{"id":3161,"date":"2023-10-04T12:13:19","date_gmt":"2023-10-04T12:13:19","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3161"},"modified":"2023-10-04T12:23:00","modified_gmt":"2023-10-04T12:23:00","slug":"week-4-ss2-second-term-further-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-4-ss2-second-term-further-mathematics-notes\/","title":{"rendered":"Week 4 – SS2 Second Term Further Mathematics Notes"},"content":{"rendered":"

\u00a0WEEK 4
\n<\/strong>TOPIC: APPLICATION OF DIFFERENTIATION ; RATE OF CHANGE, EQUATION OF MOTION , MAXIMUM AND MINIMUM POINTS AND VALUES
\n<\/strong>Rate of Change
\n<\/strong>If y = f<\/em>(x), can sometimes be interpreted as the rate at which y is changing with respect to x. if y<\/em> increases as x increases, 0, while if y<\/em> decreases as x increases, 0.
\nExample <\/strong>
\n\t\t The radius of a circle is increases a t the rate of 0.01cm\/s. find the rate at which the area is increasing when the radius of the circle is 5cm.
\nSolution
\n<\/strong>Let A<\/em>be the area of the circle of radius r<\/em>.
\nA<\/em> = 2
\n<\/sup> = 2
\nBy the chain rule:
\n =
\n = 0.01cm\/s
\nWhen r = 5cm
\n = 25 0.01
\n = 10 0.01
\n = 0.1 2<\/sup>\/s
\n = 0.3142cm2<\/sup>\/s
\nExample
\n<\/strong>Water is leaking from a hemisphere bowl of radius 20cm at the rate of 0.5cm3<\/sup>\/s. Find the rate at which the surface area of the water is decreasing when the water level is half-way from the top.
\nSolution
\n<\/strong>\"\"\/<\/p>\n

\u00a0
\n\u00a0
\n\u00a0Fig. 10 9<\/strong>
\n\t\tLet A<\/em>be the surface area of the water in the hemisphere bowl of radius r<\/em>, thenA<\/em> =r2
\n<\/sup>= = 2r
\n= =
\nV <\/em>= r 3<\/sup>
\n\t\t = 22<\/sup>
\n\t\t= =
\n =
\n\"\"\/\"\"\/ = = =
\n =
\n\t\t\t\t<\/strong>When the water level is half- way from the top r = 10cm.
\nBut = -0.5 cm3<\/sup>\/s
\n = -0.5 cm3<\/sup>\/s
\n\t\t\t= -0.5 cm2<\/sup>\/s
\nExample
\n\t\t\t\t<\/strong>Find the rate at which the volume of a spherical balloon is increasing if the surface area is increasing at the rate of 5cm2<\/sup>\/s when the radius of the spherical balloon is 4cm.
\nSolution<\/strong>
\n\t\tV<\/em>= 3
\n<\/sup> = 42
\n<\/sup> s = 42<\/sup>
\n\t\t = 8 = <\/p>\n

\u00a0MAXIMUM AND MINIMUM POINTS <\/strong>The points on a curve at which = 0, are called Stationary Points<\/strong>.
\n Stationary points fall into three major categories:<\/p>\n

    \n
  1. Those in which changes sign from positives through zero to negative. These are called maximum points.<\/strong>\n\t\t\t\t<\/li>\n
  2. Those in which changes sign from negative through zero to positive. These are called minimum points.<\/strong>\n\t\t\t\t<\/li>\n
  3. \n
    Those in which the sign of is not changed in the immediate neighborhood of the stationary points. These are called points of inflexion.<\/strong>\n\t\t\t\t\t<\/div>\n

    The terms maximum and minimum points are used in the local sense and not in the absolute sense.<\/p>\n

    \u00a0<\/li>\n<\/ol>\n

    Maximum Points
    \n<\/strong>\"\"\/
    \n\t\t\t<\/strong>
    \n\u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0Fig.10. 11 shows part of the curve y = f (x). There is a maximum at the point where x = a
    \n= \"\"\/f<\/em>‘<\/sup>(a) = 0
    \nLet us denote a point at the immediate neighborhood of a to the left by a and a pint at the immediate neighborhood of a to the right by a+<\/sup> then:
    \n At x = a, > 0
    \n\"\"\/f‘<\/sup><\/em>(a–<\/sup>) > 0
    \n At x = a+<\/sup>, < 0
    \nf<\/em>‘<\/sup> (a+<\/sup>) < 0
    \nHence for the existence of a minimum at x = a, three conditions must be satisfied:<\/p>\n

      \n
    1. \n
      f<\/em>‘<\/sup>(a) = 0 (ii) f<\/em>‘<\/sup> (a) >0\n<\/div>\n
        \n
      1. f<\/em>‘<\/sup> (a+<\/sup>)< 0\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

        Minimum Points\"\"\/
        \n\t\t\t<\/strong>
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0Fig. 10.12 shows a part of the curve y = f<\/em> (x). There is a minimum at x = b<\/em>
        \n\t\tAt x = b, f1 (b) =0
        \nAt x = b–<\/sup>, f1<\/sup> (b–<\/sup>) <0
        \nAt x = b+, <\/sup>f1<\/sup> (b+<\/sup>) <0
        \nSo a pint on a curve is a minimum at x = a.
        \nIf: (i) f<\/em>‘<\/sup> (a) = 0 (ii) f<\/em>‘ <\/sup>(a–<\/sup>) <0<\/p>\n

          \n
        1. f<\/em>‘<\/sup>(a+<\/sup>)>0\n<\/li>\n<\/ol>\n

          \"\"\/Points of Inflexion
          \n<\/strong>\"\"\/<\/p>\n

          \u00a0
          \n\u00a0
          \n\u00a0
          \n\u00a0
          \n\u00a0(a)
          \n<\/strong>–<\/sup>
          \n\t\tshows part of the curve y= f(x). We observe that
          \nf<\/em>‘<\/sup>(c) = 0
          \nf<\/em>‘<\/sup>(c–<\/sup>) < 0
          \nf<\/em>‘<\/sup>(c+<\/sup>) < 0
          \nThe point x = c<\/em> is a point of inflexion. Similarly, fig. 10.13(b) shows parts of the curve y = f<\/em> (x)
          \nWe observe that
          \nf<\/em>‘<\/sup>(d) = 0
          \nf<\/em>‘<\/sup>(d–<\/sup>) >0
          \nf ‘<\/em>(d+<\/sup>) >0
          \nThe point x = d<\/em> is a point of inflexion.
          \nA maximum point, a minimum point and a point of inflexion are all stationary points. Both maximum and minimum points are called turning points. A point of inflexion however, is not a turning point.<\/p>\n

          \u00a0Example
          \n<\/strong>Find the stationary points in each of the following curves whose equations are:<\/p>\n

            \n
          1. y = x3<\/sup>+ x2 <\/sup>-3x + 4\n<\/li>\n<\/ol>\n

            3<\/p>\n

              \n
            1. \n
              y = x4<\/sup> +4 x3<\/sup> \u2013 2x2<\/sup> \u2013 16x + 1\n<\/div>\n

              4 3\n<\/li>\n<\/ol>\n

              \u00a0Solution
              \n<\/strong><\/p>\n

                \n
              1. \n
                y = x3<\/sup> + x2<\/sup>-3x + 4\n<\/div>\n

                3
                \n = x2+ 2x \u2013 3
                \n = (x \u2013 1) (x + 3)
                \nAt the stationary points, = 0
                \n\"\"\/ (x \u2013 1) (x + 3) = 0
                \n x =1 or x = -3
                \nHence there are stationary points at x = 1 and x = -3
                \n y = x4<\/sup>+ 4 x3 <\/sup>-2x2 <\/sup>\u2013 16x + 1
                \n 4 3
                \n = x3 <\/sup>+ 4x2<\/sup> \u2013 4x \u2013 16
                \n = (x -2) (x + 2) (x + 4)
                \nAt the stationary points, = 0.
                \n (x \u2013 2) (x + 2) (x + 4) = 0
                \n x =2 or x= -2 or x = -4
                \nHence there are stationary points at x = 2, x = -2 and x = -4
                \nExample
                \n<\/strong>Find the turning points on the curve y = x4<\/sup> +
                \n 2
                \n5 x3<\/sup> -2x2 <\/sup>– 3x + 1 and distinguish between them
                \n3
                \nSolution
                \n<\/strong>y =x4<\/sup>+ 5 x3 <\/sup>\u2013 2x2 <\/sup>\u2013 3x + 1
                \n 2 3
                \n = 2x3 <\/sup>+ 5x2<\/sup> -4x \u2013 3
                \n = (2x + 1) (x-1) (x+3)
                \nAt the stationary points, =0
                \n(2x + 1) (x \u2013 1) (x + 3) = 0
                \n\"\"\/x<\/em> =, x = 1 and x<\/em> = -3 are the x-
                \nCoordinates of the stationary points.
                \n Let f<\/em>(x) = x4<\/sup>+ 5 x3<\/sup> -2x2<\/sup> \u2013 3x + 1
                \n 2 3
                \nf‘<\/sup>(x)<\/em>= 2x3 <\/sup> + 5x2 <\/sup> – 4x \u2013 3
                \n = (2x + 1) (x \u2013 1) (x + 3)
                \nLet a = = -0.5, a =-5, a+<\/sup> = -0.4
                \nThen f<\/em>‘<\/sup>(a) = 0
                \nf‘<\/sup><\/em> (a) = (-1.2+1) (-0.6 \u2013 1) (-0.6 + 3)
                \n = (-0.2) (-1.6) (2.4)>0
                \n = f1<\/sup> (a) > 0
                \n\"\"\/f ‘<\/em>(a+<\/sup>) = (-0.8 + 1) (-0.4 \u2013 1) (-0.4 + 3)
                \n = (0.2) (-1.4) (2.6) <0
                \n\"\"\/f <\/em>‘ (a+<\/sup>) < 0<\/p>\n

                \u00a0Table <\/p>\n

                \n\n\n\n\n
                \u00a0<\/td>\nf‘<\/sup>(a–<\/sup>)<\/td>\nf‘<\/sup>(a)<\/td>\nf‘<\/sup>(a+<\/sup>)<\/td>\n<\/tr>\n
                Sign<\/td>\n+<\/strong>ve
                \n \/<\/td>\n                		0
                \n —<\/td>\n                		–<\/strong>ve
                \n \\<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

                <\/strong>Hence, there is a maximum point at x =
                \nAt x = 1
                \nLet a = 1, a–<\/sup> = 0.9, a+<\/sup> = 1.1\n<\/li>\n<\/ol>\n

                f'(a–<\/sup>) = 0<\/em>
                \n\t\tf<\/em>‘ (a–<\/sup>) = (1.8 + 1) (0.9 -1) (0.9 + 3) < 0
                \nf’ <\/em>(a+<\/sup>) = (2.2 + 1) (1.1 \u2013 1) (1.1 + 4) > 0
                \nTable <\/p>\n

                \n\n\n\n\n
                \u00a0<\/td>\nf‘<\/sup> (a–<\/sup>)<\/td>\nf‘<\/sup>(a)<\/td>\nf ‘<\/sup>(a+<\/sup>)<\/td>\n<\/tr>\n
                Sign<\/td>\n-ve
                \n<\/strong> \/<\/strong><\/td>\n                		0
                \n<\/strong>–<\/strong><\/td>\n                		+ve
                \n<\/strong>\\<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

                <\/strong>Hence, there is minimum point at x = 1.
                \nAt x = -3
                \nPut a = -3, a– <\/sup>=-3.1; a+<\/sup>= -2.9
                \nf<\/em> ‘ (a) = 0
                \nf<\/em>‘ (a–<\/sup>) = (-6.2 + 1) (-3.1 \u2013 1) (-3.1 + 3) < 0
                \nf<\/em> ‘ (a+<\/sup>) = (-5.8 + 1) (-2.9 \u2013 1) 9-2.9 + 3) > 0
                \nTable <\/p>\n

                \n\n\n\n\n
                \u00a0<\/td>\nf’ (a–<\/sup>)<\/td>\nf'(a)<\/td>\nf'(a+<\/sup>)<\/td>\n<\/tr>\n
                Sign<\/td>\n–<\/strong>ve
                \n\t\t\t\t\t\t\t\t\t\t\t\t<\/strong>\/<\/strong><\/td>\n                		0
                \n–<\/strong><\/td>\n                		+<\/strong>ve
                \n\t\t\t\t\t\t\t\t\t\t\t\t<\/strong>\\<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

                <\/strong>Hence, x = -3 is a minimum point
                \nExample
                \n<\/strong> Find the stationary points on the curve y = x3<\/sup> \u2013 6x2<\/sup> + 12x \u2013 8 and distinguish between them.
                \nSolution
                \n<\/strong> y = x3<\/sup> \u2013 6x2<\/sup> + 12x \u2013 8
                \n = 3x2<\/sup> \u2013 12x = 12
                \n = 3(x2<\/sup> \u2013 4x + 4)
                \n = 3(x \u2013 2)2
                \n<\/sup>At the stationary points, = 0
                \nPut a = 2, a–<\/sup> = 1.9, a+<\/sup> = 2.1
                \nand let f<\/em>‘(x) = x3<\/sup> \u2013 6x2<\/sup> + 12x \u2013 8
                \nf ‘<\/em>(x) =3(x2 <\/sup>– 12x + 12)
                \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/em> = 3(x2 –<\/sup> 4x + 4)
                \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = (3x -2)2
                \n<\/sup>\u00a0\u00a0\u00a0\u00a0<\/sup>f ‘<\/em>(a) = 0
                \nf ‘<\/em>(a–<\/sup>) = 3(1.9 \u2013 2)2 <\/sup>> 0
                \nf ‘<\/em> (a+<\/sup>) = 3(2.1 \u2013 2)2 <\/sup>> 0
                \nTable <\/p>\n

                \n\n\n\n\n
                \u00a0<\/td>\nf’ (a–<\/sup>)<\/td>\nf'(a)<\/td>\nf'(a+<\/sup>)<\/td>\n<\/tr>\n
                Sign<\/td>\n–<\/strong>ve
                \n\t\t\t\t\t\t\t\t\t\t\t\t\t<\/strong>\/<\/strong><\/td>\n                		0
                \n–<\/strong><\/td>\n                		+<\/strong>ve
                \n\t\t\t\t\t\t\t\t\t\t\t\t\t<\/strong>\\<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

                <\/strong>Hence, there is a point of inflexion at x, =2
                \nWhen x = 2
                \ny = 8 \u2013 24 + 24 \u2013 8 = 0
                \n= \"\"\/ The point (2, 0) is a point of inflexion on the curve y = x3<\/sup> \u2013 6x2 <\/sup> + 12 \u2013 8
                \nThe second derivative test for stationary points
                \n<\/strong> We recall that a necessary condition for the existence of stationary points for the curve.
                \n\"\"\/ y = f (x) is that =. This condition is however not sufficient to determine the nature of the stationary points. We shall consider an alternative method which enables us to distinguish between the natures of stationary points.
                \n\"\"\/<\/p>\n

                  \n
                1. \"\"\/\n\t\t\t\t\t\t<\/li>\n<\/ol>\n
                    \n
                  1. \n\t\t\t\t\t<\/li>\n<\/ol>\n

                    \u00a0
                    \n\u00a0
                    \n\u00a0
                    \n\u00a0EVALUATION
                    \n<\/strong>1)<\/strong> Find the minimum and maximum points of the curve y= x3<\/sup> \u2013x-5x and sketch the curve
                    \n2) The area of a circle is increasing at the rate of 4cm2<\/sup>\/s , find the rate of change of the circumference when the radius is 6cm<\/p>\n

                    \u00a0GENERAL EVALUATION
                    \n<\/strong>1) A curve is defined by f(x) = x3<\/sup> -6x2<\/sup>-15-1 find (i) the derivative of f(x) (ii) the gradient of the curve at the point where x= 1 (iii) the minimum and maximum points
                    \n2) The distance of a particle from a starting point is S = t3<\/sup> \u2013 15t2<\/sup> +63t \u2013 40 where t = time taken in seconds, find the (i) distance of the particle from the starting point when the particle is at rest (ii) velocity when the acceleration is zero
                    \nReading Assignment : New FURTHER MATHS PROJECT 2 page 149-167<\/p>\n

                    \u00a0WEEKEND ASSIGNMENT
                    \n<\/strong>1) Find the value of x at which the function y = x2<\/sup> \u2013 7x2<\/sup> + 15x has the greatest value a) 5\/3 b) 5\/4 c) 5\/2 d) 5\/6
                    \n2) Find the values of x at the turning point of y = 2x3<\/sup> \u2013 3x2<\/sup> -12x + 8 a) 1 or 2 b) -1 or -2 c) -1 or 2 d) 1 or -2
                    \n3) Find the maximum value of the function 3x2<\/sup> \u2013x3<\/sup> a) 2 b) 4 c) 0 d) 6
                    \n4) Find the minimum value of the function f(x) = x3<\/sup> + 3x2<\/sup> \u2013 9x + 1 a) -3 b) -5 c) -4 d) 0
                    \n5) At what rate is the area of a circle changing with respect to its radius when the radius is 5cm a) 25 b 15 c) 20 d) 10<\/p>\n

                    \u00a0THEORY
                    \n<\/strong>1) The displacement of a particle is given as S = 12t \u2013 15t2<\/sup> + 4t3<\/sup> where t is the time taken . Find the velocity and acceleration of the particle after 3 seconds
                    \n2) Find the maximum and minimum points and values of the curve y = x3<\/sup> \u2013 6x2<\/sup> + 9x –<\/p>\n

                    \u00a0
                    \n\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"

                    \u00a0WEEK 4 TOPIC: APPLICATION OF DIFFERENTIATION ; RATE OF CHANGE, EQUATION OF MOTION , MAXIMUM…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,253],"tags":[],"class_list":["post-3161","post","type-post","status-publish","format-standard","hentry","category-posts","category-second-term-ss2-further-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3161","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=3161"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3161\/revisions"}],"predecessor-version":[{"id":3162,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3161\/revisions\/3162"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3161"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3161"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3161"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}