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  