Introduction to Calculus

Transcription

1 8 Introduction to Calculus TERMINOLOGY Composite function: A function of a function. One function, f (), is a composite of one function to another function, for eample g() Continuity: Describing a line or curve that is unbroken over its domain Continuous function: A function is continuous over an interval if it has no break in its graph. For every value on the graph the limit eists and equals the function value Derivative at a point: This is the gradient of a curve at a particular point Derivative function: The gradient function of a curve obtained through differentiation Differentiable function: A function which is continuous and where the gradient eists at all points on the function Differentiation: The process of finding the gradient of a tangent to a curve which is called the derivative Differentiation from first principles: The process of finding the gradient of a tangent to a curve by finding the gradient of the secant between two points and finding the limit as the secant becomes a tangent Gradient of a secant: The gradient (slope) of the line between two points that lies close together on a function Gradient of a tangent: The gradient (slope) of a line that is a tangent to the curve at a point on a function. It is the derivative of the function Rate of change: The rate at which the dependent variable changes as the independent variable changes

2 Chapter 8 Introduction to Calculus 9 INTRODUCTION CALCULUS IS A VERY IMPORTANT part of mathematics and involves the measurement of change. It can be applied to many areas such as science, economics, engineering, astronomy, sociology and medicine. We also see articles in newspapers every day that involve change: the spread of infectious diseases, population growth, inflation, unemployment, filling of our water reservoirs. For eample, this graph shows the change in crude oil production in Iran over the years. Notice that while the graph shows that production is increasing over recent years, the rate at which it is being produced seems to be slowing down. Calculus is used to look at these trends and predict what will happen in the future. There are two main branches of calculus. Differentiation is used to calculate 7,000 6,000 5,000,000,000,000,000 the rate at which two variables change in relation to one another. Anti-differentiation, or integration, is the inverse of differentiation and uses information about rates of change to go back and eamine the original variables. Integration can also be used to find areas of curved objects. Thousand Barrels per Day Crude Oil Production (Mbbl/d) Iran January 97 May 007 You will learn about integration in the HSC Course. DID YOU KNOW? Calculus comes from the Latin meaning pebble or small stone. In ancient civilisations, stones were used for counting. However, the mathematics practised by these early people was quite sophisticated. For eample, the ancient Greeks used sums of rectangles to estimate areas of curved figures. However, it wasn t until the 7 th century that there was a breakthrough in calculus when scientists were searching for ways of measuring motion of objects such as planets, pendulums and projectiles. Isaac Newton, an Englishman, discovered the main principles of calculus when he was years old. At this time an epidemic of bubonic plague closed Cambridge University where he was stuing, so many of his discoveries were made at home. He first wrote about his calculus methods, which he called fluions, in 67, but his Method of fluions was not published until 70. Gottfried Leibniz (66 76), in Germany, was also stuing the same methods and there was intense rivalry between the two countries over who was first! Search the Internet for further details on these two famous mathematicians. You can find out about the history of calculus and why it was necessary for mathematicians all those years ago to invent it. Isaac Newton

3 0 Maths In Focus Mathematics Etension Preliminary Course In this chapter you will learn about differentiation, which measures the rate of change of one variable with respect to another. Gradient Gradient of a straight line The gradient of a straight line measures its slope. You studied gradient in the last chapter. rise m run Class Discussion Remember that an increasing line has a positive gradient and a decreasing line has a negative gradient. positive negative Notice also that a horizontal line has zero gradient. Can you see why? Can you find the gradient of a vertical line? Why? Gradient plays an important part, not just in mathematics, but in many areas including science, business, medicine and engineering. It is used everywhere we want to find rates. On a graph, the gradient measures the rate of change of the dependent variable with respect to the change in the independent variable.

4 km Chapter 8 Introduction to Calculus EXAMPLES. The graph shows the average distance travelled by a car over time. Find the gradient and describe it as a rate. d 00 Hours 5 t The line is increasing so it will have a positive gradient. rise m run This means that the car is travelling at the rate of 80 km/hour.. The graph shows the number of cases of flu reported in a town over several weeks. N 5 Number of cases (00s) Weeks Find the gradient and describe it as a rate. 0 t CONTINUED

5 Maths In Focus Mathematics Etension Preliminary Course The line is decreasing so it will have a negative gradient. rise m run This means that the rate is - 50 cases/week, or the number of cases reported is decreasing by 50 cases/week. When finding the gradient of a straight line in the number plane, we think of a change in y values as changes. The gradients in the eamples above show rates of change. However, in most eamples in real life, the rate of change will vary. For eample, a car would speed up and slow down depending on where it is in relation to other cars, traffic light signals and changing speed limits. Gradient of a curve Class Discussion The two graphs show the distance that a bicycle travels over time. One is a straight line and the other is a curve. d d km 0 km t Hours Hours Is the average speed of the bicycle the same in both cases? What is different about the speed in the two graphs? How could you measure the speed in the second graph at any one time? Does it change? If so, how does it change? t

6 Chapter 8 Introduction to Calculus Here is a more general curve. What could you say about its gradient? How does it change along the curve? y Copy the graph and mark on it where the gradient is positive, negative and zero. Using what we know about the gradient of a straight line, we can see where the gradient of a curve is positive, negative or zero by drawing tangents to the curve in different places around the curve. y Notice that when the curve increases it has a positive gradient, when it decreases it has a negative gradient and when it turns around the gradient is zero. Investigation There are some ecellent computer programs that will draw tangents to a curve and then sketch the gradient curve. One of these is Geometer Sketchpad. Eplore how to sketch gradient functions using this or a similar program as you look at the eamples below.

7 Maths In Focus Mathematics Etension Preliminary Course EXAMPLES Describe the gradient of each curve.. Where the curve increases, the gradient is positive. Where it decreases, it is negative. Where it turns around, it has a zero gradient..

8 Chapter 8 Introduction to Calculus 5 Since we have a formula for finding the gradient of a straight line, we find the gradient of a curve by measuring the gradient of a tangent to the curve. EXAMPLE (a) Make an accurate sketch of y on graph paper. (b) Draw tangents to this curve at the points where -, -, -, 0,, and. (c) Find the gradient of each of these tangents. (d) Draw the graph of the gradients (the gradient function) on a number plane. (a) and (b) y There are computer programs that will draw these tangents. (c) At -, m -6 At -, m - At -, m - At 0, m 0 At, m At, m At, m 6 (d) Use the m values as the y values on this graph.

9 6 Maths In Focus Mathematics Etension Preliminary Course Drawing tangents to a curve is difficult. We can do a rough sketch of the gradient function of a curve without knowing the actual values of the gradients of the tangents. To do this, notice in the eample above that where m is positive, the gradient function is above the -ais, where m 0, the gradient function is on the -ais and where m is negative, the gradient function is below the -ais. EXAMPLES Sketch the gradient function of each curve.. First we mark in where the gradient is positive, negative and zero. Now on the gradient graph, place the points where m 0 on the -ais. These are at, and.

10 Chapter 8 Introduction to Calculus 7 To the left of, the gradient is negative, so this part of the graph will be below the -ais. Between and, the gradient is positive, so the graph will be above the -ais. Between and, the gradient is negative, so the graph will be below the -ais. To the right of, the gradient is positive, so this part of the graph will be above the -ais.. First mark in where the gradient is positive, negative and zero. CONTINUED

11 8 Maths In Focus Mathematics Etension Preliminary Course The gradient is zero at and. These points will be on the -ais. To the left of, the gradient is positive, so this part of the graph will be above the -ais. Between and, the gradient is negative, so the graph will be below the -ais. To the right of, the gradient is positive, so this part of the graph will be above the -ais. 8. Eercises Sketch the gradient function for each graph

12 Chapter 8 Introduction to Calculus Differentiation from First Principles Seeing where the gradient of a curve is positive, negative or zero is a good first step, but there are methods to find a formula for the gradient of a tangent to a curve. The process of finding the gradient of a tangent is called differentiation. The resulting function is called the derivative. Differentiability A function is called a differentiable function if the gradient of the tangent can be found. There are some graphs that are not differentiable in places. Most functions are continuous, which means that they have a smooth unbroken line or curve. However, some have a gap, or discontinuity, in the graph (e.g. hyperbola). This can be shown by an asymptote or a hole in the graph. We cannot find the gradient of a tangent to the curve at a point that doesn t eist! So the function is not differentiable at the point of discontinuity. y a This function is not differentiable at a since the curve is discontinuous at this point.

13 50 Maths In Focus Mathematics Etension Preliminary Course y b This function is not differentiable at b as the curve is discontinuous at this point. A function may be continuous but not smooth. It may have a sharp corner. Can you see why curves are not differentiable at the point where there is a corner? y c The curve is not differentiable at point c since it is not smooth at that point. A function y f( ) is differentiable at the point a if the derivative eists at that point. This can only happen if the function is continuous and smooth at a.

14 Chapter 8 Introduction to Calculus 5 EXAMPLES. Find all points where the function below is not differentiable. y B C A The function is not differentiable at points A and B since there are sharp corners and the curve is not smooth at these points. It is not differentiable at point C since the function is discontinuous at this point.. Is the function f ( ) for $ ) differentiable at all points? - for The functions f ( ) and f ( ) - are both differentiable at all points. However, we need to look at where one finishes and the other starts, at f (). For f ( ) f ] g For f ( ) - f ] g ] g- This means that both pieces of this function join up (the function is continuous). However, to be differentiable, the curve must be smooth at this point. CONTINUED

15 5 Maths In Focus Mathematics Etension Preliminary Course Sketching this function shows that it is not smooth (it has a sharp corner) so it is not differentiable at. y y - y - 8. Eercises For each function, state whether it has any points at which it is not differentiable.. y. y. y. y

16 Chapter 8 Introduction to Calculus 5 5. y 0. 5 y f ( ) 7. y f ( ) 9. f ( ) if ) + if # Z for ] [ for - # # ] \ - for y tan for 0c # # 60c. f ( ). f ( i) - cosi. g ( z) sin z 5. - y - 9 Limits To differentiate from first principles, we need to look more closely at the concept of a limit. A limit is used when we want to move as close as we can to something. Often this is to find out where a function is near a gap or discontinuous point. You saw this in Chapter 5 when looking at discontinuous graphs. In this topic, it is used when we want to move from a gradient of a line between two points to a gradient of a tangent. EXAMPLES - -. Find lim. " - lim - - ( + )( - ) lim - ( - ) lim ( + ) " " + " You did this in Chapter 5. CONTINUED

17 5 Maths In Focus Mathematics Etension Preliminary Course h - h - h. Find an epression in terms of for lim. h " 0 h h - h - h h( - h - ) lim lim h " 0 h h " 0 h lim ( - h - ) h " Find an epression in terms of for lim. " ( + - 5) lim lim " 0 " 0 lim ( + - 5) " Eercises. Evaluate + (a) lim " (b) lim " 0 - (c) lim " - t - 6 (d) lim t " t - g - (e) lim g " g (f) lim " h + h (g) lim h " 0 h (h) lim " - n - 5 (i) lim n " 5 n (j) lim " - -. Find as an epression in terms of h- h- h (a) lim h " 0 h h+ h- h (b) lim h " 0 h h- 7h+ h- h (c) lim h " 0 h h- h- h (d) lim h " 0 h h+ h- h+ h (e) lim h " 0 h h+ 5h+ 6h (f) lim h " 0 h - (g) lim " 0 - (h) lim " (i) lim " (j) lim " 0

18 Chapter 8 Introduction to Calculus 55 Differentiation as a limit y - y The formula m - is used to find the gradient of a straight line when we know two points on the line. However, when the line is a tangent to a curve, we only know one point on the line the point of contact with the curve. To differentiate from first principles, we first use the point of contact and another point close to it on the curve (this line is called a secant ) and then we move the second point closer and closer to the point of contact until they overlap and the line is at single point (the tangent ). To do this, we use a limit. If you look at a close up of a graph, you can get some idea of this concept. When the curve is magnified, two points appear to be joined by a straight line. We say the curve is locally straight. Investigation Use a graphics calculator or a computer program to sketch a curve and then zoom in on a section of the curve to see that it is locally straight. For eample, here is a parabola. 0 y -0 f () 0-0 Notice how it looks straight when we zoom in on a point on the parabola? 7.99 y.99 f () Use technology to sketch other curves and zoom in to show that they are locally straight.

19 56 Maths In Focus Mathematics Etension Preliminary Course Before using limits to find different formulae for differentiating from first principles, here are some eamples of how we can calculate an approimate value for the gradient of the tangent to a curve. By taking two points close together, as in the eample below, we find the gradient of the secant and then estimate the gradient of the tangent. y (.0, f (.0)) (, f ()) EXAMPLES. For the function f] g, find the gradient of the secant PQ where P is the point on the function where and Q is another point on the curve close to P. Choose different values for Q and use these results to estimate the gradient of the curve at P. y Q (., f(.)) P (, f())

20 Chapter 8 Introduction to Calculus 57 P ^, f() h Take different values of for point Q, for eample. Using different values of for point Q gives the results in the table. Point Q _., f ]. gi _ 0., f ] 0. gi _. 00, f ]. 00gi _ 9., f ] 9. gi _ 99., f ] 99. gi _. 999, f ]. 999gi Gradient of secant PQ (. ) ( ) m f - f (. ) ( ) m f 0 - f (. ) ( ) m f 00 - f (. ) ( ) m f 9 - f (. ) ( ) m f 99 - f (. ) ( ) m f f y - y Use m to find - the gradient of the secant. From these results, a good estimate for the gradient at P is. We can say that as approaches, the gradient approaches. f ( )- f() We can write lim. " - CONTINUED

21 58 Maths In Focus Mathematics Etension Preliminary Course. For the curve y, find the gradient of the secant AB where A is the point on the curve where 5 and point B is close to A. Find an estimate of the gradient of the curve at A by using three different values for B. A ^5, f(5) h Take three different values of for point B, for eample 9., 5. and 5.0. (a) B ^9., f( 9. ) h y - y m - f( 9. )- f( 5) (b) B ^5., f( 5. ) h y - y m - f( 5. )- f( 5) (c) B ^50., f( 50. ) h y - y m - f( 50. )- f( 5) From these results, a good estimate for the gradient at A is 0. We can say that as approaches 5, the gradient approaches 0. f ( )- f(5) We can write lim 0. " We can find a general formula for differentiating from first principles by using c rather than any particular number. We use general points Pcfc ^, ( ) h and Qf ^, ( ) h where is close to c. The gradient of the secant PQ is given by y - y m - f ( )- fc ( ) - c

22 Chapter 8 Introduction to Calculus 59 The gradient of the tangent at P is found when approaches c. We call this fl () c. f ( )- fc ( ) fl () c lim " c - c There are other versions of this formula. We can call the points Pf ^, ( ) h and Q ^ + hf, ( + h) h where h is small. y Q ( + h, f( + h)) P (, f()) Secant PQ has gradient y - y m - f ( + h) - f ( ) + h - f ( + h) - f ( ) h To find the gradient of the secant, we make h smaller as shown, so that Q becomes closer and closer to P. y Q Q ( + h, f( + h)) P (, f ()) Q Q Search the Internet using keywords differentiation from first principles, gradient of secant and tangent to find mathematical websites that show this working.

23 60 Maths In Focus Mathematics Etension Preliminary Course f ( + h) - f ( ) As h approaches 0, the gradient of the tangent becomes lim. h " 0 h We call this fl ( ). f ( + h) - f ( ) fl ( ) lim h " 0 h The symbol d is a Greek letter called delta. If we use P^, yh and Q^ +, y + h close to P where and d y are small: Gradient of secant PQ y - y m - y + - y + - As d approaches 0, the gradient of the tangent becomes lim call this. " 0 d. We lim d 0 " All of these different notations stand for the derivative, or the gradient of the tangent: d d, ( y ), ^f ( ) h, fl ( ), yl These occur because Newton, Leibniz and other mathematicians over the years have used different notation. Investigation Leibniz used where d stood for difference. Can you see why he would have used this? Use the Internet to eplore the different notations used in calculus and where they came from.

24 Chapter 8 Introduction to Calculus 6 The three formulae for differentiating from first principles all work in a similar way. EXAMPLE Differentiate from first principles to find the gradient of the tangent to the curve y + at the point where. Method : f ( )- fc ( ) fl() c lim " c - c f] g + f ] g + f ( )- fc ( ) fl() c lim " c - c f ( )- f( ) fl( ) lim " - ( + ) - lim " - - lim " - ( + )( - ) lim " - lim ( + ) + " Remember that y - is the same as f () -. Method : f ( + h) - f ( ) fl( ) lim h " 0 h f] g + f ] g + f] + hg ] + hg + When f] + hg ] + hg + + h + h + h + h + CONTINUED

25 6 Maths In Focus Mathematics Etension Preliminary Course f ( + h) - f ( ) fl( ) lim h " 0 h f( + h) - f( ) fl( ) lim h " 0 h ( h + h + ) - lim h " 0 h h + h lim h " 0 h h( + h) lim h " 0 h lim ( + h) + 0 h " 0 Method : lim d " 0 y + When y + So point ^, h lies on the curve. Substitute point ( +, + ) : + ( + ) ( + ) + lim " 0 lim ( + ) + 0 " 0 We can also use these formulae to find the derivative function generally.

26 Chapter 8 Introduction to Calculus 6 EXAMPLE Differentiate f ] g from first principles. f] g f] + hg ] + hg + 7] + hg- ^ + h + h h h - + h + h h - f] + hg - f] g ^ + h + h h - h- ^ h + h + h h h + h + 7h f ( + h) - f ( ) fl( ) lim h " 0 h h + h + 7h lim h " 0 h h( + h + 7) lim h " 0 h lim ( + h + 7) h " Try this eample using the other two formulae. 8. Eercises. (a) Find the gradient of the secant between the point ^, h and the point where.0, on the curve y +. (b) Find the gradient of the secant between ^, h and the point where on the curve. (c) Use these results to find the gradient of the tangent to the curve y + at the point ^, h.. A function f] g + has a tangent at the point ^0, h. f ( )- f() (a) Find the value of - when.. f ( )- f( ) (b) Find the value of - when.0. f ( )- f() (c) Evaluate when (d) Hence find the gradient of the tangent at the point ^, 0h.. For the function f] g -, find the derivative at point P where by selecting points near P and finding the gradient of the secant.. If f ( ), (a) find f ( + h) (b) show that f ( + h) - f ( ) h + h

27 6 Maths In Focus Mathematics Etension Preliminary Course Remember that - (c) show that f ( + h) - f ( ) h + h (d) show that fl ( ). 5. A function is given by f ( ) (a) Show that f ( + h) + h + h - 7-7h +. (b) Show that f ( + h) - f ( ) h+ h - 7h. (c) Show that f ( + h) - f ( ) + h - 7. h (d) Find fl ( ). 6. A function is given by f ( ) (a) Find f ] g. (b) Find f] + hg. (c) Find f] + hg - f] g. (d) Show that f( + h) - f( ) 5 + h. h (e) Find fl (). 7. Given the curve f ( ) - (a) find f ]- g (b) find f ]- + hg - f ]-g (c) find the gradient of the tangent to the curve at the point where For the parabola y - (a) find f ] g (b) find f] + hg - f] g (c) find fl (). 9. For the function f( ) (a) find fl ( ) (b) similarly, find the gradient of the tangent at the point ^-, -0h. 0. For the parabola y + (a) show that + + by substituting the point ^ +, y + h (b) show that + + (c) find.. Differentiate from first principles to find the gradient of the tangent to the curve (a) f] g at the point where (b) y + at the point ^6, h (c) f] g - 5 at the point where - (d) y + + at the point where (e) f] g at the point ^-, 6h.. Find the derivative function for each curve by differentiating from first principles (a) f] g (b) y + 5 (c) f] g - - (d) y (e) y (f) f] g + 5 (g) y (h) f ( ) -.. The curve y has a tangent drawn at the point ^, h. f ( )- f() (a) Evaluate when f ( )- f() (b) Evaluate when f ( )- f() (c) Evaluate when For the function f ( ) -, f ( )- f(5) (a) evaluate when

28 Chapter 8 Introduction to Calculus 65 f ( )- f( 5) (b) evaluate when (c) Use these results to find the derivative of the function at the point where Find the gradient of the tangent to the curve y at point P ^, h by finding the gradient of the secant between P and a point close to P. Short Methods of Differentiation The basic rule Remember that the gradient of a straight line y m + b is m. The tangent to the line is the line itself, so the gradient of the tangent is m everywhere along the line. y y m + b So if y m, m d ] k g k For a horizontal line in the form y k, the gradient is zero. y y k So if y k, 0 d ] k g 0

29 66 Maths In Focus Mathematics Etension Preliminary Course Investigation Differentiate from first principles: y y y Can you find a pattern? Could you predict what the result would be for n? Alternatively, you could find an approimation to the derivative of a f ( )- f ( ) function at any point by drawing the graph of y. 00. Use a graphics calculator or graphing computer software to sketch the derivative for these functions and find the equation of the derivative. Mathematicians working with differentiation from first principles discovered this pattern that enabled them to shorten differentiation considerably! For eample: When y, yl When y, yl When y, yl d ^ h n n n- You do not need to know Proof this proof. f ( ) f ( + h) ( + h) n f ( + h) - f ( ) ( + h) - n n n n- n- n- n- ^( + h) - h [( + h) + ( + h) + ( + h) + ( + h) n- n ( + h) + ] n- n- n- n- h[( + h) + ( + h) + ( + h) + ( + h) n- n ( + h) + ] f ( + h) - f ( ) fl( ) lim h " 0 h n- n- n- n- n- n- h[( + h) + ( + h) + ( + h) + ( + h) ( + h) + ] lim h " 0 h n- n- n- n- n- n- lim [( + h) + ( + h) + ( + h) + ( + h) ( + h) + ] h " 0 ( ) + ( ) + ( ) + ( ) ( ) + n n- n- n- n- n- n- n -

30 Chapter 8 Introduction to Calculus 67 EXAMPLE Differentiate f ( ) 7. fl ( ) 7 6 There are some more rules that give us short ways to differentiate functions. The first one says that if there is a constant in front of the (we call this a coefficient), then it is just multiplied with the derivative. d ^ k h kn n n- A more general way of writing this rule is: d ^ kf ( ) h kf l ( ) Proof d kf ( + h) - kf ( ) ^ kf ( ) h lim h " 0 h kf [( + h) - f ()] lim h " 0 h f ( + h) - f ( ) k lim h " 0 h kfl( ) You do not need to know this proof. EXAMPLE Find the derivative of 8. 8 If y # 8 7 7

31 68 Maths In Focus Mathematics Etension Preliminary Course Also, if there are several terms in an epression, we differentiate each one separately. We can write this as a rule: d ^ f ( ) + g ( ) h f l ( ) + g ( ) Proof You do not need to know this proof. d [ f ( + h) + g ( + h)] -[ f ( ) + g ( )] ^ f ( ) + g ( ) h lim h " 0 h f ( + h) + g ( + h) - f ( )- g ( ) lim h " 0 h f ( + h) - f ( ) + g ( + h) - g ( ) lim h " 0 h f ( + h) - f ( ) g ( + h) - g ( ) lim + G h " 0 h h f ( + h) - f ( ) g ( + h) - g ( ) lim + lim h " 0 h h " 0 h fl( ) + gl( ) EXAMPLE Differentiate +. d ( + ) + Many functions use a combination of these rules. EXAMPLES Differentiate. 7 d ] 7g 7 CONTINUED

32 Chapter 8 Introduction to Calculus 69. f ( ) fl( ) y 7 # If f( ) , evaluate fl (-) fl( ) fl (- ) 0( -) - ( - ) Differentiate + Divide by before differentiating Differentiate S rr + rrh with respect to r. We are differentiating with respect to r, so r is the variable and r and h are constants. ds r( r) + rh dr rr + rh

33 70 Maths In Focus Mathematics Etension Preliminary Course 8.5 Eercises Epand brackets before differentiating. Simplify by dividing before differentiating.. Differentiate (a) + (b) 5-9 (c) + + (d) (e) (f) (g) (h) (i) (j) - 7. Find the derivative of (a) ] + g (b) ] - g (c) ] + g] - g (d) ^ - h (e) ] + 5g^ - + h. Differentiate (a) - 6 (b) (c) ( - ) 5 (d) + (e) (f) + 5. Find fl ( ) when f ( ) If y - + 5, find when Find if y If s 5t ds - 0t, find. dt 8. Find gl ( ) given g ( ) 5 -. dv 9. Find when v 5t - 9. dt 0. If h 0t - t dh, find. dt. Given V r dv r, find. dr. If f ( ) - +, evaluate f l ().. Given f ( ) - + 5, evaluate (a) fl ( ) (b) fl (-) (c) when fl ( ) 7. If y - 7, evaluate (a) when (b) when 5. Evaluate gl ( ) when gt () t - t - t+.

34 Chapter 8 Introduction to Calculus 7 Tangents and Normals DID YOU KNOW? The word tangent comes from the Latin tangens, meaning touching. A tangent to a circle intersects it only once. However, a tangent to a curve could intersect the curve more than once. This line is a tangent to the curve at point P. A line may only intersect a curve once but not be a tangent. So a tangent to a curve is best described as the limiting position of the secant PQ as Q approaches P. Remember from earlier in the chapter that the derivative is the gradient of the tangent to a curve. is the gradient of the tangent to a curve

35 7 Maths In Focus Mathematics Etension Preliminary Course u EXAMPLES. Find the gradient of the tangent to the parabola y + at the point ^, h. + 0 At ^, h ( ) So the gradient of the tangent at ^, h is.. Find values of for which the gradient of the tangent to the curve y is equal to is the gradient of the tangent, so substitute ] - g] + g - 0, + 0 `, -. Find the equation of the tangent to the curve y at the point ^, h At ^, h ] g - 9 ] g + 7 So the gradient of the tangent at ^, h is. Equation of the tangent: y - y m _ - i y - ] - g

36 Chapter 8 Introduction to Calculus 7-6 y - or 0 - y - The normal is a straight line perpendicular to the tangent at the same point of contact with the curve. y Tangent Normal You used this rule in the If lines with gradients m and m are perpendicular, then mm - previous chapter. EXAMPLES. Find the gradient of the normal to the curve y at the point where. is the gradient of the tangent. - When # - So m The normal is perpendicular to the tangent. So mm - CONTINUED

37 7 Maths In Focus Mathematics Etension Preliminary Course m - m - So the gradient of the normal is -.. Find the equation of the normal to the curve y at the point ^-, h. is the gradient of the tangent When - ] - g + 6] -g- -5 So m -5 The normal is perpendicular to the tangent. So mm - - 5m - m 5 So the gradient of the normal is 5. Equation of the normal: y - y m _ - i y - ] -]- gg 5 5y y Eercise s. Find the gradient of the tangent to the curve (a) y - at the point where 5 (b) f ] g + - at the point ^-7, 8h (c) f] g at the point where - (d) y at the point ^-, 9h (e) y 9 at the point where (f) f] g - 7 at the point where (g) v t + t - 5 at the point where t (h) Q r - r + 8r - at the point where r (i) h t - t where t 0 (j) f] tg t 5-8t + 5t at the point where t.

38 Chapter 8 Introduction to Calculus 75. Find the gradient of the normal to the curve (a) f] g + - at the point where - (b) y at the point ^-5, 8h (c) f] g at the point where - 9 (d) y at the point ^-, -6h (e) f ] g 0 at the point where - (f) y at the point ^-7, -5h (g) A at the point where (h) f] ag a - a -6 at the point where a - (i) V h - h + 9 at the point ^9, h (j) g] g at the point where -.. Find the gradient of the (i) tangent and (ii) normal to the curve (a) y + at the point ^0, h (b) f] g 5 - at the point where - (c) y at the point where - (d) p ] g where (e) f] g - - at the point ^-6, 6h.. Find the equation of the tangent to the curve (a) y at the point ^7, h (b) f ( ) at the point ^6, h (c) y at the point ^-, -5h (d) y + at the point where (e) v t - 7t - at the point where t 5. Find the equation of the normal to the curve (a) f] g at the point ^, h (b) y at the point ^-, 7h (c) f] g 7 - at the point where 6 (d) y at the point ^-, 70h (e) y at the point where. 6. Find the equation of the (i) tangent and (ii) normal to the curve (a) f] g at the point ^, h (b) y at the point ^-, 6h 5 (c) F ] g - 5 at the point where (d) y at the point ^, -8h (e) y at the point where. 7. For the curve y - 7-5, find values of for which Find the coordinates of the point at which the curve y + has a tangent with a gradient of. 9. A function f ( ) + - has a tangent with a gradient of - 6 at point P on the curve. Find the coordinates of the point P. 0. The tangent at point P on the curve y + is parallel to the -ais. Find the coordinates of P.. Find the coordinates of point Q where the tangent to the curve y 5 - is parallel to the line 7 - y + 0.

39 76 Maths In Focus Mathematics Etension Preliminary Course. Find the coordinates of point S where the tangent to the curve y + - is perpendicular to the line + y The curve y - has a gradient of 6 at point A. (a) Find the coordinates of A. (b) Find the equation of the tangent to the curve at A.. A function h t - t + 5 has a tangent at the point where t. Find the equation of the tangent. 5. A function f] g has a tangent parallel to the line - y + 0 at point P. Find the equation of the tangent at P. Further Differentiation and Indices The basic rule for differentiating n works for any rational number n. Investigation. h (a) Show that + h - -. ( + h) (b) Hence differentiate y from first principles. (c) Differentiate y - using a short method. Do you get the same answer as (b)?. (a) Show that ( + h - )( + h + ) h. (b) Hence differentiate y from first principles. (c) Differentiate y and show that this gives the same answer as (b). We sometimes need to change a function into inde form before differentiating. EXAMPLES. Differentiate $ 7-7 # -

40 Chapter 8 Introduction to Calculus 77 7 # 7. Find the equation of the tangent to the curve y at the point where. y When y Gradient of the tangent at ^, h : Equation of the tangent: y - y m _ - i y - -] - g - + y - + or + y Eercises. Differentiate (a) - (b). (c) 6 0. (d) (e) - - (f) (g) 8 (h) - -. Find the derivative function, writing the answer without negative or fractional indices. (a) (b) 5 6 (c) (d) 5 (e) (f) 5 -

41 78 Maths In Focus Mathematics Etension Preliminary Course Note that Use inde laws to simplify first. #. 6 6 Epand brackets first. (g) (h) (i) (j) 6 +. Find the gradient of the tangent to the curve y at the point where 7.. If, find when t. t dt 5. A function is given by f ( ). Evaluate fl ( 6). 6. Find the gradient of the tangent to the curve y at the point c, m. 7. Find if y ^ + h. 8. A function f ( ) has a tangent at ^, h. Find the gradient of the tangent. 9. Find the equation of the tangent to the curve y at the point c, m Find the equation of the tangent to f ( ) 6 at the point where 9.. (a) Differentiate. (b) Hence find the gradient of the tangent to the curve y at the point where.. Find the equation of the tangent to the curve y at the point c8, m.. If the gradient of the tangent to y is at point A, find the 6 coordinates of A.. The function f ( ) has fl ( ). Evaluate. 5. The hyperbola y has two tangents with gradient -. Find 5 the coordinates of the points of contact of these tangents. This rule is also called the function of a function rule or chain rule. Composite Function Rule A composite function is a function composed of two or more other functions. For eample, 5 ^ - h is made up of a function u 5 where u -. To differentiate a composite function, we need to use the result.. du # du

42 Chapter 8 Introduction to Calculus 79 Proof Let, d y and d u be small changes in, y and u where " 0, " 0, d u " 0. du Then # du As " 0, du " 0 du lim lim # lim d " 0 d " 0 du d " 0 So u Using the definition of the derivative from first principles, this gives du #. du You do not need to learn this proof. EXAMPLES Differentiate 7. (5 + ) Let u 5 + du Then 5 7 y u 6 ` 7u du du # du 6 7u # 5 5( 5 + ) 6 Can you see a quick way of doing this question?. ( + - ) 9 Let Then ` u + - du y u 8 9u du du # du 8 9u ( 6 + ) 96 ( + )( + - ) 8 CONTINUED

43 80 Maths In Focus Mathematics Etension Preliminary Course. - - ( - ) Let u - du - y u - u du du # du - u (-) - ( -) The derivative of a composite function is the product of two derivatives. One is the derivative of the function inside the brackets. The other is the derivative of the whole function. d n n - [ f ( )] fl ( n ) [ f ( )] You do not need to know this proof. Proof Let Then ` u f( ) du fl( ) n y u n - nu du du # du n - nu # fl( ) n - fl( ) n[ f( )]

44 Chapter 8 Introduction to Calculus 8 EXAMPLES Differentiate. (8 - ) 5 n - fl ( ) $ n[ f( )] $ 5( 8 -) 0 ( 8 -). ( + 8) yl fl ( ). n[ f( )] # ( + 8) ( + 8) 0 n - 0. (6 + ) - ( 6 + ) ( 6 + ) n - yl fl( ) $ n[ f( )] 6# - ( 6 + ) - - ( 6 + ) - ( 6 + ) Eercises. Differentiate (a) ( + ) (b) ( - ) (c) ( 5 - ) (d) ( 8 + ) 6 5 (e) ( - ) 7 9 (f) (5 + 9) (g) ( - ) (h) ( + ) 8 (i) ( ) (j) ( - + ) 6 6 (k) ( - )

45 8 Maths In Focus Mathematics Etension Preliminary Course (l) ( - ) (m) ( 9) (n) ( 5 + ) (o) ( ) (p) + (q) 5 - (r) ( + ) (s) ( 7 - ) 5 (t) + (u) - (v) ( 7) 9 + (w) - + () ( + ) (y) 5 ( 7 ) -. Find the gradient of the tangent to the curve y ] - g at the point ^, h.. If f ( ) ( - ) 5, evaluate fl ().. The curve y - has a tangent with gradient at point N. Find the coordinates of N. 5. For what values of does the function f ( ) - have fl ( ) -? 9 6. Find the equation of the tangent to y ( + ) at the point where -. Product Rule Differentiating the product of two functions y uv gives the result dv u + du v Proof y uv Given that, du and d v are small changes in y, u and v. y + ( u + du)( v + dv) uv + udv + vdu + dudv ` udv + vdu + dudv ^since y uvh dv du dv u + v + du d d

46 Chapter 8 Introduction to Calculus 8 As " 0, du " 0 dv du dv lim lim < u + v + du F " 0 " 0 d d dv du dv lim < u F+ lim < v F+ lim < du F " 0 d " 0 d " 0 dv du u + v It is easier to remember this rule as yl uvl + vul. We can also write this the other way around which helps when learning the quotient rule in the net section. You do not need to know this proof. If y uv, yl ulv + vl u EXAMPLES Differentiate. ] + g] - 5g You could epand the brackets and then differentiate: ] + g] - 5g Using the product rule: y uv where u + and v -5 ul vl yl ulv + vlu ] - 5g + ] + g ] 5 + g 5 y uv where u and v ] 5 + g 5 ul 0 vl 5. ] 5 + g CONTINUED

47 8 Maths In Focus Mathematics Etension Preliminary Course We can simplify this further by factorising. yl ulv + vlu 0 ] 5 + g + 5. ] 5 + g $ 5 0 ] 5 + g + 0 ] 5 + g 5 0 ] 5 + g 6 ] ] 5 + g ] 8 + g. ( - ) 5 - Remember 5 - ] 5 - g y uv where u - and v ] 5 - g ul vl - $ ( 5 - ) yl ulv + vlu - $ 5 ] - g +- ] 5 - g ] - g ( - ) ] 5 - g ( 5 - ) $ 5 - -( - ) 5-5 ( - ) -( - ) Eercises Change this into a product before differentiating.. Differentiate (a) ] + g (b) ] - g] + g (c) ] 5 + 7g (d) ^ - h (e) ^ - h (f) ] + g 5 (g) ] - g (h) ] - g (i) ] + g] + 5g (j) ^ h ^ + h (k) (l) - 5

48 Chapter 8 Introduction to Calculus 85. Find the gradient of the tangent to the curve y ] - g at the point ^, h. 5. If f ( ) ( + )( - ), evaluate fl ( ).. Find the eact gradient of the tangent to the curve y + 5 at the point where. 5. Find the gradient of the tangent where t, given ] t - 5g] t + g. 7. Find the equation of the tangent 7 to h ( t + ) ( t - ) at the point ^, 9h. 8. Find eact values of for which the gradient of the tangent to the curve y ] + g is. 9. Given f ( ) ( - )( + ), find the equation of the tangent at the point where Find the equation of the tangent to the curve y ] - g at the point ^, h. Quotient Rule Differentiating the quotient of two functions y u v gives the result. du v - v dv u Proof u y v Given that, du and d v are small changes in y, u and v. u + du y + v + dv u + du u u ` - a v v v since y k + d v vu ( + du) uv ( + dv) - vv ( + dv) vv ( + dv) vu ( + du) - uv ( + dv) vv ( + dv) vu + vdu - uv - udv vv ( + dv) vdu - udv vv ( + dv) du dv y v u d - d d vv ( + dv) As " 0, dv " 0

49 86 Maths In Focus Mathematics Etension Preliminary Course You do not need to know this proof. R V S du dv v - u W S d d W lim lim " 0 " 0S vv ( + dv) W T X du dv v - u v ulv - vl u It is easier to remember this rule as yl. v u ulv - vlu If y v, y l v EXAMPLES Differentiate u y v where u - 5 and v 5 + ul vl 5 ulv - vlu yl v 5 ( + ) - 5 ( - 5) ( 5 + ) ( 5 + ) ( 5 + ) u y v where u and v - ul - 5 vl ulv - vl u yl v ( - 5)( - ) - ( ) ( - ) ( - ) ( - ) 5 5

50 Chapter 8 Introduction to Calculus Eercises. Differentiate (a) - (b) + 5 (c) - (d) (e) - 7 (f) (g) - (h) + - (i) (j) (k) (l) - (m) (n) + (o) (p) (q) (r) ( + 5) (s) ( - 9) 5 + (t) - (7 + ) (u) ( + ) ( - 5) (v) + + (w) () ( - 9) 5. Find the gradient of the tangent to the curve y at the point + c, m.. If f ( ) + 5 evaluate fl ( ). -. Find any values of for which the gradient of the tangent to the curve y - is equal to Given f ( ) fl ( ). 6 find if + 6. Find the equation of the tangent to the curve y at the + point c, m. 7. Find the equation of the tangent - to the curve y at the + point where. Angle Between Curves To measure the angle between two curves, measure the angle between the tangents to the curves at that point.

51 88 Maths In Focus Mathematics Etension Preliminary Course m - m tan i + mm the curves at the point of intersection. where m and m are the gradients of the tangents to EXAMPLE Find the acute angle formed at the intersection of the curves y and y ( - ). The curves intersect at the point (,). For y At (, ), ( ) ` m For y ( - ) ( - ) At (, ), ( - ) ` m - m - m tan i + mm -(-) + ( - ) i 5c 08l

52 Chapter 8 Introduction to Calculus Eercises. (a) Sketch the curves y - and y on the same set of aes. (b) Show that the curves intersect at the point Q (, 0). (c) Find the gradient of the tangent of each curve at point Q. (d) Find the acute angle at which the curves intersect at Q.. (a) Sketch the curve y and the line y 6-9 on the same set of aes. (b) Find the point P, their point of intersection. (c) Find the gradient of the curve y at P. (d) Find the acute angle between the curve and the line at P.. Find the acute angle between the curves y and y at point (,).. Find the acute angle between the curves y and y - + at their point of intersection. 5. What is the obtuse angle between the curves f ( ) - and g ( ) - at the point where they meet? 6. The curves y - and y - + intersect at two points X and Y. (a) Find the coordinates of X and Y. (b) Find the gradient of the tangent to each curve at X and Y. (c) Find the acute angle between the curves at X and Y. 7. Find the acute angle between the curve f ( ) - and the line g ( ) - at their points of intersection. 8. (a) Find the points of intersection between y and y +. (b) Find the acute angle between the curves at these points. 9. Show that the acute angle between the curves y and y - is the same at both the points of intersection. 0. Find the obtuse angles between the curves y + and y 5 - at their points of intersection.

53 90 Maths In Focus Mathematics Etension Preliminary Course Test Yourself 8. Sketch the derivative function of each graph (a) 6. Find the gradient of the tangent to the curve y at the point (-,- 0). 7. If h 60t - t, find dh when t. dt 8. Find all -values that are not differentiable on the following curves. (a) (b). Differentiate y from first principles.. Differentiate (a) (b) - + (c) ( + - ) (d) 5( - ) (e) 5 (f) 9 dv. Find if v t - t -. dt (b) (c) y 5 y Given f ( ) (- ), find the value of (a) f () (b) f ( ). 9. Differentiate (a) f] g ] + 9g 5 (b) y - (c) y ] - g (d) y 5 (e) f ( )

54 Chapter 8 Introduction to Calculus 9 0. Sketch the derivative function of the following curve. y 5. Find the equation of the tangent to the curve y that is parallel to the line y Find the gradient of the tangent to the curve y ] - g ] - g at the point where. 7. Find fl ( ) when f] g ] - g Find the equation of the tangent to the curve y at the point where Differentiate s ut + at with respect to t and find the value of t for which ds 5, u 7 and a - 0. dt. Find the equation of the tangent to the curve y at the point ^, h.. Find the point on the curve y - + at which the tangent has a gradient of. ds. Find if S r r. dr. At which points on the curve y are the tangents horizontal? 0. Find the -intercept of the tangent to - the curve y at the point where +.. Find the acute angle between the curve y and the line y + at each point of intersection.. Find the obtuse angle between the curve y and the line y 6-8 at each point of intersection. Challenge Eercise 8. If f ( ) ( - ) 5, find the value of f () and fl (). 5h +. If A, 7h - da find when h. dh. Given t 00 t +, find and find dt values of t when 0. dt. Find the equations of the tangents to the curve y ( - )( + ) at the points where the curve cuts the -ais. 5. Find the points on the curve y - 6 where the tangents are parallel to the line y -. Hence find the equations of the normals to the curve at those points.

55 9 Maths In Focus Mathematics Etension Preliminary Course 6. Find fl ( ) if f ( ) Differentiate (5 + ) ( - 9) Find the derivative of y. ( - 9) 9. If f ( ) + +, for what eact values of is fl ( ) 7? 0. Find the equation of the normal to the curve y + at the point where 8.. The tangent to the curve y a + at the point where is inclined at 5c to the -ais. Find the value of a.. The normal to the curve y + at the point where, cuts the curve again at point P. Find the coordinates of P.. Find the eact values of the - coordinates of the points on the curve y ( - - ) where the tangent is horizontal.. Find the gradient of the normal to the curve y 5 - at the point (, 8). 5. Find the equation of the tangent to the curve y at point P (, 8). Find the coordinates of point Q where this tangent meets the y -ais and calculate the eact length of PQ (a) Show that the curves y ] - g and 5 - y intersect at ^, h + (b) Find the acute angle between the curves at this point. 7. The equation of the tangent to the curve y - n + - at the point where - is given by - y - 0. Evaluate n. 8. The function f ( ) + has a tangent that makes an angle of 0c with the -ais. Find the coordinates of the point of contact for this tangent and find its equation in eact form. 9. Find all values of the function 8 f ( ) ( - )(- ) for which fl ( ) (a) Find any points at which the graph below is not differentiable. (b) Sketch the derivative function for the graph. y 90c 80c 70c 60c. Find the point of intersection between the tangents to the curve y at the points where and -.. Find the equation of the tangent to the parabola y - at the point where the tangent is perpendicular to the line + y Differentiate.. (a) Find the equations of the tangents to the parabola y at the points where the line 6-8y + 0 intersects with the parabola. (b) Show that the tangents are perpendicular.

56 Chapter 8 Introduction to Calculus 9 5. Find any values of the function f ( ) where it is not differentiable. 6. The equation of the tangent to the curve y is y a + b at the point where -. Evaluate a and b. 7. Find the eact gradient with rational denominator of the tangent to the curve y - at the point where 5. p 8. The tangent to the curve y has a gradient of - at the point where. 6 Evaluate p. 9. Find dv r when r dr and h 6 given V rr h. 0. Evaluate k if the function f ( ) - k+ has fl ( ) 8.. Find the equation of the chord joining the points of contact of the tangents to the curve y - - with gradients and -.. Find the equation of the straight line passing through ^, h and parallel to the tangent to the curve y at the point ^, h.. Find fl ( 7) as a fraction, given f ( ) +.. For the function f ( ) a+ b+ cf, ( ), fl( ) 0and fl ( ) 8 when -. Evaluate a, b and c. 5. Find the equation of the tangent to the curve S rr + rrh at the point where r ( h is a constant). 6. Differentiate (a) - ] - 5g + (b) ( ) - 7. The tangents to the curve y - + at points A and B are perpendicular to the tangent at ^, h. Find the eact values of at A and B. 8. (a) Find the equation of the normal to the curve y + - at the point P where. (b) Find the coordinates of Q, the point where the normal intersects the parabola again.