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 Differentiation & Applications T4 4 18 15 15y a x a   4 15 3y x a (iii) At Q, 2 114 15 1 3a t a a tt              3 2 24 4 15 15 3t t t t    324 12 15 4 0t t t    Using GC, 4t  or 1 2t  . Cdts of Q: 5 63,4 16aa  (iv) As ,,t x a y   . Asymptote: x = a (iv) (v) Required area = 1 22 4 1 5 1 63 1 d2 4 5 16 aa a a a t t tt                     = 4 22 4 1 1323 1 1 d640 a a t t t  = 4 22 3 1 1323 1 ln640 3 a a t t   = 221323 21 ln 4640 64aa   = 2 1113 ln 4640a   5(i) sin 2x a t cosy a t d 2 cos 2d x att  d sind y att  d sin d 2cos 2 yt xt  y x = a

River Valley High School, Mathematics Department 2023 Differentiation & Applications T5 When 4t  , 2cos 2 0t   tangent at 4t  // y-axis When 4t  , xa . Thus, the tangent to the curve at 4t  is xa . (ii) When 3t  , 3 2 ax , 2 ay , d3 d2 y x  Equation of normal: 23 22 3 aayx     At R: 23 22 3 33 4 aa x ax      Coordinates of R: 33 ,04 a  At xa , 23 22 3 32 2 3 aaya ya       Area enclosed = 1 3 2 3 3 2 2 4 3 aaa        = 2 43 3 3 .48 2a   6(i) 1 tt dxx e e dt     212 dyy t t dt    2 2 t t dy dy t dt tedxdx e dt     When t = p, 2 pdy pedx  , point P is 2(1 ,1 )pep ,

River Valley High School, Mathematics Department 2023 Differentiation & Applications T6 Equation of tangent at P is 2(1 ) 2 [ (1 )] ppy p pe x e      22 2 (1 ) (1 )p p py pe x pe e p      Since tangent passes through (1, 0), 20 2 (1) 2 2 1pppe pe p p     2 2 1 0pp    1p  (ii) 7(i)   e d e e e 1d x x x x yx y xxx          Graph is decreasing: d 0d y x   e 1 0 1 x x x    (ii)       2 2 d e 1 e 1 d e2 xx x y x x x        Graph is concave downwards: 2 2 d 0 d y x   e 2 0 2 x x x    Therefore, for graph to be decreasing and concave downwards: 12 x . (iii) gradient at  ,xy =  e1x x          2 e1 0 e e 1 e e 1 e x xx xx x yhx x x h x x h x x x x              f ( )yx x O y x =1 x O y (0,1) x =1

River Valley High School, Mathematics Department 2023 Differentiation & Applications T7     2d 2 e ed e2 xx x h xxx xx     At max/min point: d 0d h x   e 2 0 2 or 0 xxx xx    x 2- 2 2+ d d h x + 0 - x 0- 0 0- d d h x - 0 + Greatest possible h 24e 8 (i) 42 d 3 d dd x a y a t t t t  2 d d d d d d 3 y y t t x t x   At 1 2t  , Gradient of tangent at 2 1 12 3 12P  Gradient of normal at 12P At  , 8 and 2 8 ,2P x a y a a a   Equation of tangent:  128 12y a x a   14 12 3y x a Equation of normal:  2 12 8y a x a   12 98y x a  (ii) 3 14 12 3 aa att   23 32 12 1 16 16 12 1 0 11By G.C, (N.A.), 24 tt tt t     

River Valley High School, Mathematics Department 2023 Differentiation & Applications T8 When 1 , 64 , 44t x a y a   Hence the tangent cuts the curve again at  64 , 4aa (iii) At Q: 0y 140 1612 3x a x a     16 ,0Qa At R: 0y 490 12 98 6x a x a    49( ,0)6Ra Area of triangle PQR =    1 49 16 226 a a a  = 22145 units6 a 9(i)   22dd 2 4 66dd xy y xxx    dd 4 8 0dd yyx y y xxx     d 48d y x y x yx    d8 d4 y x y x y x   For tangent parallel to y-axis, 40yx 4xy Substitute 4xy into equation of curve,     224 2 4 4 66y y y y   266 66y  2 1y  1y . When 1y , 4x When 1y , 4x Coordinates are    4,1 , 4,1

River Valley High School, Mathematics Department 2023 Differentiation & Applications T9 (ii) Substitute yk into equation of the curve, 222 4 66kx k x     224 2 66 0x kx k     Considering the discriminant,   22 2 4(4) 2 66 33 1056 0 for all real values of kk k k      The line yk cuts the curve for all real values of k. 10   3 2 3 2 2 2 2 2 2 (i) 4 3 2 Differentiating wr.t. : dd12 6 3 3 dd d3 3 12 6                        d x x y y x yyx xy x y xx yx y x xy x               2 22 d 4 2 d 22     y x xy x y x x x y y x y x         3 2 3 3 Curve meets when: 4 3 2 22 1 and 1 Thus, coordinates of is 1,1      yx x x x x x xy P          