Higher. Differentiation 33

Transcription

1 hsn uknet Higher Mathematics UNIT UTCME Differentiation Contents Differentiation Introduction to Differentiation Finding the Derivative 4 Differentiating with Respect to ther Variables 8 4 Rates of Change 9 Equations of Tangents 40 6 Increasing and Decreasing Curves 44 7 Stationar Points 4 8 Determining the Nature of Stationar Points 46 9 Curve Sketching 49 0 Closed Intervals Graphs of Derivatives ptimisation 4 HSN00 This document was produced speciall for the HSNuknet website, and we require that an copies or derivative works attribute the work to Higher Still Notes For more details about the copright on these notes, please see

2 Unit Differentiation UTCME Differentiation Introduction to Differentiation From our work on Straight Lines, we saw that the gradient (or steepness ) of a line is constant However, the steepness of other curves ma not be the same at all points In order to measure the steepness of other curves, we can use lines which give an increasingl good approimation to the curve at a particular point n the curve with equation = f ( ), suppose point A has coordinates a, f ( a ) At the point B where = a + h, we have = f a + h Thus the chord AB has gradient m AB = = ( + ) f a h f a a + h a f a h f a h ( + ) If we let h get smaller and smaller, ie h 0, then B moves closer to A This means that m AB gives a better estimate of the steepness of the curve at the point A We use the notation f ( a) for the steepness of the curve when = a So f ( a + h) f ( a + h) f a + h f a f ( a) = lim h 0 h f ( a) Given a curve with equation = f ( ), an epression for f is called the derivative and the process of finding this is called differentiation It is possible to use this definition directl to find derivates, but ou will not be epected to do this Instead, we will learn rules which allow us to quickl find derivatives for certain curves f ( a) A A a a = f ( ) B a + h = f ( ) B a + h Page HSN00

3 Unit Differentiation Finding the Derivative The basic rule for differentiating f ( ) =, n R, with respect to is: If n n f ( ) = then f ( ) = n Stated simpl: the power (n) multiplies to the front of the term, and the power lowers b one (giving n ) EXAMPLES 4 Given f ( ) =, find f f ( ) = 4 Differentiate f ( ) =, 0, with respect to 4 f ( ) = For an epression of the form =, we denote the derivative with respect to b d EXAMPLE Differentiate =, 0, with respect to n d 4 = When finding the derivative of an epression with respect to, we use the notation d d EXAMPLE 4 Find the derivative of d d = Preparing to differentiate, 0, with respect to It is important that before ou differentiate, all brackets are multiplied out and there are no fractions with an term in the denominator (bottom line) For eample: = = = 4 = 4 4 = 4 Page 4 HSN00

4 Unit Differentiation EXAMPLES Differentiate with respect to, where > 0 d d ( ) = = = Given =, where 0, find d = = d = Note It is good practice to ti up our answer Terms with a coefficient For an constant a, if f = a g then f = a g Stated simpl: constant coefficients are carried through when differentiating n n So if f ( ) = a then f ( ) = an EXAMPLES A function f is defined b f ( ) = Find f f ( ) = 6 Differentiate = 4 with respect to, where 0 = 8 d 8 = Differentiate, 0, with respect to d d ( ) = = 4 Page HSN00

5 Unit Differentiation 4 Given =, > 0, find d = = 4 d = 4 Differentiating more than one term The following rule allows us to differentiate epressions with several terms If f ( ) = g ( ) + h( ) then f ( ) = g ( ) + h ( ) Stated simpl: differentiate each term separatel EXAMPLES A function f is defined for R b f ( ) = + Find f f ( ) = Differentiate d = = with respect to Note The derivative of an term (eg,, For eample: d d ( 6 ) 6, 0 = d d ) is alwas a constant = The derivative of a constant (eg, 0, π) is alwas zero For eample: d d = d 0, 0 d = Page 6 HSN00

6 Unit Differentiation Differentiating more comple epressions We will now consider more comple eamples where we will have to use several of the rules we have met EXAMPLES Differentiate =, > 0, with respect to = = = d = = Find when d = + = ( )( + ) = + = 6 = d 6 A function f is defined for 0 b f ( ) = + Find f f = + f ( ) = = 4 Differentiate d d = = = with respect to, where 0 Note You need to be confident working with indices and fractions Remember Before differentiating, the brackets must be multiplied out Page 7 HSN00

7 Unit Differentiation Differentiate d d = +, > 0, with respect to = + 6 = = 9 9 = 6 Find the derivative of ( ) 6 = + = + 6 = + 6 d = = +, > 0, with respect to Differentiating with Respect to ther Variables So far we have differentiated functions and epressions with respect to However, the rules we have been using still appl if we differentiate with respect to an other variable When modelling real-life problems we often use appropriate variable names, such as t for time and V for volume EXAMPLES Differentiate d dt t t t = 6t t with respect to t Given A( r ) = πr, find A ( r ) A( r ) = πr A ( r ) = πr When differentiating with respect to a certain variable, all other letters are treated as constants EXAMPLE Differentiate d dp p = p with respect to p Remember a a b = b Remember a b a b = + Remember π is just a constant Note Since we are differentiating with respect to p, we treat as a constant Page 8 HSN00

8 Unit Differentiation 4 Rates of Change The derivative of a function describes its rate of change This can be evaluated for specific values b substituting them into the derivative EXAMPLES Given f ( ) =, find the rate of change of f when = 4 f ( ) = 0 4 f = 0 = 0 8 = 80 Given = for 0, calculate the rate of change of when = 8 = At = 8, = d 8 = d = = = 96 = = 48 Displacement, velocit and acceleration The velocit v of an object is defined as the rate of change of displacement s with respect to time t That is: ds v = dt Also, acceleration a is defined as the rate of change of velocit with respect to time: dv a = dt EXAMPLE A ball is thrown so that its displacement s after t seconds is given b s( t ) = t t Find its velocit after seconds v ( t ) = s ( t ) = 0 t b differentiating s ( t ) = t t with respect to t Substitute t = into v( t ) : v = 0 = After seconds, the ball has velocit metres per second Page 9 HSN00

9 Unit Differentiation Equations of Tangents As we alrea know, the gradient of a straight line is constant We can determine the gradient of a curve, at a particular point, b considering a straight line which touches the curve at the point This line is called a tangent tangent The gradient of the tangent to a curve = f ( ) at = a is given b f ( a) This is the same as finding the rate of change of f at a To work out the equation of a tangent we use b = m( a) Therefore we need to know two things about the tangent: a point, of which at least one coordinate will be given; the gradient, which is calculated b differentiating and substituting in the value of at the required point EXAMPLES Find the equation of the tangent to the curve with equation at the point (, ) We know the tangent passes through (, ) = To find its equation, we need the gradient at the point where = : = = d At =, m = = 4 Now we have the point (, ) and the gradient m = 4, so we can find the equation of the tangent: b = m( a) = 4( ) = = 0 Page 40 HSN00

10 Unit Differentiation Find the equation of the tangent to the curve with equation at the point where = = We need a point on the tangent Using the given -coordinate, we can find the -coordinate of the point on the curve: = = ( ) ( ) = + = So the point is (, ) We also need the gradient at the point where = : = = d At =, m = ( ) = Now we have the point (, ) and the gradient m =, so the equation of the tangent is: b = m( a) = ( + ) + = 0 A function f is defined for > 0 b f ( ) = Find the equation of the tangent to the curve = f ( ) at P P = f ( ) We need a point on the tangent Using the given -coordinate, we can find the -coordinate of the point P: f ( ) = = = So the point is (, ) Page 4 HSN00

11 Unit Differentiation We also need the gradient at the point where = : f ( ) = f ( ) = = At =, m = = 4 4 Now we have the point, and the gradient m = 4, so the equation of the tangent is: b = m( a) ( ) = 4 = = 0 4 Find the equation of the tangent to the curve where = 8 = at the point We need a point on the tangent Using the given -coordinate, we can work out the -coordinate: = 8 = ( ) = 4 So the point is 8, 4 We also need the gradient at the point where = 8: = = = d = At = 8, m = 8 = = Now we have the point ( 8, 4) and the gradient m =, so the equation of the tangent is: b = m( a) 4 = ( + 8) = = 0 Page 4 HSN00

12 Unit Differentiation A curve has equation = + + Find the coordinates of the points on the curve where the tangent has gradient 4 The derivative gives the gradient of the tangent: d = + We want to find where this is equal to 4: + = 4 = 0 ( + )( ) = 0 = or = Now we can find the -coordinates b using the equation of the curve: = ( ) + ( ) + = + = = 6 6 So the points are (, 6 ) and ( 9 ), = + + = = 7 + = 9 8 Remember Before solving a quadratic equation ou need to rearrange to get quadratic= 0 Page 4 HSN00

13 Unit Differentiation 6 Increasing and Decreasing Curves A curve is said to be strictl increasing when 0 d > This is because when 0 d >, tangents will slope upwards from left to right since their gradients are positive This means the curve is also moving upwards, ie strictl increasing Similarl: A curve is said to be strictl decreasing when 0 d < strictl increasing 0 d > strictl decreasing strictl 0 d < increasing 0 d > EXAMPLES A curve has equation = 4 + Determine whether the curve is increasing or decreasing at = 0 = 4 + = 8 d = 8 When = 0, 8 0 d = 0 = > 0 Since > 0, the curve is increasing when 0 d = Note 0 0 < Page 44 HSN00

14 Unit Differentiation Show that the curve 4 = + + is never decreasing = + + d = ( + ) 0 Since d is never less than zero, the curve is never decreasing Remember The result of squaring an number is alwas greater than, or equal to, zero 7 Stationar Points At some points, a curve ma be neither increasing nor decreasing we sa that the curve is stationar at these points This means that the gradient of the tangent to the curve is zero at stationar points, so we can find them b solving f ( ) = 0 or = 0 d The four possible stationar points are: Turning point Horizontal point of inflection Maimum Minimum Rising Falling A stationar point s nature (tpe) is determined b the behaviour of the graph to its left and right This is often done using a nature table Page 4 HSN00

15 Unit Differentiation 8 Determining the Nature of Stationar Points To illustrate the method used to find stationar points and determine their nature, we will do this for the graph of f ( ) = Step Differentiate the function f ( ) = Step Find the stationar values b solving f ( ) = 0 f = = 0 Step Find the -coordinates of the stationar points Step 4 Write the stationar values in the top row of the nature table, with arrows leading in and out of them Step Calculate f for the values in the table, and record the results This gives the gradient at these values, so zeros confirm that stationar points eist here Step 6 Calculate f for values slightl lower and higher than the stationar values and record the sign in the second row, eg: f ( 08) > 0 so enter + in the first cell Step 7 We can now sketch the graph near the stationar points: + means the graph is increasing and means the graph is decreasing Step 8 The nature of the stationar points can then be concluded from the sketch ( ) 6 + = 0 ( 6) ( )( ) = 0 = or = f = 9 so (, 9 ) is a stat pt f = 8 so (, 8 ) is a stat pt f ( ) Graph f ( ) Graph 0 0 f ( ) Graph f ( ) Graph (, 9 ) is a ma turning point (, 8 ) is a min turning point Page 46 HSN00

16 Unit Differentiation EXAMPLES A curve has equation = Find the stationar points on the curve and determine their nature Given = , = + 9 d Stationar points eist where 0 d = : + 9 = 0 ( ) 4 + = 0 ( ) 4 + = 0 ( )( ) = 0 = 0 or = 0 = = When =, = = = 0 Therefore the point is (,0 ) Nature: d Graph So (, 0 ) is a maimum turning point, (, 4) is a minimum turning point When =, = = = 4 Therefore the point is (, 4) Page 47 HSN00

17 Unit Differentiation Find the stationar points of 4 Given = 4, = 8 d Stationar points eist where 0 d = : 8 = 0 4 ( ) = 0 4 = 0 or = 0 = 0 = When = 0, = 4( 0) ( 0) = 0 4 Therefore the point is ( 0, 0 ) Nature: 0 d Graph So ( 0, 0 ) is a rising point of inflection, 7, is a maimum turning point 8 4 = 4 and determine their nature When =, = 4 = = Therefore the point is, 8 Page 48 HSN00

18 Unit Differentiation A curve has equation = + for 0 Find the -coordinates of the stationar points on the curve and determine their nature Given = +, = d = Stationar points eist where 0 d = : = 0 = = = ± Nature: d Graph So the point where = is a maimum turning point and the point where = is a minimum turning point 9 Curve Sketching In order to sketch a curve, we first need to find the following: -ais intercepts (roots) solve = 0; -ais intercept find for 0 = ; stationar points and their nature Page 49 HSN00

19 Unit Differentiation EXAMPLE Sketch the curve with equation = -ais intercept, ie = 0: = ( 0) ( 0) = 0 Therefore the point is ( 0, 0 ) Given =, = 6 6 d Stationar points eist where 0 d = : 6 6 = 0 6 ( ) = 0 6 = 0 or = 0 = 0 When = 0, = ( 0) ( 0) = 0 Therefore the point is ( 0, 0 ) = Nature: 0 d Graph ais intercepts ie = 0: = 0 = 0 = 0 ( 0, 0) When =, ( ) = 0 or = = = 0 = Therefore the point is (, ) =, 0 ( 0, 0 ) is a maimum turning point (, ) is a minimum turning point = (, ) Page 0 HSN00

20 Unit Differentiation 0 Closed Intervals Sometimes it is necessar to restrict the part of the graph we are looking at using a closed interval (also called a restricted domain) The maimum and minimum values of a function can either be at its stationar points or at the end points of a closed interval Below is a sketch of a curve with the closed interval 6 shaded maimum 6 minimum Notice that the minimum value occurs at one of the end points in this eample It is important to check for this whenever we are dealing with a closed interval EXAMPLE A function f is defined for 4 b f ( ) = 4 + Find the maimum and minimum value of f ( ) Given f ( ) = 4 +, f ( ) = Stationar points eist where f = 0: = 0 ( ) = 0 ( )( + ) = 0 = 0 or + = 0 = = Page HSN00

21 Unit Differentiation To find coordinates of stationar points: f = 4 + = = Therefore the point is (, ) Nature: f ( ) Graph ( ) ( ) ( ) ( ) = ( ) 4 + f = = = 46 7 Therefore the point is 46 (, ) 46 (, ) 7 7 is a ma turning point (, ) is a min turning point Points at etremities of closed interval: f ( ) = ( ) ( ) 4( ) + = = Therefore the point is (, ) f 4 = = = 4, Therefore the point is Now we can make a sketch: (, 46 ) 7 ( 4, ) Note A sketch ma help ou to decide on the correct answer, but it is not required in the eam (, ) (, ) The maimum value is which occurs when = 4 The minimum value is which occurs when = Page HSN00

22 Unit Differentiation Graphs of Derivatives n n The derivative of an term is an term the power lowers b one For eample, the derivative of a cubic (where is the highest power of ) is a quadratic (where is the highest power of ) When drawing a derived graph: All stationar points of the original curve become roots (ie lie on the - ais) on the graph of the derivative Wherever the curve is strictl decreasing, the derivative is negative So the graph of the derivative will lie below the -ais it will take negative values Wherever the curve is strictl increasing, the derivative is positive So the graph of the derivative will lie above the -ais it will take positive values Quadratic Cubic Quartic dec dec inc inc dec inc inc inc dec + Linear + + Quadratic + + Cubic Page HSN00

23 Unit Differentiation EXAMPLE The curve = f ( ) shown below is a cubic It has stationar points where = and = 4 = f ( ) Sketch the graph of = f ( ) Since = f ( ) has stationar points at = and = 4, the graph of = f ( ) crosses the -ais at = and = 4 4 = f ( ) Note The curve is increasing between the stationar points so the derivative is positive there ptimisation In the section on closed intervals, we saw that it is possible to find maimum and minimum values of a function This is often useful in applications; for eample a compan ma have a function P ( ) which predicts the profit if is spent on raw materials the management would be ver interested in finding the value of which gave the maimum value of P ( ) The process of finding these optimal values is called optimisation Sometimes ou will have to find the appropriate function before ou can start optimisation Page 4 HSN00

24 Unit Differentiation EXAMPLE Small plastic tras, with open tops and square bases, are being designed The must have a volume of 08 cubic centimetres h (a) The internal length of one side of the base is centimetres, and the internal height of the tra is h centimetres (a) Show that the total internal surface area A of one tra is given b 4 A = + (b) Find the dimensions of the tra using the least amount of plastic Volume = area of base height = h We are told that the volume is 08 cm, so: Volume = 08 h = h = Let A be the surface area for a particular value of : A = + h 4 08 We have h =, so: A = = + 4 (b) The smallest amount of plastic is used when the surface area is minimised da 4 = d da Stationar points occur when 0 d = : Nature: 6 4 = 0 da d 0 + = 6 Graph = 6 So the minimum surface area occurs when = 6 For this value of : 08 h = = 6 So a length and depth of 6 cm and a height of cm uses the least amount of plastic Page HSN00

25 Unit Differentiation ptimisation with closed intervals In practical situations, there ma be bounds on the values we can use For eample, the compan from before might onl have available to spend on raw materials We would need to take this into account when optimising Recall from the section on Closed Intervals that the maimum and minimum values of a function can occur at turning points or the endpoints of a closed interval The point P lies on the graph of f = + 4, between = 0 and = 7 A triangle is formed with vertices at the origin, P and ( p,0) (a) Show that the area, A square units, of this triangle is given b A = p 6 p + p 4 (b) Find the greatest possible value of A and the corresponding value of p for which it occurs (a) The area of the triangle is A = base height = p f ( p) ( p 4) = p p + 4 = p 6 p + p = f ( ) ( p f ( p) ) P, p 7 Page 6 HSN00

26 Unit Differentiation (b) The greatest value occurs at a stationar point or an endpoint da At stationar points 0 dp = : da dp 4 p p = + = p 4 p 4 0 p 0 + = 8p + = 0 ( p )( p ) = 0 p = or p = Now evaluate A at the stationar points and endpoints: when p = 0, A = 0; when p =, when p =, when p = 7, A = 6 + = 7 ; 4 A = 6 + = ; 4 A = = 4 So the greatest possible value of A is, which occurs when p = 7 Page 7 HSN00