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 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

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