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# Parametric equation watch

1. Need help with the last part i dont get the mark scheme.
I got dy/dx=0. i get the bit about the normal being undefined but I dont get the x=tan(pi/4)
Therefore x=1 bit. What happens to y?
http://m.imgur.com/z2zjGFt,kZKYEPh
2. (Original post by Super199)
Need help with the last part i dont get the mark scheme.
I got dy/dx=0. i get the bit about the normal being undefined but I dont get the x=tan(pi/4)
Therefore x=1 bit. What happens to y?
http://m.imgur.com/z2zjGFt,kZKYEPh
It's a vertical straight line. x=1.

3. (Original post by Zacken)
It's a vertical straight line. x=1.

How did you know that? Why cant it be y=1?
4. (Original post by Super199)
How did you know that? Why cant it be y=1?
Ah not to worry got it
5. (Original post by Super199)
How did you know that? Why cant it be y=1?
Because the gradient of the normal is undefined. Does the horizontal line y=1 have an undefined gradient? (hint: no it has zero gradient), a vertical line has a undefined gradient.
6. (Original post by Zacken)
Because the gradient of the normal is undefined. Does the horizontal line y=1 have an undefined gradient? (hint: no it has zero gradient), a vertical line has a undefined gradient.
Yh cheers man got it
7. (Original post by Zacken)
Because the gradient of the normal is undefined. Does the horizontal line y=1 have an undefined gradient? (hint: no it has zero gradient), a vertical line has a undefined gradient.
Nice 'hint'.
8. (Original post by B_9710)
Nice 'hint'.
I realised after.
9. (Original post by Zacken)
It's a vertical straight line. x=1.

what package are you using for these graphs?
10. (Original post by TeeEm)
what package are you using for these graphs?
Just typed it into Wolfram Alpha and copy pasted it here
11. (Original post by Zacken)
Just typed it into Wolfram Alpha and copy pasted it here
what about the other with the circle and the triangle?
12. (Original post by TeeEm)
what about the other with the circle and the triangle?

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