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: