11-19 PROGRESSION. A level Mathematics. Pure Mathematics

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1 SSaa m m pplle e UCa ni p t ter DD iff if erfe enren tiatia tiotio nn - 9 RGRSSIN decel Slevel andmatematics level Matematics ure Matematics NW FR 07 Year/S Year

2 decel S and level Matematics Sample material bjectives fter completing tis capter ou sould be able to: find te derivative, f9() or d, of a simple function pages d use te derivative to solve problems involving gradients, tangents and normals pages identif increasing and decreasing functions pages 70 7 find te second order derivative, f 0() or _ d d, of a simple function pages 7 7 find stationar points of functions and determine teir nature pages 7 76 sketc te gradient function of a given function pages model real-life situations wit differentiation pages 79 8 is part of calculus, one of te most powerful tools in matematics. You will use differentiation in mecanics to model rates of cange, suc as speed and acceleration. ercise K Q5 rior knowledge ceck Find te gradients of tese lines. a b c (6, 6) Section 5. Write eac of tese epressions in te form n were n is a positive or negative real number. a 7 b c d 6 Sections.,.4 Find te equation of te straigt line tat passes troug: a (0, ) and (6, ) b (, 7) and (9, 4) c (0, 5) and (, 8) Section 5. 4 Find te equation of te perpendicular to te line = 5 at te point (, ). Section 5. 55

3 Capter. Gradients of curves Te gradient of a curve is constantl canging. You can use a tangent to find te gradient of a curve at an point on te curve. Te tangent to a curve at a point is te straigt line tat just touces te curve at. Te gradient of a curve at a given point is defined as te gradient of te tangent to te curve at tat point ample = Te diagram sows te curve wit equation =. Te tangent, T, to te curve at te point (, ) is sown. oint is joined to point b te cord. a Calculate te gradient of te tangent, T. b Calculate te gradient of te cord wen as coordinates: i (, 4) ii (.5,.5) iii (.,.) iv (.0,.00) v ( +, ( + ) ) c Comment on te relationsip between our answers to parts a and b. Te tangent to te curve at (, 0) as gradient, so te gradient of te curve at te point (, 0) is equal to. Te tangent just touces te curve at (, 0). It does not cut te curve at tis point, altoug it ma cut te curve at anoter point. = (, ) T a Gradient of tangent = _ = = b i Gradient of cord joining (, ) to (, 4) = 4 = ii Gradient of te cord joining (, ) to (.5,.5) =.5.5 _ = =.5 iii Gradient of te cord joining (, ) to (.,.) =.. = =. iv Gradient of te cord joining (, ) to (.0,.00) =.00.0 = =.0 v Gradient of te cord joining (, ) to ( +, ( + ) ) = ( + ) ( + ) _ = = = + Use te formula for te gradient of a straigt line between points (, ) and (, ). Section 5. Te points used are (, ) and (, ). nline plore te gradient of te cord using GeoGebra. Tis time (, ) is (, ) and (, ) is (.5,.5). Tis point is closer to (, ) tan (.,.) is. Tis gradient is closer to. is a constant. ( + ) = ( + )( + ) = + + ( + ) Tis becomes _ You can use tis formula to ceck te answers to te rest of part b. For eample, wen = 0.5, ( +, ( + ) ) = (.5,.5) and te gradient of te cord is =.5. c s gets closer to, te gradient of te cord gets closer to te gradient of te tangent at. s gets closer to zero, + gets closer to, and te gradient of te cord gets closer to te gradient of te tangent

4 Capter ercise Te diagram sows te curve wit equation =. a Cop and complete tis table sowing estimates for te gradient of te curve. -coordinate 0 stimate for gradient of curve b Write a potesis about te gradient of te curve at te point were = p. c Test our potesis b estimating te gradient of te grap at te point (.5, 0.75). 4 Te diagram sows te curve wit equation =. Te point as coordinates (0.6, 0.8). Te points B, C and D lie on te curve wit -coordinates 0.7, 0.8 and 0.9 respectivel = B C D Hint lace a ruler on te grap to approimate eac tangent. a Verif tat point lies on te curve. b Use a ruler to estimate te gradient of te curve at point. c Find te gradient of te line segments: i D ii C iii B Hint Use algebra for part c. d Comment on te relationsip between our answers to parts b and c. F is te point wit coordinates (, 9) on te curve wit equation =. a Find te gradients of te cords joining te point F to te points wit coordinates: i (4, 6) ii (.5,.5) iii (., 9.6) iv (.0, 9.060) v ( +, ( + ) ) b Wat do ou deduce about te gradient of te tangent at te point (, 9)? 4 G is te point wit coordinates (4, 6) on te curve wit equation =. a Find te gradients of te cords joining te point G to te points wit coordinates: i (5, 5) ii (4.5, 0.5) iii (4., 6.8) iv (4.0, 6.080) v (4 +, (4 + ) ) b Wat do ou deduce about te gradient of te tangent at te point (4, 6)?. Finding te derivative You can use algebra to find te eact gradient of a curve at a given point. Tis diagram sows two points, and B, tat lie on te curve wit equation = f(). = f() B You can formalise tis approac b letting te -coordinate of be 0 and te -coordinate of B be 0 +. Consider wat appens to te gradient of B as gets smaller. 0 = f() B 0 + s point B moves closer to point te gradient of cord B gets closer to te gradient of te tangent to te curve at. oint B as coordinates ( 0 +, f( 0 + )). oint as coordinates ( 0, f( 0 )). Notation represents a small cange in te value of. You can also use d to represent tis small cange. It is pronounced delta

5 Capter Te vertical distance from to B is f( 0 + ) f( 0 ). Te orizontal distance is =. So te gradient of B is f ( 0 + ) f ( 0 ) B f( 0 + ) f( 0 ) s gets smaller, te gradient of B gets closer to te gradient of te tangent to te curve at. Tis means tat te gradient of te curve at is te limit of tis epression as te value of tends to 0. You can use tis to define te gradient function. Te gradient function, or derivative, of te curve = f() is written as f9() or d d. f 9 () = lim _ f( + ) f() Te gradient function can be used to find te gradient of te curve for an value of. Using tis rule to find te derivative is called differentiating from first principles. ample Notation Te point wit coordinates ( 4, 6 ) lies on te curve wit equation =. t point te curve as gradient g. a Sow tat g = lim (8 + ). b Deduce te value of g. _ f(4 + ) f(4) a g = lim (4 + ) = lim 4 _ = lim = lim b g = 8 = lim (8 + ) lim means te limit as tends to 0. You can t evaluate te epression wen = 0, but as gets smaller te epression gets closer to a fied (or limiting) value. Use te definition of te derivative wit = 4. Te function is f() =. Remember to square everting inside te brackets. Section. Te 6 and te 6 cancel, and ou can cancel in te fraction. s te limiting value is 8, so te gradient at point is 8. ample rove, from first principles, tat te derivative of is. f() = f( + ) f() f9() = lim = lim = lim = lim = lim ( + ) () ( + + ) = lim ( + + ) s, and 0. So f9() = ercise B From first principles means tat ou ave to use te definition of te derivative. ( + ) = ( + )( + ) = ( + )( + + ) wic epands to give Factorise te numerator. n terms containing,,, etc will ave a limiting value of 0 as. For te function f() =, use te definition of te derivative to sow tat: a f9() = 4 b f9( ) = 6 c f9(0) = 0 d f9(50) = 00 f() = a Sow tat f 9 () = lim ( + ). b Hence deduce tat f9() =. Te point wit coordinates (, 8) lies on te curve wit equation =. t point te curve as gradient g. a Sow tat g = lim ( 6 + ). b Deduce te value of g. 4 Te point wit coordinates (, 4) lies on te curve wit equation = 5. Te point B also lies on te curve and as -coordinate ( + ). a Sow tat te gradient of te line segment B is given b. b Deduce te gradient of te curve at point. roblem-solving Draw a sketc sowing points and B and te cord between tem. 5 rove, from first principles, tat te derivative of 6 is 6. 6 rove, from first principles, tat te derivative of 4 is 8. (4 marks) 7 f() = a, were a is a constant. rove, from first principles, tat f9() = a. (4 marks) 60 6

6 Capter Callenge f() = a Given tat f 9 () = lim b Deduce tat f 9 () =. Differentiating n You can use te definition of te derivative to find an epression for te derivative of n were n is an number. Tis is called differentiation. For all real values of n, and for a constant a : If f() = n ten f 9 () = n n If = n ample 4 ten d d = n n If f() = a n ten f 9 () = an n d If = a n ten d = an n Find te derivative, f 9(), wen f() equals: a 6 b _ c d e 5 a f() = 6 So f9() = 6 5 b f() = So f9() = = c f() = So f9() = = d f() = = 5 So f9() = 5 4 f ( + ) f (), sow tat f 9 () = lim _ + Notation f9() and d bot represent te d derivative. You usuall use d d wen an epression is given in te form = Multipl b te power, ten subtract from te power: 6 6 = 6 5 Te new power is _ = _ _ = Section.4 You can leave our answer in tis form or write it as a fraction. You need to write te function in te form n before ou can use te rule. = + = 5 e f() = 5 = 4 So f9() = 4 5 ample 5 = 4 5 Find d wen equals: d a 7 b 4 _ c _ d 8 7 a d d = 7 = b d d = 4 = = c d d = = 6 = 6 d = 8 6 d d = = 6 5 ercise C Find f 9() given tat f() equals: e _ 6 a 7 b 8 c 4 d _ e _ 4 f g 4 i j 5 m 6 n o p 4 Find d given tat equals: d k q a b 6 9 c _ 4 d 0 _ 4 e 6 5_ 4 f 0 g 4 6 _ e = 6 = 6 ( ) = 6 d d = 6 = 9 = 9 _ 8 5 Use te laws of indices to simplif te fraction: 5 = 5 = 4 Use te rule for differentiating a n wit a = 7 and n =. Multipl b ten subtract from te power. Tis is te same as differentiating _ ten multipling te result b 4. Write te epression in te form a n. Remember a can be an number, including fractions. _ = _ Simplif te number part as muc as possible. i Hint Make sure tat te functions are in te form n before ou differentiate. l r 6 5 j

7 Capter Find te gradient of te curve wit equation = at te point were: 64 a = 4 b = 9 c = _ 4 d = Given tat = 0 and > 0, find d d.4 Differentiating quadratics ( marks) You can differentiate a function wit more tan one term b differentiating te terms one-at-a-time. Te igest power of in a quadratic function is, so te igest power of in its derivative will be. For te quadratic curve wit equation = a + b + c, te derivative is given b d d = a + b You can find tis epression for d b differentiating eac of te terms one-at-a-time: d a Differentiate a = a b = b Differentiate b 0 = b c Differentiate Te quadratic term tells ou te slope of te gradient function. ample 6 n term differentiates to give a constant. Find d given tat equals: d a + b 8 7 c a = + So d d = + b = 8 7 So d d = 8 c = So d d = 8 roblem-solving Tr rearranging unfamiliar equations into a form ou recognise. Links Te derivative is a straigt line wit gradient a. It crosses te -ais once, at te point were te quadratic curve as zero gradient. Tis is te turning point of te quadratic curve. Section 5. Constant terms disappear wen ou differentiate. Differentiate te terms one-at-a-time. Te constant term disappears wen ou differentiate. Te line = 7 would ave zero gradient is a quadratic epression wit a = 4, b = and c = 5. Te derivative is a + b = 4 = 8. 0 ample 7 Let f() = a Find te gradient of = f() at te point ( _, 0 ). b Find te coordinates of te point on te grap of = f() were te gradient is 8. c Find te gradient of = f() at te points were te curve meets te line = 4 5. a s = d = f9() = d So f9 ( ) = 4 b d = f9() = 8 8 = 8 d So = So = f() = Te point were te gradient is 8 is (, ). c = = 0 + = 0 ( )( ) = 0 So = or = t =, te gradient is 0. t =, te gradient is 8, as in part b. ercise D Find d wen equals: d a 6 + b _ + c 4 6 d e Find te gradient of te curve wit equation: Differentiate to find te gradient function. Ten substitute te -coordinate value to obtain te gradient. ut te gradient function equal to 8. Ten solve te equation ou ave obtained to give te value of. Substitute tis value of into f() to give te value of and interpret our answer in words. To find te points of intersection, set te equation of te curve equal to te equation of te line. Solve te resulting quadratic equation to find te -coordinates of te points of intersection. Section 4.4 Substitute te values of into f9() = 8 8 to give te gradients at te specified points. nline Use our calculator to ceck solutions to quadratic equations quickl. a = at te point (, ) b = + 4 at te point (, 5) c = at te point (, 5) d = + at te point (, ) e = at te point (, ) f = 4 at te point (, ) Find te -coordinate and te value of te gradient at te point wit -coordinate on te curve wit equation = +. 4 Find te coordinates of te point on te curve wit equation = were te gradient is. 65

8 Capter 5 Find te gradients of te curve = at te points and B were te curve meets te line = 4. 6 Find te gradients of te curve = at te points C and D were te curve meets te line = +. 7 f() = 8 66 a Sketc te grap of = f(). b n te same set of aes, sketc te grap of = f9(). c plain w te -coordinate of te turning point of = f() is te same as te -coordinate of te point were te grap of = f9() crosses te -ais..5 Differentiating functions wit two or more terms You can use te rule for differentiating a n to differentiate functions wit two or more terms. You need to be able to rearrange eac term into te form a n, were a is a constant and n is a real number. Ten ou can differentiate te terms one-at-a-time. If = f() ± g(), ten d = f9() ± g9(). d ample 8 Find d given tat equals: d a 4 + b + _ c + 4 a So = 4 + d d = + b = + d So d = + c = + 4 d So d = + 8 = Differentiate te terms one-at-a-time. Be careful wit te tird term. You multipl te term b _ and ten reduce te power b to get _ Ceck tat eac term is in te form a n before differentiating. ample 9 Differentiate: _ a 4 a Let = 4 = 4 Terefore d d = 8 b Let = ( + ) = 4 + Terefore d d = + c Let = ercise Differentiate: a 4 + b 5 + c 6 _ + _ + 4 Find te gradient of te curve wit equation = f() at te point were: a f() = + and is at (, 4) b f() = + and is at (, ) Find te point or points on te curve wit equation = f(), were te gradient is zero: a f() = 5 b f() = c f() = _ Differentiate: a e i = = + + = (4 + ) Terefore d d = + 4 b ( + ) c = + 4 = 4 Use te laws of indices to write te epression in te form a n. _ 4 = 4 = 4 = 4 _ Multipl out te brackets to give a polnomial function. Differentiate eac term. d f() = + 4 b c d _ ( ) f + g + _ 6 j ( + ) k ( + ) l ( ) ( 4 + ) _ press te single fraction as two separate fractions, and simplif: = Write eac term in te form a n ten differentiate. You can write te answer as a single fraction wit denominator. 67

9 Capter 5 Find te gradient of te curve wit equation = f() at te point were: a f() = ( + ) and is at (0, 0) b f() = 6 and is at (, 0) c f() = and is at ( _ 4, ) d f() = 4 and is at (, 5) _ 6 f( ) = p +, were p is a real constant and > 0. Given tat f9() =, find p, giving our answer in te form a were a is a rational number. 7 f() = ( ) 9 a Find te first terms, in ascending powers of, of te binomial epansion of f(), giving eac term in its simplest form. b If is small, so tat and iger powers can be ignored, sow tat f9() < Gradients, tangents and normals (4 marks) Hint Use te binomial epansion wit a =, b = and n = 9. Section 8. You can use te derivative to find te equation of te tangent to a curve at a given point. n te curve wit equation = f(), te gradient of te tangent at a point wit -coordinate a will be f9(a). Te tangent to te curve = f() at te point wit coordinates (a, f(a)) as equation f(a) = f9(a)( a) Links Te equation of a straigt line wit gradient m tat passes troug te point (, ) is = m( ). Section 5. Te normal to a curve at point is te straigt line troug wic is perpendicular to te tangent to te curve at. Te gradient of te normal will be _ f 9 (a ) Te normal to te curve = f() at te point wit coordinates (a, f(a)) as equation f(a ) = f 9 (a ) ( a ) Normal at = f() Tangent at ample 0 Find te equation of te tangent to te curve = + at te point (, 5). = + d d = 6 + Wen =, te gradient is. So te equation of te tangent at (, 5) is 5 = ( ) = 8 ample Find te equation of te normal to te curve wit equation = 8 at te point were = 4. = 8 = 8 d d = Wen = 4, = and gradient of curve and of tangent = 4 So gradient of normal is 4. quation of normal is = 4 ( 4) 6 = = 0 ercise F Find te equation of te tangent to te curve: First differentiate to determine te gradient function. Ten substitute for to calculate te value of te gradient of te curve and of te tangent wen =. You can now use te line equation and simplif. Write eac term in te form a n and differentiate to obtain te gradient function, wic ou can use to find te gradient at an point. Find te -coordinate wen = 4 b substituting into te equation of te curve and calculating 8 4 = 8 6 =. Find te gradient of te curve, b calculating d d = (4) _ = = 4 Gradient of normal = gradient of curve = ( = 4 4 ) Simplif b multipling bot sides b and collecting terms. nline plore te tangent and normal to te curve using GeoGebra. a = at te point (, 0) b = + at te point (, _ ) c = 4 at te point (9, ) d = at te point (, ) e = at te point (, ) f = 7 at te point (, 6) Find te equation of te normal to te curve: a = 5 at te point (6, 6) b = 8 at te point (4, ) Find te coordinates of te point were te tangent to te curve = + at te point (, 5) meets te normal to te same curve at te point (, )

10 Capter 4 Find te equations of te normals to te curve = + at te points (0, 0) and (, ), and find te coordinates of te point were tese normals meet. 5 For f() = 4 +, find te equations of te tangent and te normal at te point were = on te curve wit equation = f(). ample Find te interval on wic te function f() = + 9 is decreasing. Find f9() and put tis epression < 0. 6 Te point wit -coordinate _ lies on te curve wit equation =. Te normal to te curve at intersects te curve at points and Q. Find te coordinates of Q. (6 marks) roblem-solving Draw a sketc sowing te curve, te point and te normal. Tis will elp ou ceck tat our answer makes sense. Callenge Hint Use te discriminant to find te value of m Te line L is a tangent to te curve wit equation wen te line just touces te curve. Section.5 = 4 +. L cuts te -ais at (0, 8) and as a positive gradient. Find te equation of L in te form = m + c..7 Increasing and decreasing functions You can use te derivative to determine weter a function is increasing or decreasing on a given interval. Te function f() is increasing on te interval [a, b] if f9() > 0 for all values of suc tat a < < b. Te function f() is decreasing on te interval [a, b] if f9() < 0 for all values of suc tat a < < b. = + = 4 Te function f() = + is increasing for all real values of. Notation Te interval [a, b] is te set of all real numbers,, tat satisf a < < b. (, ) (, ) Te function f() = 4 is increasing on te interval [, 0] and decreasing on te interval [0, ]. f() = + 9 f9() = If f9() < 0 ten < 0 So ( + ) < 0 ( + )( ) < 0 So < < So f() is decreasing on te interval [, ]. ercise G Find te values of for wic f() is an increasing function, given tat f() equals: a b 4 c 5 8 d e + + f 5 + g Find te values of for wic f() is a decreasing function, given tat f() equals: a 9 b 5 c 4 d e 7 + f + 5 g _ + 9 _ ( + ) Sow tat te function f() = 4 ( + ) is decreasing for all R. 4 a Given tat te function f() = + p is increasing on te interval [, ], find one possible value for p. b State wit justification weter tis is te onl possible value for p..8 Second order derivatives Solve te inequalit b considering te tree regions <, < < and >, or b sketcing te curve wit equation = ( + )( ) Section.5 Write te answer clearl. nline plore increasing and decreasing functions using GeoGebra. You can find te rate of cange of te gradient function b differentiating a function twice. ( marks) ( mark) ample = 5 Differentiate d d = 5 Differentiate d d = 0 Sow tat te function f() = is increasing for all real values of. f() = f9() = + 4 > 0 for all real values of So + 4 > 0 for all real values of. So f() is increasing for all real values of. First differentiate to obtain te gradient function. State tat te condition for an increasing function is met. In fact f9() > 4 for all real values of. Tis is te gradient function. It describes te rate of cange of te function wit respect to. Differentiating a function = f() twice gives ou te second order derivative, f 0() or _ d d Tis is te rate of cange of te gradient function. It is called te second order derivative. It can also be written as f0(). Notation Te derivative is also called te first order derivative or first derivative. Te second order derivative is sometimes just called te second derivative. 70 7

11 Capter ample 4 Given tat = find: a d d a So b 7 = = d d = 54 8 = b d d _ d d = = ample 5 4 Given tat f() = +, find: a f9() b f 0() a f() = + = + f9() = 4 b f0() = ercise H Find d d and d wen equals: d a b Te displacement of a particle in metres at time t seconds is modelled b te function press te fraction as a negative power of. Differentiate once to get te first order derivative. Differentiate a second time to get te second order derivative. c 9 d (5 + 4)( ) e + 8 f(t) = t + _ t Te acceleration of te particle in m s is te second derivative of tis function. Find an epression for te acceleration of te particle at time t seconds. Given tat = ( ), find te value of wen d _ d = 0. 4 f() = p p + 4 Wen =, f 0() =. Find te value of p. Don t rewrite our epression for f9() as a fraction. It will be easier to differentiate again if ou leave it in tis form. Te coefficient for te second term is ( ) ( 4) = + 8 Te new power is = 5 Links Te velocit of te particle will be f9(t) and its acceleration will be f 0(t). Statistics and Mecanics, Section 7. roblem-solving Wen ou differentiate wit respect to, ou treat an oter letters as constants..9 Stationar points stationar point on a curve is an point were te curve as gradient zero. You can determine weter a stationar point is a local maimum, a local minimum or a point of inflection b looking at te gradient of te curve on eiter side. n point on te curve = f() were f9() = 0 is called a stationar point. For a small positive value : oint is a local maimum. Te origin is a point of inflection. ample 6 B oint B is a local minimum. Notation oint is called a local maimum because it is not te largest value te function can take. It is just te largest value in tat immediate vicinit. a Find te coordinates of te stationar point on te curve wit equation = 4. b B considering points on eiter side of te stationar point, determine weter it is a local maimum, a local minimum or a point of inflection. a Tpe of stationar point f9( ) f9() f9( ) Local maimum ositive 0 Negative Local minimum Negative 0 ositive oint of inflection = 4 d d = 4 Let d d = 0 Ten 4 = 0 4 = = 8 = So = 4 = 6 64 = 48 So (, 48) is a stationar point. Negative 0 Negative ositive 0 ositive Differentiate and let d d = 0. Notation Te plural of maimum is maima and te plural of minimum is minima. Solve te equation to find te value of. Substitute te value of into te original equation to find te value of. 7

12 Capter b Now consider te gradient on eiter side of (, 48). Value of Gradient Sape of curve =.9 = = wic is ve wic is +ve From te sape of te curve, te point (, 48) is a local minimum point. Make a table were ou consider a value of sligtl less tan and a value of sligtl greater tan. Calculate te gradient for eac of tese values of close to te stationar point. Deduce te sape of te curve. In some cases ou can use te second derivative, f 0(), to determine te nature of a stationar point. If a function f() as a stationar point wen = a, ten: if f 0(a). 0, te point is a local minimum if f 0(a), 0, te point is a local maimum If f 0(a) = 0, te point could be a local minimum, a local maimum or a point of inflection. You will need to look at points on eiter side to determine its nature. ample 7 a Find te coordinates of te stationar points on te curve wit equation = _ b Find d and use it to determine te nature of te stationar points. d a = d d = utting = 0 6( 4)( ) = 0 So = 4 or = Wen =, = = 7 Wen = 4, = = 0 So te stationar points are at (, 7) and (4, 0). nline plore te solution using GeoGebra Hint f 0() tells ou te rate of cange of te gradient function. Wen f9() = 0 and f 0() > 0 te gradient is increasing from a negative value to a positive value, so te stationar point is a minimum. Differentiate and put te derivative equal to zero. Solve te equation to obtain te values of for te stationar points. Substitute = and = 4 into te original equation of te curve to obtain te values of wic correspond to tese values. _ b d = 0 d _ Wen =, d = 8 wic is, 0 d So (, 7) is a local maimum point. _ Wen = 4, d = 8 wic is. 0 d So (4, 0) is a local minimum point. ample 8 Differentiate again to obtain te second derivative. Substitute = and = 4 into te second derivative epression. If te second derivative is negative ten te point is a local maimum point. If it is positive ten te point is a local minimum point. a Te curve wit equation = + 7 as stationar points at = ±a. Find te value of a. b Sketc te grap of = + 7. a = + 7 d d = + 8 = + 8 Wen d d = 0: + 8 = 0 8 = So a = 8 4 = 4 = 8 = ± b d d = + 6 = + 6 Wen =, = ( + 7 ) ( ) = 4 _ and d d = ( + 6 ( ) = 08 ) wic is negative. So te curve as a local maimum at (, 4 ). Wen =, = ( + 7 ( ) ) = 4 and _ d d = ( + 6 ) ( ) = 08 wic is positive. Write as to differentiate. Set d = 0 to determine te -coordinates of te d stationar points. You need to consider te positive and negative roots: _ ( ) 4 = _ ( ) _ ( ) _ ( ) _ ( ) = To sketc te curve, ou need to find te coordinates of te stationar points and determine teir natures. Differentiate our epression for d d to find _ d d Substitute = _ 8 _ and = into te equation of te curve to find te -coordinates of te stationar points. nline Ceck our solution using our calculator

13 Capter So te curve as a local minimum at (, 4). Te curve as an asmptote at = 0. s,. s,. ercise I 4 4 = + 7 Find te least value of te following functions: a f() = + 8 b f() = 8 c f() = 5 + Find te greatest value of te following functions: a f() = 0 5 b f() = + c f() = (6 + )( ) Find te coordinates of te points were te gradient is zero on te curves wit te given equations. stablis weter tese points are local maimum points, local minimum points or points of inflection in eac case. a = b = 9 + c = + d = ( 4 ) e = + f = + 54 g = = _ ( 6) i = 4 4 Sketc te curves wit equations given in question parts a, b, c and d, labelling an stationar points wit teir coordinates. 5 B considering te gradient on eiter side of te stationar point on te curve = +, sow tat tis point is a point of inflection. Sketc te curve = +. 6 Find te maimum value and ence te range of values for te function f() = f() = a Find te coordinates of te stationar points of f(), and determine te nature of eac. b Sketc te grap of = f(). ± as 0 so = 0 is an asmptote of te curve. Mark te coordinates of te stationar points on our sketc, and label te curve wit its equation. You could ceck d at specific points to elp wit d our sketc: Wen = _ 4, d = wic is negative. d Wen =, d = 80 wic is positive. d Hint For eac part of questions and : Find f9(). Set f9() = 0 and solve to find te value of at te stationar point. Find te corresponding value of f(). Hint Use te factor teorem wit small positive integer values of to find one factor of f9(). Section 7..0 Sketcing gradient functions You can use te features of a given function to sketc te corresponding gradient function. Tis table sows ou features of te grap of a function, = f(), and te grap of its gradient function, = f9(), at corresponding values of. = f () Maimum or minimum oint of inflection ositive gradient Negative gradient Vertical asmptote Horizontal asmptote ample 9 = f9() Cuts te -ais Touces te -ais bove te -ais Below te -ais Vertical asmptote Horizontal asmptote at te -ais Te diagram sows te curve wit equation = f(). Te curve as stationar points at (, 4) and (, 0), and cuts te -ais at (, 0). Sketc te gradient function, = f 9(), sowing te coordinates of an points were te curve cuts or meets te -ais. = fˇ() (, 4) = f() = f () = f() = f9() = f9(), ositive gradient bove -ais = Maimum Cuts -ais,, Negative gradient Below -ais = Minimum Cuts -ais > ositive gradient bove -ais Watc out Ignore an points were te curve = f() cuts te -ais. Tese will not tell ou anting about te features of te grap of = f9(). nline Use GeoGebra to eplore te ke features linking = f() and = f9()

14 Capter ample 0 Te diagram sows te curve wit equation = f(). Te curve as an asmptote at = and a turning point at (, 8). It cuts te -ais at ( 0, 0). a Sketc te grap of = f 9(). b State te equation of te asmptote of = f 9(). a b = 0 ercise 78 0 (, 8) = f() For eac grap given, sketc te grap of te corresponding gradient function on a separate set of aes. Sow te coordinates of an points were te curve cuts or meets te -ais, and give te equations of an asmptotes. a b c (6, 5) = 0 ( 9, ) (4, ) d J 8 6 = fˇ() = = 6 f() = ( + )( 4) a Sketc te grap of = f(). b n a separate set of aes, sketc te grap of = f 9(). c Sow tat f 9() = ( 4)( ). d Use te derivative to determine te eact coordinates of te points were te gradient function cuts te coordinate aes. e Draw our sketc on a separate set of aes. Te grap of = f9() will ave te same orizontal scale but will ave a different vertical scale. You don t ave enoug information to work out te coordinates of te -intercept, or te local maimum, of te grap of te gradient function. Te grap of = f() is a smoot curve so te grap of = f9() will also be a smoot curve. If = f() as an orizontal asmptotes ten te grap of = f9() will ave an asmptote at te -ais. f = 7 = 4 = 4 Hint Tis is an grap wit a positive coefficient of. Section 4.. Modelling wit differentiation You can tink of d as small cange in. It represents te rate of cange of wit respect to. d small cange in If ou replace and wit variables tat represent real-life quantities, ou can use te derivative to model lots of real-life situations involving rates of cange. Te volume of water in tis water butt is constantl canging over time. If V represents te volume of water in te water butt in litres, and t represents te time in seconds, ten ou could model V as a function of t. If V = f(t) ten dv = f9(t) would represent te rate of cange of volume V dt wit respect to time. Te units of dv would be litres per second. dt ample Given tat te volume, V cm, of an epanding spere is related to its radius, r cm, b te formula V = 4_ p r, find te rate of cange of volume wit respect to radius at te instant wen te radius is 5 cm. V = 4 π r Differentiate V wit respect to r. Remember tat dv π is a constant. = 4π r dr Wen r = 5, dv = 4π 5 Substitute r = 5. dr = 4 ( s.f.) So te rate of cange is 4 cm per cm. Interpret te answer wit units. ample large tank in te sape of a cuboid is to be made from 54 m of seet metal. Te tank as a orizontal base and no top. Te eigt of te tank is metres. Two of te opposite vertical faces are squares. a Sow tat te volume, V m, of te tank is given b V = 8 _ b Given tat can var, use differentiation to find te maimum or minimum value of V. c Justif tat te value of V ou ave found is a maimum. 79

15 Capter a Let te lengt of te tank be metres. b 80 Total area, So But So So ut = + 54 = + 54 = V = V = 54 ( ) = (54 ) V = 8 dv = 8 d dv d = 0 So = 9 0 = 8 = or But is a lengt so = Wen =, V = 8 _ = 54 8 = 6 V = 6 is a maimum or minimum value of V. c _ d V d = 4 Wen =, _ d V = 4 = d Tis is negative, so V = 6 is te maimum value of V. roblem-solving You don t know te lengt of te tank. Write it as metres to simplif our working. You could also draw a sketc to elp ou find te correct epressions for te surface area and volume of te tank. Draw a sketc. Rearrange to find in terms of. Substitute te epression for into te equation. Simplif. Differentiate V wit respect to and put dv d = 0. Rearrange to find. is a lengt so use te positive solution. Substitute te value of into te epression for V. Find te second derivative of V. _ d V < 0 so V = 6 is a maimum. d ercise K Find dθ dt were θ = t t. Find d were = pr. dr Given tat r =, find te value of dr wen t =. t dt 4 Te surface area, cm, of an epanding spere of radius r cm is given b = 4pr. Find te rate of cange of te area wit respect to te radius at te instant wen te radius is 6 cm. 5 Te displacement, s metres, of a car from a fied point at time t seconds is given b s = t + 8t. Find te rate of cange of te displacement wit respect to time at te instant wen t = 5. 6 rectangular garden is fenced on tree sides, and te ouse forms te fourt side of te rectangle. a Given tat te total lengt of te fence is 80 m sow tat te area,, of te garden is given b te formula = (80 ), were is te distance from te ouse to te end of te garden. b Given tat te area is a maimum for tis lengt of fence, find te dimensions of te enclosed garden, and te area wic is enclosed. 7 closed clinder as total surface area equal to 600p. a Sow tat te volume, V cm, of tis clinder is given b te formula V = 00pr pr, were r cm is te radius of te clinder. b Find te maimum volume of suc a clinder. 8 sector of a circle as area 00 cm. a Sow tat te perimeter of tis sector is given b te formula _ = r + 00 r, r. 00 p b Find te minimum value for te perimeter of suc a sector. 9 sape consists of a rectangular base wit a semicircular top, as sown. a Given tat te perimeter of te sape is 40 cm, sow tat its area, cm, is given b te formula = 40r r pr were r cm is te radius of te semicircle. b Hence find te maimum value for te area of te sape. 0 Te sape sown is a wire frame in te form of a large rectangle split b parallel lengts of wire into smaller equal-sized rectangles. mm a Given tat te total lengt of wire used to complete te wole frame is 5 mm, sow tat te area of mm te wole sape, mm, is given b te formula = were mm is te widt of one of te smaller rectangles. b Hence find te maimum area wic can be enclosed in tis wa. M r cm r cm N ( marks) (4 marks) (4 marks) (4 marks) 8

16 Capter Mied eercise rove, from first principles, tat te derivative of 0 is 0. (4 marks) Te point wit coordinates (, 4) lies on te curve wit equation = +. Te point B also lies on te curve and as -coordinate ( + δ ). a Sow tat te gradient of te line segment B is given b ( δ ) + δ + 6. b Deduce te gradient of te curve at point. curve is given b te equation = + +, were. 0. t te points, B and C on te curve, =, and respectivel. Find te gradient of te curve at, B and C. 4 Calculate te -coordinates of te points on te curve wit equation = 7 at wic te gradient is equal to 6. 5 Find te -coordinates of te two points on te curve wit equation = + were te gradient is. Find te corresponding -coordinates. 6 Te function f is defined b f() = + 9, [ R, Þ 0. a Find f9(). b Solve f9() = 0. 7 Given tat = 4 find d d,. 0, (4 marks) ( marks) ( marks) 8 curve as equation = _ _. a Sow tat d d = _ (4 ). ( marks) b Find te coordinates of te point on te curve were te gradient is zero. ( marks) 9 a pand ( _ )( _ + ). ( marks) b curve as equation = ( _ )( _ + ),. 0. Find d d ( marks) c Use our answer to part b to calculate te gradient of te curve at te point were = 4. ( mark) curve C as equation = a Find d in terms of. ( marks) d b Te points and Q lie on C. Te gradient of C at bot and Q is. Te -coordinate of is. i Find te -coordinate of Q. ii Find an equation for te tangent to C at, giving our answer in te form = m + c, were m and c are constants. iii If tis tangent intersects te coordinate aes at te points R and S, find te lengt of RS, giving our answer as a surd. curve as equation = 8 +, > 0. Find te equations of te tangent and te normal to te curve at te point were =. 4 Te normals to te curve = 7 + 4, at te points (0, 0) and (, 0), meet at te point N. a Find te coordinates of N. b Calculate te area of triangle N. (7 marks) 5 curve C as equation = 4 and cuts te -ais at a point. Te line L is a tangent to te curve at, and cuts te curve at te point Q. Sow tat te distance Q is 7. (7 marks) 6 Given tat = _ + 48,. 0 a Find te value of and te value of wen d d = 0. b Sow tat te value of wic ou found in part a is a minimum. 7 curve as equation = Determine, b calculation, te coordinates of te stationar points of te curve. 8 Te function f, defined for [ R,. 0, is suc tat: f9() = + a Find te value of f 0 () at = 4. b rove tat f is an increasing function. (5 marks) ( marks) (4 marks) 0 Differentiate wit respect to : + + _ + Te curve wit equation = a + b + c passes troug te point (, ). Te gradient of te curve is zero at te point (, ). Find te values of a, b and c. (5 marks) 9 curve as equation = Find te coordinates of its local maimum. (4 marks) 0 f() = a Find te coordinates of te stationar points of f(), and determine te nature of eac of tem. b Sketc te grap of = f(). 8 8

17 Capter Te diagram sows part of te curve wit equation = f(), were: _ f() = 00 50,. 0 Te curve cuts te -ais at te points and C. Te point B is te maimum point of te curve. a Find f9(). b Use our answer to part a to calculate te coordinates of B. (4 marks) Te diagram sows te part of te curve wit equation = 5 _ for wic > 0. Te point (, ) lies on te curve and is te origin. a Sow tat = _ Taking f() = _ : b Find te values of for wic f9() = 0. (4 marks) c Hence, or oterwise, find te minimum distance from to te curve, sowing tat our answer is a minimum. (4 marks) Te diagram sows part of te curve wit equation = Te curve touces te -ais at and crosses te -ais at C. Te points and B are stationar points on te curve. a Sow tat C as coordinates (, 0). ( mark) b Using calculus and sowing all our working, find te coordinates of and B. (5 marks) 4 Te motion of a damped spring is modelled using tis grap. n a separate grap, sketc te gradient function for tis model. Coose suitable labels and units for eac ais, and indicate te coordinates of an points were te gradient function crosses te orizontal ais. 5 Te volume, V cm, of a tin of radius r cm is given b te formula V = p(40r r r ). Find te positive value of r for wic dv = 0, and find te value of V wic dr corresponds to tis value of r. B C (, ) C Displacement (cm) B Time (seconds) 7 wire is bent into te plane sape BCD as sown. Sape BD is a rectangle and BCD is a semicircle wit diameter BD. Te area of te region enclosed b te wire is R m, = metres, and B = D = metres. Te total lengt of te wire is m. a Find an epression for in terms of. b rove tat R = (8 4 p). (4 marks) 8 Given tat can var, using calculus and sowing our working: c find te maimum value of R. (You do not ave to prove tat te value ou obtain is a maimum.) 8 clindrical biscuit tin as a close-fitting lid wic overlaps te tin b cm, as sown. Te radii of te tin and te lid are bot cm. Te tin and te lid are made from a tin seet of metal of area 80p cm and tere is no wastage. Te volume of te tin is V cm. a Sow tat V = p(40 ). (5 marks) Given tat can var: b use differentiation to find te positive value of for wic V is stationar. c rove tat tis value of gives a maimum value of V. ( marks) d Find tis maimum value of V. ( mark) e Determine te percentage of te seet metal used in te lid wen V is a maimum. 9 Te diagram sows an open tank for storing water, BCDF. Te sides BF and CDF are rectangles. Te triangular ends D and BCF are isosceles, and /D = /BFC = 90. Te ends D and BCF are vertical and F is orizontal. Given tat D = metres: a sow tat te area of triangle D is _ 4 m Given also tat te capacit of te container is 4000 m and tat te total area of te two triangular and two rectangular sides of te container is S m : b sow tat S = Given tat can var: c use calculus to find te minimum value of S. d Justif tat te value of S ou ave found is a minimum. D B B D Lid Tin cm cm F C (5 marks) cm ( marks) C (4 marks) (6 marks) ( marks) 6 Te total surface area, cm, of a clinder wit a fied volume of 000 cm is given b te formula = p + 000, were cm is te radius. Sow tat wen te rate of cange of te area wit respect to te radius is zero, = _ 500 p Callenge a Find te first four terms in te binomial epansion of ( + ) 7, in ascending powers of. b Hence prove, from first principles, tat te derivative of 7 is

18 Capter Summar of ke points Te gradient of a curve at a given point is defined as te gradient of te tangent to te curve at tat point. Te gradient function, or derivative, of te curve = f() is written as f9() or d d f ( + ) f() f 9 () = lim Te gradient function can be used to find te gradient of te curve for an value of. For all real values of n, and for a constant a: If f() = n ten f 9 () = n n If f() = a n ten f 9 () = an n d If = n ten d = n d n If = a n ten d = an n 4 For te quadratic curve wit equation = a + b + c, te derivative is given b d d = a + b 5 If = f() ± g(), ten d = f9() ± g9(). d 6 Te tangent to te curve = f() at te point wit coordinates (a, f(a)) as equation f(a) = f9(a)( a) 7 Te normal to te curve = f() at te point wit coordinates (a, f(a)) as equation f(a ) = _ f 9 (a ) ( a ) 8 Te function f() is increasing on te interval [a, b] if f9() > 0 for all values of suc tat a,, b. Te function f() is decreasing on te interval [a, b] if f9() < 0 for all values of suc tat a,, b. 9 Differentiating a function = f() twice gives ou te second order derivative, f 0() or _ d d 0 n point on te curve = f() were f9() = 0 is called a stationar point. For a small positive value : Tpe of stationar point f9( ) f9() f9( + ) Local maimum ositive 0 Negative Local minimum Negative 0 ositive oint of inflection Negative 0 Negative ositive 0 ositive If a function f() as a stationar point wen = a, ten: if f 0(a). 0, te point is a local minimum if f 0(a), 0, te point is a local maimum. If f 0(a) = 0, te point could be a local minimum, a local maimum or a point of inflection. You will need to look at points on eiter side to determine its nature. 86

11-19 PROGRESSION. Chapter 12 Differentiation. Unit 12 Differentiation. Edexcel A AS level and Mathematics. Pure Mathematics.

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