applications of differentiation

Transcription

1 Contents 1 applications of differentiation 1. Tangents and normals. Maima and minima 3. The Newton-Raphson method 4. Curvature 5. Differentiation of vectors 6. Comple impedance Learning outcomes In this workbook ou will learn to appl our knowledge of differentiation to solve some basic problems connected with curves. First ou will learn how to obtain the equation of the tangent line and the normal line to an point of interest on a curve. Secondl, ou will learn how to find the positions of maima and minima on a given curve. Thirdl, ou will learn how, given an approimate position of the root of a function, a better estimate of the position can be obtained using the Newton-Raphson technique. Lastl ou will learn how to characterise how sharpl a curve is turning b calculating its curvature. Time allocation You are epected to spend approimatel seven hours of independent stud on the material presented in this workbook. However, depending upon our abilit to concentrate and on our previous eperience with certain mathematical topics this time ma var considerabl. 1

2 Tangents and Normals 1.1 Introduction In this section we see how the equations of the tangent line and the normal line at a particular point on the curve = f() can be obtained. The equations of tangent and normal lines have form: = m + c, = n + d respectivel. We shall show that the product mn is 1 if these lines are to be perpendicular. The constants c, d are then obtained b using the propert that the both the normal and tangent lines and the curve pass through a common point. Prerequisites Before starting this Section ou should... Learning Outcomes After completing this Section ou should be able to... 1 be able to differentiate standard functions understand the geometrical interpretation of a derivative 3 understand the trigonometric epansions of sin(a + B), cos(a + B) obtain the condition that two given lines be perpendicular obtain the equation of the tangent line to a curve obtain the equation of the normal line to a curve

3 1. Perpendicular Lines One form for the equation of a straight line is = m + c where m and c are constants. We remember that m is the gradient of the line and its value is the tangent of the angle θ that the line makes with the positive ais. The constant c is the value obtained where the line intersects the ais. See the following diagram: = m + c c m = tan θ θ If we have a second line, with equation = n + d then, unless m = n, the two lines will intersect. These are drawn together on the following diagram. The second line makes an angle ψ with the positive ais. c = m + c θ = n + d n = tan ψ ψ A simple question to ask is what is the relation between m and n if the lines are perpendicular? If the lines are perpendicular, as shown in the net figure, the angles θ, ψ must satisf the relation: ψ θ =90 c θ ψ This is true since the angles in a triangle add up to 180. According to the figure the three angles are 90, θ and 180 ψ. Therefore 180 =90 + θ + (180 ψ) impling ψ θ =90 3 HELM (VERSION 1: April, 004): Workbook Level 1

4 In this special case that the lines are perpendicular or normal to each other the relation between the gradients m and n is easil obtained. In this deduction we use the following basic trigonometric relations and identities: sin(a B) =sin A cos B cos A sin B cos(a B) =cos A cos B + sin A sin B Now tan A = sin A cos A sin 90 =1 cos 90 =0 m = tan θ = tan(ψ 90 o ) (see previous figure) = sin(ψ 90o ) cos(ψ 90 o ) = cos ψ sin ψ = 1 tan ψ = 1 n So mn = 1 Ke Point Two straight lines = m + c, = n + d are perpendicular if m = 1 n or equivalentl mn = 1 This result assumes that neither of the lines are parallel to the ais or to the ais, as in such cases one gradient will be zero and the other infinite. Eercises Which of the following pairs of lines are perpendicular: i. = +1, = +1 ii. + 1=0, + =0 iii. =8 +3, = Answers (i) perpendicular (ii) not perpendicular (iii) perpendicular. Tangents and Normals to a curve HELM (VERSION 1: April, 004): Workbook Level 1 4

5 As we know, the relationship between an independent variable and a dependent variable is denoted b = f() As we also know, the geometrical interpretation of this relation takes the form of a curve in an plane as illustrated in the following diagram. = f() We know how to calculate a value of given a value of. We can either do this graphicall (which is inaccurate) or else use the function itself. So, at an value of 0 the corresponding value is 0 where 0 = f( 0 ) Let us eamine the curve in the neighbourhood of the point ( 0, 0 ). There are two important constructions of interest the tangent line at ( 0, 0 ) the normal line at ( 0, 0 ) These are shown in the following figure tangent line 0 θ 0 ψ normal line We note the geometricall obvious fact: the tangent and normal lines at an given point on a curve are perpendicular to each other. The curve = is drawn below. On this graph draw the tangent line and the normal line at the point ( 0 =1, 0 =1) 5 HELM (VERSION 1: April, 004): Workbook Level 1

6 1 1 1 tangent line normal line 1 θ ψ From our graph, estimate the values of θ and ψ. (You will need a protractor). θ ψ θ =63.4 o ψ = o Returning to the curve = f() : we know, from the geometrical interpretation of the derivative that d = tan θ 0 (in case ou ve forgotten, the notation on the left-hand side of this relation means evaluate d at the value 0 ) Also, in this relation, θ is the angle the tangent line to the curve = f() makes with the positive ais. This is highlighted in the following diagram: = f() θ 3. The Tangent Line to a Curve 0 d 0 = tan θ Let the equation of the tangent line to the curve = f() atthepoint ( 0, 0 )be: = m + c HELM (VERSION 1: April, 004): Workbook Level 1 6

7 where m and c are constants to be found. The line just touches the curve = f() atthe point ( 0, 0 ) so, at this point both must have the same value for the derivative. That is: m = d Since we know (in an particular case) f() and the value 0 we can readil calculate the value for m. The value of c is found b using the fact that the tangent line and the curve pass through the same point ( 0, 0 ). 0 = m 0 + c and 0 = f( 0 ) Thus m 0 + c = f( 0 ) leading to c = f( 0 ) m 0 0 Ke Point The equation of the tangent line to the curve = f() atthepoint ( 0, 0 )is = m + c where m = d and c = f( 0 ) m 0 0 Eample Find the equation of the tangent line to the curve = at the point (1,1) Solution Here f() = and 0 =1. Thus = m = d d = 0 Also c = f( 0 ) m 0 = f(1) m =1 = 1. Therefore the equation of the tangent line is = 1. (Check back to the previous guided eercise to see if this looks right ). Find the equation of the tangent line to the curve =e at the point =0. The curve and the line are displaed in the following figure: tangent line 7 HELM (VERSION 1: April, 004): Workbook Level 1

8 Specif 0 and f 0 = f() = Now obtain the values of d,f( 0 ) m 0 0 d = f( 0 ) m 0 = 0 0 =0 f() =e Now obtain the equation of the tangent line = =1 and f (0) 1(0) = e 0 0=1 0 d = +1 d =e Find the equation of the tangent line to the curve = sin 3 at the point = π and find where the tangent line intersects the ais. See the following 4 figure. tangent line Specif 0 and f 0 = f() = Now obtain the values of d,f( 0 ) m = π 4 f() =sin 3 HELM (VERSION 1: April, 004): Workbook Level 1 8

9 d = f( 0 ) m 0 = 0 =3cos 3 d d π 4 =3cos 3π = 3 4 =.1 and f ( π 4 ) ( ) mπ = sin 3π 3 π = π 4 =.37 to d.p. Now obtain the equation of the tangent line = where does the line intersect the ais? = (4+3π) = when = = 0 =1.1 to d.p. = The Normal Line to a Curve We have alread noted that, at an point ( 0, 0 )onacurve = f() the tangent and normal lines are perpendicular. Thus if the equations of the tangent and normal lines are, respectivel = m + c = n + d then m = 1 n or, equivalentl n = 1 m. We have also noted, for the tangent line m = d 0 so n can easil be obtained. To find d, we again use the fact that the normal line = n + d and the curve have a point in common: so n 0 +d=f( 0 ) leading to d = f( 0 ) n 0. 0 = n 0 +d and 0 = f( 0 ) Find the equation of the normal line to curve = sin 3 at the point = π 4. See the earlier guided eercise for the value of m. 9 HELM (VERSION 1: April, 004): Workbook Level 1

10 m = = d π 4 Hence find the value of n nm = 1 n = m = 3 nm = 1 n = 3 The equation of the normal line is = +d. 3 Now find the value of d. (Remember the normal line and the curve pass through the same point.) Now obtain the equation of the normal line. = π ) + d=sin π d= π 4 ( 3 The curve and the normal line are shown in the following figure: = normal line Find the equation of the normal line to the curve = 3 at =1. First find f(), 0, d,m,n 0 HELM (VERSION 1: April, 004): Workbook Level 1 10

11 f() = 3, 0 =1, d 1 =3 =3 1 m =3and n = 1 3 Now use the propert that the normal line = n +dand the curve = 3 pass through the point (1,1) d= 1=n +d d=1 n = = 4 3 Thus the equation of the normal line is = The curve and the normal line are shown 3 3 below: normal line Eercises 1. Find the equations of the tangent and normal lines to the following curves at the points indicated (a) = 4 + (1, 3) (b) = 1 (, ) (What would be obtained if the point was (1, 0)?) (c) = 1 (1, 1). Find the value of a (a 1) if the two curves = e and = e a are to intersect at right-angles. 11 HELM (VERSION 1: April, 004): Workbook Level 1

12 Answers. The curves will intersect at right-angles if their tangent lines, at the point of intersection, are perpendicular. Point of intersection: e = e a i.e. = a =0 The tangent line to = e a is = m + c where m = ae a =0 = a The tangent line to = e is = k + g where k = e =0 = 1 These two lines are perpendicular if a( 1) = 1 i.e. a =1. = e = e Answers 1. (a) f() = 4 + d =43 +4, d =8 =1 tangent line =8 + c. This passes through (1, 3) so 3=8+c c = 5 =8 5 normal line = 1 8 +d. This passes through (1, 3) so 3= 1 +d d= 5 = (b) f() = 1 = d 1 = 1 d = tangent line = + c. This passes through (, ) so = + c c = = + ( ) normal line = + d. This passes through so =, +d d=0 =. At (1,0) the tangent line is =1and the normal line is =0. (c) f() = 1 = 1 d 1 d = 1 =1 tangent line: = 1 + c. This passes through (1,1) so 1= 1 + c c = 1 = normal line: = +d. This passes through (1,1) so 1= +d d=3 = +3. HELM (VERSION 1: April, 004): Workbook Level 1 1