Can someone please tell me the mistake i made in this problem:
x=3cost and y=3sint
i used this equation x^2+y^2=1
(3cost)^2+(3sint)^2=9
the final answer is cos^2t +sin^2t=9(but i got 1)

- anonymous

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- anonymous

the equation for a circle is x^2 + y^2 = r^2. You have to find r yourself by plugging x and y into the equation.
\[(3\cos t)^2+(3\sin t)^2=9\cos^2t+9\sin^2t\]
\[9(\sin^2t+\cos^2t)=r^2\]
9 = r^2
r = 3
Plug r = 3 in the general equation for the circle. My preference would be to leave it as x^2 + y^2 = 9. However, you must substitute x = cost and y = sint because the radius is already factored in.

- anonymous

ty i understand it now. but now i am stuck on a similar problem you helped me previously with. i have to find when t<0 for the follwowing equations 10+t y=2t

- anonymous

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## More answers

- anonymous

* x=10+t and y=2y

- anonymous

Solve for t in both equations. Then substitute them both in t<0. Know that you have to do x and y separately. Then, solve for x and y in their respective inequalities.

- anonymous

do i have to plug in zero fpr t when i solve for their equations

- anonymous

Right away, plug in 0 for t in both equations to find what point we're talking about. You should get x = 10 and y = 2 for the point (10, 2). To derive the inequality, solve for t without plugging 0s in.

- anonymous

yah i got the answer but since its less than it is left (10,0) ?

- anonymous

hey i dont know why my response shows up 10 times

- anonymous

Yeah, to the left of (10, 0) because x<10. You're right about the point, it is at (10, 0), not (10, 2)...my fault.

- anonymous

no worries thanks again for ur help

- anonymous

i have another question if you do not mind helping. i have to find the the parametrization of the unit circle or part of it for the following problem x=cos(t^2) and y=sin(t^2)

- anonymous

This one is parametrized. Is there anything specific being asked for?

- anonymous

- anonymous

it also asks how the circle is traced ou (clockwise and counterclockwise

- anonymous

It is traced along the unit circle because if you put x and y into the equation of a circle x^2 + y^2 = r^2, then you'll get r = 1, or x^2 + y^2 = 1. The manner in which it is traced is what makes it different than just an ordinary x = cost, y = sint. It traces in a counterclockwise direction just like a normal parametrization of the unit circle. However, it traces a lot faster. Do you know what I mean by faster?

- anonymous

the graph actually moves clockwise wen t is less than zero and increases when t is greater than zero

- anonymous

Oh sorry, I didn't consider negative t values. When t is negative the x = cost and y = sint parametrization goes clockwise. However, I would think that x = cos(t^2) and y = sin(t^2) would still go counterclockwise when t is negative because the negative would just cancel itself out...

- anonymous

i mean counterclockwise when it increases

- anonymous

how did you figure out the direction

- anonymous

The ones related to the unit circle are obvious for me because I know that's the way the unit circle moves (1, 0) when t = 0, (0, 1) when t = pi/2, (-1, 0) when t = pi, etc.
However, the best way to do it is to set up a table with t, x, and y. You choose t values, and then get corresponding x's and y's. It's crucial that you pick t values in numerical order (0, 1, 2, etc.) because the order to plot the points indicates the order they come in the path. t=0 comes before t=1 and t=1 comes before t =2.

- anonymous

but for this problem the t is squared so what i do? bc if i plug in pi/2 into cos it become cos(pi(^2))/4

- anonymous

Yeah, that's a yucky value to work with. What you do is pick a t value such that t^2 = n where n is a "nice number" like pi or pi/2. Just take the square root of t to get t = sqrt(n). So, think of a nice number, then use the square root of it.

- anonymous

my teacher used another method using the derivative
if the function is increasing then the derivative>0 so itcs counterclockwise if its dcreasing it clockwise

- anonymous

Haha, I didn't even think about the derivative. Yes, that works for many unit circle problems.

- anonymous

what type of problems specifically - when dealing with squared functions

- anonymous

You have to make sure you know which direction is the "increasing" direction - the orientation of the curve - when t is increasing. For example, the orientation of the curve x = cos(-t^2) and y = sin(-t^2), the derivative is positive when moving in a clockwise direction.

- anonymous

u mean counterclockwise

- anonymous

Hmm, now I'm just confused. I guess this is why I prefer to determine orientation numerically. Yes, the example I gave wasn't a good example... you'll have to ask your teacher about derivatives...haha.

- anonymous

what happens to the values on the unit circle when you move clockwise

- anonymous

The t values? T values would be negative, though the graphing calculator freaks out when you try to use them. I generally just use positive values of t when working with parameters, unless an interval of t with negative limits is given to me.

- anonymous

Or the t values would just go down, I guess would be a better way to put it.

- anonymous

in genral on the unit circle if u move clockwise what happens to the values do they become negative

- anonymous

hey just wanted to say thank you for everything

- anonymous

Well if you chose negative values of t with increasing absolute value (-1, -2, -3), you will be plotting points moving the clockwise direction. This isn't the correct way to think about orientation, though because we plot values of t of increasing value (-3, -2, -1, etc.).
I may be beginning to steer you in a bad direction. In general, I'd say not to think too much into negative values of t besides plotting them. Plotting points does not do anything for changing orientation. The orientation of the unit circle defined as x = cost and y = sint always moves in the counter clockwise direction.
If you're confused on what the orientation is, though you can graph the set of parametric equations on a graphing calculator. It traces it out with respect to the orientation.

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