Unit 3: Partial Derivatives and Geometry of Curves

Partial Differentiation

A partial derivative of a function of several variables (e.g., f(x, y)) is its derivative with respect to one of those variables, while holding all other variables constant.

Notation

Let z = f(x, y).

• Partial derivative of f with respect to x: ∂f/∂x or fₓ

  To calculate, treat 'y' as if it were a constant (like '5') and differentiate f with respect to 'x' normally.

• Partial derivative of f with respect to y: ∂f/∂y or fᵧ

  To calculate, treat 'x' as if it were a constant and differentiate f with respect to 'y' normally.

Example: Let f(x, y) = 2x³y + sin(y) - x⁴

• ∂f/∂x = (2y) * (3x²) + 0 - 4x³ = 6x²y - 4x³ (We treated 2y as a constant multiplier and sin(y) as a constant)
• ∂f/∂y = (2x³) * (1) + cos(y) - 0 = 2x³ + cos(y) (We treated 2x³ as a constant multiplier and x⁴ as a constant)

Higher-Order Partial Derivatives

We can differentiate again to find second-order partial derivatives:

• ∂²f/∂x² = ∂/∂x (∂f/∂x) = fₓₓ
• ∂²f/∂y² = ∂/∂y (∂f/∂y) = fᵧᵧ
• ∂²f/∂x∂y = ∂/∂x (∂f/∂y) = fᵧₓ (Differentiate w.r.t. y first, then x)
• ∂²f/∂y∂x = ∂/∂y (∂f/∂x) = fₓᵧ (Differentiate w.r.t. x first, then y)

Clairaut's Theorem (Equality of Mixed Partials)

For most "well-behaved" functions (if the mixed partials are continuous), the order of differentiation does not matter.

∂²f/∂y∂x = ∂²f/∂x∂y (or fₓᵧ = fᵧₓ)

Homogeneous Functions

A function f(x, y) is a homogeneous function of degree n if, for any constant t, it satisfies the following property:

f(tx, ty) = tⁿ * f(x, y)

Alternative Form

A function f(x, y) is homogeneous of degree n if it can be written in the form:

f(x, y) = xⁿ * g(y/x) OR f(x, y) = yⁿ * h(x/y)

Example: f(x, y) = (x³ + y³) / (x - y)

• Method 1 (t-test):
  f(tx, ty) = ((tx)³ + (ty)³) / (tx - ty) = (t³x³ + t³y³) / (tx - ty)
  = [t³(x³ + y³)] / [t(x - y)] = t² * [(x³ + y³) / (x - y)]
  = t² * f(x, y)
  This is a homogeneous function of degree 2.
• Method 2 (factoring):
  f(x, y) = [x³(1 + (y/x)³)] / [x(1 - y/x)]
  = x² * [ (1 + (y/x)³) / (1 - y/x) ]
  This is in the form x² * g(y/x). Therefore, it is homogeneous of degree 2.

Euler's Theorem on Homogeneous Functions

This theorem provides a powerful relationship between a homogeneous function (of two variables) and its partial derivatives.

Statement

If f(x, y) is a homogeneous function of degree n, then: x * (∂f/∂x) + y * (∂f/∂y) = n * f

Problems Related to Euler's Theorem

Problems often ask you to "verify" the theorem, or use it to find the value of x * (∂f/∂x) + y * (∂f/∂y) without full differentiation.

Example (Deduction): If u = tan⁻¹((x³ + y³) / (x - y)), find x(∂u/∂x) + y(∂u/∂y).

1. 'u' itself is not homogeneous.
2. Let f(x, y) = tan(u) = (x³ + y³) / (x - y).
3. We already showed f(x, y) is homogeneous of degree n = 2.
4. By Euler's Theorem for 'f': x * (∂f/∂x) + y * (∂f/∂y) = n * f = 2 * f x * (∂f/∂x) + y * (∂f/∂y) = 2 * tan(u)
5. Now, find the partial derivatives of f in terms of u:
   ∂f/∂x = d/dx(tan u) = sec²(u) * (∂u/∂x)
   ∂f/∂y = d/dy(tan u) = sec²(u) * (∂u/∂y)
6. Substitute these into Euler's Theorem: x * [sec²(u) * (∂u/∂x)] + y * [sec²(u) * (∂u/∂y)] = 2 * tan(u)
7. Factor out sec²(u): sec²(u) * [ x(∂u/∂x) + y(∂u/∂y) ] = 2 * tan(u)
8. Solve for the expression we want: x(∂u/∂x) + y(∂u/∂y) = [ 2 * tan(u) ] / sec²(u) = 2 * (sin u / cos u) * (cos² u) = 2 * sin(u) * cos(u) = sin(2u)

Tangents and Normals (Cartesian)

For a curve given by y = f(x), the derivative dy/dx at a point (x₁, y₁) represents the slope of the tangent line at that point.

Let m = (dy/dx) at (x₁, y₁)

Equation of the Tangent Line

Using the point-slope form Y - y₁ = m(X - x₁):

Y - y₁ = (dy/dx) * (X - x₁)

(We use capital Y and X for the equation of the line, to distinguish from the point (x₁, y₁) on the curve).

Equation of the Normal Line

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

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

Y - y₁ = (-1/m) * (X - x₁)
(X - x₁) + m * (Y - y₁) = 0

Subtangents and Subnormals (Cartesian)

These are lengths on the x-axis related to the tangent and normal lines.

Let P(x, y) be a point on the curve. Let the tangent at P meet the x-axis at T. Let the normal at P meet the x-axis at N. Let M be the foot of the perpendicular from P to the x-axis (so M = (x, 0)).

• Length of Subtangent (TM): This is the projection of the tangent segment PT onto the x-axis.

  Length of Subtangent = | y / m | = | y / (dy/dx) |

• Length of Subnormal (MN): This is the projection of the normal segment PN onto the x-axis.

  Length of Subnormal = | y * m | = | y * (dy/dx) |

Tangents and Normals (Polar)

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

Let φ (phi) be the angle between the radius vector (from origin to P(r, θ)) and the tangent line at P.

Angle φ (phi)

The formula relating r, θ, and φ is: tan(φ) = r / (dr/dθ) = r * (dθ/dr)

Equation of Tangent and Normal (in polar coordinates)

This is less common, but the equation of the tangent line at a point (r₁, θ₁) is given by:

r₁ / r = cos(θ - θ₁) - tan(φ₁) * sin(θ - θ₁)

Where φ₁ is the value of φ at (r₁, θ₁).

The slope of the tangent line in Cartesian coordinates (m = dy/dx) can be found using:

x = r cos(θ), y = r sin(θ) dy/dx = ( (dr/dθ)sin(θ) + r*cos(θ) ) / ( (dr/dθ)cos(θ) - r*sin(θ) )

You can then use this 'm' in the Cartesian tangent/normal equations.

Subtangents and Subnormals (Polar)

These are lengths defined differently than their Cartesian counterparts.

Draw the tangent and normal lines at a point P(r, θ). 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) |