Parametric Equations and Polar Coordinates

Parametric Equations and Polar Coordinates

Parametrizations of Plane Curves In previous chapters, we have studied curves as the graphs of functions or equations involving the two variables x and y. Another way to describe a curve is by expressing both coordinates as functions of a third variable t.

Parametric Equations

t I is a parameter for the curve. is the parameter interval. If I is a closed interval, a t b, ( f a ga) ( f b gb) ( ), ( ) is the initial point of the curve and ( ), ( ) is the terminal point of the curve.

Example 1: Sketch the curve defined by the parametric equations x= t 2, y = t+ 1, < t <.

Example 2: Identify geometrically the curve in Eg. 1 (Fig. 11.2) by eliminating the parameter t and obtaining an algebraic equation in x and y.

Example 3: Graph the parametric curves ( a) x= cos t, y = sin t, 0 t 2 π. ( b) x= acos t, y = asin t, 0 t 2π

Example 4: The position P( x, y) of a particle moving in the xy plane is given by the equations and parameter interval x= t, y = t, t 0. Identify the path traced by the particle and describe the motion.

Example 5: 2 A parametrization of the graph of the function f( x) is given by x = t y = f t = t < t < 2, (),. When t 0, this parametrization gives the same path in the xy-plane as we had in Eg. 4. However, since the parameter t here can now also be negative, we obtain the left-hand part of the parabola as well; that is, we have the entire parabolic curve. = x For this parametrization, there is no starting point and no terminal point. (Fig. 11.5).

Example 6: Find a parametrization for the line through the point (a, b) having the slope, m. Solution: A Cartesian equation of the line is y b= mx ( a). If we set the parameter t = x a, we find that x = a + t and y b = mt. That is, x = a+ t, y = b+ mt, < t < parametrizes the line. This parametrization differs from the one we would obtain by the technique used in Eg. 5 when t = x. However, both parametrizations give the same line.

Example 7: Sketch and identify the path traced by the point Pxy (, ) if 1 1 x = t +, y t, t 0. t = t >

Exercise 11.1:

Calculus with Parametric Curves In this section, we apply calculus to parametric curves.

Tangents A parametrized curve x= f( t) and y= gt ( ) is differentiable at tif fand gare differentiable at t. At a point in a differentiable parametrized curve where y is also a differentiable function of x, the derivatives dy dt, dx dt, and dy dx are related by the Chain Rule: dy dy dx = dt dx dt If dx dt = 0, we may divide both sides of this equation by dx dt to solve for dy dx.

Example 1: Find the tangent to the curve π π x = sec t, y = tan t, < t < 2 2 ( ) at the point 2, 1, where t = π 4 (Fig 11.12).

Example 2: 2 d y 2 3 Find as a function of t if x = t t, y = t t. 2 dx

Length of a Parametrically Defined Curve

Using Leibniz notation: L 2 2 b dx dy = + a dt dt dt (3)

Example 4: Using the definition, find the length of the circle of radius r defined parametrically by x = rcost and y = rsin t, 0 t 2 π.

Example 5: Find the length of the astroid (Fig. 11.13) 3 3 x = cos t, y = sin t, 0 t 2 π.

Length of a Curve y = f(x) The length formula in section 6.3 is a special case of Eq. (3). Given a continuously differentiable function y = f( x), a x b, we can assign x = t as a parameter. The graph of the function f is then the curve C defined parametrically by x = t and y = f( t), a t b, a special case of what we considered before. Then, dx dy = 1 and = f ( t). dt dt From equation (1), we have dy dx giving dy dt = = f ( t), dx dt 2 dy dx + dt dt 2 [ f t ] [ f x ] 2 2 = 1 + ( ) = 1 + ( ).

Exercises 11.2:

Polar Coordinates

Definition of Polar Coordinates

Example 1: Find all the polar coordinates of the point P(2, π 6).

Corresponding coordinate pairs of P are: 2, 1, 0,, 2 6 5 2, 2, 1, 0,, 2 6 2, ± ± = + ± ± = + n n n n π π π π

Exercise: Mark down the given polar coordinates on the diagram. A(3, π 3) B( 2, π 4) C( 4, π 2) D(1, π 6) E(2, 5π 6)

Polar Equations and Graphs

If ris fixed at a constant value r= a 0, the point Pr (, θ ) will lie a units from the origin O. As θ varies over any interval of length 2 π, P then traces a circle of radius a centered at O. If θ is fixed at a constant value θ= θ and let r vary between 0 and, the point Pr (, θ ) traces the line through O that makes an angle of measure θ0 with the initial ray.

Example 2: ( a) r = 1 and r = 1 are equations for the circle of radius1 centered at O. ( b) θ = π 6, θ = 7π 6, and θ = 5π 6 are equations for the line in Fig. 11.21. Equations of the form r = a and θ= θ can be combined to define regions, segments and rays. 0

Example 3: Graph the sets of points whose polar corrdinates satisfy the following conditions: ( a) 1 r 2 and π 0 θ 2 ( b) 3 r 2 and π θ = 4 2π 5π ( c) θ (no restrictions on r) 3 6

Relating Polar and Cartesian Coordinates

The first two of these equations uniquely determine the Cartesian coordinates x and y given the polar coordinates r and θ. If x and y are given, the third equation gives two possible choices for r (a positive and a negative). ( x y) ( ) θ [ π) For each, 0, 0, there is a unique 0,2 satisfying the first two equations, each then giving a polar coordinate representation of the Cartesian point ( x, y). The other polar coordinate representations for the point can be determined from these two as in Eg. 1.

2 2 Find a polar equationfor for the circle ( 3) 9. x y + = Example 5: θ θ θ 6sin 0 6sin or 0 0 sin 6 0 6 9 9 6 Solution: 2 2 2 2 2 = = = = = + = + + r r r r r y y x y y x

Example 6: Replace the following polar equations by equivalent Cartesian equations,and identify their graphs. (a) r cosθ = 4 (b) r 2 = 4r cosθ 4 (c) r = 2cosθ sinθ

Solution: (a) (b) r cosϑ = 4 Graph : Vertical line through x = 4 on the x - axis r 2 Graph :Circle, radius 2, center ( h,k) = (2, 0). (c) r(2cosϑ sinϑ) = 4 2r cosϑ r sinϑ = 4 2x = 4r cosϑ and x = 4 x x x ( x y = 4 y = 2x 2 2 2 4x + 4x 4 = 4 Graph : Line, slope m = 2, y - intercept b = -4. + 2) y 2 2 = 4x y 2 + 4 + + y 2 = 0 y 2 = 4

Exercises 11.3:

Graphing in Polar Coordinates This section describes techniques for graphing equations in polar coordinates.

Symmetry

Slope The slope of a polar curve r = f ( θ ) is given by dy dx, not r = df / Think of graph f as the graph of parametric equations: x = r cosθ = f ( θ )cosθ, y = r sinθ = f ( θ )sinθ If f is a differentiable function of θ, then so are x and y and when dx dθ 0, we can calculate dy dx from the parametric formula : dθ dy dx = dy dx dθ = dθ = d dθ d dθ df sinθ + dθ df cosθ dθ ( f ( θ ) sinθ ) ( f ( θ ) cosθ ) f ( θ )cosθ f ( θ )sinθ 56

If then dy dx the curve r = f ( θ ) ( 0, θ ) 0 0 = = 0, and the slope equation gives: f f ( θ0 )sinθ0 ( θ )cosθ 0 f ( θ ) passes through the origin at θ = θ, 0 = tanθ 0 0

Example 1: A cardioid Graph the curve r = 1 cos θ. Solution: The curve is symmetric about the x-axis because ( r, θ) on the graph r = 1 cosθ r = 1 cos( θ ) ( r, θ ) on the graph.

Example 2: 2 Graph the curve 4cos. r Solution: The equation requires cosθ 0, so we get the entire graph by running θ from π 2 to π 2. 2 (, ) on the graph 4cos r ϑ = r = 2 = r θ 4cos( ϑ) ϑ ( r, ϑ) on the graph. { symmetric about the x axis} 2 (, ) on the graph 4cos r ϑ r = r = 2 ( ) 4cos ϑ { } ( r, ϑ) on the graph. symmetric about the origin These 2 symmetries implies symmetry about the y axis. Make a short table of values, plot the corresponding points, and use information about symmetry and tangents to guide us in connecting the points with a smooth curve. ϑ

A technique for Graphing Method 1. 1. Make a table of ( r, θ ) -values. 2. Plot the corresponding points. 3. Connect them in order of increasing. Method 2. 1. Graph r = f (θ ) in the Cartesian rθ - plane. 2. Use the Cartesian graph as a table and guide to sketch the polar coordinate graph. Method 2 is better in the sense that its drawing shows where r is positive, negative, nonexistent, and also where r is increasing/decreasing. θ

Example 3: A lemniscate Graph the lemniscate curve r 2 = sin 2θ

Exercise 11.4: