RI Applications of Differerentiation Solns

RAFFLES INSTITUTION H2 Mathematics 9758 2023 Year 6 Term 3 Revision 8 (Summary and Tutorial) Topic: Applications of Differentiation Summary for Applications of Differentiation 1 Strictly Increasing/Strictly Decreasing & Concavity IF Then f is over the interval (,)ab f( ) 0x  Strictly increasing f( ) 0x  Strictly decreasing f() 0x  Concave up f() 0x  Concave down To show “ 0 ” or “ 0 ”, we can complete the square of the expression if it is quadratic otherwise we can try to “build up the expression” from the given interval. For example to show  1f2 0x x   for 1 2x . We can rewrite the expression:  21f xx x   So the strategy here is to try to examine the signs the numerator and denominator from the given condition of 1 2x . 2 Methods for determining the nature of a stationary point Method 1: Second Derivative Test IF at xk Then  , f( )kk is f() 0x  a minimum turning point f() 0x  a maximum turning point Note that if f() 0x  then there is no conclusion about the nature of the stationary point. The first derivative test must then be used. If 1 2x , then the denominator must be “ 0 ” If 1 2x , then 21x which implies that 21 0x Therefore if the numerator is “ 0 ” and the denominator is also “ 0 ”, we can conclude that for 1 2x , f0 x 

Method 2: First Derivative Test To determine the nature of the stationary point at  , f( )kk , we need to check the signs of f( )x for x k and x k . Important note: you need to fully factorise f( )x where possible BEFORE discussing the signs of f( )x . Example: 53f( ) 2 5x xx . It is not sufficient to just differentiate and get 42f( ) 1 0 1 5x xx  . You need to factorise further to get 2 33f( ) 1 0 22xx x x        , BEFORE discussing stationary points at (0,0) , 33,322   and 33,322  making reference to the table below. x kk k kk k kk k f( )x ve 0 ve ve 0 ve ve 0 ve or ve 0 ve Nature of Stationary Point Maximum Minimum Stationary point of inflexion 3 TANGENTS AND NORMALS To find the equations of the tangent or normal at a point  , f( )kk on a curve, we need: Equation of a (straight) line f( ) ( )yk m x k   with gradient m passing through  , f( )kk .  Obtain m by finding f( )k which will the gradient of the tangent at  , f( )kk .  If the question needs equation of the normal at  , f( )kk , gradient of the normal at  , f( )kk is 1 f( )k  . When gradient at a point on a curve f( )x is parallel to the x-axis f( ) 0x  (Usually mean that the numerator of f( ) 0x  ) y-axis f( )x is undefined. (Usually mean that the denominator of f( ) 0x  )

4 MAXIMIZATION AND MINIMIZATION Some guidelines to solve problems involving maximization and minimization :  Denote each changing quantity by a variable.  Write down a formula for the quantity to be maximized or minimized.  Express the quantity to be maximized or minimized in terms of 1 variable only.  Differentiate and equate derivative to zero for stationary values.  Justify if quantity is a maximum or minimum (using 2nd or 1st derivative test).  Answer the question. Some strategies 1. Identify any right angle triangle by drawing a diagram. Use Pythagoras theorem to relate the two variables. 2. Identify similar triangles to relate the two variables. 3. Read questions carefully to identify the constant(s) used in the questions. That will reduce confusion on what variable to differentiate with respect to. 5 CONNECTED RATES OF CHANGE Some guidelines to solve questions involving rate of change:  Denote each changing quantity by a variable.  Find the equations relating the variables.  Use the chain rule to link up the derivatives.  Write down the values of the variables and the given rates of change.  Solve for the unknown rate. Some strategies 1. Write down the rate of change that is given in the question. Use the units to guide you. For example 3- 1ms is the rate of change of volume per unit second. 2. Write down the rate of change required by the question and use the chain rule to link up the derivative. For example to find d d V t and you are given d d V r and d d r t . Then using chain rule, we can get dd d dd d VV r tr t .

6 MACLAURIN SERIES Reminder: the formulas are found in MF26. If xf can be expanded as a power series for a given range of x including zero, then (3) ( ) 23f (0) f (0) f (0)f( ) f( 0 ) f ( 0 ) 2! 3! ! n nx= + x + x x x n     [in MF26] Binomial Series 2( 1) ( 1)...( 1)(1 ) 1 ... ... 2! !        n rnn nn n rxn x x x r The binomial expansion is valid for x  1, i.e. x 11 . [in MF26] N o t e t h a t w e n e e d t o h a v e “ 1 ” b e f o r e a p p l y i n g t h e f o r m u l a , s o i f w e n e e d t o e x p a n d 1 2(3 ) ,x we can do the following depending on what is required.  For ascending powers of x , we rewrite 1 11 2 22(3 ) 3 1 3 xx    before applying the binomial expansion. In this case, the expansion is valid for 13 x  (i.e. small x )  For descending powers of x , we rewrite 1 11 1 2 22 2 3(3 ) ( 3) 1xx x x     before applying the binomial expansion. In this case, the expansion is valid for 3 1x  (i.e. large x )

Small angle approximations Note that  If angle x is small, addition or subtraction to angle x may not remain small. i.e  sin x ax a  . We need to use addition formula to simplify the expression before applying the small angle approximations. o For example: sin sin cos cos sin44 4xx x    211 1 222 xx         If x is small, multiple of x is still small. o For example:   2 2cos 2 1 2 xx  For all small angles, positive or negative, we have (1) sin x x (2) 2 cos 1 2 xx (3) tan x x , where x is measured in radians.

Revision Tutorial Questions Source of Question: VJC/Promo/2018/01/Q13 1 [It is given that a sphere of radius r has surface area 24 r and volume 34 .3 r ] A touchscreen pen is made up of three parts.  The head is modelled by the curved surface of a hemisphere of radius r mm.  The body is modelled by the curved surface of a cylinder of radius r mm and length s mm.  The base is modelled by a circular disc of radius r mm. The three parts are joined together as shown in the diagram. The model is made of material of negligible thickness. (i) It is given that the volume of the model is a fixed value 3 mm ,k and the external surface area is a minimum. Use differentiation to find the values of r and s in terms of k, simplifying your answers. [7] (ii) It is given instead that the volume of the model is 31270 mm and its external surface area is 21290 mm . Show that there is only one possible value of r and find this value. [5] Solution: 1(i) [7] Volume, 23 2 22 33 krkr s r s r    External surface area is given by 22 2 2 2 22 223 3 25 3 Ar s r r krrr r k rr           Let 2 d1 0 20, 0d3 Ak rrr   3 3 10 23 3 5 rk kr     r s

23 3 2 3 2 3 3 2 3 32 3 53 5 33 55 35 3 53 5 kr s r kk s kk s kks k                     (ii) [5] Method 1 1270, 1290kA 21290 2 3 rs r  (1) 23 21270 3rs r  (2) From (1), 21290 3 2 rs r    Substitute into (2), 2 23 1290 3 21270 23 rrr r      35 645 1270 06 rr   Using GC, r = 2.0015, 14.599 or -16.601 Since 0r , r = -16.6 is rejected If 2.0015r ,   2 2 2.0015 99.57332.0015 ks    If 14.599r , 7.8364 s Hence 14.599r is rejected since s cannot be negative.