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# Stationary point of Curve watch

1. Parametric equations of a curve:

X=0.5t
Y=t^2 +1

Differentiated to 2t/0.5. Does this mean the stationary point is infinite?

Sorry if I'm being stupid I'm not the biggest fan of my teacher.

Thank you- all help is much appreciated XxX
2. dy/dt=2t+?
3. (Original post by Coconutter)
Parametric equations of a curve:

X=0.5t
Y=t^2 +t

Differentiated to 2t/0.5. Does this mean the stationary point is infinite?
No, this does not
4. Thank you for the help so far .

(Original post by BabyMaths)
dy/dt=2t+?
Sorry just realised I wrote out the original y equation wrong- it's Y=t^2 +1.
5. (Original post by ztibor)
No, this does not
Does that mean their is a stationary point? Or is there something else I need to do to show it is infinite?
6. (Original post by Coconutter)
Does that mean their is a stationary point? Or is there something else I need to do to show it is infinite?
I think you have things the wrong way round!

Your expression for dy/dx should now be 2t/(0.5) i.e. 4t. So what's the condition for a stationary point?
7. I have never come across an infinite stationary point, I don't think it makes any sense.
8. (Original post by davros)
I think you have things the wrong way round!

Your expression for dy/dx should now be 2t/(0.5) i.e. 4t. So what's the condition for a stationary point?
My expression for dy/dx is 2t/(0.5) -left it there though rather than continued to 4t. I wrote the original post out really rushed though so admittedly I've made a bit of a mess of it lol.

This is the part I'm confused about though for there to be a stationary point dy/dx= 0 so for 4t to equal t would have to be 0 but then that doesn't work with the... Ahh would that mean the stationary point is at (0,1)? Think I may have just got myself confused with the y equation .
9. (Original post by james22)
I have never come across an infinite stationary point, I don't think it makes any sense.
Sorry terrible wording in my original post- won't be writing out posts five mins before I need to leave in future lol. I mean is the gradient infinite where the curve changes direction rather than there being a stationary point.
10. (Original post by Coconutter)
My expression for dy/dx is 2t/(0.5) -left it there though rather than continued to 4t. I wrote the original post out really rushed though so admittedly I've made a bit of a mess of it lol.

This is the part I'm confused about though for there to be a stationary point dy/dx= 0 so for 4t to equal t would have to be 0 but then that doesn't work with the... Ahh would that mean the stationary point is at (0,1)? Think I may have just got myself confused with the y equation .
Yes - you got yourself confused! You want dy/dx = 0 which means t=0. Now you can substitute back to get x=0, y=1 at the stationary point.

As a double-check, if you write t in terms of x - i.e. t = 2x - then you have a really simple cartesian equation y = 4x^2 + 1, and I'm sure you know what the graph of that looks like

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