Question

find two different sets of parametric equations for the rectangular equation

Answer 1

y = 1/(x+2)

Answer 2

hey man i got x = t +2 and y = 1/t as the first set of parametric equation

Answer 3

is it right

Answer 4

and i was thinking on using a trig identity for the second one

Answer 5

hmm, that's not quite true. let's see.
First, can you you tell me the domain of \(y = \frac{1}{x+2}\)?
What are the possible values for x?

Answer 6

all real numbers except -2

Answer 7

Right, so we have to pick a function of t for x such that x will not be -2

Answer 8

but i am using a positive 2

Answer 9

Yes, but can it be that \(t + 2 = -2\)?
Is there a value for \(t\) for which it happens?

Answer 10

t=-4

Answer 11

Right, so if we set \(x = t+2\) then for \(t = -4\) we get \(x=-2\) which we cannot use. I don't see why can't we say though:
$$ x= t+2 \qquad t\neq -4 $$
Which now makes sure that x will not be -2. it seems legitimate to me

Answer 12

Ok, so say we have this, now you want to plug it into y. let's see what we get:
$$
y = \frac{1}{x+2} \implies y = \frac{1}{(t+2) + 2} = \frac{1}{t+4}
$$

Answer 13

So In that case I guess your set of equations will be:
$$ \boxed{x = t+2 \qquad y = \frac{1}{t+4}} \qquad t\neq -4$$

Answer 14

you get what I mean?

Answer 15

sort of

Answer 16

do you understand why it is not y=1/t?

Answer 17

i see why we cant use positve 2 though

Answer 18

we can use, you get the set of equations above. that will not be y=1/t it will be y=1/(t+4)

Answer 19

because t does not equal positiv 4

Answer 20

ok, I lost you, what do you mean by that?

Answer 21

no i meant -4

Answer 22

ye, t cannot be -4 because it corresponds to x = -2 which causes division by zero. x = -2 is not defined in the original equation, and it is not defined here too

Answer 23

right

Answer 24

If you want y = 1/t, then we will have to say:
$$
y = \frac{1}{x + 2} = \frac{1}{t} \qquad t \neq 0\\
x + 2 = t \\
x = t -2\\
\\
\boxed{x = t-2 \qquad y = \frac{1}{t}} \qquad t \neq 0
$$

Answer 25

so now the t does not equal 0

Answer 26

yes, we got a different set of equations, and here t cannot be 0. both the set above and this set represent the same equation, but the relation of x and y to t is different.

Answer 27

i see

Answer 28

so thats one can another be a trig identity like for example x = (sect)^2

Answer 29

it can be, but once again, we need something that will give x all the possible values it has in the original equation. if we look at \((\sec(t))^2\):
http://www.wolframalpha.com/input/?i=sec%28t%29%5E2
We see it can only be positive. We need something that can be anything but -2. we can put constraints later for the -2, but we still need the rest

Answer 30

oh

Answer 31

what about tan

Answer 32

ye we can use it, but we have to tell when tan(t) = -2

Answer 33

x = (tan(t)^2 - 2

Answer 34

well I guess we can use \(t \neq \arctan(-2) + n\pi\) or something

Answer 35

again, if you do tan(t)^2 you get rid of all the negatives:
http://www.wolframalpha.com/input/?i=tan%28t%29%5E2

Answer 36

man this is hard

Answer 37

i wonder when i am going to use this in real life

Answer 38

Lol, that's totally up to you =)
Those things have their uses, but it's up to you if you want to go in there
Anyway, tan(t) is fine, it is good enough as it is, it goes from infinity to -infinity
Now we just have to make sure to restrict t so tan(t) will never by -2
We can say:
$$
x = \tan(t) \implies y = \frac{1}{x + 2} = \frac{1}{\tan(t) + 2}\\
\tan(t) \neq -2 \implies t \neq \arctan(-2) + n\pi
$$Makes sense?

Answer 39

Just to make it clear, arctan is the inverse function of tan. sometimes it's written as \(\tan^{-1}(...)\)

Answer 40

is this stuff used in engineering?

Answer 41

so the y = arctan(-2)+npi
what does npi mean?

Answer 42

There are engineers that will not have to use it at all, and that's true for about anything.
At some point you can get specialized at something and not deal with all the rest.
But again, it has its uses and if you want you can be getting into it, it's up to you

Answer 43

I wrote \(t \neq \arctan(-2) + n\pi\) that means t cannot be arctan(-2) + npi
It doesn't say anything about y itself

Answer 44

oh my bad

Answer 45

it's actually pretty easy to understand it. let's try something a little different first.
Say you have \(y = \frac{1}{x + 2}\) and I tell you x = 4. what is y?

Answer 46

y = 1/6

Answer 47

right, and if I'd say \(x = 3\pi\)?

Answer 48

y = 1/3pi + 1/2

Answer 49

you cannot break the denominator like that, if you could then we could say:
$$\frac{1}{3} = \frac{1}{1+1+1} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = 1+1+1 = 3$$
which is wrong.

Answer 50

lol sorry 1/(3pi+2)

Answer 51

exactly. and if i say x = tan(t)?

Answer 52

1/(tan(t)+2)

Answer 53

right, not complicated at all
So whatever I set to x, I can tell what y can be. here you found y for x=tan(t)

Answer 54

right seems easy enough

Answer 55

now we just have to make sure that whatever we set to x matches the possible values of x.
For example, if I set x = 4 then I can get the y = 1/6 but in the way I lost all the possible values of x. I don't limit x, I just want to 'write it differently'

Answer 56

so I can set for example x = 2t
Now if I don't limit t at all then I know that there is a t value for every possible x value.
there is only one little problem. there is also a t value for x=-2 which is NOT a possible value.
So I have to put a limit on t. I say:
$$x \neq -2 \implies 2t \neq -2 \implies t \neq -1$$

Answer 57

it is the same for tan(t)
We can say:
$$ x = \tan(t) \implies y = \frac{1}{\tan(t) + 2} $$
But we know that x cannot be -2. you can even look at y now, you can see that the denominator cannot be 0. so we say:
$$
\tan(t) + 2 \neq 0 \implies \tan(t) \neq -2
$$To be honest, maybe it's sufficient to leave it as it is.. but we can also limit the values of t itself:
$$
\tan(t) \neq -2 \implies t \neq \arctan(-2) + n\pi
$$

Answer 58

okay

Answer 59

so that's it.. we can say our second set can be:
$$
\boxed{x = \tan(t) \qquad y = \frac{1}{\tan(t) + 2}} \qquad t \neq\arctan(-2) + n\pi
$$

Answer 60

so the npi is just any value of pi?

Answer 61

any multiple of pi, because if you look at the tan() function you can see that it is periodic.
You'll get the same value if you add or subtract \(\pi\) from the input.
That also somewhat makes sense, because \(\tan(x) = \frac{\sin(x)}{\cos(x)}\)
and for both sin() and cos(), addition of \(\pi\) to the input will swap the sign of the output.
For example
$$\cos(0) = 1 \qquad \cos(0+\pi) = \cos(\pi) = -1$$
So if you add \(\pi\) to the input then both sin() and cos() just swapped their signs, but the fraction of negative/negative is still positive, so you get the same value out of the fraction

Answer 62

thanks for that input

Answer 63

ok, so the two sets we have are good enough? =)

Answer 64

yes sir

Answer 65

Ok, good I hope it's clear