Consider the parameterization of the unit circle given by x=cos(4t^2−t), y=sin(4t^2−t) for t in (−InF, INF). Describe in words and sketch how the circle is traced out, and use this to answer the following questions.

(a) When is the parameterization tracing the circle out in a clockwise direction?
_____________
(Give your answer as a comma-separated list of intervals, for example, (0,1), (3,Inf)). Enter the word None if there are no such intervals.

(b) When is the parameterization tracing the circle out in a counter-clockwise direction?
__________

(Give your answer as a comma-separated list of intervals, for example, (0,1), (3,Inf)). Enter the word None if there are no such intervals.

C)Does the entire unit circle get traced by this parameterization?

(d) Give a time t at which the point being traced out on the circle is at (10):
t=_________

Ok, I graphed it in my graphing calculator, the circle seemed to be tracing itself for a while, but I still don't get how to write the intervals. Since it appeared to be just tracing itself clockwise, I typed in the first one (my homework is online)
(0,1),(1,0),(0,-1),(-1,0)
A message was shown saying that the left endpoint must be less than the right endpoint, so I'm not sure how to solve this one. I believe the answer for B is none, and I don't know what to put for the time, please explain this to me....

To understand the parameterization of the unit circle given by x=cos(4t^2−t), y=sin(4t^2−t), let's break it down step by step:

When t is positive, the values inside the sine and cosine functions change gradually from 0 to 1 and then back to 0, resulting in the tracing of the unit circle in the counterclockwise direction. This is because the positive values of t cause the argument inside the trigonometric functions to increase gradually.

When t is negative, the values inside the sine and cosine functions change from 0 to -1 and then back to 0, resulting in the tracing of the unit circle in the clockwise direction. This is because the negative values of t cause the argument inside the trigonometric functions to decrease gradually.

(a) When the parameterization is tracing the circle out in a clockwise direction:
The parameterization traces the circle in the clockwise direction for negative values of t. As t ranges from -Inf to 0, the circle is traced in a clockwise direction. Therefore, the interval is (-Inf, 0).

(b) When the parameterization is tracing the circle out in a counterclockwise direction:
The parameterization traces the circle in the counterclockwise direction for positive values of t. As t ranges from 0 to +Inf, the circle is traced in a counterclockwise direction. Therefore, the interval is (0, +Inf).

(c) Does the entire unit circle get traced by this parameterization?
Yes, the entire unit circle is traced by this parameterization because for all real values of t, the x and y components will vary from -1 to 1, covering all points on the unit circle.

(d) To find a time t at which the point being traced out on the circle is at (10):
The point (10) is not on the unit circle since the unit circle has a maximum radius of 1. Therefore, there is no time t at which the point (10) will be traced on the circle.

To determine the intervals when the parameterization traces the unit circle in a specific direction, we need to consider the range of the parameter t. In this case, t can take any value from negative infinity to positive infinity.

(a) To determine when the parameterization traces the unit circle in a clockwise direction, we need to identify the intervals of t where the function is decreasing as t increases.

To do this, we can find the derivative of each component of the parameterization:

dx/dt = (-sin(4t^2 - t))(8t - 1)
dy/dt = (cos(4t^2 - t))(8t - 1)

The direction of motion is determined by the signs of dx/dt and dy/dt.

For dx/dt, we need to find when (-sin(4t^2 - t))(8t - 1) < 0.
For dy/dt, we need to find when (cos(4t^2 - t))(8t - 1) < 0.

Simplifying the inequalities, we can determine the intervals where the parameterization traces the unit circle in a clockwise direction.

(b) To determine when the parameterization traces the unit circle in a counter-clockwise direction, we need to identify the intervals of t where the function is increasing as t increases.

Using the same process as in part (a), we can find the intervals where the signs of dx/dt and dy/dt indicate a counter-clockwise direction.

(c) To determine if the entire unit circle gets traced by this parameterization, we need to consider the entire range of t from negative infinity to positive infinity. By looking at a graph of the parameterization, if the curve goes through all the points on the unit circle, then the entire unit circle is traced.

(d) To find a time t at which the point being traced is at (10), you need to solve the equations x(t) = 10 and y(t) = 10 simultaneously. However, in this case, since the unit circle has a radius of 1, there are no t-values where the point on the unit circle is at (10).

It's important to carefully analyze the equations and consider the behavior of the parameterization to determine the intervals and if the entire unit circle is traced.

To understand how the unit circle is traced out by the given parameterization, let's analyze the behavior of the variables x and y in the equations x = cos(4t^2 - t) and y = sin(4t^2 - t).

The parameterization defines x and y as functions of t. As t varies from negative infinity to positive infinity, the values of x and y change, and the point (x, y) traces the unit circle.

(a) To determine when the parameterization traces the circle in a clockwise direction, we need to find the intervals where the angle 4t^2 - t is decreasing. Recall that the standard unit circle is traversed in a counterclockwise direction. So, if we want to go around it clockwise, we need to traverse it in the opposite direction.

To find these intervals, we can differentiate the angle function 4t^2 - t with respect to t and look for intervals where the derivative is negative. Differentiating 4t^2 - t gives us 8t - 1.

Setting 8t - 1 < 0, we solve for t:
8t < 1
t < 1/8

Therefore, the parameterization is tracing the unit circle clockwise for t in the interval (-∞, 1/8).

(b) On the other hand, to find when the parameterization traces the circle in a counterclockwise direction, we look for intervals where the angle function 4t^2 - t is increasing. In this case, we want to traverse the standard unit circle in its original direction.

So, we need to find the intervals where the derivative 8t - 1 is positive. Solve 8t - 1 > 0:
8t > 1
t > 1/8

Therefore, the parameterization is tracing the unit circle counterclockwise for t in the interval (1/8, ∞).

(c) Yes, the entire unit circle is traced by this parameterization since the values of t in the interval (-∞, ∞) cover the entire real number line. This means that the parameterization will trace all points on the unit circle multiple times as t changes.

(d) To find a specific time t when the point being traced out on the circle is at (1, 0), which corresponds to (10) as written in the question, we need to equate the values of x and y to their corresponding values at (1, 0).

For x = 1, we have cos(4t^2 - t) = 1.
Solving this equation, we find that 4t^2 - t = 0.
Factorizing, we have t(4t - 1) = 0.

From this equation, we can see that either t = 0 or 4t - 1 = 0. Solving for t in the second equation, we find t = 1/4.

So, the point (1, 0) is traced when t = 0 or t = 1/4.

In summary:

(a) The parameterization traces the circle clockwise for t in the interval (-∞, 1/8).
(b) The parameterization does not trace the circle counterclockwise in any interval, so the answer is None.
(c) Yes, the entire unit circle is traced by this parameterization.
(d) The point (1, 0) is traced at times t = 0 and t = 1/4.

Consider the parameterization of the unit circle given by x = cos(ln(2t)), y = sin(ln(2t)) for t in (0,∞). De- scribe in words and sketch how the circle is traced out, and use this to answer the following questions.

SOLUTION
Let h(t) = ln(2t). Then h′(t) = 1 . The particle is moving
t
counterclockwise when h′(t) > 0, that is, when t is in (0,∞).
Any other values of t, t ≤ 0, are not in the domain of the func- tion, so the particle is never moving clockwise.
The full circle is traced out if h(t) = ln(2t) produces values over one period of the functions sin(t) and cos(t). In this case h(t) produces all the needed values, so the full circle must be produced.
To find a point where the parameterization is tracing out the point (1,0), we are looking for a t such that h(t) = 0. Such a value is t = 0.5.
Correct Answers:
• None
• (0,infinity) •A
• 0.5