# Parametric equations

• Jun 25th 2010, 10:38 PM
hooke
Parametric equations
A curve has parametric equations :

$x=t^3-\mu t$ , where $\mu$ is a constant

$y=\frac{t^2}{1+t^2}$

(1) Show that the curve is symmetrical about the y-axis

(2) For $\mu>0$ ,find in terms of $\mu$ the coordinates of the point where the curve intersects itself.
• Jun 26th 2010, 02:16 AM
Ackbeet
Ok, what ideas have you had so far?
• Jun 26th 2010, 02:17 AM
mr fantastic
Quote:

Originally Posted by hooke
A curve has parametric equations :

$x=t^3-\mu t$ , where $\mu$ is a constant

$y=\frac{t^2}{1+t^2}$

(1) Show that the curve is symmetrical about the y-axis

(2) For $\mu>0$ ,find in terms of $\mu$ the coordinates of the point where the curve intersects itself.

(1) x --> -x => t --> -t and t --> -t => y --> y therefore the function is even (symmetric about y-axis).

(2) You are trying to find a crunode: Crunode - Wikipedia, the free encyclopedia.

Note that crunodes occur at points where dy/dx has the indeterminant form 0/0. So get dy/dx, find the value of t for which dy/dx is indeterminant and then substitute this value of t into x = x(t) and y = y(t). (You know in advance that the point is a crunode but you should test it anyway).
• Jun 26th 2010, 02:28 AM
hooke
Quote:

Originally Posted by mr fantastic
(1) x --> -x => t --> -t and t --> -t => y --> y therefore the function is even (symmetric about y-axis).

(2) You are trying to find a crunode: Crunode - Wikipedia, the free encyclopedia.

Note that crunodes occur at points where dy/dx has the indeterminant form 0/0. So get dy/dx, find the value of t for which dy/dx is indeterminant and then substitute this value of t into x = x(t) and y = y(t). (You know in advance that the point is a crunode but you should test it anyway).

thank you sir,why is the gradient of tangent 0/0 at the crunode? I only see two different tangents but not a vertical tangent.

just out of curiousity, is there an algebraic or non-calculus solution to part (2)?
• Jun 26th 2010, 02:42 AM
mr fantastic
Quote:

Originally Posted by hooke
thank you sir,why is the gradient of tangent 0/0 at the crunode? I only see two different tangents but not a vertical tangent.

just out of curiousity, is there an algebraic or non-calculus solution to part (2)?

0/0 does not mean vertical tangent. It means indeterminant (which makes perfect sense if you go back and read the link again).

Not that I'm aware of.
• Jun 26th 2010, 02:58 AM
hooke
Quote:

Originally Posted by mr fantastic
0/0 does not mean vertical tangent. It means indeterminant (which makes perfect sense if you go back and read the link again).

Not that I'm aware of.

i have read it a few times but i do not understand how can a point with 2 different tangents ended up with an indeterminant.
• Jun 26th 2010, 07:54 AM
HallsofIvy
Well, what else can it be? If the derivative were not indeterminant, then it would have one "determinant" value and that would be the slope of the tangent line.

But mr fantastic was answering a different question, "why is 0/0 not "infinite?"(giving a vertical tangent). The answer to that is that limits such as $\lim_{x\to 0}\frac{x}{x}= 1$, $\lim_{x\to 0} \frac{2x}{x}= 2$ all involve fractions of the form "0/0" but different limits and so different "slopes". To have an infinite limit your fraction must be of the form "a/0" where a is not 0.
• Jun 28th 2010, 03:21 AM
hooke
Quote:

Originally Posted by hooke
A curve has parametric equations :

$x=t^3-\mu t$ , where $\mu$ is a constant

$y=\frac{t^2}{1+t^2}$

(1) Show that the curve is symmetrical about the y-axis

(2) For $\mu>0$ ,find in terms of $\mu$ the coordinates of the point where the curve intersects itself.

This is my work,

$\frac{dy}{dx}=\frac{2t}{(1+t^2)^2(3t^2-\mu)}$

so i am supposed to find a value of t which gives 0/0 but that seems impossible..

If $t=\pm\sqrt{\frac{\mu}{3}}$, it only makes the denominator 0 and if t=0, only the numerator is 0

??
• Jun 28th 2010, 03:58 PM
mr fantastic
Quote:

Originally Posted by hooke
This is my work,

$\frac{dy}{dx}=\frac{2t}{(1+t^2)^2(3t^2-\mu)}$

so i am supposed to find a value of t which gives 0/0 but that seems impossible..

If $t=\pm\sqrt{\frac{\mu}{3}}$, it only makes the denominator 0 and if t=0, only the numerator is 0

??

Perhaps there's a misprint in the question .... (I had half-suspected as much earlier because if the curve is symmetric about the y-axis I would have expected the crunode to lie on the y-axis ....)