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PARAMETRIC EQUATIONS Sometimes the equation of a curve is given by expressing the coordinates x and y as functions of a third variable (usually t), called a parameter. Using t as parameter enables us to refer to a particular point on quite complex curves (that we’ve not met so far) Example 1 Plot the curve given parametrically by the equations x = t2 – 4 and y = t3 – 4t −3 t −2 −1 use t from −3 to 3 0 1 2 3 t –4 t3 – 4t 2 4 y 3 2 1 x –6 –4 –2 2 –1 –2 –3 –4 What is the Cartesian equation connecting x and y ? Use substitution by making t the subject of either parametric equation. 4 6 Working : So the Cartesian equation reads y 2 x 2 ( x 4) Some curves to plot 1. use t from −3 to 3 x = t2 + 1 and y = t + 2 −3 t −2 −1 0 1 2 3 t2 + 1 t+ 2 4 y 3 2 1 x −2 −1 1 2 −1 −2 Find the Cartesian equation of the loci above 3 4 5 6 7 8 x = t2 2. −3 t use t from −3 to 3 and y = t3 −2 −1 0 1 2 3 2 t t3 30 y 25 20 15 10 5 x −2 −1 1 2 3 4 5 6 7 8 9 −5 −10 −15 −20 −25 −30 3. Find the value of the parameter and the other coordinate on the curve x=t and y = 2 t at y 3 2 [2] 4. Show that the parametric equations (i) x = 1 + 2t and y = 2 + 3t (ii) x= 1 2t 3 and y= t 2t 3 Both represent the same straight line. (Form the same cartesian equations) [6] 10 Answers x = t2 + 1 1. and −3 10 −1 t 2 t +1 t+ 2 use t from −3 to 3 y=t+2 −2 5 0 6 −1 2 1 0 1 2 1 2 3 2 5 4 3 10 5 2 4 8 3 9 27 y 5 4 3 2 1 x –2 –1 1 2 3 4 5 6 –1 –2 –3 –4 Cartesian equation : x = t2 2. and −3 9 – 27 t 2 t t3 6 x = y2 – 4y + 5 use t from −3 to 3 y = t3 −2 4 -8 −1 1 –1 0 0 0 1 1 1 y 4 2 x –2 –1 1 –2 –4 –6 Cartesian equation : y2 = x 3 2 3 4 5 6 3. t = x = 4 3 4. Both should give the Cartesian equation 3x – 2y + 1= 0 PARAMETRIC DIFFERENTIATION – SOME EXAMPLES x = f (t) Find the derivative y = g(t) dy g (t ) dx f (t ) dy for each of the following parametric equations, dx 1. x t 2 , y 2t 2. x t2, y t3 3. x sin t , y cos 2t 4. x e t , y tet 5. x sin 3 , y cos 3 Answers: 1. 1 t 2. 3t 2 3. – 4 sin t 4. 1 + t 5. cot Further examples to work through 6. Find an expression for dy for each of the parametrically defined curves ; dx (a) x 4t 2 1, y t 3 t (b) x t2 , 7. Find the gradient at t = 2, on the parabola x 3 sin , y cos 2 y 2t 8. Find the coordinates of the points on the parametrically defined curve where the gradient has a value of 1. Give each answer using 2dp. x 2t 3 t y 2t 2 1 I have included the curve for you to interpret the solution graphically. 4 y 3 2 1 x –4 –3 –2 –1 1 –1 –2 2 3 4 9. Find the equation of the normal at ( –8, 4) to the curve which is given parametrically by x t3 , y t2 . Draw the normal on the curve below. Label the normal with the straight line equation. 4 y 3 2 1 x –9 –8 –7 –6 –5 –4 –3 –2 –1 1 2 3 4 5 6 7 8 –1 –2 Note : The gradient is not defined when t = 0 because the tangent at the origin is the y axis. The point where t = 0 is called a cusp. 9