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September 29th, 2010, 02:40 PM  #1 
Newbie Joined: Oct 2009 Posts: 20 Thanks: 0  Multivariable Calculus: parametric curves problem
Consider the parametric curves C1: , t?R and C2 : , ,t?R. Find all lines that are tangent to both of the parametric curves. * Some clues would be highly appreciative. Thanks

October 3rd, 2010, 07:06 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Multivariable Calculus: parametric curves problem
The way I went about solving this problem was to eliminate the parameter t to get Cartesian equations for the two curves. C1: x1 = y1² or y1 = ±x1^(½) C2: y2 = ¼x2² Now, we require dy1/dx1 = dy2/dx2 = (y2  y1)/(x2  x1) (a) dy1/dx1 = ±½x1^(½) (b) dy2/dx2 = ½x2 (c) (y2  y1)/(x2  x1) = (¼x2²  ±x1^(½))/(x2  x1) Setting (a) and (b) equal, we find ±½x1^(½) = ½x2 (d) x2 = ±x1^(½) Next, setting (c) and (a) equal, and using (d) to eliminate x2 we have (¼(±x1^(½))²  ±x1^(½))/(±x1^(½)  x1) = ±½x1^(½) (¼x1^(1)  ±x1^(½)) = ±½x1^(½)(±x1^(½)  x1) (¼x1^(1)  ±x1^(½)) = ½x1^(1) ± ½x1^(½) Multiply through by 4x1. 1  ±4x1^(3/2) = 2  ±2x1^(3/2) ±2x1^(3/2) = 1 x1^(3/2) = 1/(±2) Here, we see the only real solution for x1 occurs when the plus/minus is minus (which we will use hereafter) and is: x1 = (1/2)^(2/3) = 2^(?) x2 = x1^(½) = 2^(?) y1 = x1^(½) = 2^(?) y2 = ¼x2² = ¼2^(?) = 2^(4/3) Let's check to see if the slopes given by (a), (b) and (c) are equivalent. (a) ½x1^(½) = ½(2^(?))^(½) = ½(2^(?)) = 2^(?) (b) ½x2 = ½(2^(?)) = 2^(?) (c) (y2  y1)/(x2  x1) = [2^(4/3)  2^(?)]/[2^(?)  2^(?)] = [2^(4/3)(1 + 2)]/[2^(?)(2 + 1)] = 2^(?) Thus, the one tangent line to both curves is given by: y  2^(4/3) = 2^(?)[x + 2^(?)] or y + 2^(?) = 2^(?)[x  2^(?)] both reduce to (had to check them both to make sure ) y = 2^(?)x  2^(4/3) 
October 4th, 2010, 04:17 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,968 Thanks: 2217 
Tangent to C1: (y  t) = (1/(2t))(x  t²) Tangent to C2: (y  u²) = u(x  2u) For these to coincide, u = 1/(2t) and t/2 = u², so and Hence the common tangent has equation (as per the previous post). 
October 4th, 2010, 10:06 AM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Multivariable Calculus: parametric curves problem
Wow! What a difference knowing what you're doing makes! I truly appreciate seeing how to do this problem parametrically. It makes sense now that I see it, but I just didn't know how. 

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