Functions and Models 1. Parametric Curves 1.7 3 Parametric Curves Imagine that a particle moves along the curve C shown in Figure 1. It is impossible.

Presentation on theme: "Functions and Models 1. Parametric Curves 1.7 3 Parametric Curves Imagine that a particle moves along the curve C shown in Figure 1. It is impossible."— Presentation transcript:

1 Functions and Models 1

2 Parametric Curves 1.7

3 3 Parametric Curves Imagine that a particle moves along the curve C shown in Figure 1. It is impossible to describe C by an equation of the form y = f (x) because C fails the Vertical Line Test. But the x- and y-coordinates of the particle are functions of time and so we can write x = f (t) and y = g(t). Figure 1

4 4 Parametric Curves Such a pair of equations is often a convenient way of describing a curve and gives rise to the following definition. Suppose that x and y are both given as functions of a third variable t (called a parameter) by the equations x = f (t) y = g(t) (called parametric equations). Each value of t determines a point (x, y), which we can plot in a coordinate plane. As t varies, the point (x, y) = (f (t), g(t)) varies and traces out a curve C, which we call a parametric curve.

5 5 Parametric Curves The parameter t does not necessarily represent time and, in fact, we could use a letter other than t for the parameter. But in many applications of parametric curves, t does denote time and therefore we can interpret (x, y) = (f (t), g(t)) as the position of a particle at time t.

6 6 Example 1 – Graphing a Parametric Curve Sketch and identify the curve defined by the parametric equations x = t 2 – 2t y = t + 1 Solution: Each value of t gives a point on the curve, as shown in the table. For instance, if t = 0, then x = 0, and y = 1 so the corresponding point is (0, 1).

7 7 Example 1 – Solution In Figure 2 we plot the points (x, y) determined by several values of the parameter t and we join them to produce a curve. A particle whose position is given by the parametric equations moves along the curve in the direction of the arrows as t increases. cont’d Figure 2

8 8 Example 1 – Solution Notice that the consecutive points marked on the curve appear at equal time intervals but not at equal distances. That is because the particle slows down and then speeds up as t increases. It appears from Figure 2 that the curve traced out by the particle may be a parabola. This can be confirmed by eliminating the parameter t as follows. We obtain t = y – 1 from the second equation and substitute into the first equation. cont’d

9 9 Example 1 – Solution This gives x = t 2 – 2t = (y – 1) 2 – 2(y – 1) = y 2 – 4y + 3 and so the curve represented by the given parametric equations is the parabola x = y 2 – 4y + 3. cont’d

10 10 Parametric Curves No restriction was placed on the parameter t in Example 1, so we assumed that t could be any real number. But sometimes we restrict t to lie in a finite interval. For instance, the parametric curve x = t 2 – 2t y = t + 1 0  t  4 shown in Figure 3 is the part of the parabola in Example 1 that starts at the point (0, 1) and ends at the point (8, 5). Figure 3

11 11 Parametric Curves The arrowhead indicates the direction in which the curve is traced as t increases from 0 to 4. In general, the curve with parametric equations x = f (t) y = g(t) a  t  b has initial point (f (a), g(a)) and terminal point (f (b), g(b)).

12 12 Graphing Devices

13 13 Graphing Devices Most graphing calculators and computer graphing programs can be used to graph curves defined by parametric equations. In fact, it’s instructive to watch a parametric curve being drawn by a graphing calculator because the points are plotted in order as the corresponding parameter values increase.

14 14 Example 6 – Graphing x as a Function of y Use a graphing device to graph the curve x = y 4 – 3y 2. Solution: If we let the parameter be t = y, then we have the equations x = t 4 – 3t 2 y = t Using these parametric equations to graph the curve, we obtain Figure 9. Figure 9

15 15 Example 6 – Solution It would be possible to solve the given equation (x = y 4 – 3y 2 ) for y as four functions of x and graph them individually, but the parametric equations provide a much easier method. cont’d

16 16 Graphing Devices In general, if we need to graph an equation of the form x = g(y), we can use the parametric equations x = g(t) y = t Notice also that curves with equations y = f (x) (the ones we are most familiar with—graphs of functions) can also be regarded as curves with parametric equations x = t y = f (t)

17 17 Graphing Devices Graphing devices are particularly useful when sketching complicated curves. For instance, the curves shown in Figures 10, 11, and 12 would be virtually impossible to produce by hand. Figure 12 Figure 11 Figure 10 x = t + 2 sin 2t y = t + 2 cos 5t x = 1.5 cos t – cos 30t y = 1.5 sin t – sin 30t x = sin(t + cos 100t) y = cos(t + sin 100t)

18 18 Graphing Devices One of the most important uses of parametric curves is in computer-aided design (CAD). We will investigate special parametric curves, called Bézier curves, that are used extensively in manufacturing, especially in the automotive industry. These curves are also employed in specifying the shapes of letters and other symbols in laser printers.

19 19 The Cycloid

20 20 Example 7 – Deriving Parametric Equations for a Cycloid The curve traced out by a point P on the circumference of a circle as the circle rolls along a straight line is called a cycloid (see Figure 13). If the circle has radius r and rolls along the x-axis and if one position of P is the origin, find parametric equations for the cycloid. Figure 13

21 21 Example 7 – Solution We choose as parameter the angle of rotation  of the circle (  = 0 when P is at the origin). Suppose the circle has rotated through  radians. Because the circle has been in contact with the line, we see from Figure 14 that the distance it has rolled from the origin is | OT | = arc PT = r  Therefore the center of the circle is C(r , r). Figure 14

22 22 Example 7 – Solution Let the coordinates of P be (x, y). Then from Figure 14 we see that x = | OT | – | PQ | = r  – r sin  = r (  – sin  ) y = | TC| – | QC | = r – r cos  = r (1 – cos  ) cont’d

23 23 Example 7 – Solution Therefore parametric equations of the cycloid are x = r (  – sin  ) y = r (1 – cos  )   One arch of the cycloid comes from one rotation of the circle and so is described by 0    2 . Although Equations 1 were derived from Figure 14, which illustrates the case where 0 <  <  /2, it can be seen that these equations are still valid for other values of . Figure 14 cont’d

24 24 Example 7 – Solution Although it is possible to eliminate the parameter  from Equations 1, the resulting Cartesian equation in x and y is very complicated and not as convenient to work with as the parametric equations. cont’d

25 25 The Cycloid One of the first people to study the cycloid was Galileo, who proposed that bridges be built in the shape of cycloids and who tried to find the area under one arch of a cycloid. Later this curve arose in connection with the brachistochrone problem: Find the curve along which a particle will slide in the shortest time (under the influence of gravity) from a point A to a lower point B not directly beneath A.

26 26 The Cycloid The Swiss mathematician John Bernoulli, who posed this problem in 1696, showed that among all possible curves that join A to B, as in Figure 15, the particle will take the least time sliding from A to B if the curve is part of an inverted arch of a cycloid. Figure 15

27 27 The Cycloid The Dutch physicist Huygens had already shown that the cycloid is also the solution to the tautochrone problem; that is, no matter where a particle P is placed on an inverted cycloid, it takes the same time to slide to the bottom (see Figure 16). Figure 16

28 28 The Cycloid Huygens proposed that pendulum clocks (which he invented) should swing in cycloidal arcs because then the pendulum would take the same time to make a complete oscillation whether it swings through a wide or a small arc.