RI Differentiation C6B Tut Sect A (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 ___________________________________ Tutorial 6B: Applications of Differentiation Page 1 of 4 Tutorial 6B: Applications of Differentiation Section A (Basic Questions) 1 9740/2009/02/Q1 (ii), (iii) A curve C has parametric equations 2 4xt t , 32 .yt t The tangent to the curve at the point P where 2t is denoted by l. (i) Find the cartesian equation of l. [3] (ii) The tangent l meets C again at the point Q. Use a non-calculator method to find the coordinates of Q. [4] (i) 2 4xt t , 32yt t d 24d x tt  2d 32d y ttt  2d32 d2 4 yt t xt   At 2,t  2d3 ( 2 ) 2 ( 2 ) 1 6 2d2 ( 2 ) 4 8 y x    224 ( 2 ) 1 2x  , 32221 2y Equation of tangent to the curve at 2t , is 12 2( 12) 2 12 yx yx    (ii) When the tangent l meets C again at the point Q, we have 21 2yx -----------------------(1) and 2 4xt t , 32yt t ----------(2) Substitute (2) into (1) : 32 2 32 2 2( 4 ) 12 81 20 (2 ) ( 6 ) 0 (2 ) (2 ) (3 ) 0 2 (n.a.) or 3 tt t t tt t tt t ttt tt            When 3,t  2(3 ) 4 (3 ) 3x     and 32(3 ) (3 ) 1 8y    i.e. coordinates of Q are ( 3, 18).

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ ________________________________ Tutorial 6B: Applications of Differentiation Page 2 of 4 2 JPJC Promo 9758/2021/Q8(a) A closed container is constructed using a sheet of metal with area 100 cm2. The container comprises 2 shapes, a cone and a cylinder. The slant height, l, of the cone is 10 cm. Given that the cylinder has height h, and radius r, and the height of the cone is kh, where k is a positive constant. (i) show that r rrh 2 10100 2  , [1] (ii) use differentiation to show that the exact maximum volu me of the container is given that 12 5 0 0132 7Vk  cm3, proving that it is a maximum. [6] [Volume of Cone = hr 2 3 1 , Curved Surface Area of Cone = rl ] (i) 2 2 2 (10) 100 100 10 (Shown).2 rr h r rrh r      (ii)   2 22 2 32 2 1 1 100 10 133 2 1 1 100 1023 d1 1 100 3 20d2 3 d 0d 20 400 4(3)(100) 20 40 10 10(NA as 0) or 2( 3) 6 3 rrVr k h r h k r r Vk r r r V kr rr V r rr                      

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ ________________________________ Tutorial 6B: Applications of Differentiation Page 3 of 4 To verify 10 3r  gives maximum V: Method 1: 2nd derivative test  2 2 d1 1 6 20 0 since 0d3 10 is max when 3 V kr rr Vr       Method 2: 1st derivative test   2d1 3 1 1 0 1 100 3 20 1 10d2 3 2 3 3 V kr r k r rr            Since k>0 and r>0, 1 103 k  and  10 0 r r 10 3    10 3 10 3    10 3 r  >0 0 <0 d d V r >0 0 <0 10 is max when 3Vr  1 1000 1000 1000 2500 1Max 1 1 (Shown).2 3 32 79 2 7 3Vk k             

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ ________________________________ Tutorial 6B: Applications of Differentiation Page 4 of 4 3 ASRJC Promo 9758/2021/Q4 A curve C has equation 32 2 23 3 0 .yx yx   (i) Find d d y x in terms of x and y. [2] (ii) Find the equation of the normal to the curve at the point P(2, 3). [2] (iii) Given that C meets the y-axis at the point R and the normal in (ii) meets the y-axis at the point A, find the area of triangle APR in the form 3ab , where a and b are integers to be determined. [3] (i) 32 2 23 3 0yx yx   Differentiating with respect to x, 22 22 2 2 dd3( 2 4 ) 6 0dd d(3 4 ) 2 6 d d26 d3 4 yyyy x y xxx yyx y yx x yyx x yx y        (ii) Gradient of tangent to curve at P(2, 3) = 2 2 2(3) 6(2) 23(3) 4(2)(3)   Gradient of normal to curve at P = 1 2 Equation of normal at P: 1 2 1 2 3( 2 ) 4 yx yx       (iii) When x = 0, normal cuts y-axis at A(0, 4). C: 32 2 23 3 0yx yx   When x = 0, 3y = 3  3 3y C meets y-axis at R(0, 3 3 ). Area of triangle APR = 31 2( 4 3 )2  = 343 , where a = 4, b = 3.