TMJC Term 4 Revision Chapter 7 and 8 Differentiation Solutions

Term 4 Revision Topical Quick Check: Differentiation and its Applications TMJC 2024 Page 1 of 11 JC1 H2 Mathematics (9758) Term 4 Revision Topical Quick Check Chapter 7 Differentiation Chapter 8 Applications of Differentiation 1 SAJC Promo 9758/2022/Q3b / JPJC Promo 9758/2022/Q2aiii Differentiate the following with respect to x. (i) 2 2ln 1 x x   + [2] (ii) 1sec3 sin 2xx − [3] Solutions (i) ( ) 2 2 2 2ln ln 2 ln ln 1 1 1ln 2 ln ln 1 2 x xx x xx  = + − + + = + − + ( ) 2 2 2 2 d 2 d 1ln ln 2 ln ln 1d d 2 1 11 (2 )2( 1) 1 1 x xxxx x xxx x xx   = + − +   + =− + =− + (ii) 1d sec3 sin 2d xxx − ( ) ( ) 1 2 1sec3 2 sin 2 3sec3 tan 3 12 x x x x x −=+ − 1 2 2sec3 3tan 3 sin 2 14 x x x x −=+  −

Term 4 Revision Topical Quick Check: Differentiation and its Applications TMJC 2024 Page 2 of 11 2 DHS Promo 9758/2022/Q1b The graph of f ( )yx= has a maximum turning point at (2,3) and passes through the origin. The lines 1x=− and 2y= are asymptotes to the graph, as shown in the diagram below. Sketch the graph of f '( ),yx= showing clearly the axial intercepts and the asymptotes. [3] Solutions x y (2,3) O ( )fyx =

Term 4 Revision Topical Quick Check: Differentiation and its Applications TMJC 2024 Page 3 of 11 Revision Guide Page 3 Let’s Try Now. 3 MI PU2 P1 Promo 9758/2022/Q7 A curve C has parametric equations 1cos 2 , sin , for 0 2xy    = =   . (i) Show that d1 d 4sin y x =− . [3] (ii) Sketch C, showing clearly the features of the curve at the points where 0 = and 1 2= . [2] (iii) The tangent to the curve C at the point where p = is parallel to the line 2 0.yx+= Find the equation of this tangent. [4] (iv) The tangent from part (iii) meets the x-axis at P and the y-axis at Q. Find the area of the triangle OPQ. [3] Tangents and Normals The equation of the tangent and normal at any point ( )00,xy on a curve ( )fyx= is given by: Tangent ( )00y y m x x− = − where d d ym x= is the gradient of tangent at the point ( )00,xy Normal ( )00 1y y x x m− =− − Note: • Tangent parallel to x-axis d 0d y x= • Tangent parallel to y-axis d d y x is undefined, i.e. DENOMINATOR of d d y x is ZERO.

Term 4 Revision Topical Quick Check: Differentiation and its Applications TMJC 2024 Page 4 of 11 Solutions (i) cos 2 sin dd 2sin 2 cosdd d d d dd d cos 2sin 2 cos 4sin cos 1 4sin xy xy y y xx          == =− = = − = − =− (ii) (iii) 120 2y x y x+ =  =− Tangent parallel to 20yx+= means d1 d2 y x =− when p = . 11 4sin 2 1sin 2 6 p p p  − =− = = 11When , cos , sin6 3 2 6 2p x y  = = = = = Equation of tangent: 1 1 1 2 2 2 13 24 yx yx − =− −  =− +

Term 4 Revision Topical Quick Check: Differentiation and its Applications TMJC 2024 Page 5 of 11 (iv) 3When 0, 4 1 3 3When 0, 2 4 2 xy y x x == = =  = 33 ,0 and 0, 24PQ           1 3 3Area of Triangle 242 9 16 OPQ=   =

Term 4 Revision Topical Quick Check: Differentiation and its Applications TMJC 2024 Page 6 of 11 4 RVHS Promo 9758/2022/Q11(i)-(iii) To ease food security concerns, the government is building tents at the community garden plots to allow urban farming to take place more efficiently among residents. The organizing committee is building a tent consisting of two rectangular pieces on the ro of and two rectangular vertical sides as shown in the diagram below. The base of the tent has sides x m by y m, and covers a total floor area of 40 m2. The vertical sides of the tent are 4 m tall, and the roof adds another 0.01y2 m to the overall height of the tent. (Diagram not drawn to scale) (i) The total external surface area of the tent is denoted by A m2. Show that A is given by 2 168 40 1 25Ax x= + + . [3] (ii) Suppose that A has a stationary value at some x, show that x satisfies the equation 6425 16 256 0xx+ − = . [3] (iii) The design team decides that the material for the tent costs $3.10 per m 2, estimate the minimum total cost of the material for the whole tent. [3] Solutions (i) Floor area 4040xy y x=  = ( ) 2 22 2 22 2 42 22 2 Area 2(4 ) 2 0.01 ....(*) 2 1 40 4082 100 2 256 4008 2 ....(**) 400 168 2 1 25 168 40 1 25 yA x x y xx xx xx xx xx xx x x = + +   = +  +   = + +   = + +      = + + 4 m x m y m 0.01y2 m

Term 4 Revision Topical Quick Check: Differentiation and its Applications TMJC 2024 Page 7 of 11 (ii) 2 168 40 1 25Ax x= + + 1/ 2 23 d 1 16 328 40 1d 2 25 25 A x x x −     = + + −         3 2 d 1288d 1651 25 A x x x =− + Let d 0d A x = 3 2 12880 1651 25x x −= + 23 16 16125 5xx+= 26 16 256125 25xx+= 6425 16 256 0xx+ − = (Shown) (iii) Solving 6425 16 256 0xx+ − = using GC Rejecting all negative and complex roots. 1.4063x= Using GC, 2 2 d 15.67 0d A x = when x = 1.4063 Hence A is minimum when 1.4063x= Minimum Cost = ( ) ( ) ( ) 2 163.10 8 1.4063 40 1 25 1.4063   + +  = $177.53

Term 4 Revision Topical Quick Check: Differentiation and its Applications TMJC 2024 Page 8 of 11 Revision Guide Page 3 Let’s Try Now. 5 MJC Prelim 9740/2008/P1/Q13 A piece of wire of length d units is cut into two pieces. One piece is bent to form a circle of radius r units, and the other piece is bent to form a regular hexagon. Prove that, as r varies, the sum of the areas enclosed by the two shapes is a minimum when the radius of the circle is approximately 0.076d units. [7] Solutions Length of circle = 2πr Perimeter of hexagon = d − 2 πr ( ) 2 2 22 total area 12 πππ 6 sin 2 6 3 3π 2π 24 A drr r d r = −    =+          = + − ( )( ) ( ) d3 2π 2π 2πd 12 π32π 2π 6 A r d rr r d r = + − − = − − r Maxima / Minima 1. Draw a clear diagram with all the given information included. 2. Denote each changing quantity (weight, volume, radius, length etc.) by a variable. 3. Construct the equation(s) relating the variables. 4. Express quantity to be maximised/minimised in terms of a single variable, say x (if there are 2 variables, express 1 in terms of another). 5. Using differentiation, find the stationary point(s). 6. Use 1st or 2nd derivative test to determine/prove nature of the stationary point.

Term 4 Revision Topical Quick Check: Differentiation and its Applications TMJC 2024 Page 9 of 11 ( ) ( ) dAt stationary point, 0d π32π 2π 0 6 π 3 32 36 3 26 π3 A r r d r dr dr =  − − =  + = = + 22 2 d π32π0d3 A r = +  Alternative method (First Derivative) r ( ) 3 26 π3 d −    + ( ) 3 26 π3 d   + ( ) 3 26 π3 d −    + d d A r − 0 + Slope Thus, the sum of the areas enclosed by the two shapes is a minimum when the radius of the circle is approximately 0.076d units.

Term 4 Revision Topical Quick Check: Differentiation and its Applications TMJC 2024 Page 10 of 11 Revision Guide Page 3 Let’s Try Now. 6 JPJC Promo 9758/2022/Q5 The diagram shows a V -shaped tank with dimensions = 4 mL , = 0.6 mW and = 0.7 mH . The tank is initially empty. Water is pumped into the tank at a rate of 0.0025 m3/s. At any instant from the start of water flowing into the tank, the water in the tank has a depth of y m and a su