Parametric Curves


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Parametric Curves

Let f and g be continuous functions defined on an interval I of the real line. Then the set of all points with coordinates (x,y), where x = f(t) and y = g(t) for some number t in I, is called a parametric curve in the plane, and the variable t is called the parameter.

The axis, or real line, of the parameter t is considered to be distinct from the coordinate axes of the plane of the curve. The parameter in a parametric curve frequently represents time.

The curve is the path of the moving point. Because f and g are continuous, the curve has no breaks in it.

Note that a parametric curve has a direction assigned to it, indicated usually by an arrowhead. The direction is usually corresponding to increasing values of the parameter, so when plotting the coordinates in order of t (negative values of t to positive values of t), if the graph’s direction moves right or counterclockwise, the arrowhead should point right or counterclockwise, and if the graph’s direction moves left or clockwise, the arrowhead should point left or clockwise.

When the curve starts and ends at the same point, it is called a closed curve.

General Plane Curves: Parametrization
A set S of points in the plane is called a plane curve if S is the parametric curve x = f(t) and y = g(t), for some continuous functions f and g and interval I.

When a plane curve is discussed, it is only dealing with the set of points and not any particular parametrization (set of parametric equations) of the curve. Therefore, a plane curve has no distinguished direction.

Plane Curve Examples: Involute of a Circle
A string is wound around a fixed circle. One end is unwound in such a way that the part of the string not lying on the circle is extended in a straight line. The curve that traces the free end of the string is called an involute of a circle.

Suppose the circle has the equation x² + y² = a², and suppose the end of the string begins at the point A = (a,0). At some subsequent time, let P be the position of the end of the string, where the unwound string is shown in red, and let T be the point where the string leaves the circle. Then PT is tangent to the circle at T. The path of P will be parametrized in terms of the angle TOA, denoted as t.

Let points R on OA and S on TR be shown in the above graph. Then:

Since the string does not stretch or slip on the circle:

TP = at = arcTA

Since a tangent to a circle is perpendicular to the radial line to the point of contact:

angle OTP = π/2

Since they are similar triangles:

If P has the coordinates (x,y), then:

x = OR + SP = acost + atsint
y = RT - ST = asint - atcost

The above is the parametric equations of the involute in terms of t.

Plane Curve Examples: Cycloid
If a circle rolls without slipping along a straight line, the path created by a point fixed on the circle is called a cycloid.

Suppose this straight line is the x-axis, that the circle has the radius a, and it lies above the straight line. Additionally, the point whose motion is being followed is originally at the origin O.

The point has moved to position P when the circle has rolled through an angle of t. Additionally, it is tangent to the line at T.

Since no slipping occurs:

segment OT = arcPT = rθ = at

Let PQ be perpendicular to TC. If P has the coordinates (x,y), then:

Therefore, the following equations are the parametric equations of the cycloid:

x = a(t - sint)
y = a(1 - cost)

Ellipses and Their Points
Suppose an ellipse had the parametric equations x = 3cost and y = sint (0 ≤ t ≤ 2π). Since cost’s amplitude is greater than sint’s amplitude, the radius is unequal and so it will not result in a circle.

The following is a graph of the ellipse:

The points P on an ellipse can be found using two circles with radii of 3 and 1, and interpreting the parameter as an angle.

Note that between the points [-3,3] on the x-axis is called the major axis, and between the points [-1,1] on the y-axis is called the minor axis.

Note that the parametric equations x = cost and y = sint is the equation of a circle with radius 1.

Sketch the following parametric curve, indicating its direction with an arrowhead. Then eliminate the parameter and provide an ordinary equation in x and y whose graph contains the parametric curve.

x = 3cost, y = 3sint (0 ≤ t ≤ 3π/2)

If t = 0, then x = 3cos0 = 3 and y = 3sin0 = 0, making the coordinate (3,0).
If t = π/2, then x = 0 and y = 3, making the coordinate (0,3).
If t = π, then x = -3 and y = 0, making the coordinate (-3,0).
If t = 3π/3, then x = 0 and y = -3, making the coordinate (0,-3).
Because of the domain of t, t is not tested for 2π.

Since the amplitude of cost is equal to the amplitude of sint, the parametric curve is a circle with a restricted domain of t.

The direction is counterclockwise, because as t increases from 0 to 3π/2, the values move in a counterclockwise direction from 0, making the arrow point counterclockwise.

Therefore:

To find an ordinary equation in x and y whose graph contains the parametric curve, consider the following:

If x² + y² = r², where r = 3, then:

(3cost)² + (3sint)² = 9cos²t + 9sin²t = 9

Therefore, the ordinary equation in x and y whose graph contains the parametric curve is x² + y² = 9.

Sketch the following parametric curve, indicating its direction with an arrowhead. Then eliminate the parameter and provide an ordinary equation in x and y whose graph contains the parametric curve.

x = acoss, y = bsins (0 ≤ s ≤ 2π), where a > b > 0.

If s = 0, then x = acos0 = a and y = bsin0 = 0, making the coordinate (a,0).
If s = π/2, then x = 0 and y = b, making the coordinate (0,b).
If s = π, then x = -a and y = 0, making the coordinate (-a,0).
If s = 3π/3, then x = 0 and y = -b, making the coordinate (0,-b).
If s = 2π, then x = a and y = 0, making the coordinate (a,0).

Since a > b, the amplitude of coss is greater than the amplitude of sins, and therefore the parametric curve is an ellipse. Additionally, since the coordinates when s = 0 and s = 2π are the same, the ellipse starts and ends at the same point, causing the parametric curve to be a closed curve.

The direction is counterclockwise, because as t increases from 0 to 2π, the values move in a counterclockwise direction from 0, making the arrow point counterclockwise.

Therefore:

To find an ordinary equation in x and y whose graph contains the parametric curve, consider the following:

Since coss = x/a, and sins = y/b, and cos²s + sin²s = 1, then:

cos²s + sin²s = 1
(x/a)² + (y/b)² = 1
x²/a² + y²/b² = 1

Therefore, the ordinary equation in x and y whose graph contains the parametric curve is x²/a² + y²/b² = 1.

Sketch the following parametric curve, indicating its direction with an arrowhead. Then eliminate the parameter and provide an ordinary equation in x and y whose graph contains the parametric curve.

x = t, y = 1 - t (0 ≤ t ≤ 1)

Constructing a table of values:

The direction is right, because as t increases from 0 to 1, the values move to the right of 0, making the arrow point right.

Therefore:

To find an ordinary equation in x and y whose graph contains the parametric curve, eliminate the parameter t:

Since x = t, then:

y = 1 - t = 1 - x

Therefore, the ordinary equation in x and y whose graph contains the parametric curve is y = -x + 1.

Sketch the following parametric curve, indicating its direction with an arrowhead. Then eliminate the parameter and provide an ordinary equation in x and y whose graph contains the parametric curve.

x = 1 + 2t, y = t² (-∞ < t < ∞)

Constructing a table of values:

The direction is right, because as t increases from ∞ to ∞, the values move to the right of 0, making the arrow point right.

Therefore:

To find an ordinary equation in x and y whose graph contains the parametric curve, eliminate the parameter t:

Since x = 1 + 2t:

Therefore, the ordinary equation in x and y whose graph contains the parametric curve is the following:

Sketch the following parametric curve, indicating its direction with an arrowhead. Then eliminate the parameter and provide an ordinary equation in x and y whose graph contains the parametric curve.

x = 1/t, y = t - 1 (0 < t < 4)

Constructing a table of values:

The direction is left, because as t increases from 0 to 4, the values move to the left of 0, making the arrow point left.

Therefore:

To find an ordinary equation in x and y whose graph contains the parametric curve, eliminate the parameter t:

Since x = 1/t:

Therefore, the ordinary equation in x and y whose graph contains the parametric curve is the following:

Sketch the following parametric curve, indicating its direction with an arrowhead. Then eliminate the parameter and provide an ordinary equation in x and y whose graph contains the parametric curve.

x = 3sin2t, y = 3cos2t (0 ≤ t ≤ π/3)

Constructing a table of values:

Since the amplitude of cos2t is equal to the amplitude of sin2t, the parametric curve is a circle with a restricted domain of t.

The direction is clockwise, because as t increases from 0 to π/3, the values move in a clockwise direction from 0, making the arrow point clockwise.

Therefore:

To find an ordinary equation in x and y whose graph contains the parametric curve, consider the following:

If x² + y² = r², where r = 3, then:

(3cos2t)² + (3sin2t)² = 9cos²2t + 9sin²2t = 9

Therefore, the ordinary equation in x and y whose graph contains the parametric curve is x² + y² = 9.

Sketch the following parametric curve, indicating its direction with an arrowhead. Then eliminate the parameter and provide an ordinary equation in x and y whose graph contains the parametric curve.

x = 3sinπt, y = 4cosπt (-1 ≤ t ≤ 1)

Constructing a table of values:

Since cosπt’s amplitude > sinπt’s amplitude, the parametric curve is an ellipse. Additionally, since the coordinates when t = -1 and t = 1 are the same, the ellipse starts and ends at the same point, causing the parametric curve to be a closed curve.

The direction is clockwise, because as t increases from -1 to 1, the values move in a clockwise direction from -1, making the arrow point clockwise.

Therefore:

To find an ordinary equation in x and y whose graph contains the parametric curve, consider the following:

Since sinπt = x/3, and cosπt = y/4, and cos²πt + sin²πt = 1, then:

cos²πt + sin²πt = 1
(y/4)² + (x/3)² = 1
y²/16 + x²/9 = 1

Therefore, the ordinary equation in x and y whose graph contains the parametric curve is y²/16 + x²/9 = 1.

Sketch the following parametric curve, indicating its direction with an arrowhead. Then eliminate the parameter and provide an ordinary equation in x and y whose graph contains the parametric curve.

x = cos³t, y = sin³t (0 ≤ t ≤ 2π)

Constructing a table of values:

Since the amplitude of cos2t is equal to the amplitude of sin2t, the parametric curve would be a circle. However, since the trigonometric functions are cubed, it will not be in the shape of a circle.

Additionally, since the coordinates when t = 0 and t = 2π are the same, the parametric curve starts and ends at the same point, causing the parametric curve to be a closed curve.

The direction is counterclockwise, because as t increases from 0 to 2π, the values move in a counterclockwise direction from 0, making the arrow point counterclockwise.

Therefore:

To find an ordinary equation in x and y whose graph contains the parametric curve, consider the following:

Therefore, the ordinary equation in x and y whose graph contains the parametric curve is the following:

Find a parametrization of the parabola y = x² using the slope of the tangent line at the general point as the parameter.

The slope m of the tangent line at the general point of y is the derivative of y, which is 2x. Therefore:

m = 2x

Since this is a parametrization, m is the parameter and not the coordinate value, therefore:

m = 2x
x = m/2

Using the original function to find y:

y = x² = (m/2)² = m²/4

Therefore, the parametrization of the parabola y = x² is x = m/2 and y = m²/4.