Applications of Differentiation

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1 Contents 1 Applications of Differentiation 1.1 Tangents and Normals 1. Maima and Minima The Newton-Raphson Method Curvature Differentiation of Vectors Case Stud: Comple Impedance 60 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.

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 are often written as = m + c, = n + d respectivel. We shall show that the product of their gradients m and n is such that mn is 1 which is the condition for perpendicularit. Prerequisites Before starting this Section ou should... Learning Outcomes On completion ou should be able to... be able to differentiate standard functions understand the geometrical interpretation of a derivative know the trigonometric epansions of sin(a + B), cos(a + B) obtain the condition that two given lines are perpendicular obtain the equation of the tangent line to a curve obtain the equation of the normal line to a curve HELM (008): Workbook 1: Applications of Differentiation

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 Figure 1: c = m + c m = tan θ θ Figure 1 If we have a second line, with equation = n + d then, unless m = n, the two lines will intersect at one point. These are drawn together in Figure. The second line makes an angle ψ with the positive -ais. c = m + c θ = n + d n = tan ψ ψ Figure 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 Figure, the angles θ and ψ must satisf the relation: ψ θ = 90 c θ ψ Figure HELM (008): Section 1.1: Tangents and Normals

4 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 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 Figure ) = sin(ψ 90o ) cos(ψ 90 o ) = cos ψ sin ψ = 1 tan ψ = 1 n So mn = 1 Ke Point 1 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. Eercise Which of the following pairs of lines are perpendicular? (a) = + 1, = + 1 (b) + 1 = 0, + = 0 (c) = 8 +, = (a) perpendicular (b) not perpendicular (c) perpendicular 4 HELM (008): Workbook 1: Applications of Differentiation

5 . Tangents and normals to a curve 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 Figure 4. = f() Figure 4 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 ). constructions of interest the tangent line at ( 0, 0 ) the normal line at ( 0, 0 ) These are shown in Figure 5. There are two important tangent line 0 θ 0 ψ normal line Figure 5 We note the geometricall obvious fact: the tangent and normal lines at an given point on a curve are perpendicular to each other. HELM (008): Section 1.1: Tangents and Normals 5

6 Task The curve = is drawn below. On this graph draw the tangent line and the normal line at the point ( 0 = 1, 0 = 1): 1 1 tangent line 1 θ 1 ψ normal line From our graph, estimate the values of θ and ψ in degrees. (You will need a protractor.) θ ψ θ 6.4 o ψ 15.4 o Returning to the curve = f() : we know, from the geometrical interpretation of the derivative that d = tan θ (the notation d means evaluate d at the value = 0) Here θ is the angle the tangent line to the curve = f() makes with the positive -ais. This is highlighted in Figure 6: = f() d = tan θ θ 0 Figure 6 6 HELM (008): Workbook 1: Applications of Differentiation

7 . The tangent line to a curve Let the equation of the tangent line to the curve = f() at the point ( 0, 0 ) be: = m + c where m and c are constants to be found. The line just touches the curve = f() at the 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 Ke Point The equation of the tangent line to the curve = f() at the point ( 0, 0 ) is = m + c where m = d and c = f( 0 ) m 0 Alternativel, the equation is 0 = m( 0 ) where m = d and 0 = f( 0 ) Eample 1 Find the equation of the tangent line to the curve = at the point (1,1). Solution Method 1 Here f() = and 0 = 1 thus d = m = d = Also c = f( 0 ) m 0 = f(1) m = 1 = 1. The tangent line has equation = 1. Method 0 = f( 0 ) = f(1) = 1 = 1 The tangent line has equation 1 = ( 1) = 1 HELM (008): Section 1.1: Tangents and Normals 7

8 Task 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 First specif 0 and f: 0 = f() = 0 = 0 f() = e Now obtain the values of d and f( 0 ) m 0 : d = f( 0 ) m 0 = d = e d = 1 and f (0) 1(0) = e 0 0 = 1 0 Now obtain the equation of the tangent line: = = HELM (008): Workbook 1: Applications of Differentiation

9 Task Find the equation of the tangent line to the curve = sin at the point = π 4 and find where the tangent line intersects the -ais. See the following figure: tangent line First specif 0 and f: 0 = f() = 0 = π 4 f() = sin Now obtain the values of d and f( 0 ) m 0 correct to d.p.: d = f( 0 ) m 0 = d = cos d ( π ) 4 and f mπ 4 = sin π 4 Now obtain the equation of the tangent line: = π 4 = cos π 4 = =.1 ( ) π 4 = 1 + π 4 =.7 to d.p. = (4 + π) so = (to d.p.) HELM (008): Section 1.1: Tangents and Normals 9

10 Where does the line intersect the -ais? = When = = 0 = 1.1 to d.p. 4. The normal line to a curve We have alread noted that, at an point ( 0, 0 ) on a curve = 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 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: 0 = n 0 + d and 0 = f( 0 ) so n 0 + d = f( 0 ) leading to d = f( 0 ) n 0. Task Find the equation of the normal line to curve = sin at the point = π 4. [The equation of the tangent line was found in the previous Task.] First find the value of m: m = d π 4 = m = Hence find the value of n: n = nm = 1 n = 10 HELM (008): Workbook 1: Applications of Differentiation

The equation of the normal line is = + d. Now find the value of d to d.p.. (Remember the normal line must pass through the curve at the point = π 4.) ( π ) + d = sin π 4 4 d = 1 π 4 0.

11 The equation of the normal line is = + d. Now find the value of d to d.p.. (Remember the normal line must pass through the curve at the point = π 4.) ( π ) + d = sin π 4 4 d = 1 π Now obtain the equation of the normal line to d.p.: = = The curve and the normal line are shown in the following figure: normal line Task Find the equation of the normal line to the curve = at = 1. First find f(), 0, d, m, n: f() =, 0 = 1, 1 d = = m = and n = 1 1 HELM (008): Section 1.1: Tangents and Normals 11

12 Now use the propert that the normal line = n + d and the curve = pass through the point (1, 1) to find d and so obtain the equation of the normal line: d = = 1 = n + d d = 1 n = = 4. Thus the equation of the normal line is = The curve and the normal line through (1, 1) are shown 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, ) (b) = 1, (, ) What would be obtained if the point was (1, 0)? (c) = 1/, (1, 1). Find the value of a if the two curves = e and = e a are to intersect at right-angles. 1 HELM (008): Workbook 1: Applications of Differentiation

13 s 1. (a) f() = 4 + d = 4 + 4, d = 8 =1 tangent line = 8 + c. This passes through (1, ) so = 8 5 normal line = d. This passes through (1, ) so = (b) f() = 1 d = 1 d = 1 = ( ) tangent line = + c. This passes through, so = + ( ) normal line = + d. This passes through, so =. At (1, 0) the tangent line is = 1 and the gradient is infinite (the line is vertical), and the normal line is = 0. (c) f() = 1 d = 1 1 d = 1 =1 tangent line: = 1 + c. This passes through (1, 1) so = normal line: = + d. This passes through (1, 1) so = +.. 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 (a = 1 not sensible) The tangent line to = e a is = m + c where The tangent line to = e is = k + g where m = ae a =0 = a k = e =0 = 1 These two lines are perpendicular if a( 1) = 1 i.e. a = 1. = e = e HELM (008): Section 1.1: Tangents and Normals 1

2.3. One-to-One and Inverse Functions. Introduction. Prerequisites. Learning Outcomes

2.3. One-to-One and Inverse Functions. Introduction. Prerequisites. Learning Outcomes One-to-One and Inverse Functions 2. Introduction In this Section we eamine more terminolog associated with functions. We eplain one-to-one and man-to-one functions and show how the rule associated with