TMJC H2 Chapter 8 Applications of Differentiation Discussion Solutions 2024

Chapter 8 Applications of Differentiation TMJC 2024 Page 1 of 24 H2 Mathematics (9758) Chapter 8 Applications of Differentiation Discussion Solutions Level 1 1 2012 ACJC JC1 Promo/9 (modified) The diagram below shows a rectangle of height x m and width y m inscribed in an equilateral triangle of side a m. (i) Show that  3 .2x a y  (ii) Hence find the maximum area of the rectangle. Q1 Solution (i)     3 1 2tan 30 1 2 (shown)2 3x a y x a y x a y       (ii) Let area of rectangle be 2mA Then   2 2 3 2 2 3 3A xy a y y ay y     3 3d d 2 A y ya   For maximum area, 3d 0d 2 3 0 2 A ay ay y     2 2 3d 0d A y   Hence A is maximum when .2 ay Maximum 2 2 23 3 3 m2 2 2 2 8 a aA a a            

Chapter 8 Applications of Differentiation TMJC 2024 Page 2 of 24 2 2018(9758)/I/7 A curve C has equation 2 2 2 2 4 1 2 x y x xy   . (i) Show that 2d 2 d 2 16 y x y x xy y   . [3] The points P and Q on C each have x-coordinate 1. The tangents to C at P and Q meet at the point N. (ii) Find the exact coordinates of N. [6] Q2 Suggested Solution (i)   2 2 2 2 2 2 2 2 2 2 2 4 1 2 2 8 1 8 x y x xy x y x xy x y xy         Differentiate w.r.t x,   2 2 d d2 16 2 d d d 2 Shownd 2 16 y yx y y xy x x y x y x xy y      (ii) Substitute 1x into  1 , 2 2 2 2 8 1 9 1 1 3 y y y y      2 1When , 3 12d 3 1 1d 2 163 3 17 54 y y x                  2 1When , 3 12d 3 1 1d 2 163 3 17 54 y y x                 Therefore, the equation of tangents at 1x are    1 17 1 23 54y x        and    1 17 1 33 54y x    . Solving equation (2) and (3) simultaneously    3 2 , Cross multiply first Recall: Equation of tangent is    0 0y y m x x  

Chapter 8 Applications of Differentiation TMJC 2024 Page 3 of 24  2 17 13 27 1 17 x x     0y  1 coordinates of is ,0 . 17N       3 Specimen Paper (9758)/I/1 A circular ink-blot is expanding such that the rate of change of its diameter D with respect to time t is 0.25cm/s. Find the rate of change of both the circumference and the area of the circle with respect to t when the radius of the circle is 1.5cm. Give your answers correct to 4 decimal places. Q3 Suggested Solution Let the radius, circumference and area of the circle be r cm, C cm and A cm2 respectively. d 0.25d d d and when 1.5d d 2 Differentiate w.r.t : d d d d 0.25 0.785 To find 4 (to 4 d Given: .p.) : D t C A rt t C r D t C D t t            The rate of increase of circumference is 0.7854cm/s. 2 2 2π π π 2 4 d d d d d d π 0.252 π 8 D Dr A A D t D t D A D            When   1.5, 2 2 1.5 3r D r    ,     π 3d d 8 1.1781 to 4 d.p. A t   The rate of increase of area is 1.1781cm2/s. Strategy: Use Implicit differentiation Look carefully at the requirement of the question Look carefully at the requirement of the question

Chapter 8 Applications of Differentiation TMJC 2024 Page 4 of 24 Level 2 4 2009(9740)/II/1 (modified) A curve C has parametric equations 2 3 24 , .x t t y t t    (i) Find the coordinates of the x-intercepts and sketch the curve for 2 1 t   . You do not need to label the coordinates of the turning point(s). [3] The tangent to the curve at the point P where 2t  is denoted by l. (ii) Find the Cartesian equation of l. [3] (iii) The tangent l meets C again at the point Q. Use a non-calculator method to find the coordinates of Q. [4] (iv) Determine the acute angle between the tangent l and the line 3y x  . [2] Q4 Suggested Solution (i)   3 2 20, 0 1 0 or 1 When 1, 3, 0 When 1, 0, 0 y t t t t t t t x y t x y               Therefore the coordinates of the x-intercepts are    3,0 and 0,0 . GC Keystrokes to sketch parametric equations: y x Label coordinates of end-point When 1, 5, 2t x y   Label coordinates of end-point When 2, 4, 4t x y    0,0 Label coordinates of x-intercept   3 2 20, 0 1 0 or 1 When 1, 3, 0 When 1, 0, 0 y t t t t t t t x y t x y              

Chapter 8 Applications of Differentiation TMJC 2024 Page 5 of 24 (ii)   2 2 2 3 2 d d 2 4 3 2d d d d d dd d 3 2 2 4 When 2, 2 4 2 12, 2 2 12 d 2,d x y t t tt t y y t xx t t t t t x y y x                Equation of the tangent, l is 12 2( 12) 2 12 y x y x      (iii) Curve C: . 2 4x t t  and 3 2y t t  --------(1) Tangent line l: 2 12y x  --------- (2) To find the point of intersection Q between tangent line l and curve C, we sub (1) into (2) and attempt to solve for the values of t: Sub. 2 4x t t  and 3 2y t t  into l:   3 2 2 2 4 12t t t t           3 2 2 2 8 12 0 2 6 0 2 3 0 t t t t t t t t            2 or 3 (rejected, this is point ) t t P    When 3t  , 3,x 18y Coordinates of Q:  3, 18  Comments: The tangent l meets C again at the point Q means point Q is the point of intersection between tangent line l and curve C. Recall: Equation of tangent is    0 0y y m x x   Always reject with a reason. Note: “non-calculator method” means cannot use GC and get 3t  . You have to solve for the values of t algebraically, i.e. using factorisation. (You can use GC to guide/check your factorisation) Note: Since the line is tangent to the curve at P (when t = 2), it will imply that t = 2 is a solution of the polynomial (hence t – 2 is a factor)

Chapter 8 Applications of Differentiation TMJC 2024 Page 6 of 24 (iv) Let  be the acute angle between the two lines. 1 opp risetan gradient of lineadj run tan 2 tan 2 A A A        1 opp risetan gradient of lineadj run tan 1 tan 1 B B B          1 1 exterior angle of sum of 2 interior oppo site angles of tan 2 tan 1 18.4 A B           Alternatively, using MF26      1 tan tan tan tan 1 tan tan 2 1 1 1 2 1 3 1tan 18.43 A B A B A B             α B A x y y = 2x − 12 y = x + 3 O Concept used: tan Gradient of line  where θ is the angle between the line and the x-axis Strategy to solve this question: Draw diagram α

Chapter 8 Applications of Differentiation TMJC 2024 Page 7 of 24 5 2017(9758)/II/1 (modified) A curve C has parametric equations 3x t , 2y t . (i) The tangent at the point 3 ,2P pp      on C meets the x-axis at D and the y-axis at E . The point F is the midpoint of DE . Find a cartesian equation of the curve traced by F as p varies. [5] (ii) Show that the area of triangle ODE is independent of p, where O is the origin. Q5 Suggested Solution (i) 2 2 d 3 d , 2d d d 2 d 3 x y t t t y tx    At point P, 2d 2 d 3t p yt p p x     Equation of tangent at point P: 22 32 3y p p x p         When 0, 4 . 0, 4 6 6When 0, . ,0 x y p E p y x D p p           Coordinates of F are 6 0 0 4 3, , 2 .2 2 pp pp              Any point on the curved traced by F will have 3 , 2x y pp  To obtain the cartestian equation, subst 3p x into 2y p Cartesian equation traced by F : 6y x (ii) Coordinates of D and E are 6 ,0p      and  0, 4p respectively. Area of triangle ODE = 1 6 4 122 pp      which is independent of p. (shown) Find d d y x in terms of the parameter t first. OE and OD are perpendicular to each other, thus the area of triangle ODE can be found using   1 base height2 Recall: Equation of tangent is    0 0y y m x x  

Chapter 8 Applications of Differentiation TMJC 2024 Page 8 of 24 6 2016(9740)/II/1 Water i