Axioms Posted October 18, 2011 Posted October 18, 2011 I know to find the slope of parametric equations you find dy/dx. If you are given the parametric equations and you find an answer in terms of t for the gradient and you are given points in terms of x and y. E.G. Find the tangent of x= 1 + lnt, y= t^2 +2 ; (1,3) I get dy/dx = 2t^2 Now the text book is not very clear... What should I do with the 2t^2 and what should I do with this to find the equation of the tangent? Must I find what t is equal to at that given point or should I eliminate the parameter? at y = 3 3 = t^2 +2 therefore t = 1 or t = -1 (but t cannot = -1 because of the terms of x for real numbers) therefore you substitute t = 1 into dy/dx and find the gradient to be 2 I then get y - 3 = 2(x -1) and get an answer of y=2x +1 I am not sure if its right and if the reasoning is correct or not
DrRocket Posted October 18, 2011 Posted October 18, 2011 (edited) I know to find the slope of parametric equations you find dy/dx. That is one way. A more geometric approach involves finding the tangent vector to the parameterized curve. If [math]\gamma (t) = (x(t), y(t)) [/math] is your curve then the tangent vector (velocity if [math]\gamma[/math] represents position) is just [math]\gamma '(t) = (x'(t),y'(t))[/math]. The slope of the tangent line at some point [math]\gamma (t_0)[/math] is then just [math]\frac{y'(t_0)}{x'(t_0)}[/math] This generalizes immediately to curves in 3 or more dimensions, except that "slope" of a tangent is not a defined notion. Edited October 18, 2011 by DrRocket
Axioms Posted October 18, 2011 Author Posted October 18, 2011 Is that not essentially the same thing though? You still need to define t from the given points and then substitute.
DrRocket Posted October 18, 2011 Posted October 18, 2011 Is that not essentially the same thing though? You still need to define t from the given points and then substitute. No, you simply need to know the point as a function of t, then differentiate. You do not need to find y as a function of x. In fact the parameterized curve need not be the graph of a function -- for instance it can have closed loops.
Axioms Posted October 18, 2011 Author Posted October 18, 2011 Ok that makes sense. Thanks makes it simple
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