8 Differential Calculus 1 Introduction

Transcription

1 8 Differential Calculus Introduction The ideas that are the basis for calculus have been with us for a ver long time. Between 5 BC and 5 BC, Greek mathematicians were working on problems that would find their absolute solution with the invention of calculus. However, the main developments were much more recent; it was not until the th centur that major progress was made b mathematicians such as Fermat, Roberval and Cavalieri. In the 7th centur, calculus as it is now known was developed b Sir Isaac Newton and Gottfried Wilhelm von Leibniz. Sir Isaac Newton famousl discovered gravit when an apple fell on his head. Sir Isaac Newton Gottfried Wilhelm von Leibniz Consider the graph of a quadratic, cubic or trigonometric function. Differential calculus is a branch of mathematics that is concerned with rate of change. In a graph, the rate of change is the gradient. Although linear functions have a constant gradient, most functions have changing gradients. Being able to find a pattern for the gradient of curves is the aim of differentiation. Differentiation is the process used to find rate of change. The gradient of a straight line is constant. For eample, in the diagram below, the gradient. (, ) 8

2 8 Differential Calculus Introduction However, when a curve is considered, it is obvious that the gradient is constantl changing. D B F C A E The sections AB, CD, EF have positive gradient (the function is increasing) and sections BC, DE have negative gradient (the function is decreasing).the question we need to answer is: how do we measure the gradient of a curve? 8. Differentiation b first principles We know gradient, and one method of finding the gradient of a straight line is to use. B P A Consider the coordinates (, f()) and h, f h the gap between the - coordinates is h. This can be used to find an approimation for the gradient at P as seen in the diagram. Gradient f h f h f h f h This is calculating the gradient of the chord AB shown in the diagram. As the chord becomes smaller, the end-points of the curve are getting closer together, and h becomes smaller. bviousl, this approimation becomes more accurate as h becomes smaller. Finall h becomes close to zero and the chord s length becomes so small that it can be considered to be the same as the point P. 8

3 8 Differential Calculus Introduction The gradient of a function, known as the derivative, with notation f lim hs f h f h is defined: The notation lim means the limit as h tends to zero. This is the value to which the hs epression converges as h becomes infinitesimall small. f The idea of a limit is similar to sum of a infinite series met in Chapter and also to horizontal asmptotes in Chapter. Eample Find the derivative of f. f h h h h f h f h h h h f h f h h h h h f h f So f lim (as h S ) hs h Hence at an point on the curve, the gradient is given b. This process is known as differentiation b first principles. Eample Find the derivative of f. f h h h h h f h f h h h h h h f h f h h h h h h h f h f So f lim (as h h S ) hs h Eample Find the derivative of f 7. f h 7 h 7 7h f h f 7 7h 7 7h f h f h 7h h 7 f h f So f lim 7 hs h 85

4 8 Differential Calculus Introduction Eample Find the derivative of f 5. f h 5 f h f 5 5 f h f h h So f lim hs f h f h What happens in a sum or difference of a set of functions? Consider the sum of the previous three eamples, i.e. f 7 5. Eample Find the derivative of f 7 5. f h h 7 h 5 h h h 7 7h 5 f h f h h h 7 7h h h h 7h f h f h h h h 7h h h h 7 f h f So f lim 7 (as h h S ) hs h This demonstrates that differentiation of a function containing a number of terms can be differentiated term b term. Eercise Find f using the method of differentiation from first principles: f 5 f 8 f 5 f 7 f 8 f 5 f f f f f f 8 f 8 f 5 f

5 8. Differentiation using a rule Looking at the patterns in Eercise, it should be obvious that for: 8 Differential Calculus Introduction f a n f an n Multipl b the power and subtract from the power This rule can be used to perform differentiation. In particular, notice that f a gives f a. a Unless specificall required, differentiation b first principles is not used the above rule makes the process much shorter and easier. This is no surprise the gradient of a linear function is constant. Also, f k gives f. The gradient of a horizontal line is zero. k k Differential calculus was developed b two mathematicians, Isaac Newton and Gottfried Leibniz. There are two commonl used notations: Functional / Newtonian notation Derivative f f Geometrical / Leibniz notation d d Either notation can be used, and both will appear in questions. 87

6 8 Differential Calculus Introduction Eample Differentiate 5. Simplifing, 5 As with first principles, we can differentiate a sum b differentiating term b term. d d 5# 5 5 Eample Find the derivative of f. f f Eample Differentiate 5. Sometimes it is necessar to simplif the function before differentiating. d d # # 5 Eample Find g () for g. Here we are evaluating the derivative when. First differentiate: g Then substitute : g # 7 So the gradient of g() at is 7. 88

7 8 Differential Calculus Introduction Eample Find the coordinates of the points where the gradient is f 8 7. for Here we are finding the points on the curve where the derivative is First differentiate: f 8. Then solve the equation f 8 or At, 55 and at, 5 So the coordinates required are, 55 and, 5. Eercise Differentiate these functions. a f b f c d f 5 e f f f f 7 g f h f 8 i f j f 5 k 5 l f 5 5 m 5 7 n o p q 5 r 5 Find f () for f. Find g () for g. Find the gradient of when. 5 Find the gradient of when. Find the coordinates of the point where the gradient is for f. 7 Find the coordinates of the points where the gradient is for f 8. 8

8 8 Differential Calculus Introduction 8. Gradient of a tangent A tangent is a straight line that touches a curve (or circle) at one point. Differentiation can be used to find the value of the gradient at an particular point on the curve. At this instant the value of the gradient of the curve is the same as the gradient of the tangent to the curve at that point. Finding the gradient using a graphing calculator Using a graphing calculator, the value of the gradient at an point can be calculated. For eample, for, at This is helpful, especiall for checking answers. However, we often need the derivative function and so need to differentiate b hand. The calculator can onl find the gradient using a numerical process and is unable to differentiate algebraicall. Tangents and normals The gradient at a point is the same as the gradient of the tangent to the curve at that point. ften it is necessar to find the equation of the tangent to the curve. Method for finding the equation of a tangent. Differentiate the function.. Substitute the required value to find the gradient.. Find the -coordinate (if not given).. Find the equation of the tangent using this gradient and the point of contact using m.

9 8 Differential Calculus Introduction Eample Find the equation of the tangent to at. Differentiating, d d and so at, d d The point of contact is when, and so, i.e., Using m, the equation of the tangent is 7 The normal to a curve is also a straight line. The normal to the curve is perpendicular to the curve at the point of contact (therefore it is perpendicular to the tangent). Finding the equation of a normal to a curve is a ver similar process to finding the tangent. Method for finding the equation of a normal. Differentiate the function.. Substitute the required value to find the gradient.. Find the gradient of the perpendicular using m m.. Find the -coordinate (if not given). 5. Find the equation of the normal using this gradient and the point of contact using m. Eample Find the equation of the tangent, and the equation of the normal, to at. d Using the method,. d At, d. d At,.

10 8 Differential Calculus Introduction So the equation of the tangent is The equation of the normal uses the same point but the gradient is different. Using m the gradient of the normal is. m, Using m the equation of the normal is To find the perpendicular gradient turn the fraction upside down and change the sign. Eample Find the equation of the tangent, and the equation of the normal, to where the curve crosses the -ais. The curve crosses the -ais when. So, i.e. (, ) Differentiating, At d d., d d. Using m,. the equation of the tangent is Then the gradient of the normal will be. Using m the equation of the normal is Eercise Find the equation of the tangent and the equation of the normal to: a at b at c at d at e at f at The curve meets the -ais at A and the -ais at B. Find the equation of the tangents at A and B. Find the equation of the normal to at. The tangent at P(, ) to the curve meets the curve again at Q. Find the coordinates of Q. 5 Find the equation of the tangent to at.

11 8 Differential Calculus Introduction Find the equations of the tangents to the curve at the points where the curve cuts the -ais. Find the point of intersection of these tangents. 7 Find the equations of the tangents to the curve 5 at the points of intersection of the curve and the line 5. 8 Find the equation of the normal to, which has a gradient of. Find the equations for the tangents at the points where the curves and meet. For 7, find the equation of the tangent at. For, find the equation of the normal to the curve at. Now find the area of the triangle formed between these two lines and the -ais. Tangents on a graphing calculator It is possible to draw the tangent to a curve using a graphing calculator. To find the equation of the tangent to at, the calculator can draw the tangent and provide the equation of the tangent. 8. Stationar points The gradient of a curve is constantl changing. In some regions, the function is increasing, in others it is decreasing, and at other points it is stationar. B C A D At points A, B, C and D, the tangent to the curve is horizontal and so the gradient is zero. These points are known as stationar points. ften these points are ver important to find, particularl when functions are used to model real-life situations.

12 8 Differential Calculus Introduction For eample, a stone is thrown and its height, in metres, is given b t. h(t) ht t t, t h t t and so h t when t i.e. t So the maimum height of the stone is given b h metres, which is the point where the gradient is zero. We have met this concept before as maimum and minimum turning points in Chapters, and, and these are in fact eamples of stationar points. Stationar points are when d. d Stationar points are coordinate points. The -coordinate is when the stationar point occurs. The -coordinate is the stationar value. Note that the maimum turning point is not necessaril the maimum value of that function. Although it is the maimum value in that region (a local maimum) there ma be greater values. For eample, for the cubic function the greatest value is not a turning point as it tends to infinit in the positive -direction. Method for finding stationar points. Differentiate the function. d. Solve the equation. d. Find the -coordinate of each stationar point. Eample Find the stationar points of 7 5. d Differentiating, d 5 When d d, 5

13 8 Differential Calculus Introduction So 5 or 5 When 5, 7 So the stationar points are and when, 5 7 5, 7 and 5, 7. Determining the nature of stationar points There are four possible tpes of stationar point. Maimum Minimum turning Rising point of Falling point of turning point point infleion infleion r r A stationar point of infleion is when the sign of the gradient does not change either side of the stationar point. There are two methods for testing the nature of stationar points. Method Using the signs of f () Here the gradient immediatel before and after the stationar point is eamined. This is best demonstrated b eample. Eample Find the stationar points of 5 and determine their nature. Using the steps of the method suggested above, d. d d. d or 5

14 8 Differential Calculus Introduction. When, 5 8 When, 5 Therefore the coordinates of the stationar points are, 8 and,. To find the nature of the stationar points, we can eamine the gradient before and after and using a table of signs. This means the negative side of r This means the positive side of r d d Shape We can choose values either side of the stationar point to test the gradient either side of the stationar point. This is the meaning of the notation and. means taking a value just on the positive side of, that is slightl higher than. means taking a value just on the negative side of, that is slightl lower than. It is important to be careful of an vertical asmptotes that create a discontinuit. d So for, could be used and so. What d is important is whether this is positive or negative. The brackets are both negative and so the gradient is positive. A similar process with, sa,,, fills in the above table. This provides the shape of the curve around each stationar value and hence the nature of each stationar point. So, 8 is a maimum turning point and, is a minimum turning point. Strictl these should be known as a local maimum and a local minimum as the are not necessaril the maimum or minimum values of the function these would be called the global maimum or minimum. Eample Find the stationar point for d. d d. when d and determine its nature.

15 8 Differential Calculus Introduction. When,, i.e. (, ). d d Shape So the stationar point (, ) is a rising point of infleion. Method Using the sign of f () When a function is differentiated a second time, the rate of change of the gradient of the function is found. This is known as the concavit of the function. The two notations used here for the second derivative are: f in functional notation and This Leibniz notation arises from differentiating This is d d d d d d. d d in Leibniz notation. d d again. For a section of curve, if the gradient is increasing then it is said to be concave up. The curve is getting less steep in this section, i.e. it is becoming less negative and so is increasing. r Similarl, if the gradient is decreasing it is said to be concave down. Looking at the sign of d d can help us determine the nature of stationar points. Consider a minimum turning point: d At the turning point,, although the gradient is zero, the gradient is increasing d d (moving from negative to positive) and so is positive. d 7

16 8 Differential Calculus Introduction Consider a maimum turning point: d At the turning point,, although the gradient is zero, the gradient is decreasing d d (moving from positive to negative) and so is negative. d At a point of infleion, d d is zero. d This table summarizes the nature of stationar points in relation d and d d. Maimum Minimum turning Rising point of Falling point of turning point point infleion infleion d d d d This method is often considered more powerful than method (when the functions become more complicated). For eamination purposes, it is alwas best to use the second derivative to test nature. However, note that for stationar points of infleion, it is still necessar to use a table of signs. Although the table above is true, it is unfortunatel not the whole picture. A positive or d negative answer for provides a conclusive answer to the nature of a stationar point. d 8

17 8 Differential Calculus Introduction d is not quite as helpful. In most cases, this will mean that there is a stationar d point of infleion. However, this needs to be tested using a table of signs as it is possible that it will in fact be a minimum or maimum turning point. A table of signs is also required to determine whether a stationar point of infleion is rising or falling. See the second eample below for further clarification. Eample Find the stationar points of,. d. d. This is stationar when So d. d or. When,, i.e., and when,, i.e. (, ). To test the nature using the second derivative, d d 8 8 At, d d turning point. 8 8 and since this is negative, this is a maimum At, d d 8 8 point. and since this is positive, this is a minimum turning So stationar points are,, a local maimum, and (, ), a local minimum. Eample Find the stationar point(s) of. d. d. Stationar when d, d So

18 8 Differential Calculus Introduction. When,, i.e. (, ).. To test the nature using the second derivative, d d At, d d d As for this stationar point, no assumptions can be made about its d nature and so a table of signs is needed. d d Shape Hence (, ) is a minimum turning point. This can be verified with a calculator. This is an eceptional case, which does not often occur. However, be aware of this anomal. Eercise Find the stationar points and determine their nature using a table of signs. a b c d f 8 7 f 5 e f Find the stationar points and determine their nature using the second derivative. a b c d 8 5 f 5 e f 5

19 8 Differential Calculus Introduction Find the stationar points and determine their nature using either method. a b c d e f 5 f 5 5 Find the distance between the turning points of the graph of. 8.5 Points of infleion The concavit of a function is determined b the second derivative. f 7 f Concave up Concave down So what happens when f? We know that when f and f, there is a stationar point normall a stationar point of infleion (with the eceptions as previousl discussed). In fact, apart from the previousl noted eceptions, whenever f it is known as a point of infleion. The tpe met so far are stationar points of infleion when the gradient is also zero (also known as horizontal points of infleion). However, consider the curve: A Looking at the gradient between the turning points, it is constantl changing, the curve becoming steeper and then less steep as it approaches B. So the rate of change of gradient is negative (concave down) around A and then positive (concave up) around B. Clearl the rate of change of gradient d must be zero at some point between A d and B. This is the steepest part of the curve between A and B, and it is this point that is known as a point of infleion. This is clearl not stationar. So a point of infleion can now be defined to be a point where the concavit of the graph changes sign. B If d, d there is a point of infleion:

20 8 Differential Calculus Introduction If d, it is a stationar d point d If, it is a non-stationar d point of infleion (assuming a change in concavit) e.g. Anomalous case Method for finding points of infleion d. Differentiate the function twice to find d. d. Solve the equation. d. Find the -coordinate of each point. d. Test the concavit around this point, i.e. must change sign. d Eample Find the points of infleion of the curve whether the are stationar. f 5 5 and determine. f 5 5 f. For points of infleion, So f or or ;. f. f f.

21 8 Differential Calculus Introduction. f f There is a change in concavit (the sign of the second derivative changes) around each point. So each of these three points is a point of infleion. To test whether each point is stationar, consider f f. Hence provides a non-stationar point of infleion. f Hence also provides a non-stationar point of infleion. f 5 5 Hence provides a stationar point of infleion. The three points of infleion are,., (, ) and This can be verified on a calculator.,.. Eercise 5 Eercise 5 Find the points of infleion for the following functions and determine whether the are stationar. a b f 5 f 7 c i 8

22 8 Differential Calculus Introduction ii iii 5 f a b c Make a general statement about quadratic functions. Find the points of infleion for the following functions and determine whether the are stationar. a b c d f 7 8 f a b c d Make a general statement about cubic functions. Find the points of infleion for the following functions and determine whether the are stationar. a b c d f f 5 5 Find the equation of the tangent to at the point of infleion. 5 For the graph of 5, find the distance between the point of infleion and the root. 8. Curve sketching Bringing together knowledge of functions, polnomials and differentiation, it is now possible to identif all the important features of a function and hence sketch its curve. The important features of a graph are: Vertical asmptotes (where the function is not defined) This is usuall when the denominator is zero. Intercepts These are when and. Stationar points and points of infleion d d Determine when and when. d d Behaviour as S;q This provides horizontal and oblique asmptotes. With the eception of oblique asmptotes, all of the necessar concepts have been met in this chapter, and in Chapters,, and.

23 8 Differential Calculus Introduction blique asmptotes In Chapter we met horizontal asmptotes. These occur where true for oblique asmptotes. Consider the function. It is clear that as S;q, the becomes negligible and so Hence is a horizontal asmptote for this function. S;q. S. This is also Now consider the function 5. 5 In a similar wa, the fractional part tends to zero as S;q and so S. This means that is an oblique asmptote (also known as a slant asmptote). Method for sketching a function. Find the vertical asmptotes (where the function is not defined).. If it is an improper rational function (degree of numerator degree of denominator), divide algebraicall to produce a proper rational function.. Consider what happens for ver large positive and negative values of. This will provide horizontal and oblique asmptotes.. Find the intercepts with the aes. These are when and. 5. Find the stationar points and points of infleion (and their nature). d Determine when and when d. d d. Sketch the curve, ensuring that all of the above important points are annotated on the graph. Eample Find the asmptotes of 7. Clearl the function is not defined at and so this is a vertical asmptote. 5 Dividing,

24 8 Differential Calculus Introduction Hence 5 and so as S;q, S 5. Therefore 5 is an oblique asmptote. This is clear from the graph: Eample Sketch the graph of identifing all asmptotes, intercepts, stationar, points, and non-horizontal points of infleion. There is a vertical asmptote at Dividing, This gives. So is an oblique asmptote. Putting and gives an intercept at (,). There are no other roots. Preparing for differentiation, d Differentiating, d d Stationar when, i.e. when d or or Substituting into the original function provides the coordinates (, ) and,. For the nature of these stationar points,

25 8 Differential Calculus Introduction d d At, d d 7 At, d so, d so (, ) is a local minimum turning point. is a local maimum point. The notation means for all. So H, means for all real values of. As d d H, there are no non-horizontal points of infleion. (, ) All of these features can be checked using a graphing calculator, if it is available. In some cases, an eamination question ma epect the use of a graphing calculator to find some of these important points, particularl stationar points. For eample, the calculator provides this graph for the above function: In fact, it is possible to be asked to sketch a graph that would be difficult without use of the calculator. Consider the net eample. Eample In order to graph this function using a calculator, we need to rearrange into a form. 7

26 8 Differential Calculus Introduction ; C Clearl this function is not defined for a large section (in fact 8 8 ). The oblique asmptotes are not immediatel obvious. Rearranging the equation to give ; B makes it clearer. As S;q, S; S; Eercise Find all asmptotes (vertical and non-vertical) for these functions. 5 f f f f 8 Sketch the graphs of these functions, including asmptotes, stationar points and intercepts. 5 8

27 8 Differential Calculus Introduction Sketching the graph of the derived function Given the graph of the original function it is sometimes useful to consider the graph of its derivative. For eample, non-horizontal points of infleion now become obvious from the graph of the derived function, since the become stationar points. Horizontal points of infleion are alread stationar. If the original function is known, then it is straightforward to sketch the graph of the derived function. This can be done b: a) finding the derivative and sketching it b) using a graphing calculator to sketch the derived function. Eample Sketch the derivative of 5. Using method a) Differentiating gives d d

28 8 Differential Calculus Introduction Using method b) These methods are possible onl if the function is known. If the function is not known, the gradient of the graph needs to be eamined. Eample Sketch the graph of the derivative of this graph. (, ) (, ) Note where the graph is increasing, stationar and decreasing. Stationar points on the original graph become roots of the derived function regions are above the -ais -ais d d. (, ) d d 7, So the graph becomes d, d increasing and decreasing regions are below the (, ) With polnomial functions, the degree of a derived function is alwas one less than the original function.

29 8 Differential Calculus Introduction We can consider this process in reverse, and draw a possible graph of the original function, given the graph of the derived function. In order to do this note that:. roots of the derived function are stationar points on the original graph. stationar points on the derived function graph are points of infleion on the original graph. Eample Given this graph of the derived function the original function f. f, sketch a possible graph of f() 5 We can see that there are roots on this graph and hence stationar points on the original at,, and 5. It is helpful to consider the gradient of the original to be able to draw a curve. S S S 5 S f' So a possible graph of the original function 5 f is We cannot determine the -values of the points on the curve but we can be certain of the shape of it. This will be covered in further detail in Chapter. Sketching the reciprocal function Sometimes we are asked to sketch the graph of a reciprocal function, i.e. f. If f() is known and a calculator is used, this is eas. However, if it is not known, then we need to consider the following points.

30 8 Differential Calculus Introduction. At f, Hence roots on the original graph become vertical f Sq. asmptotes on the reciprocal graph.. At a vertical asmptote, f Sq. Hence vertical asmptotes on f() f become roots on the reciprocal graph.. Maimum turning points become minimum turning points and minimum turning points become maimum turning points. The -value of the turning point stas the same but the -value is reciprocated.. If f() is above the -ais, is also above the -ais, and if f() is below the f -ais, is also below the -ais. f We will now demonstrate this b eample. Eample If f sin, p draw f. sin csc This is the standard wa of drawing the curves of sec, csc and cot. Eample For the following graph of f, draw the graph of f. f() (, ) (, ) (, ) (, )

31 8 Differential Calculus Introduction Eercise 7 Sketch the graph of the derived function of the following: a b c d e f 7 g h i 8 Sketch the graph of the derived function of the following: a b Linear (, 8) Quadratic (, ) c Quadratic d Cubic (, ) e (, ) Cubic f (, 5) Cubic 5 (, ) g (a, b) Cubic h (, 7) Quartic (c, d) (, ) (, )

32 8 Differential Calculus Introduction i Quartic j (, 7) Fifth degree (, ) (, ) (, ) (, ) Sketch a possible graph of the original function f, given the derived function graph f in each case. a b c d (, ) 7 5 For the following functions, draw the graph of f a f b f c f 5 d f 5 e f e f f ln g f cos, p h f 5 For the following graphs of f, sketch the graph of the reciprocal function f. a b (, )

33 M C 7 M C 7 M C 7 M C 7 M C 7 M C 7 M C 7 M CE 8 5 M CE 8 5 M CE 8 5 M CE 8 5 M CE 8 5 M CE 8 5 M CE 8 5 M+ % M+ % + M+ % + + M+ % + M+ % + M+ % + M+ % + X = X = N X = X = X = X = X = 8 Differential Calculus Introduction c d e (, 7) f (, 8) (, ) (, 5) (, ) 7 Review eercise N N N N N N Differentiate f 5 using first principles. For f find f., d Given that find., d A function is defined as f Find values of for which the. function is increasing. 5 Given that f 5 and g, find h fg. Hence find h. Find the positive value of for which the gradient of the tangent is for. Hence find the equation of the tangent at this point. 7 Sketch the graph of the derivative for the graph below. (, 7) (, ) 5

34 M C 7 M C 7 M C 7 M C 7 M C 7 M C 7 M CE 8 5 M CE 8 5 M CE 8 5 M CE 8 5 M CE 8 5 M CE 8 5 M+ % + M+ % + M+ % + M+ % + M+ % + M+ % + X = X = X = X = X = X = 8 Differential Calculus Introduction N N N 8 Find the stationar points of and investigate their nature. Find the equation of the tangent to 5 at and find the equation of the normal to at. Find the point where these lines cross. Sketch the graph of including all asmptotes, stationar, points and intercepts. N Find the equations of all the asmptotes of the graph of 5 5 [IB Nov p Q] N N The line is a tangent to the curve a b at the point (, 7). Find the values of a and b. [IB Nov p Q7] For the following graphs of f, draw the graphs of f. a (, ) b (, 7) (, ) (, 7)