CALCULUS APPLICATIONS OF DIFFERENTIATION LESSON PLAN. C3 Topic Overview

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1 CALCULUS C3 Topic Overview C3 APPLICATIONS OF DIFFERENTIATION Differentiation can be used to investigate the behaviour of a function, to find regions where the value of a function is increasing or decreasing and to find maimum and minimum values. illustrates how the technique of differentiation is applied to solving the problem of determining the maimum and minimum values of polynomial functions. assumes you know the basic rules of differentiation given in Topic C and can solve quadratic equations [Ref: Algebra A3 Quadratic Equations]. LESSON PLAN Lesson No. LESSON TITLE Page Eamples SAQs 1 C3.1 Maimum and Minimum Values C3.1 C3. Maimum and Minimum Values C3. Appendi C3.3 Solutions to the Self-Assessment Questions 11 Page 1

2 CALCULUS: DIFFERENTIATION C3.1 Maimum and Minimum Values 1 C3.1 MAXIMUM AND MINIMUM VALUES There are some situations where we need to investigate maimum or minimum values of a particular quantity. For eample, we might wish to find: the load resistance for which the power dissipated is a maimum in an electrical circuit the point at which the bending moment of a cantilever beam is a maimum in a mechanical system the dimensions of a cylindrical hot water tank for which heat loss is a minimum. In these applications, we may be able to graph one variable against the other over a limited range of values and then find the point on the graph at which a maimum or minimum point occurs. However, in certain situations, we can also use calculus techniques to find the eact maimum or minimum values. STATIONARY POINTS A stationary point on a curve is a point at which the gradient is zero, i.e. the tangent line is horizontal at the point. The graph may have a local maimum, local minimum or a point of inflection at a stationary point. In Topic C1.3., the derivative f '() of a function was defined as the gradient of the curve y = f() at a particular value of. Hence, a given point (a, f(a)) on the curve y = f() is a stationary point if f '(a) = 0. A. MAXIMUM TURNING POINT A stationary point is a local maimum turning point if the gradient f '() changes sign from positive to negative at = a. f() Positive Negative a a a a Sign of f '() 0 a a means a value just less than = a means a value just more than = a Page

3 CALCULUS: DIFFERENTIATION C3.1 Maimum and Minimum Values 1 B. MINIMUM TURNING POINT A stationary point is a local minimum turning point if the gradient f '() changes sign from negative to positive at = a. f() Negative Positive a a a a Sign of f '() 0 C. POINT OF INFLECTION A stationary point is a point of inflection if the gradient f '() does not change sign at = a. f() Positive Positive a a a a Sign of f '() 0 or f() Negative Negative a a a a Sign of f '() 0 Page 3

4 CALCULUS: DIFFERENTIATION C3.1 Maimum and Minimum Values 1 SUMMARY Stationary points on a curve occur where f '() = 0. The nature of the stationary points can be deduced by observing changes in the sign of f '() in the vicinity of the stationary points or by eamining the sign of f "() at the stationary points. This is eplained below: The Sign of the First Derivative f '() The first derivative f '() measures the rate of change of f () with respect to. A f '() = 0 B Slope: f '() > 0 f() is increasing f '() < 0 f() is decreasing Turning points (f '() = 0): A is a local maimum B is a local minimum The Sign of the Second Derivative f "() The second derivative f "() measures the rate of change of f'() with respect to. At a maimum turning point, the gradient is decreasing since it changes from ve just before the turning point to ve just after the turning point. At a minimum turning point, the gradient is increasing since it changes from ve just before the turning point to ve just after the turning point. Hence, at a stationary point: f "() < 0 Maimum turning point f "() > 0 Minimum turning point Unfortunately, the use of the second derivative to determine the nature of a stationary point does not help if f "() = 0 at the stationary point. In such cases, the point can be a maimum, minimum or a point of inflection. If f "() = 0 at a stationary point, then it is necessary to investigate the changes of sign in the first derivative f '() to determine the nature of the stationary point (as shown in Eample 3.3). Page 4

5 CALCULUS: DIFFERENTIATION C3.1 Maimum and Minimum Values 1 EXAMPLE 3.1 Find the position and nature of the stationary point on the graph of the function y = 4. Hence sketch the curve. Solution: Step 1: Find the stationary point Differentiate: f() = 4 f ' () = f " () = At the stationary point(s): f ' () = 0 = 0 = 0 When = 0, y = 4 = 4 Thus, there is one stationary point at (0, 4) on the curve y = 4 Step : Determine the nature of the stationary point From Step 1, f " () = At the stationary point, where = 0 f " (0) = < 0 Local Maimum The sign of the second derivative is negative, hence (0, 4) is a maimum turning point. Step 3: Sketch the curve Before sketching the graph of a function, first determine certain critical values. These are: the points where the graph cuts the ais and the yais the values of y = f() over a range of values of the stationary points (maimum, minimum, points of inflection) the slope of the curve between and beyond the stationary point(s). On the ais, y = 0 4 = 0 = 4 = ± Hence, the graph cuts the ais at points (, 0) and (, 0) On the yais, = 0 y = 4 = 4 0 = 4 Hence, the graph cuts the yais at the point (0, 4) Page 5

6 CALCULUS: DIFFERENTIATION C3.1 Maimum and Minimum Values 1 Tabulate the values of f() = 4 over the range = 4 to = 4. Also determine the sign of f '() at these values (the numerical values of f '() are not required). Observe that the sign of f '() changes from ve just before and ve just after the stationary point at (0, 4). This confirms that point (0,4) is a maimum turning point f() = f '() = 0 Slope of the curve Maimum Turning Point Graph of f() = 4 f() SAQ C3.1 Find the position and nature of the stationary point on the graph of the following functions. Hence sketch the curve. (a) y = 4 (b) y = 4 1 Page 6

7 CALCULUS: DIFFERENTIATION C3. Maimum and Minimum Values C3. MAXIMUM AND MINIMUM VALUES The following eamples involve cubic functions and require more work to determine the stationary point(s). EXAMPLE 3. Find the position and nature of the stationary point(s) on the graph of the function y = (1 )(1 ). Hence sketch the curve. Solution: Step 1: Find the stationary point(s) Differentiate: f() = (1 )(1 ) At the stationary point(s): = (1 ) = 3 f '() = 1 3 f " () = 6 f'() = = ± When = 0.577, y = 3 = When = 0.577, y = 3 = = 0 = 1 = Thus, there are stationary points at (0.58, 0.38) and (0.58, 0.38) 1 3 Step : Determine the nature of the stationary points From Step 1, f " () = 6 At = 0.58 f " (0.58) = 3.48 < 0 Local maimum At = 0.58 f " (0.58) = 3.48 > 0 Local minimum Hence, there is a local maimum turning point at (0.58, 0.38) and a local minimum turning point at (0.58, 0.38) Page 7

8 CALCULUS: DIFFERENTIATION C3. Maimum and Minimum Values Step 3: Sketch the curve First, find the points where the graph cuts the ais and the yais. On the ais, y = 0 (1 )(1 ) = 0 Solving this equation gives = 0, 1, 1 Hence, the graph cuts the ais at points (1, 0), (0, 0) and (1, 0) On the yais, = 0 y = (1 )(1 ) = 0 Hence, the graph cuts the yais at the point (0,0) Tabulate the values of f() = (1 )(1 ) over the range = 1.5 to = 1.5. Observe changes in the sign of f ' () which confirm that there is a minimum turning point at (0.58,0.38) and a maimum turning point at (0.58, 0.38) f() = (1)(1) f '() = Slope of the curve Minimum Turning Point Maimum Turning Point Graph of f()= (1 )(1 ) f() Page 8

9 CALCULUS: DIFFERENTIATION C3. Maimum and Minimum Values EXAMPLE 3.3 Find the position and nature of the stationary point(s) on the graph of the function y = ( 1)( 5 7). Hence sketch the curve. Solution: Step 1: Find the stationary point(s) Differentiate: f() = ( 1)( 5 7) = f '() = f " () = 6 1 At the stationary point(s): f '() = = = 0 ( ) = 0 = [Ref: A3.1 Factorisation Method] When =, y = ( 1)( 5 7) = ( 1)( 5() 7) = 1 Thus, there is a stationary point at (, 1) Step : Determine the nature of the stationary point From Step 1, f " () = 6 1 At = f "() = 6() 1 = 0 Inconclusive Since the sign of the second derivative is zero at the point (, 1), it is necessary to investigate the changes of sign in the first derivative f '() to determine the nature of the stationary point. Tabulate the values of f() = ( 1)( 5 7) over the range = 0 to = 4. Observe that the sign of f ' () indicates that there is a point of inflection at (, 1) f() = ( 1)( 5 7) f'() = Slope of the curve Point of Inflection Page 9

10 CALCULUS: DIFFERENTIATION C3. Maimum and Minimum Values Step 3: Sketch the curve Determine points where the graph cuts and ais and the yais. On the ais, y = 0 ( 1)( 5 7) = 0 1 = 0 or 5 7 = 0 The discriminant of the quadratic 5 7 is b 4ac = 5 8 = 3 < 0 and so 5 7 = 0 has no real roots. [Ref: A3. Formula Method (The Discriminant)] Thus, the graph cuts the ais at one point (1, 0) On the yais, = 0 y = ( 1)( 5 7) = 7 Thus, the graph cuts the yais at the point (0, 7) Graph of f() = ( 1)( 5 7) f() SAQ C3. Find the position and nature of the stationary point(s) on the graph of the following functions. Hence sketch the curve. (a) y = ( 1)( ) (b) y = ( 1)( ) (c) y = 3 8 Page 10

11 CALCULUS: DIFFERENTIATION C3.3 APPENDIX C3.3 SOLUTIONS TO SELF-ASSESSMENT QUESTIONS SAQ C3.1 (a) f() = 4 At the stationary points: f ' () = 4= 0 =, y = 4 At =, f " () = > 0 Local minimum at (, 4) On the ais, y = 4 = ( 4 ) = 0 = 0 or = 4 On the yais, y = 4 = 0 Graph of y = 4 (b) f() = 4 1 At the stationary points: f ' () = 4= 0 =, y = 3 At =, f " () = < 0 Local maimum at (, 3) On the ais 4 1 = = 0 Using the quadratic roots formula: ± = (1) 4(1)(1) 4 ± 1 = = 3.73 or 0.68 Thus, the graph cuts the ais at points (3.73, 0) and (0.7, 0) and on the yais, y = 4 1 = 1 4 [Ref: Algebra A3.] Graph of y = Page 11

12 CALCULUS: DIFFERENTIATION C3.3 APPENDIX SAQ C3. (a) f() = (1)() = 3 At the stationary points: f '() = 3 = 0 ± 4 4 = 6 = 0.55 or 1. y = 0.63 or.11 using quadratics roots formula At = 0.55, f " () = 6 = > 0 Local minimum at (0.55, 0.63) At = 1., f " () = 6 = < 0 Local maimum at (1.,.11) On the ais, y = (1)() = 0 = 0, 1, Graph of y = (1)() (b) f() = (1)() = At the stationary points: f '() = 3 6 = 0 3( ) = 0 = 0, y = 4, 0 At = 0, f " () = 6 6 = > 0 Local minimum at (0, 4) At =, f " () = 6 6 = < 0 Local maimum at (, 0) On the ais, y = (1)() = 0 = 1, On the yais, y = (1)() = 4 Graph of y = (1)() Page 1

13 CALCULUS: DIFFERENTIATION C3.3 APPENDIX (c) f() = 3 8 At the stationary points: f '() = 3 = 0 = 0, y = 8 At = 0, f " () = 6 = 0 Inconclusive Eamine sign of f '() = 3 Table of values of f() = 3 8 over the range = 3 to = f() = f '() = 3 0 Slope of the curve Point of Inflection On the ais, y = 3 8 = 0 3 = 8 = On the yais, y = 8 Graph of y = Page 13

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1 x Unit 1. Calculus Topic 4: Increasing and decreasing functions: turning points In topic 4 we continue with straightforward derivatives and integrals: Locate turning points where f () = 0. Determine the