DHS Differentiation & Its Applications (9758) (Revision Solutions)

Y5 Topical Revision Package 1 4. Differentiation and its Applications (solutions) 1 4. Differentiation and its Applications (solutions) (I) Tangents and Normals (Normal/Implicit Differentiation) 1 (a) (b) ( ) ( )( ) ( )( ) 1 1 212 221 12 Let sin 3 3sin (3 ) d 1 9 sin (3 ) 319 d sin 3 sin 3 1 9 xy x x x y x xxx x x x x − − − −− = −  −−−= = − ( )( ) cos 2 3 3 Taking ln on both sides, cos 2 ln ln 3ln Differentiate w.r.t. , xyx xy x x x = = = ( )( ) ( ) ( ) ( ) ( ) cos2 3 3 3 3 1d 32sin2 ln cos2 ........... (1) d At , 1d 3Subst into (1): 2sin2 ln cos2 d yxy x yx x xy y y x −+= = = ⇒= −+= ππ ππ ππ π ππ 2d 3d y x π∴= 2 (i) (ii) (iii) ( )de e e e 1d x x xxyyx x xx − − −−= ⇒ = −= − Graph is decreasing: d 0d y x < ( )e 1 0 1x xx− −< ⇒ > [Also accept 1x≥ ] ( ) ( ) ( ) 2 2 d e 1e 1 e 2 d xx xy xx x −− −= −− − = − Graph is concave downwards: 2 2 d 0 d y x < ( )e 2 0 2x xx− −< ⇒ < Therefore, for graph to be decreasing and concave downwards: 12 x<< . [Also accept 12 x≤< , 12 x<≤ , or 12 x≤≤ ] Consider gradient at ( ),ab = ( )e1a a− − : ( ) ( ) e10 e e1 a aa hb aa ha a a − −− − = −− −= − − Since e aba −= ( ) 2e e1 eaa aha a a a−− −= − −= Note There is no need to simplify d d y x expression. We need the corresponding y-coordinate to evaluate the gradient to the curve at x π= . 1x> 2x< Alternative Tangent at ( ),ab : ( )( )e2ayb a xa −−= − − Subst ( )0,Rh into tangent and e aba −= : ( ) 2e e1 eaa aha a a a−− −= − −=

Y5 Topical Revision Package 1 4. Differentiation and its Applications (solutions) 2 ( )2d 2e e e 2d a aah aa a aa − −−=−= − At max/min point: d 0d h a = ( )e2 0 2 or 0 aaa aa − −= = = a 2 – 2 2+ d d h a + 0 – Slope a 0 – 0 0+ d d h a – 0 + Slope Alternative : Second derivative Test ( ) 2 2 2 ed 24 d 0 ah aa a − = −+ < when 2a= Greatest possible h 24e−= at 2.a= 3. (i) (ii) f( )yx= strictly increasing ⇒ f '( )x > 0 f( )yx= concave downwards ⇒ f ''( ) 0x < [so the gradient function f '( )x decreases] We observe from f '( )yx= graph that 1x> (ans) For stationary points on graph of f( )yx= , Need f '( ) 0x = ⇒ 3, 0x=− x (−3)− −3 (−3)+ f '( )x −ve 0 −ve tang Point of inflexion at x = −3 x 0− 0 0+ f '( )x −ve 0 +ve tang Minimum point at x = 0. [We observe the signs from f '( )yx= graph] (iii) Alternative : From the graph, greatest h occurs when d d y x is most negative (i.e. where d d y x is min, in this case). Thus need 2 2 d d 02y x x=⇒= Greatest possible 22 2 42e eh −= = x y 1 −3 −1.5 x =1

Y5 Topical Revision Package 1 4. Differentiation and its Applications (solutions) 3 4 (i) (ii) (iii) Differentiating 111 xya+= w.r.t. x , 22 1 1d 0d y x yx−− = 2 2 d d yy xx=− Since 0, 0,xy≠≠ 2 0x > and 2 0y > . It follows that d 0d y x < and thus y is a decreasing function. From (i), d 0d y x < at all points ( ),xy . Thus there are no points on the curve such that d 0d y x = There are no stationary points on the curve. Gradient at ( )2 ,2aa ( ) ( ) 2 2 2 1 2 a a = −= − Equation of tangent at ( )2 ,2aa : ( )2 12ya xa−= −− i.e. 4y xa= −+ Solving 111 4 xya y xa  +=  = −+ 111 4x ax a+= − 22 440x ax a⇒− + = ( ) 2 20xa⇒− = 2xa⇒= , which is the point where we construct the tangent With no other intersection points, the tangent does not meet the curve again. 5(a) (b) ( ) ( ) g( ) 2 g( ) g( )f () t a ne f ' () s e c e g ' () ex xxx xx= ⇒= 21 1 2 3ef '(5) sec (e ) ( 3)e 9.81cos (e) −= −= = − ( )22 22 21d (2 1)(2 ) (2) 21d ( 21 ) ( 21 ) xxx y x xxy xx x x −−−= ⇒= =− −− Function is increasing ( ) 2 21d 0 0,d (2 1) xxy xx −⇒ >⇒ > − where 1 2x≠ Thus inequality reduces to ( 1) 0xx −> 0 or 1xx∴< > [Accept 0 or 1xx≤≥ ] Implicit differentiation right away!

Y5 Topical Revision Package 1 4. Differentiation and its Applications (solutions) 4 6(i) (ii) (iii) 223 4 2 20x xy y− + −= Differentiate with respect to x : dd6 4 22 0 0dd yyxx y y xx   − + + −=    dd32 22 0dd d 32 (shown)d 22 yyxx yy xx y xy x xy − −+ = −= − For tangents to the curve parallel to x-axis, d 0d y x = 32 022 xy xy − =− 3 2yx⇒= Solving 223 4 2 20 3 2 x xy y yx  − + −= = , 2 22 333 4 2 20 3 4022x xx x x− + −=⇒ −= 2 ,3 3 xy= = 2 ,3 3 xy= −= − The points are 2 ,3 3   and 2 ,3 3 −− At ( )0,1P , ( ) ( ) ( ) ( ) 30 21d 1d 20 21 y x −= =− Gradient of normal 1=− Equation of normal at P: ( )11 0yx−= − − i.e. 1yx= − Solving 223 4 2 20 1 x xy y yx  − + −=  = − , ( ) ( ) 223 4 1 21 2 0x xx x− − + − −= Such points ( ),xy must satisfy 2 conditions: 1. d 0d y x = 2. Lie on the curve 223 4 2 20x xy y− + −=

Y5 Topical Revision Package 1 4. Differentiation and its Applications (solutions) 5 ( ) 29 80 980 8 or 0 9 xx xx x −= −= = At point Q, 8 9x= Area of triangle OPQ = ( )18129   = 4 9 7(i) (ii) +=xya , for 0, 0xy>> , a positive constant Differentiate with respect to x : 1 1d 0d22 y xxy += d d yy x x =− ≠ 0 since 0y≠ Hence C has no stationary points. As 0→x , d d y x →−∞ . The tangent to C approaches the line x = 0 (the y-axis). 8(i) (ii) (iii) 2(4 ) 16 48xy y−+= ----- (1) Differentiate with respect to x : [ ] dd2(4 ) 4 16 0 dd dd8(4 ) 2(4 ) 16 0 dd d2(4 ) 16 8(4 ) d d 4(4 ) d (4 ) 8 yyxy xx yyxy xy xx yxy xyx y xy x xy − −+ = −− − + = −− = − −= −− Tangent // x-axis d 04 0d y xyx⇒ =⇒ −= Substitute into eqn (1) : 20 16 48 3yy+ = ⇒= ⇒ eqn of tangent is 3.y= Coordinates of P = ( )3 4 ,3 Tangent // y-axis d d y x⇒ is undefined (4 ) 8 0 4 8xy y x⇒ − −=∴= − Substitute into eqn (1) : ( ) 2 78 16 4 8 48 4xx+ −= ⇒= Eqn of tangent is 7.4x= Coordinates Q = ( )7 4 ,1− P O Note: Wherever the y-coord point Q, triangle OPQ has base of length 1 and height 8 9 . Q 1 Same concept as Q6(ii).

Y5 Topical Revision Package 1 4. Differentiation and its Applications (solutions) 6 (iv) Coordinates ( )7 4 ,3R Area of triangle PQR = [ ] 731 2 443 ( 1) ( ) 2−− − = units2 [Note that triangle PQR is a right angled triangle!!!] (II) Tangents and Normals (Parametric Equations) 9(i) (ii) dd e sin e cos ; e cos e ( sin )dd tt t txy tt t ttt −−= + = − +− d e cos e ( sin ) d e sin e cos tt tt ytt x tt −−− +−= + 2e (cos sin ) ee (sin cos ) t t t tt tt − −−+= =−+ Gradient of normal at tp= is 2 2 1 ee p p−−=− . Equation of normal is 2e cos e ( e sin )p ppy px p−−= − 2e ( e sin ) e cospp pyx p p −= −+ For π 2p= , equation of normal becomes π π 2e( e)yx= − At 0x= , 3 2ey π =− . At 0y= , 2ex π = . Area of Triangle OAB = 3 222211 e e e unit22 ππ π   =      10(i) (ii) d d d 11 2d dd 2 yyt x t x tt=×= ×= At P, d1 d y xp= and gradient of normal = p− Equation of normal to C : 22 ()y p px p−= −− , i.e. 3 2y px p p= −++ For 2p= , equation of normal becomes 2 8 2(2) 2 12 yx yx = − ++ = −+ When this normal cuts C, 22 2 12tt= −+ 2 60tt +−= ( )( 2) 3 0tt− += 2 or 3tt= =− At P, 2t = . Hence, at Q, t = 3− . ( )9, 6Q≡− Solve simultaneous eqn { 2 ,2 2 12 xty t yx = = = −+

Y5 Topical Revision Package 1 4. Differentiation and its Applications (solution