Parametric equations difficulty

[ksdhart]

I'm having difficulties with parametric equations. I try to solve this problem but I end up with an impossible solution. The text of the problem is:

53) Consider the Cartesian equation x=sqrt(1-y^2), -1 <= y <= 1, of a right semicircle of radius 1. Find a set of parametric equations and interpret the motion of a particle moving along this semicircle by using the parameter x = sin(t)

Given that I have two equations of x equals something, I set them equal. That gives me sin(t) = sqrt(1-y^2). I squared both sides to get sin^2(t) = 1 - y^2. Then I rearranged the terms and got y^2 = 1 - sin^2(t) or y^2 = cos^2(t). Thus, y = cos(t).

Now, the problem I have comes when I replace y with cos(t) in the inequality I was given. That leaves me with -1 <= cos(t) <= 1. Taking the inverse cosine of all parts leaves me with pi <= t <= 0, which has no solutions, as t cannot be greater than pi and less than zero.

The only thing I can think of is that when taking the inverse cosine, the <= signs become >= signs. If that's the case, then the inequality works, but it doesn't make any sense to me why that would happen.

 

[Ishuda]

I'm having difficulties with parametric equations. I try to solve this problem but I end up with an impossible solution. The text of the problem is:


Given that I have two equations of x equals something, I set them equal. That gives me sin(t) = sqrt(1-y^2). I squared both sides to get sin^2(t) = 1 - y^2. Then I rearranged the terms and got y^2 = 1 - sin^2(t) or y^2 = cos^2(t). Thus, y = cos(t).

Now, the problem I have comes when I replace y with cos(t) in the inequality I was given. That leaves me with -1 <= cos(t) <= 1. Taking the inverse cosine of all parts leaves me with pi <= t <= 0, which has no solutions, as t cannot be greater than pi and less than zero.

The only thing I can think of is that when taking the inverse cosine, the <= signs become >= signs. If that's the case, then the inequality works, but it doesn't make any sense to me why that would happen.

The short answer is that 'yes, the inequality signs have to be turned around because the derivative of the cosine is negative in that open interval'. See below for the start of a long explanation.

We have
-1 <= cos(t) <= 1
is true of all (real valued) t. So taking the arccos won't do any good in one sense.

In another sense, it does help since the
arccos(-1) = \(\displaystyle \pi\) + 2 m \(\displaystyle \pi\)
where m is any integer. Similarly,
arccos(1) = 2 n \(\displaystyle \pi\)
where n is any integer. So if we want a restriction on t we want to choose n as a function of m so that we can use those formulas. If we let n = m +1, that will do, i.e.
\(\displaystyle \pi\) + 2 m \(\displaystyle \pi\) <= t <= 2 n \(\displaystyle \pi\) = 2 (m+1) \(\displaystyle \pi\) = 2 \(\displaystyle \pi\) + 2 m \(\displaystyle \pi\),
i.e.
\(\displaystyle \pi\) + 2 m \(\displaystyle \pi\) <= t <= 2 \(\displaystyle \pi\) + 2 m \(\displaystyle \pi\).

We now need to choose an m so that sin(t) is positive in the interval. Well sin(t) is positive in the intervals
2 p \(\displaystyle \pi\) <= t <= \(\displaystyle \pi\) + 2 p \(\displaystyle \pi\)
where p is some integer

Now consider neither, one, or both of m and p are even. How about odd? As a hint, you will wind up with a basic interval plus the cyclic pieces repeated every 2\(\displaystyle \pi\) and the basic interval can be taken as what you derived, i.e. [0,\(\displaystyle \pi\)]

 

[Ishuda]

First, let me say I get carried away sometimes and the above is a step by step process to arrive at the answer which, if you are familiar with sines and cosines, can be arrived at (almost) immediately to get the answer. The process is about the same as yours.

The reasoning I would use on a test in determining the Domain of t would be something like this: Consider the behaviors of sin(t) and cos(t) over a complete cycle. If t is between 0 and \(\displaystyle \pi\), cos(t) covers the complete range between 1 and -1. Also, sin(t) is positive in this region. This region for t will work. If t is between \(\displaystyle \pi\) and 2\(\displaystyle \pi\), cos(t) also covers the complete range between 1 and -1 but sin(t) is negative. Thus, this region won't work. Since the sine and cosine are cyclic with period two \(\displaystyle \pi\), we have
y = cos(t)
x = sin(t)
for
2m\(\displaystyle \pi\) <= t <= \(\displaystyle \pi\) + 2m\(\displaystyle \pi\)

Depending on just what I thought the instructor/test wanted, I would give my answer as one obtained for m = 0.

 

[HallsofIvy]

I'm having difficulties with parametric equations. I try to solve this problem but I end up with an impossible solution. The text of the problem is:


Given that I have two equations of x equals something, I set them equal. That gives me sin(t) = sqrt(1-y^2). I squared both sides to get sin^2(t) = 1 - y^2. Then I rearranged the terms and got y^2 = 1 - sin^2(t) or y^2 = cos^2(t). Thus, y = cos(t).

Now, the problem I have comes when I replace y with cos(t) in the inequality I was given. That leaves me with -1 <= cos(t) <= 1. Taking the inverse cosine of all parts leaves me with pi <= t <= 0, which has no solutions, as t cannot be greater than pi and less than zero.

No, it doesn't. cosine is a periodic function with period \(\displaystyle 2\pi\). cos(t)= -1 for \(\displaystyle t= -\pi\) also. So take \(\displaystyle -\pi\le t\le 0\). Or you can say that \(\displaystyle cos(t)= 1\) for \(\displaystyle t= 2\pi\). Then \(\displaystyle -1\le cos(t)\le 1\) for \(\displaystyle \pi \le x\le 2\pi\).

The only thing I can think of is that when taking the inverse cosine, the <= signs become >= signs. If that's the case, then the inequality works, but it doesn't make any sense to me why that would happen.