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

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1 - 9 RGRSSIN Sample Unit Chapter decel S level and Mathematics level Mathematics ure Mathematics NW FR 07 Year /S

2 decel S and level Mathematics Sample material bjectives fter completing this chapter ou should be able to: Find the derivative, f9() or d, of a simple function pages d Use the derivative to solve problems involving gradients, tangents and normals pages Identif increasing and decreasing functions pages 70 7 Find the second order derivative, f 0() or _ d d, of a simple function pages 7 7 Find stationar points of functions and determine their nature pages Sketch the gradient function of a given function pages Model real-life situations with differentiation pages 79 8 is part of calculus, one of the most powerful tools in mathematics. You will use differentiation in mechanics to model rates of change, such as speed and acceleration. ercise K Q5 rior knowledge check Find the gradients of these lines. a b c (6, 6) Section 5. Write each of these epressions in the form n where n is a positive or negative real number. a 3 7 b 3 c 3 d 6 Sections.,.4 3 Find the equation of the straight line that passes through: a (0, ) and (6, ) b (3, 7) and (9, 4) c (0, 5) and (, 8) Section 5. 4 Find the equation of the perpendicular to the line = 5 at the point (, ). Section

3 Chapter. Gradients of curves The gradient of a curve is constantl changing. You can use a tangent to find the gradient of a curve at an point on the curve. The tangent to a curve at a point is the straight line that just touches the curve at. The gradient of a curve at a given point is defined as the gradient of the tangent to the curve at that point ample = The diagram shows the curve with equation R =. The tangent, T, to the curve at the point (, ) is shown. oint is joined to point b the chord. a Calculate the gradient of the tangent, T. b Calculate the gradient of the chord when has coordinates: i (, 4) DR ii (.5,.5) iii (.,.) iv (.0,.00) v ( + h, ( + h) ) c Comment on the relationship between our answers to parts a and b. R The tangent to the curve at (, 0) has gradient, so the gradient of the curve at the point (, 0) is equal to. The tangent just touches the curve at (, 0). It does not cut the curve at this point, although it ma cut the curve at another point. = (, ) T a Gradient of tangent = _ = 3 = b i Gradient of chord joining (, ) to (, 4) = 4 = 3 ii Gradient of the chord joining (, ) to (.5,.5) =.5.5 _ = =.5 iii Gradient of the chord joining (, ) to (.,.) =.. = =. iv Gradient of the chord joining (, ) to (.0,.00) =.00.0 = =.0 v Gradient of the chord joining (, ) to ( + h, ( + h) ) = ( + h) ( + h) _ = + h + h + h h + h = h = + h Use the formula for the gradient of a straight line between points (, ) and (, ). Section 5. The points used are (, ) and (, 3). nline plore the gradient of the chord using GeoGebra. This time (, ) is (, ) and (, ) is (.5,.5). This point is closer to (, ) than (.,.) is. This gradient is closer to. DRF RFT h is a constant. ( + h) = ( + h)( + h) = + h + h h( + h) This becomes _ h You can use this formula to confirm the answers to questions i to iv. For eample, when h = 0.5, ( + h, ( + h) ) = (.5,.5) and the gradient of the chord is =.5. c s gets closer to, the gradient of the chord gets closer to the gradient of the tangent at. s h gets closer to zero, + h gets closer to, so the gradient of the chord gets closer to the gradient of the tangent

4 Chapter ercise The diagram shows the curve with equation =. a Cop and complete this table showing estimates for the gradient of the curve. -coordinate 0 3 stimate for gradient of curve b Write a hpothesis about the gradient of the curve at the point where = p. c Test our hpothesis b estimating the gradient of the graph at the point (.5, 0.75) lace a ruler on the graph approimate each tangent. T The diagram shows the curve with equation = F.. The point has coordinates (0.6, 0.8). FTto FTHint The points B, C and D lie on the curve with -coordinates 0.7, 0.8 and 0.9 respectivel. R = B C 0.6 D 0.4 F F _ F F =. _F a Verif that point lies on the curve. b Use a ruler to estimate the gradient of the curve at point. c Find the gradient of the line segments: i D ii C iii B Hint Use algebra for part c. d Comment on the relationship between our answers to parts b and c. 3 F is the point with coordinates (3, 9) on the curve with equation =. a Find the gradients of the chords joining the point F to the points with coordinates: i (4, 6) ii (3.5,.5) iii (3., 9.6) iv (3.0, 9.060) v (3 + h, (3 + h) ) b What do ou deduce about the gradient of the tangent at the point (3, 9)? 4 G is the point with coordinates (4, 6) on the curve with equation =. a Find the gradients of the chords joining the point G to the points with coordinates: i (5, 5) ii (4.5, 0.5) iii (4., 6.8) iv (4.0, 6.080) v (4 + h, (4 + h) ) b What do ou deduce about the gradient of the tangent at the point (4, 6)?. Finding the derivative You can use algebra to find the eact gradient of a curve at a given point. This diagram shows two points, and B, that lie on the curve with equation = f(). = f() B You can formalise this approach b letting the -coordinate of be 0 and the -coordinate of B be 0 + h. Consider what happens Dpproach enden to the gradient of B as h gets smaller. 0 = f() B 0 + h s point B moves closer to point the gradient of chord B gets closer to the gradient of the tangent to the curve at. DR oint B has coordinates ( 0 + h, f( 0 + h)). oint has coordinates ( 0, f( 0 )). Notation h represents a small change in the value of. You can also use d to represent this small change. It is pronounced delta

5 Chapter The vertical distance from to B is f( 0 + h) f( 0 ). The horizontal distance is 0 + h 0 = h. So the gradient of B is f ( 0 + h) f ( 0 ) h h B f( 0 + h) f( 0 ) s h gets smaller, the gradient of B gets closer to the gradient of the tangent to the curve at. This means that the gradient of the curve at is the limit of this epression as the value of h tends to 0. You can use this to define the gradient function. The gradient function, or derivative, of the curve = f() is written as f9() or d d. f 9 () = lim _ f( + h) f() h 0 h The gradient function can be used to find the gradient of the curve for an value of. Using this rule to find the derivative is called differentiating from first principles. ample The point with coordinates (4, 6) lies on the curve with equation =. t point the curve has gradient g.. a Show that g = lim (8 + h) ). h 0 b Deduce the value of g.. _ f(4 + h) ) f(4) a g = lim h 0 h (4 + h) = lim 4 h 0 h _ 6 + 8h + h = lim 6 h 0 h 8h + h = lim h 0 h b g = 8 = lim (8 + h) h 0 DR FTh 4 4 DDR Notation lim means the h 0 limit as h tends to 0. You can t evaluate the epression when h = 0, but as h gets smaller the epression gets closer to a fied (or limiting) value. Th T= e FTeval FTthe FTfie Use the definition of the derivative with = 4. The function is f() =. Remember to square everthing inside the brackets. Section.3 The 6 and the 6 cancel, and ou can cancel h in the fraction. s h 0 the limiting value is 8, so the gradient at point is 8. / / / ample 3 rove, from first principles, that the derivative of 3 is 3. f() = 3 f( + h) f() f9() = lim h 0 h _ ( + h) = lim 3 () 3 h 0 h = lim h 0 = lim h 0 = lim h h + 3h + h 3 3 _ h 3 h + 3h + h 3 h h(3 + 3h + h ) h = lim h 0 (3 + 3h + h ) s h 0, 3h 0 and h 0. So f9() = 3 From first principles means that ou have to use the definition of the derivative. You are starting our proof with a known definition, so this is an eample of a proof b deduction. ( + h) 3 = ( + h)( + h) = ( + h)( + h + h ) which epands to give h + 3h + h 3 n terms containing h, h, h 3, etc will have a limiting value of 0 as h 0. ercise B For the function f() =, use the definition of the derivative to show that: a f9() = 4 b f9( 3) = 6 6 c f9(0) = 0 d f9(50) = 00 f() = a Show that t f9 9( 9 () = lim ( + h). b Hence deduce that f9() =. h 0 3 The point coordinatesdr with coordinates (, 8) lies on the curve with equation = 3. t point the curve has grd gradient g. a Show that g = lim D). ( 6h + h ). b Deduce the value of g. h 0 4 The point with coordinates (, 4) lies on the curve with equation = 3 5. The point B also lies on the curve and has -coordinate ( + h). a Show that the gradient of the line segment B is given b h 3h. b Deduce the gradient of the curve at point. Factorise the numerator. (DR DR 0DR (DR +DR 0 roblem-solving Draw a sketch showing points and B and the chord between them. 5 rove, from first principles, that the derivative of 6 is 6. 6 rove, from first principles, that the derivative of 4 is 8. (4 marks) 7 f() = a, where a is a constant. rove, from first principles, that f9() = a. (4 marks) 60 6

6 Chapter Challenge f() = a Given that f 9 () = lim h 0 b Deduce that f 9 () =.3 Differentiating n You can use the definition of the derivative to find an epression for the derivative of n where n is an number. This is called differentiation. For all real values of n, and for a constant a : If f() = n then f 9 () = n n If = n ample 4 then d d = n n If f() = a n then f 9 () = an n d If = a n then d = an n Find the derivative, f 9(), when f() equals: a 6 b _ d 3 c e a f() = 6 So f9() = 6 5 b f() = So f9() = = c f() = So f9() = 3 = 3 d f() = 3 = 5 So f9() = 5 4 f ( + h) f () h, show that f 9 () = lim h 0 _ + h f9t Notation f9() and d both represent the d derivative. You usuall use d d T9( T( T) when an epression is given in the form = FTder FTNota FTf 5 Multipl b the power, then subtract from the power: 6 6 = 6 5 The new power is _ = _ _ = Section.4 You can leave our answer in this form or write it as a fraction. You need to write the function in the form n before ou can use the rule. 3 = + 3 = 5 e f() = 5 = 4 So f9() = 4 5 ample 5 = 4 5 Find d when equals: d a 7 3 b 4 _ c 3 _ d a d d = 7 33 = b d d = 4 = = c d d = 3 3 = 6 3 = 6 e _ 36 3 d = Write the epression in the form a n. Remember a be an number, including fractions. d d = = 6 5 3_ = _ e = 36 3 = 6 ( 3 ) = d d = 6 3 = 9 = 9 Simplif the number part as much as possible. Hint Make sure that the ercise C functions are in the form n before ou differentiate. Find f 9() given that f() DRcan equals: a 7 b 8 c 4 d _ 3 e _ 4 f g 3 h 4 i j 5 m 3 6 n 3 o p 4 Find d given that equals: d k q 3 a 3 b 6 9 c _ 4 d 0 _ 4 e 6 5_ 4 f 0 g 4 6 _ 3 3 h _ 8 5 Use the laws of indices to simplif the fraction: 5 = 5 = 4 Use the rule for differentiating a n with a = 7 and n = 3. Multipl b 3 then subtract from the power. This is the same as differentiating _ then multipling the result b 4. af f(d f(d)d)dr i l r j

7 Chapter / 3 Find the gradient of the curve with equation = 3 at the point where: 64 a = 4 b = 9 c = _ 4 d = Given that 3 = 0 and > 0, find d d.4 Differentiating quadratics ( marks) You can differentiate a function with more than one term b differentiating the terms one-at-a-time. The highest power of in a quadratic function is, so the highest power of in its derivative will be. For the quadratic curve with equation Links The derivative is a straight line with = a + b + c, the derivative is given b gradient a. It crosses the -ais once, at the d curve has zero d = a + b point where the quadratic gradient. This is the turning point of the quadratic curve. Section 5. Ttur You can find this epression for d b differentiating each of the terms one-at-a-time: d a Differentiate a = a b = b Differentiate entiate b 0 = b c Differentiate 0 n term differentiates constantft to give a constant. b = 8 7 So d d = 8 c = So d d = 8 3 bf e F roblem-solving Tr rearranging unfamiliar equations into a form ou recognise. The quadratic term tells Constant terms ou the slope of the disappear when gradient function. ou differentiate. ample 6 Rentiat 7D7D Find d given that equals: d a + 3 b D 7 c Db DD8 a = + 3 Differentiate the terms one-at-a-time. So d d = + 3 The constant term disappears when ou differentiate. The line = 7 would have zero gradient is a quadratic epression with a = 4, b = 3 and c = 5. The derivative is a + b = 4 3 = 8 3. ample 7 Let f() = a Find the gradient of = f() at the point ( _, 0 ). b Find the coordinates of the point on the graph of = f() where the gradient is 8. c Find the gradient of = f() at the points where the curve meets the line = 4 5. a s = d = f9() = d So f9 ( ) = 4 b d = f9() = 8 8 = 8 d So = So = f() = 3 The point where the gradient is 8 is (, 3). c = = = 0 ( )( ) = 0 So = or = t =, the gradient is 0. t =, the gradient is 8, as in part b. ercise D b.r Find d when equals: d a b _ + c 4 6 d e Find the gradient of the curve with equation: Differentiate to find the gradient function. Then substitute the -coordinate value to obtain the gradient. ut the gradient function equal to 8. Then solve the equation ou have obtained to give the value of. Substitute this value of into f() to give the value of and interpret our answer in words. To find the points T of intersection, set the equation of the curve equal to the equation of the line. Solve the resulting quadratic equation find the -coordinates of the points of intersection. Section 4.4 Substitute the values of into f9() = 8 8 to give the gradients at the specified points. nline Use our calculator to check Rto Rinters solutions to quadratic equations quickl. als:dals: DRF.RFT T a = 3 at the point (, ) b = + 4 at the point (, 5) c = at the point (, 5) d = + 3 at the point (, ) e = 3 at the point (, ) f = 4 at the point (, ) 3 Find the -coordinate and the value of the gradient at the point with -coordinate on the curve with equation = Find the coordinates of the point on the curve with equation = where the gradient is 3. 65

8 Chapter 5 Find the gradients of the curve = at the points and B where the curve meets the line = 4. 6 Find the gradients of the curve = at the points C and D where the curve meets the line = f() = 8 66 a Sketch the graph of = f(). b n the same set of aes, sketch the graph of = f9(). c plain wh the -coordinate of the turning point of = f() is the same as the -coordinate of the point where the graph of = f9() crosses the -ais..5 Differentiating functions with two or more terms You can use the rule for differentiating a n to differentiate iate functions with two or more terms. You need to be able to rearrange each term into the form a n, where a is a constant and n is a real number. Then ou can differentiate the terms one-at-a-time.. If = f() ± g(), then d = f9() ± g9(). d ample 8 Find d given that equals: d a b 3 DR DR_ + _ _ 3 _ c + 4 Rc DR a = d So d = + Differentiate the terms one-at-a-time. b = 3 + Be careful with the third term. You multipl the d So d = 3 + term b _ and then reduce the power b to get _ c = d So d = = FTa is FTis a D+ DD Check that each term is in the form a n before differentiating. ample 9 Differentiate: _ a 4 a Let = 4 = 4 Therefore d d = Write each term in the form a n then = + 4 differentiate. 3 = 4 You can write the answer as a single fraction with 3 denominator 3. ercise Differentiate: a 4 + b c 6 3_ + _ + 4 Find the gradient of the DRdiffere curve with equation = f() at the point where: a f() = and is at (, 4) b f() = 3 + and is at (, 3) 3 Find the point or points on the curve with equation = f(), where the gradient is zero: a f() = 5 b f() = c f() = 3_ Differentiate: a e i Therefore d d = 8 3 b Let = 3 (3 + ) = Therefore d d = c Let = = = b 3 (3 + ) c = 3 (4 + ) hed Use the laws of indices to write the epression in the form a n. _ 4 = 4 = 4 = 4 _ Multipl out the brackets to give a polnomial function. Differentiate each term. d f() = b c 3 d _ ( ) f 3 + g + 3 _ h 3 6 j ( + ) k 3 ( + ) l (3 ) ( 4 + ) _ press the single fraction as two separate fractions, and simplif: = hedrf RFT 67

9 Chapter / 5 Find the gradient of the curve with equation = f() at the point where: a f() = ( + ) and is at (0, 0) b f() = 6 and is at (3, 0) c f() = and is at ( _ 4, ) d f() = 3 4 and is at (, 5) _ 6 f( ) = p +, where p is a real constant and > 0. Given that f9() = 3, find p, giving our answer in the form a where a is a rational number. (4 marks) 7 f() = ( ) 9 a Find the first 3 terms, in ascending powers of, of the binomial epansion of f(), giving each term in its Hint Use the binomial simplest form. epansion with a =, b = b If is small, so that and higher powers can be ignored, n = 9. Section 8.3 show that f9() < FTep FTand.6 Gradients, tangents ts and normals You can use the derivative to find the equation of the tangent to a curve at a given point. n the curve with equation = f(), the gradient of the tangent at a point with -coordinate a will be f9(a). The tangent to the curve DR) = f() at the Links The equation of a straight line with point with coordinates (a, f(a)) has gradient m that passes through the point ( equation, ) is = m( ). Section 5. f(a) = f9(a)( a) ada)d) DRat DD)(D D D The normal to a curve at point is the straight line through which is perpendicular to the tangent to the curve at. The gradient of the normal will be _ f 9 (a ) The normal to the curve = f() at the point with coordinates (a, f(a)) has equation f(a ) = f 9 (a ) ( a ) FTn DR =DR f(dr f(adr DR) a,, Normal at = f() Tangent at ample 0 Find the equation of the tangent to the curve = at the point (3, 5). = d d = When = 3, the gradient is. So the equation of the tangent at (3, 5) is 5 = ( 3) = 8 ample Find the equation of the normal to the curve with equation = 8 3 at the point where = 4. = 8 3 = 8 3 d d = 3 When = 4, = and gradient of curve and of tangent = 3 4 So gradient of normal is quation of normal is = 4 3 ( 4) 3 6 = = 0 ercise F Find the equation of the tangent to the curve: First differentiate to determine the gradient function. Then substitute for to calculate the value of the gradient of the curve and of the tangent when = 3. You can now use the line equation and simplif. Write each term in the form a n and differentiate to obtain the gradient function, which ou can use to find the gradient at an point. Find the -coordinate when = 4 b substituting into the equation of the curve and calculating F = 8 6 =. Find the gradient of the curve, b calculating d d = 3 (4) _ = 3 = 3 4 Gradient of normal = gradient of curve = _ 3_ ( = ) Simplif b multipling both sides b 3 and collecting terms. F= 8 8FT nline plore the tangent and normal to the curve using GeoGebra. a = at the point (, 0) b = + at the point (, _ ) c = 4 at the point (9, ) d = at the point (, ) e = at the point (, ) f = 7 at the point (, 6) Find the equation of the normal to the curve: a = 5 at the point (6, 6) b = 8 at the point (4, ) 3 Find the coordinates of the point where the tangent to the curve = + at the point (, 5) meets the normal to the same curve at the point (, )

10 Chapter 4 Find the equations of the normals to the curve = + 3 at the points (0, 0) and (, ), and find the coordinates of the point where these normals meet. 5 For f() = 4 +, find the equations of the tangent and the normal at the point where = on the curve with equation = f(). ample 3 Find the interval on which the function f() = is decreasing. Find f9() and put this epression < 0. / 6 The point with -coordinate _ lies on the curve with equation =. The normal to the curve at intersects the curve at points and Q. Find the coordinates of Q. (6 marks) roblem-solving Draw a sketch showing the curve, the point and the normal. This will help ou check that our answer makes sense. f() = f9() = If f9() < 0 then < 0 So 3( + 3) < 0 Solve the inequalit b considering the three regions < 3, 3 < < and >, or b sketching the curve with equation = 3( + 3)( ) Section 3.5 Challenge Hint Use the discriminant to find the value of m The line L is a tangent to the curve with equation when the line just touches the curve. Section.5 = 4 +. L cuts the -ais at (0, 8) and has a positive gradient. Find the equation of L in the form = m + c..7 Increasing and decreasing functions You can use the derivative to determine whether a function is increasing or decreasing on a given interval. The function f() is increasing on the interval [a, b] if f9() > 0 for all values of such that a < < b. The function f() is decreasing on the interval [a, b] if f9() < 0 for all values of such that a < < b. = 3 + = 4 The function f() = 3 + is increasing for all real values of. a,f (, ) (, ) The function f() = 4 is increasing on the interval [, 0] and decreasing on the interval [0, ]. Notation The interval [a, b] is the set of all real numbers,, that satisf a < < b. / / 3( + 3)( ) < 0 So 3 < < So f() is decreasing on the interval [ 3, ]. ercise G Write the answer clearl. nline plore increasing and decreasing functions using GeoGebra. Find the values of for which f() is an increasing function, given that f() equals: a b 4 3 c 5 8 d e f g 4 + h FT R cf F4 Find the values of for which f() is a decreasing function, given that f() equals: a 9 b 5 c 4 d 3 3 e f + 5 g _ + 9 _ h ( + 3) 3 Show that the function f() = 4 ( + 3) is decreasing for all R. 4 a Given that t the function DR+ f() = + p is increasing on the interval [, ], find one possible value for p.. ( marks) b State with justification whether this is the onl possible value for p. ( mark).8 Second order derivatives pdr p.dr DR DR + You can find the rate of change of the gradient function b differentiating a function twice. ample = 5 3 Differentiate d d = 5 Differentiate d d = 30 Show that the function f() = is increasing for all real values of. f() = f9() = > 0 for all real values of So > 0 for all real values of. So f() is increasing for all real values of. First differentiate to obtain the gradient function. State that the condition for an increasing function is met. In fact f9() > 4 for all real values of. This is the gradient function. It describes the rate of change of the function with respect to. Differentiating a function = f() twice gives ou the second order derivative, f 0() or _ d d This is the rate of change of the gradient function. It is called the second order derivative. It can also be written as f0(). Notation The derivative is also called the first order derivative or first derivative. The second order derivative is sometimes just called the second derivative. 70 7

11 Chapter ample 4 Given that = find: a d d a So b 7 = = d d = = b d d _ d d = = ample 5 a f() = Given that f() = 3 +, find: a f9() b f 0() = 3 + f9() = b f0() = ercise H Find d d and d d a when equals: b The displacement of a particle in metres at time t seconds is modelled b the function press the fraction as a negative power of. Differentiate once to get the first order derivative. Differentiate a second time to get the second order r derivative. c 9 3 d (5 + 4)(3 ) e f(t) = t + _ t The acceleration of the particle in m s is the second derivative of this function. Find an epression for the acceleration of the particle at time t seconds. 3 Given that = ( 3) 3, find the value of when d _ d = 0. 4 f() = p 3 3p + 4 When =, f 0() =. Find the 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 this form. The coefficient for the second term is ndr DRF RFT ( ) 3 ( 4) = The new power is 3 = 5 Links The velocit of the particle will be f9(t) and its acceleration will be f 0(t). Statistics and Mechanics Year, Section 6. roblem-solving When ou differentiate with respect to, ou treat an other letters as constants..9 Stationar points stationar point on a curve is an point where the curve has gradient zero. You can determine whether a stationar point is a local maimum, a local minimum or a point of inflection b looking at the gradient of the curve on either side. n point on the curve = f() where f9() = 0 is called a stationar point. For a small positive value h: oint is a local maimum. Notation oint is called a local maimum because it The origin is not the largest value the is a point of function can take. It is just the inflection. RB B oint B largest value in that immediate is a local vicinit. minimum. ample 6 a Find the coordinates stationr of the stationar point on the curve with equation = 4 3. b B considering points on eitherd either side of the stationar point, determine whether it is a local maimum, a local minimum or a point of inflection. a Tpe of stationar point f9( h) f9() f9( h) Local maimum ositive 0 Negative Local minimum Negative 0 ositive oint of inflection = 4 3 d d = 43 3 Let d d = 0 Then = = 3 3 = 8 = So = 4 3 = 6 64 = 48 So (, 48) is a stationar point. Negative 0 Negative ositive 0 ositive F nimumd Differentiate and let d d = 0. Notation The plural of maimum is maima and the plural of minimum is minima. Solve the equation to find the value of. Substitute the value of into the original equation to find the value of. 73

12 Chapter b Now consider the gradient on either side of (, 48). Value of Gradient Shape of curve =.9 = = which is ve which is +ve Make a table where ou consider a value of slightl less than and a value of slightl greater than. Calculate the gradient for each of these values of close to the stationar point. Deduce the shape of the curve. _ b d = 30 d _ When =, d = 8 which is, 0 d So (, 7) is a local maimum point. _ When = 4, d = 8 which is. 0 d So (4, 0) is a local minimum point. Differentiate again to obtain the second derivative. Substitute = and = 4 into the second derivative epression. If the second derivative is negative then the point is a local maimum point. If it is positive then the point is a local minimum point. From the shape of the curve, the point (, 48) is a local minimum point. In some cases ou can use the second derivative, f 0(), to determine ermine the nature of a stationar point. If a function f() has a stationar point when = a,, then: f 0() tells ou the rate if f 0(a). 0, the point is a local minimum of change of the gradient function. When f9() = 0 if f 0(a), 0, the point is a local maimum Tf and f 0() > 0 the gradient is If f 0(a) = 0, the point could be a local l minimum, a local increasing from a negative maimum or a point of inflection. You will need to look at value to a positive value, FTof FTHint points on either side to determine its nature. so the stationar point is a minimum. ample 7 a Find the coordinates of the stationar points on the curve with equation = _ b Find d and use it to determine the nature of the stationar points. d a = d d = utting = 0 6( 4)( ) = 0 +DR 6DR D+ So = 4 or = When =, = = 7 When = 4, = = 0 So the stationar points are at (, 7) and (4, 0). nline plore the solution using GeoGebra. DRF RFT Differentiate and put the derivative equal to zero. Solve the equation to obtain the values of for the stationar points. Substitute = and = 4 into the original equation of the curve to obtain the values of which correspond to these values. ample 8 a The curve with equation = has stationar points at = ±a. a Find the value of a. b Sketch the graph of = a = d d = + 8 = + 8 When d d = 0: + 8 = 0 8 = So a = = 4 = 8 ± 3 = ± 3 _ b d d = 3 D + 6 = + D 6 When = 3, = D D= D+ D 3D3 ( ) ( 3 ) 3 = 4 _ and d d = ( ) 3 ( 3 ) = 08 which is negative. So the curve has a local maimum at ( 3, 4 ). When = 3, = _ ( ) ( 3 ) 3 = 4 and _ d d = ( ) 3 ( 3 ) = 08 which is positive. Write differentiate. as determine the -coordinates of the d stationar FT FT Set FTto d points. F Fd F = 0 FT You need to consider the positive and negative roots: _ ( 3 ) 4 = _ ( 3 ) _ ( 3 ) _ ( 3 ) _ ( 3 ) = DRF DR RFT To sketch the curve, ou need to find the coordinates of the stationar points and determine their natures. Differentiate our epression for d d to find _ d d Substitute = _ 3 8 _ and = 3 into the equation of the curve to find the -coordinates of the stationar points. nline Check our solution using our calculator

13 Chapter So the curve has a local minimum at ( 3, 4). The curve has an asmptote at = 0. s,. s, = ercise I Hint For each part of Find the least value of the following functions: questions and : a f() = + 8 b f() = f() = 5 Find f9(). 8 c + Set f9() = 0 and solve Find the greatest value of the to find the value of a f() = 0 5 b f() at the stationar point. f() = (6 + )( ) Find the corresponding 3 Find the coordinates F5 F DR following functions: = 3 + c gradient is zero value of f(). of the points where the on the curves with the given equations. stablish whether these points are local maimum points, local minimum points or points of inflection in each case. a = DRb b = 9 + c = 3 + d = ( e = + f = 4 3) + 54 g = 3 h = _ ( 6) i = 4 4 Sketch the curves with equations given in question 3 parts a, b, c and d, labelling an stationar points with their coordinates. F) F= F = 5 R R Rc f( 5 B considering the gradient on either side of the stationar point on the curve = , show that this point is a point of inflection. Sketch the curve = Find the maimum value and hence the range of values for the function f() = f() = a Find the coordinates of the stationar points of f(), and determine the nature of each. b Sketch the graph of = f(). ± as 0 so = 0 is an asmptote of the curve. Mark the coordinates of the stationar points on our sketch, and label the curve with its equation. You could check d at specific points to help with d our sketch: When = _ 4, d = which is negative. d When =, d = 80 which is positive. d Hint Use the factor theorem with small positive integer values of to find one factor of f9(). Section 7..0 Sketching gradient functions You can use the features of a given function to sketch the corresponding gradient function. This table shows ou features of the graph of a function, = f(), and the graph 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 the -ais Touches the -ais bove the -ais Below the -ais Vertical asmptote Horizontal asmptote at the -ais s The diagram shows the curve with equation = f(). The curve has stationar points at (, 4) and (, 0), and cuts the -ais at ( 3, 0). Sketch the gradient function, = f 9(), showing the coordinates of an points where the curve cuts or meets the -ais. = fˇ() (, 4) = f() = f () = f() 3 = f() = f9(), ositive gradient bove -ais = Maimum Cuts -ais,, Negative gradient Below -ais = Minimum Cuts -ais > ositive gradient bove -ais Watch out Ignore an points where the curve = f() cuts the -ais. These will not tell ou anthing about the features of the graph of = f9(). nline Use GeoGebra to eplore the ke features linking = f() and = f9()

14 Chapter ample 0. Modelling with differentiation The diagram shows the curve with equation = f(). The curve has an asmptote at = and a turning point at ( 3, 8). It cuts the -ais at ( 0, 0). a Sketch the graph of = f 9(). b State the equation of the asmptote of = f 9(). 0 ( 3, 8) = f() You can think of d as small change in. It represents the rate of change of with respect to. d small change in If ou replace and with variables that represent real-life quantities, ou can use the derivative to model lots of real-life situations involving rates of change. a Draw our sketch on a separate set of aes. The graph of = f9() will have the same horizontal scale but will have a different vertical scale. = fˇ() 3 You don t have enough information to work out the coordinates of the -intercept, or the local maimum, of the graph of the gradient function. The graph of = f() is a smooth curve so the graph of = f9() also be a smooth curve. b = 0 T) f() an horizontal asmptotes then of = f9() will have an asmptote at -ais. FTf( FTf9 FT9( FT( FT) FT) will If = has the graph the ercise J For each graph given, sketch the graph of the corresponding gradient function on a separate set of aes. Show the coordinates ordinates of an points F-a F-a where the curve cuts or meets the -ais, and give the equations of an asmptotes. a b c (6, 5) = 0 ( 9, ) (4, 3) d 8 6 DRbRF DD8D8 DRb e FT = f( FT= f f = 7 = 4 V The volume of water in this water butt is constantl changing over time. If V represents the volume of water in the water butt in litres, and t represents the time in seconds, then model V as a function of t. If V = f(t) then dv ou could = f9(t) would represent the rate of change of volume dt with respect to time. The units of dv V would be litres per second. dtt ample DR Given that the volume, V cm 3, of an epanding sphere is related to its radius, r cm, b the formula V = 4_ 3 p r3, find the rate of change of volume with respect to radius at the instant when the radius is 5 cm. V = 4 3 π r3 Differentiate V with respect to r. Remember that dv π is a constant. = 4π r dr When r = 5, dv = 4π 5 Substitute r = 5. dr DD5 = 34 (3 s.f.) So the rate of change is 34 cm 3 per cm. Interpret the answer with units. 5DR = 3 = 6 f() = ( + )( 4) a Sketch the graph of = f(). b n a separate set of aes, sketch the graph of = f 9(). c Show that f 9() = ( 4)(3 ). d Use the derivative to determine the eact coordinates of the points where the gradient function cuts the coordinate aes. = 4 Hint This is an 3 graph with a positive coefficient of 3. Section 4. ample large tank in the shape of a cuboid is to be made from 54 m of sheet metal. The tank has a horizontal base and no top. The height of the tank is metres. Two opposite vertical faces are squares. a Show that the volume, V m 3, of the tank is given b V = 8 _ 3 3 b Given that can var, use differentiation to find the maimum or minimum value of V. c Justif that the value of V ou have found is a maimum

15 Chapter a Let the length of the tank be metres. b 80 Total area, So But So So ut = = = 3 V = V = 54 ( 3 ) = 3 (54 ) V = 8 dv = 8 d dv d = 0 So = 9 0 = 8 = 3 3 or But is a length so = 3 When = 3, V = 8D8 D 3D3 D _ 3 33 = 54 8 = 36 V = 36 is a maimum or minimum value of V. c _ d V d = 4 When = 3, _ d V = 4 3 = d This is negative, so V = 36 is the maimum value of V. roblem-solving You don t know the length of the tank. Write it as metres to simplif our working. You could also draw a sketch to help ou find the correct epressions for the surface area and volume of the tank. Draw a sketch. Rearrange to find in terms of. Substitute the epression for into the equation. Simplif. Differentiate V with respect to and put dv d = 0. Rearrange to find. is a length so use the positive solution. DRF D =DR 3DR = Substitute the value of into the epression for V. Find the second derivative of V. _ d V < 0 so V = 36 is a maimum. d / / ercise K Find dθ dt where θ = t 3t. Find d where = pr. dr 3 Given that r =, find the value of dr when t = 3. t dt 4 The surface area, cm, of an epanding sphere of radius r cm is given b = 4pr. Find the rate of change of the area with respect to the radius at the instant when the radius is 6 cm. 5 The displacement, s metres, of a car from a fied point at time t seconds is given b s = t + 8t. Find the rate of change of the displacement with respect to time at the instant when t = 5. 6 rectangular garden is fenced on three sides, and the house forms the fourth side of the rectangle. a Given that the total length of the fence is 80 m, show that the area,,, of the garden is given b the formula = (80 ), where is the distance from the house to the end of the garden. b Given that the area is a maimum for this length of fence, find the dimensions of the enclosed garden, and the area which is enclosed. T, 7 closed clinder has total surface area equal F. to 600p. a Show that the volume, V cm 3, of this clinder is given b the formula V = 300pr pr 3, where r cm is the radius of the clinder. b Find the maimum volume of such a clinder. 8 sector of a circle has area 00 cm. a Show that the perimeter bd of this sector is given b the formula M N = r _ + 00 r, r. _ 00 p r cm b Find the minimum value for the perimeter. 9 shape consists sts of a rectangular base with a semicircular top, as shown. a Given that the of the shape is 40 cm, show that its r cm area, cm, is given DR_ Dperimeter b the formula = 40r r pr where r cm is the radius of the semicircle. ( marks) b Hence find the maimum value for the area of the shape. (4 marks) 0 The shape shown is a wire frame in the form of a large rectangle split b parallel lengths of wire into smaller equal-sized rectangles. mm a Given that the total length of wire used to complete the whole frame is 5 mm, show that the area of mm the whole shape, mm, is given b the formula = where mm is the width of one of the smaller rectangles. b Hence find the maimum area which can be enclosed in this wa. (4 marks) (4 marks) 8

16 Chapter / / Mied eercise rove, from first principles, that the derivative of 0 is 0. (4 marks) The point with coordinates (, 4) lies on the curve with equation = The point B also lies on the curve and has -coordinate ( + δ ). a Show that the gradient of the line segment B is given b ( δ ) + 3δ + 6. b Deduce the gradient of the curve at point. 3 curve is given b the equation = , where. 0. t the points, B and C on the curve, =, and 3 respectivel. Find the gradient of the curve at, B and C. 4 Calculate the -coordinates of the points on the curve with equation = 7 3 at which the gradient is equal to 6. 5 Find the -coordinates of the two points on the curve with equation = 3 + where the gradient is. Find the corresponding -coordinates. (4 marks) 6 The function f is defined b f() = + 9 9, [ R,, Þ 0. a Find f9(). ( marks) b Solve 3DR DR f9() = 0. ( marks) 7 Given that = 3 4,. 0, find d d 8 curve has equation = _ 3_. d a Show that d 3 =D= 3 D D_D_ D D D(4 ). ( marks) D b Find the coordinates of the point on the curve where the gradient is zero. ( marks) ated(4d 9 a pand ( 3_ )( _ R,FT =DR DR = + ). ( marks) b curve has equation = ( 3_ )( _ + ),. 0. Find d d c Use our answer to part b to calculate the gradient of the curve at the point where = 4. ( marks) ( mark) / / / / curve C has equation = a Find d in terms of. ( marks) d b The points and Q lie on C. The gradient of C at both and Q is. The -coordinate of is 3. i Find the -coordinate of Q. ii Find an equation for the tangent to C at, giving our answer in the form = m + c, where m and c are constants. iii If this tangent intersects the coordinate aes at the points R and S, find the length of RS, giving our answer as a surd. 3 curve has equation = 8 + 3, > 0. Find the equations of the tangent and the normal to the curve at the point where =. 4 The normals to the curve = , at the points FT 0) and (, 0), meet at the point N. a Find the coordinates of N. (7 marks) b Calculate the area of triangle N. 5 curve C has equation = 3 FT(0, 4 and cuts the -ais at a point. The line L is a tangent to the curve at,, and cuts the curve at the point Q. Show that the distance Q is 7. (7 marks) 6 Given that = 3_ + 48 od3d,. 0 a find the value of and the value of when d d = 0. (5 marks) b show that the value of which ou found in part a is a minimum. ( marks) 7 curve has equation ationdation D =D= D Determine, b calculation, the coordinates of the stationar points of the curve. 8 The function f, defined for [ R,. 0, is such that: f9() = + a Find the value of f 0 () at = 4. b rove that f is an increasing function. (4 marks) / 0 Differentiate with respect to : _ + The curve with equation = a + b + c passes through the point (, ). The gradient of the curve is zero at the point (, ). Find the values of a, b and c. (5 marks) 9 curve has equation = Find the coordinates of its local maimum. (4 marks) 0 f() = a Find the coordinates of the stationar points of f(), and determine the nature of each of them. b Sketch the graph of = f(). 8 83

17 Chapter / The diagram shows part of the curve with equation = f(), where: _ f() = 00 50,. 0 The curve cuts the -ais at the points and C. The point B is the maimum point of the curve. a Find f9(). b Use our answer to part a to calculate the coordinates of B. (4 marks) The diagram shows the part of the curve with equation = 5 _ for which > 0. The point (, ) lies on the curve and is the origin. a Show that = _ Taking f() = _ : b Find the values of for which f9() = 0. (4 marks) c Hence, or otherwise, find the minimum distance from to the curve, showing that our answer is a minimum. (4 marks) 3 The diagram shows part of the curve with equation = 3 spdr The curve touches the -ais at and crosses the -ais at C. The points and B are stationar points on the curve. a Show that C has coordinates (3, 0). ( mark) b Using calculus culus and showing all our working, find the coordinates DRB of and B. (5 marks) 4 The motion of a damped spring is modelled using this graph. n a separate graph, sketch the gradient function for this model. Choose suitable labels and units for each ais, and indicate the coordinates of an points where the gradient function crosses the horizontal ais. DR 5 The volume, V cm 3, of a tin of radius r cm is given b the formula V = p(40r r r 3 ). Find the positive value of r for which dv = 0, and find the value of V which dr corresponds to this value of r. Displacement (cm) B B C (, ) C Time (seconds) / / 7 wire is bent into the plane shape BCD as shown. Shape BD is a rectangle and BCD is a semicircle with diameter BD. The area of the region enclosed b the wire is R m, = metres, and B = D = metres. The total length of the wire is m. a Find an epression for in terms of. b rove that R = (8 4 p). (4 marks) 8 Given that can var, using calculus and showing our working: c find the maimum value of R. (You do not have to prove that the value ou obtain is a maimum.) (5 marks) 8 clindrical biscuit tin has a close-fitting lid which overlaps the tin cm b cm, as shown. The radii of the tin and the lid are both cm. cm The tin and the lid are made from a thin sheet of metal of area Lid 80p cm and there is no wastage. The volume of the tin is V cm 3. cm a Show that V = p(40 3 ). (5 marks) Given that can var: b use differentiation to find the positive value of for which V is stationar. Tin c rove that this value of gives a maimum F. value of V. ( marks) d Find this maimum value of V.. ( mark) e Determine the percentage of the sheet metal used in the lid when V is a maimum. ( marks) B C 9 The diagram shows s an open tank for storing water, BCDF. The sides BF and CDF are rectangles. The triangular ar ends D and are isosceles, D and /D ends D F and BCF are horizontal. Given that D = a show that the area DRBCF DRa F D = /BFC = 90. The anddr FDR vertical and F is metres:dmetres: FDD odo of triangle D is _ 4 m Given also that the capacit of the container is 4000 m 3 and that the total area of the two triangular and two rectangular sides of the container is S m : FR DRF b show that S = Given that can var: c use calculus to find the minimum value of S. d justif that the value of S ou have found is a minimum. CDFR B D C (4 marks) (6 marks) ( marks) 6 The total surface area, cm, of a clinder with a fied volume of 000 cm 3 is given b the formula = p + 000, where cm is the radius. Show that when the rate of change of the area with respect to the radius is zero, 3 = _ 500 p Challenge a Find the first four terms in the binomial epansion of ( + h) 7, in ascending powers of h. b Hence prove, from first principles, that the derivative of 7 is

18 Chapter Summar of ke points The gradient of a curve at a given point is defined as the gradient of the tangent to the curve at that point. The gradient function, or derivative, of the curve = f() is written as f9() or d d f ( + h) f() f 9 () = lim h 0 h The gradient function can be used to find the gradient of the curve for an value of. 3 For all real values of n, and for a constant a: If f() = n then f 9 () = n n If f() = a n then f 9 () = an n d If = n then d = n d n If = a n then d = an n 4 For the quadratic curve with equation = a + b + c,, the derivative is given b d d = a + b 5 If = f() ± g(), then d = f9() ± g9(). d 6 The tangent to the curve = f() at the point with coordinates (a, f(a)) has equation f(a) = f9(a)( a) 7 The normal to the curve = f() at the point with coordinates (a, f(a)) has equation f(a ) = _ f 9 (a ) ( a ) 8 The function f() is increasing on the interval [a, b] if f9() > 0 for all values of such that a,, b. The function f() is decreasing on the interval [a, b] if f9() < 0 for all values of such that a,, b.. 9 Differentiating f() twice gives ou the second order derivative, f 0() or _ d functiondr DR a function = d 0 n point on the curve rved= DRon = f() where f9() = 0 is called a stationar point. For a small positive value h: Tpe of stationar point f9( h) f9() f9( + h) Local maimum ositive 0 Negative Local minimum Negative 0 ositive oint of inflection Negative 0 Negative ositive 0 ositive If a function f() has a stationar point when = a, then: if f 0(a). 0, the point is a local minimum if f 0(a), 0, the point is a local maimum. If f 0(a) = 0, the point could be a local minimum, a local maimum or a point of inflection. You will need to look at points on either side to determine its nature. 86

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