Knowlet

Unit 4: Geometry of Curves

Tangents and Normals (Cartesian Form)

For a curve given by y = f(x), the derivative dy/dx at a point (x₀, y₀) represents the slope of the tangent line at that point.

Let m = (dy/dx) at (x₀, y₀)

Equation of the Tangent Line

Using the point-slope form y - y₁ = m(x - x₁):

y - y₀ = m * (x - x₀)
y - y₀ = (dy/dx) at (x₀,y₀) * (x - x₀)

Equation of the Normal Line

The normal line is perpendicular to the tangent line at the point of tangency. Its slope is the negative reciprocal of the tangent's slope.

Slope of normal = -1 / m = -1 / (dy/dx)

y - y₀ = (-1/m) * (x - x₀)
(y - y₀) * m = -(x - x₀) (x - x₀) + m * (y - y₀) = 0

Special Cases:

  • If dy/dx = 0 (horizontal tangent), the normal line is vertical (x = x₀).
  • If dy/dx is undefined (vertical tangent, dx/dy = 0), the tangent line is vertical (x = x₀) and the normal line is horizontal (y = y₀).

Tangent at the Origin

If a curve passes through the origin (0, 0), we can find the equation of the tangent line(s) at the origin by a simple method.

Method:

  1. Ensure the curve's equation f(x, y) = 0 passes through (0, 0).
  2. Find the lowest degree terms in the equation.
  3. Set these lowest degree terms equal to zero. This gives the equation of the tangent(s) at the origin.

Examples:

  • Curve: y = x² + 2x

    Rewrite as y - 2x - x² = 0. Lowest degree terms are (y - 2x) (degree 1). Set to zero: y - 2x = 0, or y = 2x. This is the single tangent at the origin.

  • Curve: y² = x³ + x²

    Rewrite as y² - x² - x³ = 0. Lowest degree terms are (y² - x²) (degree 2). Set to zero: y² - x² = 0, which means (y - x)(y + x) = 0. This gives two tangents at the origin: y = x and y = -x. (The origin is a "node" or crossing-point for this curve).

  • Curve: y² = x³

    Rewrite as y² - x³ = 0. Lowest degree term is y² (degree 2). Set to zero: y² = 0, or y = 0 (the x-axis). (This indicates a "cusp" at the origin).

Angle of Intersection of Curves

The angle of intersection between two curves, f(x) and g(x), at a point (x₀, y₀) is defined as the acute angle between their tangent lines at that point.

Steps:

  1. Find the point(s) of intersection (x₀, y₀) by solving the two equations simultaneously.
  2. Find the slope of the first curve at (x₀, y₀): m₁ = f'(x₀)
  3. Find the slope of the second curve at (x₀, y₀): m₂ = g'(x₀)
  4. If θ is the angle between the tangents, use the formula:
    tan(θ) = | (m₁ - m₂) / (1 + m₁m₂) |
  5. Solve for θ: θ = tan⁻¹( | (m₁ - m₂) / (1 + m₁m₂) | )

Special Cases:

  • If m₁ = m₂, then tan(θ) = 0, so θ = 0. The curves touch tangentially.
  • If m₁m₂ = -1, then the denominator is zero, so tan(θ) is undefined. This means θ = 90° (or π/2). The curves are orthogonal (intersect at a right angle).

Derivative of Arc-Length (Cartesian Form)

Let s be the arc length of a curve y = f(x), measured from some fixed point.

We can consider a small segment of the arc, ds, as the hypotenuse of a right triangle with sides dx and dy.

By the Pythagorean theorem:

(ds)² = (dx)² + (dy)²

Dividing by (dx)²:

(ds/dx)² = 1 + (dy/dx)²
Derivative of arc length w.r.t x: ds/dx = √(1 + (dy/dx)²)

Dividing by (dy)²:

(ds/dy)² = (dx/dy)² + 1
Derivative of arc length w.r.t y: ds/dy = √(1 + (dx/dy)²)

Polar Coordinates: Angle between Radius Vector and Tangent

For a curve given in polar coordinates as r = f(θ):

  • The radius vector is the line segment from the origin (pole) to the point P(r, θ) on the curve.
  • The tangent line is the tangent to the curve at P(r, θ).
  • Let φ (phi) be the angle between the radius vector and the tangent line.
The formula relating r, θ, and φ is: tan(φ) = r / (dr/dθ)
tan(φ) = r * (dθ/dr)
Exam Tip: This formula is extremely important for polar curves. A common mistake is flipping the fraction. Remember it as tan(φ) = r / (derivative of r w.r.t θ).

Logarithmic trick: Sometimes it's easier to take the log first.
If ln(r) = ln(f(θ)), then differentiating w.r.t θ gives:
(1/r) * (dr/dθ) = f'(θ) / f(θ)
This can be rearranged to 1 / tan(φ) = cot(φ) = (1/r) * (dr/dθ).

Derivative of Arc-Length (Polar Form)

For a polar curve r = f(θ), we can find a similar relation for the arc length s.

The relationship is (ds)² = (dr)² + (r dθ)².

Dividing by (dθ)²:

(ds/dθ)² = (dr/dθ)² + r²
Derivative of arc length w.r.t θ: ds/dθ = √(r² + (dr/dθ)²)

Dividing by (dr)²:

(ds/dr)² = 1 + r²(dθ/dr)²
Derivative of arc length w.r.t r: ds/dr = √(1 + r²(dθ/dr)²)

Polar Subtangent and Subnormal

Similar to the Cartesian versions, these are lengths related to the tangent and normal lines, but defined relative to the radius vector.

Draw the tangent and normal lines at a point P(r, θ) on the curve. Draw a line through the origin (pole) O perpendicular to the radius vector OP. This new line intersects the tangent at T and the normal at N.

  • Polar Subtangent (OT): The length of the segment from the pole (O) to the intersection (T) of the perpendicular line and the tangent line.
    Length of Polar Subtangent = | r² * (dθ/dr) | = | r * tan(φ) |
  • Polar Subnormal (ON): The length of the segment from the pole (O) to the intersection (N) of the perpendicular line and the normal line.
    Length of Polar Subnormal = | dr/dθ | = | r * cot(φ) |

Radius of Curvature (Cartesian and Polar)

Curvature (κ) measures how sharply a curve bends. The radius of curvature (ρ) is the reciprocal of the curvature (ρ = 1/κ). It is the radius of the "best-fit" circle (called the osculating circle) that is tangent to the curve at a given point.

Cartesian Form for y = f(x)

ρ = [ 1 + (dy/dx)² ]³ᐟ² / | d²y/dx² |
ρ = [ 1 + (y')² ]³ᐟ² / | y'' |
Common Mistake: Forgetting the absolute value in the denominator. The radius ρ is a length, so it must be non-negative.

Parametric Form for x = f(t), y = g(t)

Let x' = dx/dt, y' = dy/dt, x'' = d²x/dt², y'' = d²y/dt².

ρ = [ (x')² + (y')² ]³ᐟ² / | x'y'' - y'x'' |

Polar Form for r = f(θ)

Let r' = dr/dθ and r'' = d²r/dθ².

ρ = [ r² + (r')² ]³ᐟ² / | r² + 2(r')² - r * r'' |

This formula is complex and used less often, but you should be familiar with it. The Cartesian formula is the most important one to memorize.

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