SAJC Chapter 6 Apps of Differentiation (Tutor) 2022

SAJC 2022 JC 2 H1 Mathematics Page 1 of 39 Chapter 6 (Pure Mathematics) Applications of Differentiation Objectives At the end of the chapter, you should be able to: (a) interpret f′(x) > 0, f′(x) = 0 and f′(x) < 0 graphically; (b) Determine the maximum and minimum points (local maxima a nd minima) and stationary points of inflexion analytically, in simple cases, using the first derivative test; [Note: The use of the second derivative test is not required.] (c) locate the maximum and minimum points with the help of a graphic calculator; (d) find and use the equations of tangents to a curve; (e) interpret the derivative as an instantaneous rate of change of a physical quantity, and use the concept to solve a variety of maxima and minima problems and problems involving connected rates of change. Content 6.1 Some Notation and Terminology 6.1.1 Gradient of a straight line 6.1.2 Equation of a straight line 6.1.3 Gradient of perpendicular lines 6.1.4 Gradient of a curve 6.2 Equation of Tangents to a curve 6.3 Increasing and Decreasing Functions 6.4 Stationary Points 6.4.1 Tests for types of Stationary Point 6.5 Maxima and Minima Problems 6.6 Rate of Change References New Syllabus Additional Mathematics (8th Edition), Shinglee Publishers Pte Ltd. Relevant Resources • http://mathinsite.bmth.ac.uk/applet/difffns/difffns.html (Applet to explore graphically the differentiation (1st and 2nd) of various functions)

SAJC 2022 JC 2 H1 Mathematics Page 2 of 39 Introduction There are many applications of differentiation in science and engineering. Differentiation is also used in analysis of finance and economics. One important application of differentiation is in the area of optimisation, which means finding the condition for a maximum (or minimum) to occur. This is important in business e.g cost reduction, profit increase and engineering e.g maximum strength, minimum cost. Example : Solar Two sustainable energy project in California A power tower produces electricity from sunlight by focusing thousand s of sun -tracking mirrors, called heliostats, on a single receiver sitting on top of a tower. The receiver captures the thermal energy of the sun and stores it in tanks of molten salt (to the right of the tower) at temperatures greater than 500 degrees centigrade. When electricity is needed, the energy in the molten salt is used to create steam, which drives a conventional electricity-generating turbine (to the left of the tower). Differentiation is used to maximise the efficiency of the process. Source: http://www.intmath.com/Calculus/Calculus-intro.php 6.1 Some Notation and Terminology y 2y 1y 1x 2x x 22( , )Q x y 11( , )P x y 0

SAJC 2022 JC 2 H1 Mathematics Page 3 of 39 6.1.1 Gradient of a straight line The gradient of a straight line from point 11( , )P x y to 22( , )Q x y is defined as 21 21 yy xx − − . i.e. The gradient of a straight line is the ratio of the change in y-coordinate to that of the x-coordinate between two points on the line. Note: 1. The gradient of a straight line is constant throughout. 2. There are 4 different types of straight lines. gradient 0= positive gradient gradient is undefined negative gradient 6.1.2 Equation of a straight line There are two methods of forming an equation of a straight line. Method 1 y mx c=+ where m is the gradient of the line and c is the y-intercept of the line. Method 2 ( )11y y m x x− = − where m is the gradient of the line and ( )11,xy is a point on the line. Note: Method 1 should be used when the gradient and y-intercept of the line are known and method 2 should be used when the gradient and one point on the line are known.

SAJC 2022 JC 2 H1 Mathematics Page 4 of 39 Example 1 Find the equation of line l that has (a) a gradient of 2 and passes the point ( )0,3 (b) a gradient of 3 and passes the point ( )3,5 Solution: (a) Equation of line l, 23yx=+ (b) Equation of line l, ( )5 3 3 3 9 5 34 yx yx yx − = − = − + =− 6.1.3 Gradient of perpendicular lines Product of the gradients of two lines that are perpendicular to each other is -1. i.e. Lines 11y m x c=+ and line 22y m x c=+ are perpendicular to each other if and only if 12 1mm =− Note: 1 2 1m m=− and 2 1 1m m=− Hence two lines are perpendicular to each other  the gradient of one is negative reciprocal of the other. Example 2 Find the equation of the line that is perpendicular to the line 23yx=+ and passes through the point ( )4,2 . Solution: Gradient of the line 1 2=− Equation of line l, ( )124 2 1 222 1 42 yx yx yx − =− − =− + + =− +

SAJC 2022 JC 2 H1 Mathematics Page 5 of 39 6.1.4 Gradient of a curve Given any curve ( )fyx= , The line touching the curve at point P is called the tangent to the curve at point P. i.e. line AB. The line passing through P and perpendicular to the tangent at P is known as the normal to the curve at point P. i.e. line CD. Note: 1. The gradient of a curve is not constant throughout. 2. The gradient of the curve at any point is defined as the gradient of the tangent to the curve at that point. Example 3 The curve 2 2y ax bx= + + has gradient of 2 at the point ( )1,2 , find the value of a and b. Solution: ( ) 2 21y ax bx= + + substitute ( )1,2 into (1), ( ) 22 02 ab ab = + + += 2dy ax bdx =+ 1 2x dy dx = = ( ) ( ) 2 1 2 2 2 3 ab ab += += (3) – (2), 2a= 2b =− A B C D ( )fyx= P A C

SAJC 2022 JC 2 H1 Mathematics Page 6 of 39 6.2 Equation of Tangents to a curve For any curve ( )fyx= , gradient at point ( , )ab is ( )f' a or d d xa y x = . Equation of tangent at ( , )ab is ( )( )f'y b a x a− = − Example 4 Find the equation of the tangent of the curve 22 5 6y x x= − + at the point where it crosses the y-axis. Solution: 22 5 6 d 45d y x x y xx = − + =− When the curve crosses the y-axis, 0x= . 6y= . At point ( )0,6 , Gradient of tangent ( ) 0 d d 4 0 5 5 x y x == =− =− Equation of tangent at ( )0,6 , 56yx=− + Normal Tangent x y 0 f ( )yx= ( ), ab

SAJC 2022 JC 2 H1 Mathematics Page 7 of 39 6.2.1 Using GC to Draw a Tangent Line in the Function Mode We can verify the answer to Example 4 using GC as shown below. Step Screenshot Note 1. Enter the equation of graph of 2 1Y = 2 5 6xx−+ 2. Press p and change window setting 3. Press s 4. Press y¼< 5. Key in ‘0’ (since 0 is the value of x where tangent is to be drawn) By default, the value of x is 0. You can enter other value of your choice.

SAJC 2022 JC 2 H1 Mathematics Page 8 of 39 6. Press Í The tangent line and its equation are displayed. From GC, the equation of tangent at ( )0,6 is 56yx=− + . Exercise 1 1. The equation of a curve is .12 xxy += Find the equation of the tangent to the curve at .2=x [Ans: 7 14yx=+ ] Solution: 2 12dy dx x=− At 2x= , ( ) 1922 22y = + = , 2 172 42 dy dx = − = . Gradient of tangent when 2x= is 7 4 Equation of tangent when 2x= , ( )97 224 7 7 9 4 2 2 7 14 yx yx x − = − = − + =+ 2. Find the equation of the tangent to the curve 3)2( −= xy at the point ( )3,1 . Calculate the coordinates of the point where this tangent meets the curve again. [Ans: 38yx=− , ( )0, 8− ] Solution:

SAJC 2022 JC 2 H1 Mathematics Page 9 of 39 ( ) 2 32dy xdx =− . At )1,3( , ( ) 2 3 3 2 3dy dx = − = . Equation of the tangent at )1,3( , ( )1 3 3 38 yx yx − = − =− When tangent meets curve, ( ) ( ) 3 32 2 3 8 2 6 9 0 6 9 0 xx x x x x x x − = − − + = − + = ( ) 2 30 0 or 3 ( ) xx x x