7.Differentiation Solutions 2023 (RVHS)

River Valley High School, Mathematics Department 2023 Differentiation & Applications T1 Differentiation: (Classified to 3 key sub-topics Tangents and Normals/ Rate of Change & Maxima Problems / Maclaurin’s series) Solutions Tangents & Normals 1 1 23 2 2 4 1 1 2cos 11 2 1 d dx x x x xx                  ln sinxt cos sin dx t dt t cotyt 2cosecdy tdt  Therefore 2 1 sin sin cos dy t dx t t  2 2sin costt 2 sin 2t (proved) 11Gradient of normal sin 222 t sin 2 1t 2 24tt     Therefore 21ln sin ln ln 24 2 2x        1y 11Equation of normal: 1 ln 2 22yx         11 ln 2 122yx     PQ = RQ  normal at R is the reflection of the above normal in the x-axis  equation of normal at R is  11 ln 2 122yx        11 ln 2 122yx     2 (i) 22 d 1 1 d 1 x t tt   and 22 d 11 d 1 y t tt     2d d d 21ddd y y x tttx    Since y = ax is an oblique asymptote,  as x, the gradient of the curve approaches a. But x  t0  2d 2 1 1d y tx    ,  a = 1 (proved)

River Valley High School, Mathematics Department 2023 Differentiation & Applications T2 (ii) As t , 0 2x  and 0 2y  . As t , 0 2x  and 0 2y  . From the diagram, For y = kx not to intersect the curve, 1 or 1kk  (iii) Eqn of normal at t = 1,         1 4 1 31 4 3 1 1 44 y x yx            To find the points of intersection,         11 1 113 tan 1 tan 1 44 4 2 tan 4 0 2 tttt tt                Using GC, t = - 8.41 O y x 1.72 1.72 y = x O y x 1.72 1.72 y = x y = kx

River Valley High School, Mathematics Department 2023 Differentiation & Applications T3 3(i) (ii) e xyx d ( e ) e e (1 )d        x x xy xxx At P , d, e , e (1 ) d aa yx a y a a x     Equation of tangent to the curve at P is e e (1 )( ) e (1 )( ) e aa aa y a a x a y a x a a            At Q , 0, x y h e (1 )(0 ) eaah a a a     2e ( 1)( ) e ea a aa a a a      2d 2 e ( e ) e (2 )d a a ah a a a aa         For stationary values of h , d 0d h a  0a (N.A. since 0a ) or 2a 2 2 2 d d h a 0 0 0 Tangent Maximum value of 24eh 4 (i) 2 111,x a y a t t t              , a > 0 2 d d xa t t  3 d2 1d y at t   2 2 33 d d d 2 2 11d d d y y t t atx t x a tt                  (ii) When 1 2t  , 9 d 15,, 2 d 4 ayx a y x   . Eqn of tangent:  9 15 24 ay x a  

River Valley High School, Mathematics Department 2023 Differe