1. PARAMETRIC EQUATIONS

2. An equation which is given as a relationship between x and y only is a Cartesian equation. Sometimes it is more convenient to express x and y in terms of a third variable, often t. e.g. x = t 2 – 3t 1 t1 t y = t + These are called parametric equations. A point on the curve can be found by choosing a value of t. For example, if we let t = 2, the point found is: ( –2, 5 2 )

3. Example 1: Find the cartesian equation of the curves with parametric equations: a) x = t + 2 ; y = t 2 – 5 b) x = 1 + sin t ; y = cos t – 4 a) Since x = t + 2t = x – 2 Substitute in y = t 2 – 5 y = ( x – 2 ) 2 – 5 y = x 2 – 4x – 1 b) Using the identity: sin 2 t + cos 2 t = 1 ( x – 1 ) 2 + ( y + 4 ) 2 = 1 If possible, make t the subject of one of the equations. If the parametric equations involve trigonometric functions, look for an appropriate identity.

4. 1t21t2 t2 t2 – x = 1t21t2 t2 t2 + y =... (1)... (2) (1) + (2): x + y = t Multiply (2) by 2t 2 :2t 2 y = t 3 + 2 2(x + y) 2 y = ( x + y ) 3 + 2 ( x + y ) 2 ( 2y – ( x + y ) ) = 2 Example 2: Find the cartesian equation of the curves with parametric equations: 1t21t2 t2 t2 – x = 1t21t2 t2 t2 + y = ; ( x + y ) 2 ( y – x ) = 2 Substitute in (3):... (3)

5. To find the gradient of a curve with parametric equations the chain rule is used. i.e. d yd xd yd x = dy d t × d td xd td x Example 1: Find d yd xd yd x for the curve with parametric equations: x = t 2 + t ; y = t 3 – 2t d xd td xd t = 2t + 1 3t 2 – 2 dy d t = 3t 2 – 2 × 1 2t + 1 = 3t 2 – 2 2t + 1 d yd xd yd x =

6. Example 3: Find the gradient of the curve with parametric equations x = t 2 ; y = 5 + 6t at the point ( 4, – 7 ) d yd xd yd x = dy d t × d td xd td x Using: 6 × 1 2 t = 3 t3 t At the point ( 4, – 7 ), t = – 2 d yd xd yd x = 3232 – Note, care is needed here. From the x co-ordinate t 2 = 4, it would be easy to deduce t = 2. d xd td xd t = 2t d yd td yd t = 6 d yd xd yd x =

7. Example 4: Find the equation of the tangent to the curve with parametric equations x = 4t ; y = t 3 + 1 at the point where t = 2. dy d t = d xd td xd t = d yd xd yd x = 3t 2 4 1414 × At t = 2, d yd xd yd x = 3(4) × 1414 = 3 y = (2) 3 + 1 =9 x = 4(2) = 8 The equation of the tangent is:y – 9 = 3 ( x – 8 ) y = 3x – 15 d yd xd yd x = dy d t × d td xd td x Using:

8. Example 5: Find the equation of the tangent to the curve with parametric equations x = t 2 ; y = t 3 at the point P where t = 2. The tangent intersects the curve again at the point Q as shown. Find the coordinates of Q. x y P Q ( 1, – 1 ) d yd xd yd x = × 1 2t 3t 2 At t = 2, d yd xd yd x = 3 P is ( 4, 8 ) The equation of the tangent is: y – 8 = 3 ( x – 4 ) y = 3x – 4... (1) For intersection at Q: sub. original equation in (1): t 3 = 3t 2 – 4 t 3 – 3t 2 + 4 = 0 ( t – 2 )( t 2 – t – 2 ) = 0 Hence at Q, t = – 1, ( t – 2 )( t – 2 )( t + 1 ) = 0 so Q is

9. Summary of key points: This PowerPoint produced by R.Collins ; Updated Nov. 2012 If possible, make t the subject of one of the equations. If the parametric equations involve trigonometric functions, look for an appropriate identity. i.e. d yd xd yd x = dy d t × d td xd td x To find the gradient of a curve with parametric equations, the chain rule is used: To find the cartesian equation of a curve given in parametric form: