6803449 - Thus we have 2 t t 2-2 =-2 = ⇒ t 2 + t-2 = 0...

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March 3, 2011 1 Solution The curve is given by x = t 3 - 6 t,y = 3 t 2 ....(1). To obtain the gradiant of the tangent we need to find dy dx . Now dx dt = d dt ( t 3 - 6 t ) = 3 t 2 - 6 Again dy dt = d dt (3 t 2 ) = 6 t Thus dy dx = 6 t 3 t 2 - 6 = 2 t t 2 - 2 The tangent line is parallel to the line (-t,2t).The cartesian equation of the line is given by y = - 2 x The gradiant of the tangent is then equal to the gradiant of the line=-2
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Unformatted text preview: Thus we have 2 t t 2-2 =-2 = ⇒ t 2 + t-2 = 0 Solving the above equation we get t =-2 , 1. Putting the values of t in (1),we get the required points as (4,12),(-5,3). 1...
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This note was uploaded on 11/12/2011 for the course M 408M taught by Professor Gilbert during the Fall '07 term at University of Texas.

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