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Published byFelix Horn Modified over 9 years ago
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PARAMETRIC EQUATIONS DIFFERENTIATION WJEC PAST PAPER PROBLEM (OLD P3) JUNE 2003
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PAST PAPER P3 JUNE 2003 A curve has parametric equations Show that the tangent to the curve at the point P, whose parameter is p, has equation:
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First find the gradient of the tangent at the point where the parameter is p
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where the parameter is p we simply replace t with p This is the gradient of the TANGENT at the required point with parameter p
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The equation of the tangent is found using the standard equation of a straight line: Where t=p The equation of the tangent is
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So the equation of the TANGENT can be simplified to:
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THE QUESTION CONTINUES TO SAY: The tangent meets the x axis at A. Find the least value of the length OA, where O is the origin. When a line crosses the x axis we have the y coordinate as zero. USE y=0 in the equation of the tangent that we have just found.
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When y=0 This will be a least value when cos p=1(the most that cos p can be) Because the most denominator gives the least fraction. Of course the x coordinate IS the distance OA
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CONCLUDE BY SAYING: The least value of the distance OA is 2
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