 My Math Forum Multivariable Calculus: parametric curves problem
 User Name Remember Me? Password

 Calculus Calculus Math Forum

 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 = y1� or y1 = �x1^(�) C2: y2 = �x2� Now, we require dy1/dx1 = dy2/dx2 = (y2 - y1)/(x2 - x1) (a) dy1/dx1 = ��x1^(-�) (b) dy2/dx2 = �x2 (c) (y2 - y1)/(x2 - x1) = (�x2� - �x1^(�))/(x2 - x1) Setting (a) and (b) equal, we find ��x1^(-�) = �x2 (d) x2 = �x1^(-�) Next, setting (c) and (a) equal, and using (d) to eliminate x2 we have (�(�x1^(-�))� - �x1^(�))/(�x1^(-�) - x1) = ��x1^(-�) (�x1^(-1) - �x1^(�)) = ��x1^(-�)(�x1^(-�) - x1) (�x1^(-1) - �x1^(�)) = �x1^(-1) � -�x1^(�) 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^(-�) = -2^(?) y1 = -x1^(�) = -2^(-?) y2 = �x2� = �2^(?) = 2^(-4/3) Let's check to see if the slopes given by (a), (b) and (c) are equivalent. (a) -�x1^(-�) = -�(2^(-?))^(-�) = -�(2^(?)) = -2^(-?) (b) �x2 = �(-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 - t�) Tangent to C2: (y - u�) = u(x - 2u) For these to coincide, u = 1/(2t) and t/2 = -u�, 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.  Tags calculus, curves, multivariable, parametric, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post maluita659 Calculus 7 February 15th, 2014 01:15 PM math221 Calculus 5 March 24th, 2013 04:52 PM math221 Calculus 0 March 19th, 2013 04:33 PM bach71 Calculus 2 March 9th, 2013 05:05 PM Robertinho! Linear Algebra 1 September 29th, 2011 12:31 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top       