Laboratory 25
Parametric Equations and Polar Coordinates

Introduction

    Certain curves in 2D (or 3D) cannot be described as being the graph of a function because they do not pass the vertical line test. This is the case for the circle with equation x^2 + y^2 = 1, for example. However, in many cases the graph may be described using parametric equations. For the circle, set x = cos(t),   y = sin(t),   0≤ t ≤2π, then x^2 + y^2 = cos^2(t) + sin^2(t) = 1, so the point (cos(t), sin(t)), 0≤t≤2π, always lies on the circle. Furthermore, as t increases from 0 to 2π, this point moves at a constant rate in an anticlockwise direction around the circle (starting at (1, 0), and returning to the same point). We say that the equations x = cos(t), y = sin(t) are parametric equations for the circle x^2 + y^2 = 1.
           Generally, if x and y are given in terms of a third variable t∈[a, b], say x = f(t), y = g(t), then as t varies, the point (f(t), g(t)) traces a curve in the plane called a parametric curve. The corresponding equations are called parametric equations of the curve. A given curve may be parametrized in many different ways: for example, x = cos(2t), y = sin(2t), 0≤t≤π and x = cos(t^(1/2)), y = sin(t^(1/2)), 0≤t≤ (2π)^(1/2), are different parametrizations of the circle x^2 + y^2 = 1.

Example 25.1, The Cycloid

    The equations of some curves are more easily represented using parametric equations. This is the case with the cycloid, which is the locus of a point on the circumference of a rolling circle. One arch of the cycloid is paramtrized by the equations
                                       x(t) = r(t - sin(t)),              y(t) = r(1 - cos(t)),    0≤t≤2π .
where r is the radius of the circle. To plot this curve for r = 1 we use the ParametricPlot command (the AspectRatio->Automatic command will give the same scale on both axes).

Clear[x, y] x[t_] := t - Sin[t] ; y[t_] := 1 - Cos[t]

ParametricPlot[{x[t], y[t]}, {t, 0, 2Pi}, AspectRatio -> Automatic] ;

[Graphics:HTMLFiles/index_24.gif]

Experiment 25.1a

    Plot each of the following curves using the given parametrizations (use the AspectRatio->Automatic command to obtain an accurate representation).

    1.  x = t^3 - 2t, y = t^2 - t, -2≤t≤2.
    
    2.  x = sin(4t), y = cos(3t), 0≤t≤2π.
    
    3.  x = sin(t + sin t), y = cos (t + cos t ), 0≤t≤2π.
    

Experiment 25.1b

    Write down parametric equations for the ellipse  x^2/9 + y^2/4 = 1, and use them to plot its graph.

Example 25.2, Tangents, Arc Length and Areas using Parametric Curves

Area

The area bounded by the curve y = f(x), the x-axis and the lines x = c, x = d, is given by the formula  A = ∫_c^df(x) x, when f(x) ≥0. Suppose the curve is parametrized by x(t), y(t), a≤t≤b, then using the substitution rule for integration, this area can be written

               &nbs ...      A = ∫_c^df(x) x = ∫_a^by(t) x ' (t) t .

We use this formula to calculate the area under one arch of the cycloid x(t) = r(t - sin(t)) , y(t) = r(1 - cos(t)).

Clear[x, y] x[t_] := r (t - Sin[t]) ; y[t_] := r (1 - Cos[t])

Integrate[y[t] * x '[t], {t, 0, 2Pi}]

3 π r^2

If the parametrization traces the curve from right to left, it will be necessary to take the absolute value of your answer. In addition, you shoud calculuate the area of those parts of the graph that lie beneath the x-axis separately.

Arc Length

    Given a parametric curve x(t), y(t), a≤t≤b, the usual arc length formula is easily modified to give the length of the arc as
                                                           L = ∫_a^b (((d x)/(d t))^2 + ((d y)/(d t))^2)^(1/2) t.

As we have seen, the corresponding indefinite integrals often cannot be evaluated in terms of elementary functions. This is not the case for the cycloid x(t) = (t - sin(t)),      y(t) = (1 - cos(t)). It has arc length from t = 0 to t = s obtained as follows (the Assumptions → s > 0 command is used to tell Mathematica that s is positive):

Clear[x, y] x[t_] := t - Sin[t] ; y[t_] := 1 - Cos[t]

L[s_] := Integrate[Sqrt[(x '[t])^2 + (y '[t])^2], {t, 0, s}, Assumptionss>0]

L[s]

4 - 2 (2 - 2 Cos[s])^(1/2) Cot[s/2]

The length of one arch of the cycloid is:

L[2Pi]

8

Experiment 25.2

    Find the arc length and area enclosed by each of the following parametrized curves (use a numerical approximation where appropriate). In each case, first give a graph of the curve.

    1.  x = 3 cos(t), y = 4 sin(t), 0≤t≤2π .
    
    2.  x = cos^3(t), y = sin^3(t), 0≤t≤2π.
    
    

Example 25.3, Polar Coordinates

            If a point P in the plane has cartesian coordinates (x, y) and (r, θ) are its polar coordinates, then FormBox[RowBox[{x, =, RowBox[{r, , cos(θ), Cell[]}]}], TraditionalForm]and y = r sin(θ), where θ is the angle OP makes with the x-axis, positive when measured in the counter clockwise direction, and negative in the clockwise direction. r is the distance from P to the origin O, but we can allow r to be negative by agreeing that (-r, θ) and (r, θ) lie on the same line through the origin O, but on opposite sides of the origin. Clearly x^2 + y^2 = 1, and θ = arctan(y/x).

To graph the cardiod FormBox[Cell[TextData[Cell[BoxData[r(θ) = 1 + sin(θ)]]]], TraditionalForm], we first need to load a Graphics package:

<<Graphics`Graphics`

r[θ_] := 1 + Sin[θ]

PolarPlot[r[θ], {θ, 0, 2π}] ;

[Graphics:HTMLFiles/index_73.gif]

Experiment 25.3

    Graph the following polar curves using PolarPlot (make sure the Graphics package is loaded).  You may find it easier to use t in place of θ, but θ can be obtained with the use of the Palette.
    1.  r = cos(θ/2) + cos(θ/3) .
    
    2.  r = sin(5/3θ).
    
    Choose the parameter interval to make sure you produce a closed curve.

Example 25.4, Areas and Arc Length for Polar Graphs

Areas

    Using the formula for the area of a sector of a circle, A = 1/2r^2θ (where r is the radius and θ is the angle subtended, measured in radians), we obtain the formula for the area enclosed by a polar curve r = f(θ) and the rays θ = a, θ = b:

               &nbs ... bsp;  A = 1/2∫_a^br^2θ = 1/2∫_a^b (f(θ))^2θ .

For example, let us calculate the area enclosed by the 3 leafed rose  r(θ) = cos(3θ). The main difficulty is calculating the limits of integration:

r[θ_] := Cos[3θ]

PolarPlot[r[θ], {θ, 0, 2π}] ;

[Graphics:HTMLFiles/index_88.gif]

We find the area of one leaf of the rose. Notice that cos(3θ) = 0 when 3θ = ± π/2, ± 3π/2 etc.  This occurs when θ = ± π/6, ± π/2 etc. To see that these rays give rise to the correct limits of integration, let us calculate the slope of the curve at the origin:

                                     =

since f(θ) = 0 at the origin.  We see for example that when θ = π/6, the slope is tan(π/6), so that the ray θ = π/6 is tangent to the curve at the origin and hence gives rise to one of the limits of integration. Consequently, the area of one leaf of the curve is:

Integrate[1/2 (r[θ])^2, {θ, -π/6, π/6}]

π/12

To find the area enclosed by the inner loop of the curve r = 1 + 2sin(θ), we proceed as follows:

r[θ_] := 1 + 2 * Sin[θ]

PolarPlot[r[θ], {θ, 0, 2Pi}] ;

[Graphics:HTMLFiles/index_101.gif]

Notice that for  7π/6≤θ≤11π/6, r is negative. In this case, as we mentioned earlier, the point (r, θ) is plotted as (-r, θ + π)  (e.g., r(3π/2) = 1 + sin(3π/2) = 1 - 2 = -1, so (-1, 3π/2) is plotted as (1, π/2)).

Now r(θ) = 1 + 2 sin(θ) = 0 when sin(θ) = -1/2, or when θ = -π/6, 7π/6, 11π/6 etc. It is clear that the limits of integration are 7π/6 and 11π/6.

Integrate[1/2 (r[θ])^2, {θ, 7π/6, 11π/6}]

-(3 3^(1/2))/2 + π

Arc Length

Suppose tath a polar curve is defined by r = f(θ), then x = f(θ) cos(θ) and y = f(θ) sin(θ), and we see that d x/d θ = f ' (θ) cos (θ - f(θ) sin(θ) = d r/d θ cos(θ) - r sin(θ) and similarly d y/d θ = d r/d θ sin(θ) + r cos(θ) .

Using the identity cos^2(θ) + sin^2(θ) = 1, it is now easy to check that  (d x/d θ)^2 + (d y/d θ)^2 = r^2 + (d r/d θ)^2.
Thus our earlier formula for arc length becomes   L = ∫_a^b (r^2 + ((d r)/(d θ))^2)^(1/2) θ for curves given in terms of a polar equation.

Let us calculate the length of arc of one leaf of the curve r(θ) = cos(3θ) (we need to use NIntegrate in this case):

r[θ_] := Cos[3θ]

NIntegrate[Sqrt[(r[θ])^2 + (r '[θ])^2], {θ, -π/6, π/6}]

2.22748

For the curve r(θ) = 1 + 2 sin(θ), to find its full arc length we use the interval 0≤θ≤2π:

Clear[r, θ] ; r[θ_] := 1 + 2 * Sin[θ]

NIntegrate[Sqrt[(r[θ])^2 + (r '[θ])^2], {θ, 0, 2π}]

13.3649

Experiment 25.4

    Find the area enclosed and the arc lengths of each of the following closed curves (give the graph in each case using PolarPlot).  
    1.  r = 4 sin(3θ), 0≤θ≤2π.
    
    2.  r = 2cos(θ) - sin^2(θ), 0≤θ≤π.

Credits

These laboratories were created by Raouf Boules, Geoff Goodson, Ohoe Kim and Mike O'Leary for use in the Calculus courses of Towson University. Commercial use is prohibited without permission of the authors. Non-commercial use is permitted, provided this credit section is retained.

Created by Mathematica  (August 23, 2004)