TMJC H2 Chapter 8 Applications of Differentiation Learning Package 2024

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Chapter 8 Applications of Differentiation TMJC 2024 Page 1 of 18 . Normal at A Tangent at A ( )fyx= x y 00( , )A x y R S P Q H2 Mathematics (9758) Chapter 8 Applications of Differentiation Core Concept Notes Success Criteria: Surface Learning Deep Learning Transfer Learning  Differentiate an equation of a curve defined implicitly to find gradient of tangents to a curve  Differentiate parametric equations to find gradient of tangents to a curve  Using relationship 1 gradient of tangent − to find gradient of normal to a curve  Find equations of tangents and normal to a curve  Use Chain Rule d d d d d d y y x t x t  = to link up the derivatives and rate given  Solve local maxima and minima problems  Solve connected rates of change problems §1 Tangents and Normals Given A ( )00,xy is a point on a curve, then a) the line PQ touching the curve at A is called the tangent to the curve at A and b) the line RS perpendicular to the tangent at A is called the normal to the curve at A. Recall: Gradient of curve at A = Gradient of the tangent at A 0 d d xx y x = =

Chapter 8 Applications of Differentiation TMJC 2024 Page 2 of 18 The gradient of the tangent and normal at 00( , )xy are respectively given by: • Gradient of tangent, 0 d d xx ym x = = • Gradient of normal 1 m=− Recall: The product of gradients of perpendicular lines is −1 The equation of the tangent and normal at any point ( )00,xy on a curve f ( )yx= is given by: Equation of tangent ( )00y y m x x− = − where 0 d d xx ym x = = Equation of normal ( )00 1y y x x m− = − − Example 1 The equation of a curve is 2 51y x x= + − . Find the equations of the (i) tangent and (ii) normal at the point (2, 13). Solution: (i) Non-GC approach: d 25d y xx =+ 2 d 9d x y x = = Equation of tangent: ( )13 9 2yx− = − 95yx=− GC approach to get equation of tangent directly: Step 1 Press !. Enter the equation of the graph 2 51y x x= + − under 1y Press % to see the graph Step 2 To draw/ obtain the tangent press `p and select 5: Tangent( Check using GC Gradient of Tangent = d d y x

Chapter 8 Applications of Differentiation TMJC 2024 Page 3 of 18 Using GC, Equation of tangent: 95yx=− Step 3 At the graph screen, key in the x-coordinate of the point required e.g. 2 and press e (ii) Gradient of normal at (2, 13) 1 9=− Equation of normal at (2, 13): ( )113 2 9 1 119 99 yx yx − = − − = − + Note: Equation of normal cannot be found directly from GC Conditions for using GC Approach: 1. There are no unknown constants in the equation of the curve. 2. y can be explicitly expressed in terms of x. Gradient of Normal = 1 d d y x −

Chapter 8 Applications of Differentiation TMJC 2024 Page 4 of 18 Example 2 (Curve is defined implicitly) [2015 JJC Promo/5 (modified)] A curve has equation 223 4 2 2 0x xy y− + − = . (i) Show that d 3 2 d 2 2 y x y x x y −= − . [3] (ii) Find the equation of tangent to the curve at the point P with coordinates ( )0,1 . [2] (iii) The normal to the curve at the point P meets the curve again at point Q. Find the area of triangle OPQ, where O is the origin. [4] Solution: (i) 223 4 2 2 0x xy y− + − = Differentiate w.r.t. x dd6 4 2 2 0 0dd yyx x y y xx    − + + − =     dd3 2 2 2 0dd d 3 2 (shown)d 2 2 yyx x y y xx y x y x x y − − + = −= − (ii) At ( )0,1P , ( ) ( ) ( ) ( ) 3 0 2 1d 1d 2 0 2 1 y x −== − Equation of tangent at P: ( )1 1 0 1 yx yx − = − =+ (iii) At ( )0,1P , gradient of normal 1=− Equation of normal at P: ( )1 1 0 1 yx yx − = − − =− Since the normal to the curve at point P meets the curve again at point Q, substitute 1yx=− into equation of the curve: ( ) ( ) ( ) 22 2 3 4 1 2 1 2 0 9 8 0 9 8 0 x x x x xx xx − − + − − = −= −= 8 or 0 9xx== (point P) At point Q, 81,.99xy== Area of triangle OPQ = ( )18129   = 4 9 units2 y x Draw diagram to help with better visualisation.

Chapter 8 Applications of Differentiation TMJC 2024 Page 5 of 18 Example 3 (Curve is defined parametrically) [2015 PJC Prelim/2/4 (modified)] The parametric equations of a curve are 2xt= , 2y t t=− . (i) The point P on the curve has parameter p. Show that the equation of the tangent at P is ( ) 22 2 1py p x p= − − . [3] (ii) The tangent at P meets the x- and y- axes at the points Q and R respectively. Find, in terms of p, the coordinates of Q and R. [2] (iii) Find the equation of the tangent at the point ( )4,6 and determine if this tangent meets the curve again. [4] Solution: (i) 22 dd 2 2 1dd x t y t t xy tttt = = − = = − d 2 1 d2 yt xt −= At the point P, tp= ( ) ( ) 22 21 2 py p p x p p −− − = − ( ) ( )( ) 3 2 22 2 2 2 1py p p p x p− − = − − ( ) 3 2 3 22 2 2 2 1 2py p p p x p p− + = − − + ( ) 22 2 1 (shown)py p x p= − − (ii) ( ) 22 2 1py p x p= − − At Q, 0y = ( ) 20 2 1 p x p= − − 2 21 px p= − Coordinates of Q are 2 ,021 p p   − At R, 0x = 22py p=− ,02 pyp= −  Coordinates of R are 0, 2 p − Chain Rule: d d d d 1xx dd d d d d y y t y xx t x t t == Remember to change parameter t to p at point P to find gradient and equation of tangent/normal at point P. Note: No t.

Chapter 8 Applications of Differentiation TMJC 2024 Page 6 of 18 (iii) At (4, 6), ( ) ( )( ) 2 22 rejected, 4, 2 are not the given coordinates 4 2 or 2 (2) 2 2 ( 2) 26 xt tt yy == = = −  = −  = − − − == Hence 2.tp= = − From (i), equation of tangent: ( ) 22 2 1py p x p= − − At 2tp= = − , Equation of tangent: ( ) ( )( ) ( ) 2 2 2 2 2 1 2yx− = − − − − 4 5 4yx=+ To determine if the tangent meets the curve again, Substitute 2xt= , 2y t t=− into 4 5 4yx=+ ( ) ( ) 22 2 2 2 4 4 5 4 40 20 t tt t t t t −+ ++ = = = + =− Since there is only one solution for t, which is the given point ( )4,6 , the tangent at ( )4,6 does not meet the curve again. Discussion: 1. Write down the gradient of a tangent that is parallel to the x-axis. 0 2. Given that the tangent is parallel to the y-axis, what can you say about the gradient of normal? 0 Tangent intersects curve again?

Chapter 8 Applications of Differentiation TMJC 2024 Page 7 of 18 Note: Tangent parallel to x-axis d 0d y x= Tangent parallel to y-axis d d y x is undefined Example 3 (Extension) The parametric equations of a curve are 2xt= , 2y t t=− . (iv) Find the equation of the tangent to the curve that is parallel to the x-axis. (v) Find the equation of the tangent to the curve that is parallel to the y-axis. Solution: (iv) From (i), 22 d 2 1, , . d2x ytyt xtt t −= = − = Since tangent is parallel to x-axis, 21 02 2 1 0 1 . d 0 2 d y x t t t t − = = = − = When 1 ,2t = 2 1 1 1 .2 2 4y = = − −  The equation of the tangent to the curve that is parallel to the x-axis is 1 .4y =− (v) Since tangent is parallel to y-axis, d d y x is undefined 21 is undefined2 20 0. t t t t − = = When 0,t = 2 0.0x = = The equation of the tangent to the curve that is parallel to the y-axis is 0.x = y x 0 tangent