<p> A– LEVEL TOPIC REVIEW unit C4 parametric functions</p><p>1. Eliminate the parameter t from the following parametric equations to find the equation in Cartesian form. State also the range of values that can be taken by x and y, assuming t . a) x=2 t - 1, y = t 2 + 2 (3 marks) b) x=2sin t , y = 3cos t (3 marks)</p><p>2. Write the parametric equations x=sin t , y = 2cos 2 t , 0＃ t 2p in Cartesian form. Hence sketch the curve, indicating the coordinates of any axis intercepts, and of the end points of the curve. (7 marks)</p><p>1 1 3. a) Find the gradient of the curve x= t +, y = t - , expressing your answer in terms of the t t parameter t. (3 marks) b) Hence find the equation of the tangent to the curve at the point where t = 2, giving your answer in the form ax+ by + c = 0 , where a, b and c are integers. (5 marks)</p><p>4. The equation of a curve is given as x=1 + t2 , y = t 3 - 3 t in parametric form. dy d2 y a) Find and in terms of t. (6 marks) dx dx2 b) Find the equation of the normal to the curve at the point (5, 2), giving your answer in the form ax+ by + c = 0 , where a, b and c are integers. (4 marks) c) Find the coordinates of the stationary points on the curve, and determine their natures. (6 marks)</p><p>5. The diagram shows the curve with parametric equations y x=ln t , y = t2 - 4 t + 3, t > 0 . a) State the equation of the asymptote of the curve. (1 mark) x b) Find the coordinates of the points where the curve C with parametric equations cuts the x-axis. (3 marks) c) Hence show that the area of the finite region bounded by C and the x-axis is 4- 3ln 3. (4 marks)</p><p>1 6. The curve given parametrically by x=tan t , y = sin t , 0 � t p y 2 B is sketched in the diagram. A 1 1 At the points A and B, t = 6 p and t = 3 p respectively. Find the R area of the region R, giving your answer in surd form. (5 marks) x [Hint : it will be useful to remember from unit C3 that if f (x )= sec x , f ( x ) = sec x tan x ]. A– LEVEL TOPIC REVIEW : ANSWERS</p><p> unit C4 parametric functions</p><p>1. a) t=1 ( x + 1) M1 d2 y 3+ 3 t -2 2 2 2 M1A1 2 = 1 2 dx2 t y=4 ( x + 1) + 2 A1 d2 y 3 3 x纬 , y 2 A1 = + A1 dx24 t 4 t 3 2 2 b) sint+ cos t = 1 M1 9 b) t = 2, m = 4 M1A1 骣x2 骣 y 2 + = 1 A1 4 琪 琪 y-2 = -9 ( x - 5) M1 12– 12 桫2 桫 3 1– 1 4x+ 9 y - 38 = 0 -2＃x 2, - 3 ＃ y 3 A1 A1 dy c) = 0� t 1 M1A1 dx 2. y=2( 1 - 2sin2 t) M1 t =1� (2, 2) M1 2 y=2 - 4 x A1 d2 y =3 + 3 > 0 minimum A1 y G1 dx2 4 4 2 t = -1 (2, 2) 1 M1 2 – 1 1 x d y 3 3 – 1 = - - < 0 maximum A1 dx2 4 4 – 2 (0, 2)- 1 , 0 1 , 0 intercepts : ( 2) ( 2 ) M1A1 5. a) x� マ  t0 = y 3 A1 end points :(- 1, - 2) (1, - 2) M1A1 b) t2 -4 t + 3 = 0� t = 1, t 3 M1A1 (0, 0) and (ln3, 0) A1 dy1 dx 1 3 3. a) =1 + ; = 1 - M1A1 2 1 dt t2 dt t 2 c) (t-4 t + 3) dt M1 t 1 1 1+ 2 2 3 dy t dy t +1 1 2 = or = A1 轾t-4 t + 3ln t A1 1 2 臌2 1 dx 1- dx t -1 t 2 9 1 ( 2-12 + 3ln 3) -( 2 - 4 + 0) M1 b) x=5, y = 3 M1 2 2 3ln 3- 4� magnitude - 4 3ln 3 A1 5 m = 3 A1</p><p>1p 3 5 5 3 y-2 = 3( x - 2 ) M1 2 6. sint sec t dt M1A1 5x- 3 y - 8 = 0 1p M1A1 6</p><p>1p 3 dy dx sect tan t dt M1 =3t2 - 3; = 2 t 4. a) M1A1 1p dt dt 6</p><p>2 1p 2 dy3 t - 3 dy 3 3 -1 3 sect 1 = 2 - = or =2t - 2 t A1 [ ] p M1A1 dx2 t dx 6 3</p>