Points Where a Parametric Equation Crosses ItselfDate: 10/12/2001 at 12:59:21 From: Jessica Stover Subject: Find all points where a parametric equation crosses itself I am supposed to find all the points where the function crosses itself of the parametric equation: x=t(9-t^2) and y=(t^2-1)(t^2-9) on the interval -4<t<4. I don't even know where to start. Could you help me, please? Thank you, Jessica Stover Date: 10/12/2001 at 16:04:44 From: Doctor Rob Subject: Re: Find all points where a parametric equation crosses itself Thanks for writing to Ask Dr. Math, Jessica. One way to proceed is to call the two times that have the same x and y values t and T. Then you have the equations t*(9-t^2) = T*(9-T^2), (t^2-1)*(t^2-9) = (T^2-1)*(T^2-9). These you need to solve for t and T, subject to the condition that t - T is nonzero. One way to solve these equations is to eliminate one of the variables, say t, from the two equations, to give you one equation in the unknown T, and then solve that. The equation you get has degree 6, and factors into two linear factors times a biquadratic factor. This allows you to solve for the roots explicitly. There seem to be 6 real roots, all in the indicated interval. Once you know the values of T, you can find the values of x and y. Good luck! - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 10/15/2001 at 20:04:09 From: jstover Subject: Re: Find all points where a parametric equation crosses itself Dr. Math, Thank you for all of your help, but I was wondering if you could go through the steps for solving for T. I cannot figure it out or come up with 6 answers when I do. Thank you again. Jessica Stover Date: 10/16/2001 at 10:00:56 From: Doctor Rob Subject: Re: Find all points where a parametric equation crosses itself Thanks for writing back, Jessica. Expand both equations (1) -t^3 + 9*t = -T^3 + 9*T, (2) t^4 - 10*t^2 + 9 = T^4 - 10*T^2 + 9, Add t times the (1) to (2): -t^2 + 9 = T^4 - t*T^3 - 10*T^2 + 9*t*T + 9. (3) -t^2 = T^4 - t*T^3 - 10*T^2 + 9*t*T. Now add -t times (3) to (1): (4) 9*t = -t*T^4 + (t^2-1)*T^3 + 10*t*T^2 - 9*(t^2-1)*T. Solve (3) for t^2: (5) t^2 = -T^4 + t*T^3 + 10*T^2 - 9*t*T. Substitute (5) in (4): 0 = -t*T^4 + ([-T^4 + t*T^3 + 10*T^2 - 9*t*T]-1)*T^3 + 10*t*T^2 - 9*([-T^4 + t*T^3 + 10*T^2 - 9*t*T]-1)*T - 9*t, (6) 0 = t*(T^6-19*T^4+91*T^2-9) + (-T^7-19T^5+91*T^3-9*T). Factor this into (7) 0 = (t-T)*(T^6-19*T^4+91*T^2-9). Since t - T is nonzero, that factor can be dropped. Thus (8) 0 = T^6 - 19*T^4 + 91*T^2 - 9. This is the 6th degree equation for T. It can be further factored as follows: 0 = (T^2 - 9)*(T^4 - 10*T^2 + 1), (9) 0 = (T - 3)*(T + 3)*(T^4 - 10*T^2 + 1). The first two factors give (10) T = 3, (11) T = -3. The last factor gives (12) T^2 = 5 + 2*sqrt(6), (13) T^2 = 5 - 2*sqrt(6), which in turn yield (14) T = sqrt(5+2*sqrt[6]), (15) T = -sqrt(5+2*sqrt[6]), (16) T = sqrt(5-2*sqrt[6]), (17) T = -sqrt(5-2*sqrt[6]). Thus (10), (11), and (14)-(17) are the six possible values of T. I leave it to you to find the corresponding values of t (which must be chosen from the same list of 6 values), subject to t - T being nonzero. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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