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PARAMETRIC EQUATIONS DIFFERENTIATION

Solve a curve tangent problem from WJEC Past Paper P3 June 2003 using differentiation with parametric equations. Find the gradient of the tangent at point P, equation of tangent, and least value of length OA.

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PARAMETRIC EQUATIONS DIFFERENTIATION

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  1. PARAMETRIC EQUATIONS DIFFERENTIATION WJEC PAST PAPER PROBLEM (OLD P3) JUNE 2003

  2. 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:

  3. First find the gradient of the tangent at the point where the parameter is p

  4. 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

  5. The equation of the tangent is found using the standard equation of a straight line: Where t=p The equation of the tangent is

  6. So the equation of the TANGENT can be simplified to:

  7. 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.

  8. When y=0 Of course the x coordinate IS the distance OA 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.

  9. CONCLUDE BY SAYING: The least value of the distance OA is 2

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