17.3. Parametric Curves. Introduction. Prerequisites. Learning Outcomes

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1 Parametric Curves 17.3 Introduction In this section we eamine et another wa of defining curves - the parametric description. We shall see that this is, in some was, far more useful than either the Cartesian description or the polar form. Although we shall onl stud planar curves (curves ling in a plane) the parametric description can be easil generalised to the description of spatial curves which twist and turn in three dimensional space. Prerequisites Before starting this Section ou should... Learning Outcomes After completing this Section ou should be able to... 1 be familiar with Cartesian coordinates be familiar with trigonometric functions and how to manipulate then 3 be able to differentiate simple functions 4 be able to locate turning points and distinguish between maima and minima. sketch planar curves given in parametric form understand that the same curve can be described using man different parametrisations recognise some conics given in parametric form

2 1. Parametric Curves In this section we eplore the use of a parameter t in the description of curves. We shall see that it has some advantages over the more usual Cartesian description. We start with a simple eample. Eample Plot the curve =cos t =3sin t }{{} parametric equations of the curve 0 t π }{{ } parameter range Solution The approach to sketching the curve is straightforward. We simpl give the parameter t various values as it ranges through 0 π and, for each value of t, calculate corresponding values of (, ) which are then plotted on a Cartesian plane. The value of t and the corresponding values of, are recorded in the following table: t 0 π π 3π 4π π 6π 7π 8π 9π 10π Plotting the (, ) coordinates gives the curve in Figure 1. 3 t = π t = 7π t = 3π 3π Figure 1. The curve resembles part of an ellipse. This can be verified b eliminating t from the parametric equations to obtain an epression involving, onl. If we divide the first parametric equation b and the second b 3, square both and add we obtain ( ) ( ) + = cos t + sin t 1 3 HELM (VERSION 1: March 18, 04): Workbook Level 1

3 i.e =1 which we easil recognise as an ellipse whose major-ais is the ais. Also, as t ranges from 0 π it is clear from =cos t that decreases from 0 and, from =3sin t, that increases from 0 3. We conclude that the parametric equations =cos t, =3sin t together with the parametric range 0 t π/ describe that part of the ellipse + =1in 4 9 the positive quadrant. On the curve in Figure 1 we have used an arrow to indicate the direction that we move along the curve as t increases from its initial value 0. Plot the curve = t +1 =t 3 0 t 1 Do ou recognise this curve as a conic section? First construct a table of (, ) values as t ranges from 0 +1 Your solution t t Now plot the points on a Cartesian plane Your solution 3 HELM (VERSION 1: March 18, 04): Workbook Level 1

4 0 1 1 t = Now eliminate the t variable from = t +1, =t 3toobtain the form of the curve. Your solution = 4 1 which is the equation of a parabola. Eample Sketch the curve = t +1 =t t 1 HELM (VERSION 1: March 18, 04): Workbook Level 1 4

5 Solution This is ver similar to the previous guided eercise (ecept for t 4 replacing t in the epression for and t replacing t in the epression for. The corresponding table of values is t t = We see that this is identical to the curve drawn previousl. This is confirmed b eliminating the t parameter from the epressions defining,. Here t = 1so =( 1) 3 which is the same as obtained in the guided eercise. The main difference is that particular values of t locate (in general) different (, ) points on the curve. We conclude that a given curve in the plane can have man (in fact infinitel man) parametric descriptions. Show that the two parametric descriptions (a) = cos t = sin t 0 t π (b) = t = 1 t 0 t 1 describe the same curve. Your solution Eliminate t from the equations in (a) + = cos t + sin t =1 5 HELM (VERSION 1: March 18, 04): Workbook Level 1

6 Your solution Eliminate t from the equations in (b) = 1 =1 or + =1 i.e. both parametric parametric descriptions represent (part of) the circle centred at the origin of radius 1.. General Parametric Form We will assume that an curve in the plane ma be written in parametric form: = g(t) } {{ = h(t) } t 0 t t 1 }{{} parametric equations of the curve parameter range in which g(t), h(t) are given functions of t and the parameter t ranges over the values t 0 t 1. As we give values to t within this range then corresponding values of, are calculated from = g(t), = h(t) which can then be plotted on an plane. In Workbook 1, section 3, we discovered how to obtain the derivative d from a knowledge of d the parametric derivatives d d and.wefound dt dt d d = d dt d dt and d d = ( d dt d d dt dt ( d ) 3 dt ) d dt Note that derivatives with respect to the parameter t are often denoted b a dot d dt ẋ d dt ẏ d dt ẍ etc so that d d = ẏ ẋ and d ẋÿ ẏẍ = d ẋ 3 Knowledge of the derivative is sometimes useful in curve sketching. Eample Sketch the curve = t 3 +3t +t =3 t t 3 t 1. HELM (VERSION 1: March 18, 04): Workbook Level 1 6

7 Solution = t 3 +3t +t = t(t +3t +)=t(t + )(t +1) =3 t t = (t +t 3) = (t + 3)(t 1) so that vanishes when, 1, and vanishes when t = 3, 1. We easil calculate the values of, at these and at other values of t. t We see t = and give rise to the same coordinate values for (, ). This represents a double-point in the curve which is one where the curve crosses itself. Now d dt =3t +6t +, d dt = t d d = (1 + t) 3t +6t + so there is a turning point when t = 1. The reader is urged to calculate d and to show that d this is negative when t = 1 (i.e. = 0, = 4)indicating a maimum when. (The reader should check that vertical tangents occur at t = 0.43, 1.47 (to d.p.)). We can now make a reasonable sketch of the curve; t = 0.5 t = 1 t = 1.5 t =, 0(double point) t =.5.5 t increasing t = 3 6 t = Standard Forms of Conic Sections in Parametric Form We have seen above that, given a curve in the plane, there is no unique wa of representing it in parametric form. However, for some commonl occurring curves, particularl the conics, there are accepted standard parametric equations. The Parabola 7 HELM (VERSION 1: March 18, 04): Workbook Level 1

8 The standard parametric equations for a parabola are: = at =at (the range for the parameter ( t is usuall omitted). Clearl, we have t = a and b eliminating t= a or =4a which we recognise as the standard Cartesian description of a parabola. 1 t.3) 4a ) (As an illustration, the sketch shows the curve with a = and t = t =1 1 3 t = 1 The Ellipse Here, the standard equations are = a cos t = b sin t Again, eliminating t (dividing the first equation b a, the second b b, squaring and adding) we have ( ( ) + = cos a) t + sin t 1 b or, in more familiar form: a + b =1. If we choose the range for the parameter t as 0 t 7π the following segment of the ellipse is 4 obtained. t = π t = 3π b t = π 4 4 3π 4 π 4 t = π a t = 5π 4 t = 7π 4 3 t = π Here we note that (ecept in the special case when a = b, giving a circle) the parameter t is not the angle that the radial line makes with the the positive ais. HELM (VERSION 1: March 18, 04): Workbook Level 1 8

9 The Hperbola The standard equations are = a cosh t = b sinh t In this case, to eliminate t we use the identit cosh t sinh t =1giving rise to the equation of the hperbola in Cartesian form: a b =1. In the following curve we have chosen a parameter range 1 t. t = t = 0.5 t =1.5 t = 1 t =1.5 To obtain the complete curve the parameter range <t< must be used. These parametric equations onl give the right-hand branch of the hperbola. To obtain the left-hand branch we would use = a cosh t = b sinh t Eercises 1. In the followng eamples sketch the given parametric curves, Also, eliminate the parameter to give the Cartesian equation in and. (a) = t, = t 0 t 1 (b) = t, = t +1 0 t (c) = = t 0 <t<3 t (d) =3sin πt =4cos πt 1 t 0.5. Find the tangent line to the parametric curve = t t t +t at the point where t =1. 3. For each of the following curves epressed in parametric form obtain epressions for d d and d. Use this information to help make a sketch of these curves. d (a) = t t, = t 4t (b) = t 3 3t, = t t 9 HELM (VERSION 1: March 18, 04): Workbook Level 1

10 1. (a) = 1 1 (b) =3 1 t =3 (c) = ( +)= t = 0.5 (d) =1 t = 1. d dt =t +1 d dt =t 1 d d t +1 = t 1 when t =1then d d =3 when t =1 =0, = tangent line is =3 + HELM (VERSION 1: March 18, 04): Workbook Level 1 10

11 3. (a) d dt =t 4 d dt =t d dt = d dt = d d = t 4 t = t t 1 d [(t ) (t 4)] = = d 8(t 1) 3 1 (t 1) 3 4 t = t =4 8 (b) =(t )(t +t +1)=(t )(t +1) =(t + 1)(t ) d dt =t 1 d dt =3t 3 d dt = d dt =6t t = 1, d d = [(3t 3) (t 1)6t] = 6t +6t 6 (3t 3) 3 7(t 1) 3 11 HELM (VERSION 1: March 18, 04): Workbook Level 1

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