# Thread: simple parameter trig equations HELP !

1. ## simple parameter trig equations HELP !

just realli confused with parameterss especially with trig equations

a) x = 2costheta y=2sintheta 0<theta<pi

b) x=cos t, y=cos t 0< t < 2pi

c) x=2cos theta, y = root 3 sin theta 0<theta<2pi

d) x = 2 cos t, y = sin t , 0< theta < pi

e) 2t/[(1+t)squared], y = [(1-t)squared]/[(1+t)squared]

i have the answers to these questions i jst dont haf the solution and i dont no how to get there
im also realli confused about the restrictions

any help is GREATLY appreciated
thankyouu !

2. Originally Posted by THSKluv
just realli confused with parameterss especially with trig equations

a) x = 2costheta y=2sintheta 0<theta<pi Mr F says: Square each of them and then add together.

b) x=cos t, y=cos t 0< t < 2pi Mr F says: Square each of them and then add together.

c) x=2cos theta, y = root 3 sin theta 0<theta<2pi Mr F says: ${\color{red} x = \cos t}$ and ${\color{red} \frac{y}{\sqrt{3}} = \sin t}$. Square each of these and then add together.

d) x = 2 cos t, y = sin t , 0< theta < pi Mr F says: ${\color{red} \frac{x}{2} = \cos t}$ and ${\color{red} y = \sin t}$. Square each of these and then add together.

e) 2t/[(1+t)squared], y = [(1-t)squared]/[(1+t)squared]

i have the answers to these questions i jst dont haf the solution and i dont no how to get there
im also realli confused about the restrictions

any help is GREATLY appreciated
thankyouu !
e) $x = \frac{2t}{(1 + t)^2} \Rightarrow 2x = \frac{4t}{(1 + t)^2}$ .... (1)

$y = \frac{(1 - t)^2}{(1 + t)^2} = \frac{1 - 2t + t^2}{(1 + t)^2}$ .... (2)

(1) + (2): $2x + y = \frac{4t}{(1 + t)^2} + \frac{1 - 2t + t^2}{(1 + t)^2} = \frac{1 + 2t + t^2}{(1 + t)^2} = 1$.

3. thankyouu !!!

4. uhm just a question
for a) , the answer in my textbook says root (4 - x squared) except when i square and add together i get x squared plus y squared = 4
is there a difference??

5. Originally Posted by THSKluv
uhm just a question
for a) , the answer in my textbook says root (4 - x squared) except when i square and add together i get x squared plus y squared = 4
is there a difference??
Yes.

The path lies on the circle $x^2 + y^2 = 4$. But there is a restriction on the values of x and y since $0 < \theta < \pi$. This restiction means that $-2 < x < 2$ and $0 < y < 2$.

So you only want the top half of this circle, that is, $y = \sqrt{4 - x^2}$.

6. Hello, THSKluv!

If we are to graph these functions, it's more work . . .

$a)\;\;x \:=\: 2\cos\theta\qquad y\:=\:2\sin\theta\qquad 0\le\theta\le\pi$
We have: . $\begin{array}{ccc}\dfrac{x}{2} &=& \cos\theta \\ \\[-3mm] \dfrac{y}{2} &=& \sin\theta \end{array}$

Square: . $\begin{array}{ccc}\left(\dfrac{x}{2}\right)^2 &=& \cos^2\!\theta \\ \\[-3mm] \left(\dfrac{y}{2}\right)^2 &=& \sin^2\!\theta \end{array}$

$\text{Add: }\;\frac{x^2}{4} + \frac{y^2}{4} \;=\;\underbrace{\cos^2\!\theta + \sin^2\!\theta}_{\text{This is 1}} \qquad \Rightarrow\qquad \frac{x^2}{4} + \frac{y^2}{4} \;=\;1$

Hence, we have: . $x^2+y^2 \:=\:4$

This is a circle: center (0,0), radius 2.

Because of the domain, $[0, \pi]$, the graph is the upper semicircle.

$b)\;\;x\:=\:\cos t \qquad y\:=\:\cos t\qquad 0\le t \le 2\pi$
We see that: . $y \:=\:x$

This is a straight line, through the origin, slope 1.

The domain, $[0, 2\pi]$, makes for an interesting graph.

When $t = 0\!:\;\;x \:=\:y\:=\:\cos 0 \:=\:1$
. . The graph starts at (1,1).

When $t = 2\pi\!:\;\;x \:=\:y\:=\:\cos2\pi \:=\:1$
. . The graph ends at (1,1) . . . same point!

What happens in-between?

When $t = \tfrac{\pi}{2}\!:\;\;x\:=\:y\:=\:\cos\tfrac{\pi}{2 } \;=\;0$
. . The graph is at the Origin.
When $t = \pi\!:\;\;x\:=\:y\:=\:\cos\pi \:=\:-1$
. . The graph is at (-1,-1).
When $t = \tfrac{3\pi}{2}\!:\;\;x\:=\:y\:=\:\cos\tfrac{3\pi} {2} \:=\:0$
. . The graph is at the Origin.

The graph looks like this:
Code:
                |     P
|     *(1,1)
|   *
| *
- - - - - - + - - - - -
* |
*   |
(-1,-1)*     |
Q     |

As $t$ goes from 0 to $\pi$, line segment $\overrightarrow{PQ}$ is plotted.

As $t$ goes from $\pi$ to $2\pi$, line segment $\overrightarrow{QP}$ is plotted.

$c)\;\;x\:=\:2\cos\theta \qquad y\:=\:\sqrt{3}\sin\theta \qquad 0< \theta < 2\pi$
Same procedure as (a) . . .

We have: . $\begin{array}{ccc}\dfrac{x}{2} &=&\cos\theta \\ \\[-3mm]\dfrac{y}{\sqrt{3}} &=& \sin\theta \end{array}$

Square: . $\begin{array}{ccc}\left(\dfrac{x}{2}\right)^2 &=& \cos^2\!\theta \\ \\[-3mm] \left(\dfrac{y}{\sqrt{3}}\right)^2 &=& \sin^2\!\theta \end{array}$

$\text{Add: }\;\frac{x^2}{4} + \frac{y^2}{3} \:=\:\underbrace{\cos^2\!\theta + \sin^2\!\theta}_{\text{This is 1}} \qquad\Rightarrow \qquad \frac{x^2}{4} + \frac{y^2}{3} \:=\:1$

This is an ellipse: center (0,0), semimajor axis 2, semiminor axis √3.

Because of the domain, $[0,2\pi]$, we have the entire ellipse.

$d)\;\;x \:=\:2\cos t\quad y\:=\:\sin t\quad 0< \theta < \pi$
Same procedure as c).

$e)\;\;x\:=\:\frac{2t}{(1+t)^2} \quad y \:=\:\frac{(1-t)^2}{(1+t)^2}$

mr fantastic is correct . . . The equation is: . $2x + y \:=\:1$
. . This is a line with intecepts: $\left(\tfrac{1}{2},0\right),\;(0, 1)$

But I am having difficulty determining the extent of this graph.

Since no domain is mentioned, I assume: $t \in (-\infty,\infty)$

Then I did some "eyeball" limits . . .

If $t = \text{-}\infty\!:\;\;x \:=\:\frac{\text{-}\infty}{\infty^2} \:=\:0, \quad y \:=\:\frac{\infty^2}{\infty^2} \:=\:1 \quad\Rightarrow\quad (0,1)$

If $t = \text{-}1\!:\;\text{ unde{f}ined.}$

If $t = 0\!:\;\; x \:=\:0,\;y \:=\:1 \quad\Rightarrow\quad (0,1)$

If $t = 1\!:\;\;x \:=\:\tfrac{1}{2},\;y \:=\:0 \quad\Rightarrow\quad \left(\tfrac{1}{2},0\right)$
If $t = +\infty\!:\;\;x \:=\:\frac{\infty}{\infty^2} \:=\:0, \quad y \:=\:\frac{\infty^2}{\infty^2} \:=\:1 \quad\Rightarrow\quad (0,1)$

I have most of it figured out, but there are some fuzzy areas.

7. Originally Posted by Soroban
[snip]
mr fantastic is correct . . . The equation is: . $2x + y \:=\:1$
. . This is a line with intecepts: $\left(\tfrac{1}{2},0\right),\;(0, 1)$

But I am having difficulty determining the extent of this graph.

Since no domain is mentioned, I assume: $t \in (-\infty,\infty)$

Then I did some "eyeball" limits . . .

If $t = \text{-}\infty\!:\;\;x \:=\:\frac{\text{-}\infty}{\infty^2} \:=\:0, \quad y \:=\:\frac{\infty^2}{\infty^2} \:=\:1 \quad\Rightarrow\quad (0,1)$

If $t = \text{-}1\!:\;\text{ unde{f}ined.}$

If $t = 0\!:\;\; x \:=\:0,\;y \:=\:1 \quad\Rightarrow\quad (0,1)$

If $t = 1\!:\;\;x \:=\:\tfrac{1}{2},\;y \:=\:0 \quad\Rightarrow\quad \left(\tfrac{1}{2},0\right)$
If $t = +\infty\!:\;\;x \:=\:\frac{\infty}{\infty^2} \:=\:0, \quad y \:=\:\frac{\infty^2}{\infty^2} \:=\:1 \quad\Rightarrow\quad (0,1)$

I have most of it figured out, but there are some fuzzy areas.
The graph of $x = \frac{2t}{(t + 1)^2}$ has a vertical asymptote at t = -1, a horizontal asymptote x = 0, passes through (0, 0) and has a turning point at $t = 1 \Rightarrow x = \frac{1}{2}$. So the range of this graph is $\left(-\infty , \, \frac{1}{2}\right]$. Hence the domain of the parametric curve is $-\infty < x \leq \frac{1}{2}$.
Similar analysis can be applied to the graph of $y = \frac{(1 - t)^2}{(1 + t)^2}$. It has a horizontal asymptote y = 1, vertical asymptote at t = -1, turning point at $t = 1 \Rightarrow y = 0$ and y-intercept at y = 1. The range of this graph is $[0, +\infty)$. Hence the range of the parametric curve is $0 \leq y < +\infty$.
Hence the path is the part of line 2x + y = 1 over the domain $x \leq \frac{1}{2}$.