hazemm Posted November 8, 2008 Report Share Posted November 8, 2008 (edited) 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 November 8, 2008 by hazemm Reply Link to post Share on other sites More sharing options...
Max Posted November 8, 2008 Report Share Posted November 8, 2008 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 + 3Now 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 + 3y = (x² - 4x) + 3...y = (x - 2)² -1You 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) + 2Hope this explains it Which task is this btw. ? Reply Link to post Share on other sites More sharing options...
hazemm Posted November 8, 2008 Author Report Share Posted November 8, 2008 I thought it was something along those line. Thanks alot for clarifying everything.Its task 1 and its about parametric curves and involves calculus, if that any help. Reply Link to post Share on other sites More sharing options...
hazemm Posted November 10, 2008 Author Report Share Posted November 10, 2008 (edited) 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 November 10, 2008 by hazemm Reply Link to post Share on other sites More sharing options...
Max Posted November 10, 2008 Report Share Posted November 10, 2008 The general equation for a circle with center point (h, k) and radius r is as follows:(x - h)² + (y - k)² = r²Hope this helps If you're still stuck I'll try to work it out later. Reply Link to post Share on other sites More sharing options...
hazemm Posted November 10, 2008 Author Report Share Posted November 10, 2008 The general equation for a circle with center point (h, k) and radius r is as follows:(x - h)² + (y - k)² = r²Hope this helps 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... Reply Link to post Share on other sites More sharing options...
hazemm Posted November 10, 2008 Author Report Share Posted November 10, 2008 would u happen to know how to differentiate y=±√(1-x ²) because i think im doing it wrong. thanks again bro Reply Link to post Share on other sites More sharing options...
Max Posted November 10, 2008 Report Share Posted November 10, 2008 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 broy1 = (1-x²)0.5==>y1' = 0.5(1-x²)-0.5 * -2xy1' = -x(1-x²)-0.5andy2 = -(1-x²)0.5==>y2' = -0.5(1-x²)-0.5 * -2xy2' = x(1-x²)-0.5That should be it. Reply Link to post Share on other sites More sharing options...
hazemm Posted November 10, 2008 Author Report Share Posted November 10, 2008 Yeah, looks correct to me.y1 = (1-x²)0.5==>y1' = 0.5(1-x²)-0.5 * -2xy1' = -x(1-x²)-0.5andy2 = -(1-x²)0.5==>y2' = -0.5(1-x²)-0.5 * -2xy2' = x(1-x²)-0.5That should be it.yea its right. you used chain rule right?thanks again man Reply Link to post Share on other sites More sharing options...
Max Posted November 11, 2008 Report Share Posted November 11, 2008 you used chain rule right?Yeah I did, no problem! Reply Link to post Share on other sites More sharing options...
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