}}{3}, \frac23\right)$

Do you have any other alternatives to explaining $\displaystyle P(-t)$?

The others I find straightforward, at least so far.

The whole P(t) stuff is confusing. Here are two explanations I found in a text:

Quote:

The mapping of t into P(t) can be described by positioning a copy of the real line vertically with t = 0 coinciding with the point (1,0) on the unit circle. For any real number t, the point P(t) is obtained by "wrapping" the line around the circle and marking as P(t) the point on the circle that corresponds to the position of t.

From this I understand, whatever number t is, $\displaystyle \left(-\frac{\sqrt{5}}{3}, \frac23\right)$, you place it on the circle. In this case, the quadrant is 2, so the P(t) goes there. I understand no more.

Quote:

The reference number for each of the points is t, so the coordinates of each of the points have the same magnitude as those of P(t). However, the coordinates might be negative.

Understanding - The values will remain the same, except the signs might change.

Re: Find coordinates of point P(t)

For part b), think of the original angle t landing you in the second quadrant, beginning at (1,0). Now, if you begin there again, but movie in the negative direction the same distance along the unit circle, this time you will end up in the third quadrant. Make sense?

Re: Find coordinates of point P(t)

Quote:

Originally Posted by

**MarkFL** For part b), think of the original angle t landing you in the second quadrant, beginning at (1,0). Now, **if you begin there again**, but move in the negative direction the same distance along the unit circle, this time you will end up in the third quadrant. Make sense?

Are we beginning at (1,0)?

If so, I get it from the perspective of the angle being formed looking the same, just moving in different directions -- it's an obtuse angle.

What exactly is $\displaystyle -t$ supposed to do? In the same way moving $\displaystyle \pi$ from $\displaystyle t$ suggests we are moving by half a circle counterclockwise.

Re: Find coordinates of point P(t)

Yes, beginning at (1,0) but moving in the opposite direction.