05 2016 - 2017 H2 Maths Differentiation Applications Tutorial Solutions BMQ2 3 4 and PQ5 6 7 (LMS)

National Junior College Mathematics Department 2016 Differentiation Applications Page 1 of 5 National Junior College 2016 – 2017 H2 Mathematics Differentiation Applications Tutorial Solutions Basic Mastery Questions 2. 222 3 5x xy y   . Differentiate implicitly wrt x, we get:   dd4 3 3 2 0 dd d 2 3 3 4d d 3 4 .d 2 3 yyx y x y xx y y x y xx y y x x y x          At (4, 3), gradient of tangent is         3 3 4 4d7 .d 2 3 3 4 6 y x   Hence equation of tangent is:  734 6 75 .63 yx yx     Gradient of normal, 16 .d 7 d y x   Hence equation of normal is:  634 7 6 45 .77 yx yx      3. 22 8 4 6 4 0x x y y xy      . Differentiate implicitly wrt to x, we get:   d d d2 8 2 4 6 6 0d d d d 6 2 4 6 2 4d d 8 2 6 .d 6 2 4 y y yx y y x x x x y x y x yx y x y x x y               For tangent to be parallel to x-axis,

National Junior College Mathematics Department 2016 Differentiation Applications Page 2 of 5 d 0d 8 2 6 06 2 4 8 2 6 0 4 3 . (1) y x xy xy xy xy        Substituting (1) into 22 8 4 6 4 0x x y y xy      , 22(4 3 ) 8(4 3 ) 4 6(4 3 ) 4 0 y y y y yy          2 (4 3 )(4 3 8 6 ) 4 4 0 y y y yy         2 22 (4 3 )( 4 3 ) ( 2) 0 ( 2) (3 4) 0 ( 2 3 4)( 2 3 4) 0 ( 2 2)(2 3) 0 y y y yy y y y y yy                      31 or 2yy   11 or respectively.2xy   Hence coordinates of the points are   131,1 and , 22  . 4. 5 secxa  , 3 tanya  where ππ 22    . 2dd 5 sec tan , 3 secdd xy aa    . 2 d d 3 sec 3d .dd 5 sec tan 5sin d y ya xxa           When the normal is parallel to yx ,

National Junior College Mathematics Department 2016 Differentiation Applications Page 3 of 5 1 1 1d d d 1d 3 15sin 3sin . 5 y x y x         Hence, 1 3 255 sec sin , 54x a a       1 393 tan sin . 54y a a       The point on the curve where the normal is parallel to the line yx is 25 9,44aa  . Practice Questions 5. 33 2.x xy y k   --- (1) Differentiate implicitly wrt x, we get:     22 22 2 2 dd3 6 0 dd d63 d d3 .d6 yyx y x y xx yx y x y x y x y x x y          Since C has a tangent which is parallel to the y-axis, the normal at the point of contact of the tangent with C is parallel to the x-axis, i.e. 2 2 2 1 0d d 6 03 6. y x xy xy xy     Substitute 26xy into (1), we get:     32 2 3 6 3 3 63 6 6 2 216 6 2 216 4 0 (shown). y y y y k y y y k y y k             Hence       2 3 4 4 4 216 .2 216 ky    For real values of 3y ,     2 4 4 216 0 1 (shown).54 k k   When 6x , 2 2 66 1 1 or 1. y y y     Hence from (1), when 1y ,        33 6 6 1 2 1 220,k       And when 1y ,        33 6 6 1 2 1 212.k        

National Junior College Mathematics Department 2016 Differentiation Applications Page 4 of 5 6. 23,.x t y t d 2d x tt  , 2d 3d y tt  . 2 d d 3 3 d .dd 2 2 d y y t t t xxt t      Then equation of tangent is   32 33 3 3 2 2 2 3 3 2 3 0 (proven). ----(1) ty t x t y t tx t y tx t          (i) For tangents that pa ss through  ,XY , 32 3 0.Y tX t   Since the equation of the tangent passing through  ,XY is a cubic equation in t, there are at most only 3 real roots for t, hence there cannot be more than 3 tangents passing through  ,XY . [Note that each value of t results in one tangent equation when substituted into (1).] (ii) Equation of tangent at 2t :     3 2 3 2 2 0 2 6 8 0 3 4 0. yx yx yx          Since the tangent meets the curve again at tu , substitute 2xu and 3yu into 3 4 0yx   ,       32 2 2 3 4 0 1 4 4 0 1 2 0 uu u u u uu           Hence 1u . 7(i) 21xt , 1 21y t  . d 2d x t  ,   2 d2 d 21 y t t   .   2 d d1 d 0.dd 21 d y y t xx t t       Therefore curve shows a decreasing function. (ii) At 1t , 3x and 1y . At 5 8t  , 1 4x and 4 9y . Hence, gradient of chord     4 1 49 1 934    . For tangent to be parallel to chord,     2 2 d4 d9 1 4 9 219421 y x t t        Since   2 2 1 0,t t is undefined, and hence there is no tangent to the curve parallel to the chord. (iii) Gradient of normal to the curve is given by:   21 21d d ty x    At 1 2t  , 0x , 1 2y , 1 4d d y x  . Hence equation of normal to the curve at 1 2t  is:  1 402 2 8 1. yx yx    

National Junior College Mathematics Department 2016 Differentiation Applications Page 5 of 5 Since the normal meets the curve again,       2 2 12 8 2 1 121 2 8 4 1 2 1 32 2 9 0 19 02 16 19 or .2 16 tt tt tt tt tt                        At 9 16t  , 17 8x and 8y . Hence the normal to the curve at 1 2t  meets the curve again at 17 ,88  . (iv) 140 4x 1 8x  . Coordinates of P are 1 ,08  . Therefore area of triangle OPQ is 1 1 1 2 1 38 22     .