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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: $x=t^2$ , $y=t$ t?R and C2 : $x=2t$ , $y=t^2$ ,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 $t= -1/\sqrt[\small3]2$ and $u= -\sqrt[\small3]{2}/2.$ Hence the common tangent has equation $^{y= -\frac{\sqrt[\small3]{2}}{2}x\,-\,\frac{\sqrt[\small3]{4}}{4}}$ (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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