# parametric equations of a curve

• Jun 30th 2012, 05:34 AM
Stuck Man
parametric equations of a curve
A curve: x=(1+t)^0.5, y=(1-t)^-0.5

After plotting the curve I don't see any turning points. I can see a vertical asymptote at x=2^0.5.

I have calculated that there is a stationary point at t=-1 and the point (0,2^-0.5). I don't see it on the graph though. At that point the second derivative is 0.25*(2^0.5) and hence greater than 0 and there is a minimum turning point.

The answers give the point (0,2^0.5) and says it is a minimum end point.
• Jun 30th 2012, 06:25 AM
tom@ballooncalculus
Re: parametric equations of a curve
Quote:

Originally Posted by Stuck Man
I don't see it on the graph though.

The point, or it looking like a starionary point? It's there.

Quote:

Originally Posted by Stuck Man
The answers give the point (0,2^0.5)

Must be a typo.

Quote:

Originally Posted by Stuck Man
and says it is a minimum end point.

i.e. would be a minimum turning point if it wasn't an end point.
• Jun 30th 2012, 06:36 AM
Stuck Man
Re: parametric equations of a curve
Left of the supposed stationary point the graph slope down slightly as if there is a horizontal asymptote. The point is not an end point.
• Jun 30th 2012, 06:40 AM
Stuck Man
Re: parametric equations of a curve
• Jun 30th 2012, 06:55 AM
earboth
Re: parametric equations of a curve
Quote:

Originally Posted by Stuck Man
A curve: x=(1+t)^0.5, y=(1-t)^-0.5

After plotting the curve I don't see any turning points. I can see a vertical asymptote at x=2^0.5.

I have calculated that there is a stationary point at t=-1 and the point (0,2^-0.5). I don't see it on the graph though. At that point the second derivative is 0.25*(2^0.5) and hence greater than 0 and there is a minimum turning point.

The answers give the point (0,2^0.5) and says it is a minimum end point.

1. You certainely have determined that $\displaystyle -1 \leq t < 1 ~\implies~ \left\lbrace \begin{array}{rcl}0\leq x \leq \sqrt{2} \\ y \ge \frac1{\sqrt{2}}\end{array}\right.$

2. Your curve is the right branch (drawn in red) of the graph of $\displaystyle \displaystyle{f(x)=\frac1{\sqrt{2-x^2}}}$ (drawn in green)

3. Obviously the left end point is a minimum with a horizontal tangent.
• Jun 30th 2012, 07:00 AM
Plato
Re: parametric equations of a curve
You can see that $\displaystyle \left( {0,\frac{1}{{\sqrt 2 }}} \right)$ is a minimum.
• Jun 30th 2012, 07:07 AM
Stuck Man
Re: parametric equations of a curve
My graphing program is displaying an incorrect line on the left of x=0. It is GraphCalc from 2003. Is there any others you know about?
• Jun 30th 2012, 09:21 AM
tom@ballooncalculus
Re: parametric equations of a curve