# Change in domain as variable changes in Parametric Equations.

1. Dec 20, 2013

### Lebombo

This question is from the parametric equations chapter of my calc book.

I am given x =$\frac{1}{\sqrt{t+1}}$

and y = $\frac{t}{t+1}$, for (x> -1)

Solving the x(t) for t, we get $\frac{1- x^{2}}{x^{2}}$

Eliminating the parameters by substitution, we get y = $1 - x^{2}$ for (x > 0)

My question is, what process is used to determine the change in domain?

(Please see attached photo to see the exact content my question has stemmed from)

Regards

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Last edited: Dec 20, 2013
2. Dec 20, 2013

### gopher_p

You're given $x=\frac{1}{\sqrt{t+1}}$. Regardless of what $t$ is, $x$ is going to be positive. The domain isn't changing. It's just being restated in terms of $x$ instead of $t$. This makes sense considering that the entire exercise is done to eliminate the parameter $t$.

3. Dec 20, 2013

### Lebombo

The only thing I can think of is to say:

if t > -1, and t = $\frac{1 - x^{2}}{x^{2}}$,

then by substitution:

$\frac{1 - x^{2}}{x^{2}}$ > -1

1 > 0 ...which is a true statement but doesn't tell me anything about the domain in terms of x.

4. Dec 20, 2013

### Staff: Mentor

Actually, it does. Your starting inequality is defined only if x â‰  0. So you're tacitly assuming that x â‰  0 when you multiply both sides of the inequality by it. Because you ended with an inequality that is true for all x (other than 0), you can conclude that the domain for your parametric equation is all real x other than 0.

5. Dec 20, 2013

### Lebombo

My apologies, but your reply has happened to have gone over my head a bit.

"Your starting inequality is defined only if x â‰  0."

Just so we're on the same page, the starting inequality you are referring to is: $\frac{1 - x^{2}}{x^{2}}$ > -1 ?

If so, then yes, I agree and understand that this inequality is only defined if x â‰ 0

"Because you ended with an inequality that is true for all x (other than 0),"

Is this the end inequality being referred to:

1 > 0 ?

If so, then I don't understand how you conclude true of all x **(other than zero)**

I do not know how any restrictions can be set on an inequality or equation if there are no variables present in the in equality. If no variables are present in an inequality, the inequality is either true or false. 1 is greater than 0 is a true statement and restrictions on a statement like that do not make any sense to me.

Additionally, I may be misreading your reply on another point as well. The book says the ending restriction is x > 0, not just x â‰  0. I'm not clear on whether or not this has been taken into account in your reply.

Again, my apologies that this just not make sense to me just yet.

6. Dec 21, 2013

### Spacemoss

The original parametric equation was defined as having the restriction that t > -1. In this case, the resulting graph is a downward parabola, and t > -1 draws the downward right hand side of the parabola.

When you put the equation into a rectangular equation it has the form $$y=1-x^2$$

If you allow all x in the rectangular form it draws the whole parabola.

All they are asking it what interval of x draws the same picture the parametric form drew. So, for the two pictures to match now, x>0.

P.S. In the original, the right half of the downward parabola is drawn from right to left as t increases, but never quite hits the y-axis, that is, x never quite hits 0. So for the rectangular domain to be the same it has to x>0, not â‰¥0.

Last edited: Dec 21, 2013
7. Dec 21, 2013

### Staff: Mentor

Yes. Sorry if I wasn't clear.
Yes again.
Because that inequality, 1 > 0, doesn't depend on x at all. The first inequality (the one you started with) is true for all real x other than 0. The last inequality, which doesn't involve x at all, is true for all real x. However, we still need to consider that the first inequality had a restriction, and this restriction should be carried through to the 1 > 0 inequality.
It's because you arrived at the final inequality (1 > 0) from an inequality that had a restriction. The two inequalities are equivalent if they have the same solution set, which we can force by removing a single value from consideration.
My comments came from your question about t = (1 - x2)/x2, with t > -1 and nothing else. In a previous post, gopher_p noted that because x = $\frac 1 {\sqrt{t + 1}}$, x has to be positive. If you combine this restriction with the one I already mentioned (x â‰  0), you get x > 0.
No problem. That's what we do here!