Polar Curves and Tangents

In this section, we will explore the fascinating world of polar curves and how to find their tangents. Unlike Cartesian coordinates (x, y) which define a point by its horizontal and vertical distances from an origin, polar coordinates (r, θ) define a point by its distance (r) from a fixed point (the pole) and the angle (θ) that the line connecting the pole to the point makes with a fixed direction (the polar axis). This coordinate system is particularly useful for describing curves that are circular or spiral in nature.

Understanding Polar Coordinates

The relationship between Cartesian coordinates (x, y) and polar coordinates (r, θ) is given by:

• x = r cos(θ)
• y = r sin(θ)

And conversely:

• r² = x² + y² => r = ±√(x² + y²)
• tan(θ) = y/x

The pole is the origin (0,0) in Cartesian coordinates. The polar axis is typically the positive x-axis.

Equations of Polar Curves

Polar curves are defined by an equation relating 'r' and 'θ'. Some common examples include:

• Circles:
• r = a (a circle centered at the pole with radius 'a')
• r = 2a cos(θ) (a circle with diameter '2a' passing through the pole, centered at (a, 0) in Cartesian)
• r = 2a sin(θ) (a circle with diameter '2a' passing through the pole, centered at (0, a) in Cartesian)
• Cardioids:
• r = a(1 + cos(θ)) or r = a(1 - cos(θ))
• r = a(1 + sin(θ)) or r = a(1 - sin(θ))
• Lemniscates:
• r² = a² cos(2θ)
• r² = a² sin(2θ)
• Spirals:
• r = aθ (Archimedean spiral)

Finding the Tangent to a Polar Curve

To find the tangent to a polar curve r = f(θ), we first convert the polar equation to parametric Cartesian equations:

x = r cos(θ) = f(θ) cos(θ)

y = r sin(θ) = f(θ) sin(θ)

The slope of the tangent line in Cartesian coordinates is given by dy/dx. Using the chain rule for parametric equations:

dy/dx = (dy/dθ) / (dx/dθ)

Calculating dy/dθ and dx/dθ

We need to differentiate x and y with respect to θ. Using the product rule:

dx/dθ = d/dθ [f(θ) cos(θ)] = f'(θ) cos(θ) - f(θ) sin(θ)

dy/dθ = d/dθ [f(θ) sin(θ)] = f'(θ) sin(θ) + f(θ) cos(θ)

So, the slope of the tangent is:

dy/dx = [f'(θ) sin(θ) + f(θ) cos(θ)] / [f'(θ) cos(θ) - f(θ) sin(θ)]

If the polar curve is given by r = f(θ), we can also express this using dr/dθ:

dy/dx = [r sin(θ) + (dr/dθ) cos(θ)] / [r cos(θ) - (dr/dθ) sin(θ)]

This formula gives the slope of the tangent at any point (r, θ) on the curve.

Special Cases and Interpretation

The slope dy/dx can be undefined or zero in certain situations, indicating vertical or horizontal tangents, respectively.

Horizontal Tangents

A horizontal tangent occurs when dy/dθ = 0 and dx/dθ ≠ 0.

This means: r sin(θ) + (dr/dθ) cos(θ) = 0

Or, if r ≠ 0: sin(θ) + (r'/r) cos(θ) = 0, where r' = dr/dθ.

This can be rewritten as: r' / r = -tan(θ)

Vertical Tangents

A vertical tangent occurs when dx/dθ = 0 and dy/dθ ≠ 0.

This means: r cos(θ) - (dr/dθ) sin(θ) = 0

Or, if r ≠ 0: cos(θ) - (r'/r) sin(θ) = 0.

This can be rewritten as: r' / r = cot(θ)

We must also consider cases where r = 0. If r = 0 at θ = θ₀, then the tangent is vertical if the limit of dy/dx as θ → θ₀ is infinite, and horizontal if the limit is zero. This often corresponds to the pole.

Example 1: Tangent to a Cardioid

Find the slope of the tangent to the cardioid r = 1 + cos(θ) at θ = π/2.

Here, f(θ) = r = 1 + cos(θ).

First, find dr/dθ:

dr/dθ = -sin(θ).

Now, calculate the values at θ = π/2:

r = 1 + cos(π/2) = 1 + 0 = 1

dr/dθ = -sin(π/2) = -1

The slope dy/dx is:

dy/dx = [r sin(θ) + (dr/dθ) cos(θ)] / [r cos(θ) - (dr/dθ) sin(θ)]

Substitute the values at θ = π/2:

dy/dx = [1 * sin(π/2) + (-1) * cos(π/2)] / [1 * cos(π/2) - (-1) * sin(π/2)]

dy/dx = [1 * 1 + (-1) * 0] / [1 * 0 - (-1) * 1]

dy/dx = [1 + 0] / [0 + 1] = 1/1 = 1

The slope of the tangent at θ = π/2 is 1.

Example 2: Horizontal and Vertical Tangents of a Cardioid

Find the horizontal and vertical tangents for the cardioid r = 1 + cos(θ).

We have r = 1 + cos(θ) and dr/dθ = -sin(θ).

Horizontal Tangents:

Set dy/dθ = 0, which means r sin(θ) + (dr/dθ) cos(θ) = 0.

(1 + cos(θ)) sin(θ) + (-sin(θ)) cos(θ) = 0

sin(θ) + cos(θ)sin(θ) - sin(θ)cos(θ) = 0

sin(θ) = 0

This occurs at θ = 0 and θ = π.

At θ = 0: r = 1 + cos(0) = 2. The point is (2, 0) in polar, which is (2, 0) in Cartesian. The tangent is horizontal.

At θ = π: r = 1 + cos(π) = 0. The point is the pole. When r=0, we need to check limits. However, sin(θ) = 0 at θ=π means dy/dθ = 0. We must check dx/dθ at θ=π. dx/dθ = r cos(θ) - (dr/dθ) sin(θ) = 0*(-1) - (0)*(0) = 0. Since both are zero, we need a more careful analysis using the limit of dy/dx, or observe the graph. For r=1+cos(θ), the cusp is at the pole, and it has a horizontal tangent there.

Vertical Tangents:

Set dx/dθ = 0, which means r cos(θ) - (dr/dθ) sin(θ) = 0.

(1 + cos(θ)) cos(θ) - (-sin(θ)) sin(θ) = 0

cos(θ) + cos²(θ) + sin²(θ) = 0

cos(θ) + 1 = 0

cos(θ) = -1

This occurs at θ = π.

At θ = π: r = 0. As noted above, this is the cusp at the pole. The tangent is vertical.

So, the cardioid r = 1 + cos(θ) has a horizontal tangent at (2, 0) and a vertical tangent at the pole (r=0, θ=π).

Angle of Tangent with Polar Axis

Sometimes, instead of the slope dy/dx, we are interested in the angle that the tangent line makes with the polar axis. Let this angle be denoted by ψ (psi). Then, tan(ψ) = dy/dx.

We also know that in polar coordinates, the angle of the tangent line with the radius vector (the line from the pole to the point) is given by tan(φ) = r / (dr/dθ).

The angle ψ that the tangent makes with the polar axis is related to the angle of the radius vector θ and the angle φ by:

ψ = θ + φ

Therefore, tan(ψ) = tan(θ + φ) = [tan(θ) + tan(φ)] / [1 - tan(θ) tan(φ)]

Substituting tan(φ) = r / (dr/dθ):

tan(ψ) = [tan(θ) + r / (dr/dθ)] / [1 - tan(θ) * (r / (dr/dθ))]

Multiply numerator and denominator by dr/dθ:

tan(ψ) = [tan(θ) (dr/dθ) + r] / [(dr/dθ) - tan(θ) r]

Substitute tan(θ) = sin(θ)/cos(θ):

tan(ψ) = [(sin(θ)/cos(θ)) (dr/dθ) + r] / [(dr/dθ) - (sin(θ)/cos(θ)) r]

Multiply numerator and denominator by cos(θ):

tan(ψ) = [sin(θ) (dr/dθ) + r cos(θ)] / [(dr/dθ) cos(θ) - r sin(θ)]

Notice that this is exactly the expression for dy/dx we derived earlier!

So, tan(ψ) = dy/dx = [r sin(θ) + (dr/dθ) cos(θ)] / [r cos(θ) - (dr/dθ) sin(θ)]

Subtangents and Subnormals in Polar Coordinates

While subtangents and subnormals are more commonly discussed in Cartesian coordinates, analogous concepts can be considered. The length of the subtangent is the projection of the tangent segment from the point of tangency to the x-axis onto the x-axis. The length of the subnormal is the projection of the normal segment from the point of tangency to the x-axis onto the x-axis.

In polar coordinates, the concept is less direct. However, we can consider the projection of the tangent vector onto the radial and transverse directions. The transverse direction is perpendicular to the radius vector, and the radial direction is along the radius vector.

The angle φ between the radius vector and the tangent is given by tan(φ) = r / (dr/dθ).

The length of the subtangent (along the radius vector) can be considered as r tan(φ) = r * (r / (dr/dθ)) = r² / (dr/dθ).

The length of the subnormal (along the radius vector) can be considered as r / tan(φ) = r * (dr/dθ) / r = dr/dθ.

Example 3: Lemniscate Tangents

Consider the lemniscate r² = a² cos(2θ).

To find derivatives, it's often easier to differentiate implicitly or convert to Cartesian coordinates. Let's use implicit differentiation with respect to θ.

2r (dr/dθ) = a² (-sin(2θ) * 2) = -2a² sin(2θ)

So, dr/dθ = -a² sin(2θ) / r.

Now, let's find the slope dy/dx:

dy/dx = [r sin(θ) + (dr/dθ) cos(θ)] / [r cos(θ) - (dr/dθ) sin(θ)]

Substitute dr/dθ:

dy/dx = [r sin(θ) + (-a² sin(2θ) / r) cos(θ)] / [r cos(θ) - (-a² sin(2θ) / r) sin(θ)]

Multiply numerator and denominator by r:

dy/dx = [r² sin(θ) - a² sin(2θ) cos(θ)] / [r² cos(θ) + a² sin(2θ) sin(θ)]

Substitute r² = a² cos(2θ):

dy/dx = [a² cos(2θ) sin(θ) - a² sin(2θ) cos(θ)] / [a² cos(2θ) cos(θ) + a² sin(2θ) sin(θ)]

Factor out a²:

dy/dx = [cos(2θ) sin(θ) - sin(2θ) cos(θ)] / [cos(2θ) cos(θ) + sin(2θ) sin(θ)]

Using trigonometric identities sin(A-B) = sinAcosB - cosAsinB and cos(A-B) = cosAcosB + sinAsinB:

dy/dx = sin(θ - 2θ) / cos(θ - 2θ) = sin(-θ) / cos(-θ) = -sin(θ) / cos(θ) = -tan(θ).

This is a remarkably simple result for the slope of the tangent to a lemniscate!

Let's find the points where tangents might be horizontal or vertical. For dy/dx = -tan(θ) to be zero (horizontal tangent), we need tan(θ) = 0, which means θ = 0 or θ = π. However, the domain for r² = a² cos(2θ) requires cos(2θ) ≥ 0. This means -π/2 + nπ ≤ 2θ ≤ π/2 + nπ, or -π/4 + nπ/2 ≤ θ ≤ π/4 + nπ/2. So, θ=0 is in the domain, giving a horizontal tangent. At θ=0, r² = a² cos(0) = a², so r = ±a. The points are (a, 0) and (-a, 0) in Cartesian.

For dy/dx = -tan(θ) to be undefined (vertical tangent), we need tan(θ) to be undefined, which means θ = π/2 or θ = 3π/2. These angles are not in the primary domain of the lemniscate where cos(2θ) ≥ 0. However, if we consider the full curve, we can see vertical tangents. The points where cos(2θ)=0 are 2θ = ±π/2, ±3π/2, etc., so θ = ±π/4, ±3π/4, etc. At these angles, r=0 (the pole). Vertical tangents occur at the pole for lemniscates.

Summary of Tangent Properties

When analyzing tangents to polar curves, remember to:

• Convert to parametric form or use the direct formula for dy/dx.
• Calculate r and dr/dθ at the point of interest.
• Substitute these values into the dy/dx formula.
• For horizontal tangents, set the numerator (dy/dθ) to zero.
• For vertical tangents, set the denominator (dx/dθ) to zero.
• Always check points where r = 0, as these can be singular points (cusps, self-intersections) with special tangent behavior.
• Consider the domain of the polar function, especially for curves like lemniscates where r² must be non-negative.

Understanding the geometry of polar curves, such as their symmetry and the locations of loops or cusps, can greatly aid in predicting and verifying tangent behavior.