flsweetheart422 31 Report post Posted November 30, 2008 (edited) I have no idea how to solve this problem. I have the final answer, but don't know how to get there. I think I need to use implicit differentiation, but I'm not sure, and I'm quite confused. Any help would be greatly appreciated. Find the equation of the tangent line to the curve is x^2+y^2 = 169 at the point (5, -12). A. 5y-12x=-120B. 5x-12y=119C. 5x-12y=169D. 12x+5y=0E. 12x+5y=169I have been told that the final answer is (B)5x-12y=119, but none of my friends seem to be able to explain it. Edited November 30, 2008 by flsweetheart422 Share this post Link to post Share on other sites
Abu 278 Report post Posted November 30, 2008 2x + 2y(dy/dx) = 0(dy/dx) = -2x/2y = -x/y(Just to confirm I'm not rusty: http://www.mathway.com/answer.aspx?p=ccal?...y/dx?p=0?p=?p=)Gradient of tangent line = 5/12Edit: http://www.webmath.com/cgi-bin/equline.cgi...k=equline2.htmlTherefore: y=5/12x-169/12Multiply through by 12 and you should get your answer. Share this post Link to post Share on other sites
SharkSpider 35 Report post Posted December 1, 2008 Well, you've got a circle with radius 13 that has no translation from the origin.The key to implicit differentiation is that you take d/dx of every term. Then, you need to use the chain rule on all terms that contain y, so that d/dx of y^2 is 2y times dy/dx.Remember that with equations of y = something, differentiating gives you the slope of the tangent line, and that really, all you're doing is implicitly differentiating, because d/dx of y is 1 times dy/dx, which is what you get when you differentiate normal functions.So, differentiating...x^2 + y^2 = 169 gives:2x + 2y(dy/dx) = 0, so dy/dx = -x/y.Or, 5/12.y - b = m(x - a) For point-intercept format.So:y - (-12) = 5/12(x - 5)y = 5x/12 - 25/12 - 1212y = 5x - 25 - 14412y - 5x = -1695x-12y=169So the answer they gave you was wrong. The answer should be C. Share this post Link to post Share on other sites
flsweetheart422 31 Report post Posted December 1, 2008 Thanks guys! I appreciate it! Share this post Link to post Share on other sites
rofler 0 Report post Posted December 11, 2008 You don't really need calculus at all for this question. Any tangent line to a circle will be perpendicular to the line through the point of tangency and the center of the circle. This line passes through the center $(0,0)$ and the point of tangency, $(5,-12)$ so will have equation $y=-(12/5)x$.Then, the line of tangency is perpendicular to this line, so has slope $5/12$, and passes through $(5,-12)$ so has equation $y=(5/12)(x-5)-12$ or $5x-12y=169$. Share this post Link to post Share on other sites