17.3. Parametric Curves. Introduction. Prerequisites. Learning Outcomes

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1 Parametric Curves 7.3 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 On completion ou should be able to... HELM (08): Section 7.3: Parametric Curves be familiar with Cartesian coordinates be familiar with trigonometric and hperbolic functions and be able to manipulate them be able to differentiate simple functions be able to locate turning points and distinguish between maima and minima. sketch planar curves given in parametric form understand how the same curve can be described using different parameterisations recognise some conics given in parametric form 33

2 . Parametric curves Here 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 6 Plot the curve = cos t }{{ = 3 sin 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 Plotting the (, ) coordinates gives the curve in Figure 6. 5π 6π 7π 8π 9π 0π 3 t = π t = 7π t = 3π 3π Figure 6 The curve in Figure 6 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 ( ) + ( 3 ) = cos t + sin t i.e = which we easil recognise as an ellipse whose major-ais is the -ais. Also, as t ranges from 0 π = cos t decreases from 0, and = 3 sin t increases from 0 3. We conclude that the 34 HELM (08): Workbook 7: Conics and Polar Coordinates

3 parametric equations = cos t, = 3 sin t together with the parametric range 0 t π describe that part of the ellipse 4 + = in the positive quadrant. On the curve in Figure 6 we have 9 used an arrow to indicate the direction that we move along the curve as t increases from its initial value 0. Task Plot the curve = t + = t 3 0 t Do ou recognise this curve as a conic section? First construct a table of (, ) values as t ranges from 0 : Your solution t t Now plot the points on a Cartesian plane: Your solution 0 t = Now eliminate the t-variable from = t +, = t 3 to obtain the form of the curve: HELM (08): Section 7.3: Parametric Curves 35

4 Your solution = 4 which is the equation of a parabola. Eample 7 Sketch the curve = t + = t t Solution This is ver similar to the previous Task (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 = Figure 7 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 = so = ( ) 3 which is the same as obtained in the last Task. The main difference is that particular values of t locate (in general) different (, ) points on the curve for the two parametric representations. We conclude that a given curve in the plane can have man (in fact infinitel man) parametric descriptions. 36 HELM (08): Workbook 7: Conics and Polar Coordinates

5 Task Show that the two parametric representations below describe the same curve. (a) = cos t = sin t 0 t π (b) = t = t 0 t Eliminate t from the parametric equations in (a): Your solution + = cos t + sin t = Eliminate t from the parametric equations in (b): Your solution = = or + = What do ou conclude? Your solution Both parametric descriptions represent (part of) a circle centred at the origin of radius.. General parametric form We will assume that an curve in the plane ma be written in parametric form: = g(t) = h(t) }{{} parametric equations of the curve t 0 t t }{{} parameter range in which g(t), h(t) are given functions of t and the parameter t ranges over the values t 0 t. 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 derivatives d.3, we discovered how to obtain the derivative d d d d = d d d and. We found and d d = ( d d d ) d from a knowledge of the parametric ( ) 3 d HELM (08): Section 7.3: Parametric Curves 37

6 Note that derivatives with respect to the parameter t are often denoted b a dot: d ẋ so that d ẏ d ẍ etc d d = ẏ d ẋÿ ẏẍ and = ẋ d ẋ 3 Knowledge of the derivative is sometimes useful in curve sketching. Eample 8 Sketch the curve = t 3 + 3t + t = 3 t t 3 t. Solution = t 3 + 3t + t = t(t + )(t + ) = 3 t t = (t + 3)(t ) so that = 0 when,, and = 0 when t = 3,. We calculate the values of, at various values of t: t We see that 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 = 3t + 6t +, d = t d d = ( + t) 3t + 6t + so there is a turning point when t =. The reader is urged to calculate d and to show that d this is negative when t = (i.e. at = 0, = 4) indicating a maimum when. (The reader should check that vertical tangents occur at t = 0.43 and t =.47, to d.p.) We can now make a reasonable sketch of the curve: t = 0.5 t = t =.5 t =, 0(double point) t =.5.5 t increasing t = 3 6 t = 6 Figure 8 38 HELM (08): Workbook 7: Conics and Polar Coordinates

7 3. 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 The standard parametric equations for a parabola are: = at = at Clearl, we have t = ( ) and b eliminating t we get = a or = 4a which we recognise a 4a as the standard Cartesian description of a parabola. As an illustration, Figure 9 shows the curve with a = and t.3 t = t = 3 t = Figure 9 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 t + sin t or, in more familiar form: a b a + b =. If we choose the range for t as 0 t 7π 4 the following segment of the ellipse is 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 Figure 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 (08): Section 7.3: Parametric Curves 39

8 In the stud of the orbits of planets and satellites it is often preferable to use plane polar coordinates (r, θ) to treat the problem. In these coordinates an ellipse has an equation of the form r = A + B cos θ, with A and B positive numbers such that B < A. Not onl is there a difference in the equations on passing from Cartesian to polar coordinates; there is also a change in the origin of coordinates. The polar coordinate equation is using a focus of the ellipse as the origin. In the Cartesian description the foci are two points at +e along the -ais, where e obes the equation e = a b, if we assume that a < b i.e. we choose the long ais of the ellipse as the -ais. This problem gives some practice at algebraic manipulation and also indicates some shortcuts which can be made once the mathematics of the ellipse has been understood. Eample 9 An ellipse is described in plane polar coordinates b the equation r = + cos θ Convert the equation to Cartesian form. [Hint: remember that = r cos θ.] Solution Multipling the given equation b r and then using = r cos θ gives the results = r + so that r = We now square the second equation, remembering that r = +. We now have 4( + ) = ( ) = + so that = We now recall the method of completing the square, which allows us to set 3 + = 3( + 3 ) 9 ) Putting this result into the equation and collecting terms leads to the final result ( + 3 ) a + b = with a = 3 and b = 3. This is the standard Cartesian form for the equation of an ellipse but we must remember that we started from a polar equation with a focus of the ellipse as origin. The presence of the term + 3 in the equation above actuall tells us that the focus being used as origin was a distance of 3 to the right of the centre of the ellipse at = 0. The preceding piece of algebra was necessar in order to convince us that the original equation in plane polar coordinates does represent an ellipse. However, now that we are convinced of this we can go back and tr to etract information in a more speed wa from the equation in its original (r, θ) form. 40 HELM (08): Workbook 7: Conics and Polar Coordinates

9 Solution (contd.) Tr setting θ = 0 and θ = π in the equation r = + cos θ We find that at θ = 0 we have r = 3 while at θ = π we have r =. These r values correspond to the two ends of the ellipse, so the long ais has a total length + 3 = 4 3. This tells us that a = 3, eactl as found b our longer algebraic derivation. We can further deduce that the focus acting as origin must be at a distance of from the centre of the ellipse in order to lead to the two r values 3 at θ = 0 and θ = π. If we now use the equation e = a b mentioned earlier then we find that 9 = 4 9 b, so that b =, as obtained b our length algebra. 3 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 giving rise to the equation of the hperbola in Cartesian form: a b =. In Figure we have chosen a parameter range t. t = t =0 t = 0.5 t = t =0.5 t = t =.5 Figure 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 HELM (08): Section 7.3: Parametric Curves 4

10 Eercises. In the following sketch the given parametric curves. Also, eliminate the parameter to give the Cartesian equation in and. (a) = t, = t 0 t (b) = t, = t + 0 t (c) = t = t 0 < t < 3 (d) = 3 sin πt = 4 cos πt t 0.5. Find the tangent line to the parametric curve = t t t + t at the point where t =. 3. For each of the following curves epressed in parametric form obtain epressions for d d and d and use this information to help make a sketch. d (a) = t t, = t 4t (b) = t 3 3t, = t t s. (a) = (b) = 3 t = 3 (c) = ( + ) = (d) = t =.5 d d. = t + = t d = t + d t when t = = 0, = d when t = then = 3 d tangent line is = HELM (08): Workbook 7: Conics and Polar Coordinates

11 3. (a) d d = t 4 = t d d = = d d = t 4 t = t t d d = [(t ) (t 4)] 8(t ) 3 = (t ) 3 t = t = (b) = (t )(t + t + ) = (t )(t + ) = (t + )(t ) d d = t = 3t 3 d d = = 6t d = [(3t 3) (t )6t] = 6t + 6t 6 d (3t 3) 3 7(t ) 3 t =, HELM (08): Section 7.3: Parametric Curves 43