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Math SL IA type1: parametric curves


hazemm

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Hey,

Im in the middle of my IA and im stuck on part of the task. From the points

(24,-3)

(15,-2)

(8,-1)

(3,0)

(0,1)

(-1,2)

(0,3)

(3,4)

(8,5)

im supposed to give an equation that expresses the relationship between x and y. I know that if these points are plotted they look like a horizontal parabola I just dont know what the equation for this type of parabola is, since its not a function. Anyways, if you can help me figure out an equation that links x and y it would be helpful.

thanks

Edited by hazemm
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Simple :) Make the x the y and the y the x coordinates - like this:

(-3, 24)

(-2, 15)

(-1, 8)

etc.

Graph it and use the quadratic regression on your calculator to find its equation. I was bored, so I quickly did it for you:

y = x² - 4x + 3

Now to get an equation for the horizontal parabola you replace x with y and solve for y. I quickly did this for you:

y = x² - 4x + 3

y = (x² - 4x) + 3

...

y = (x - 2)² -1

You replace x with y and you get the equation for your horizontal parabola:

x = (y - 2)² - 1.

If you want to graph it, solve it for y:

y = +/- sqrt(x+1) + 2

Hope this explains it :)

Which task is this btw. ?

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hey,

im supposed to generate points x and y such that x=cos(t) and y=sin(t) for different values of t(-4 to 4).

the points are as followed:

(-0.65,0.76)

(-0.99,-0.14)

(-0.42,-0.91)

(0.54 ,-0.84)

(1,0)

(0.54,0.84)

(-0.42,0.91)

(-0.99,0.14)

(-0.65,-0.76)

once i got these points i graphed them and it gave me a unit circle. now im supposed to find the equation which relates x and y, which is were im stuck at.

for this i got x=+or-cos(sin^-1(y)), but when i graph it gives me an incompelte circle...does any one know a better equation to relate x and y, and if you can explain it, it would be a great help.

cheers

Edited by hazemm
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The general equation for a circle with center point (h, k) and radius r is as follows:

(x - h)² + (y - k)² = r²

Hope this helps :D

If you're still stuck I'll try to work it out later.

yea man helped alot, in the end i got x²+y²=1 then i made y the subject of the equation which gave me

y=±√(1-x ²) would that be correct...

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y=±√(1-x ²) would that be correct...

Yeah, looks correct to me.

would u happen to know how to differentiate y=±√(1-x ²) because i think im doing it wrong. thanks again bro

y1 = (1-x²)0.5

==>

y1' = 0.5(1-x²)-0.5 * -2x

y1' = -x(1-x²)-0.5

and

y2 = -(1-x²)0.5

==>

y2' = -0.5(1-x²)-0.5 * -2x

y2' = x(1-x²)-0.5

That should be it.

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