Unit 5: Partial Derivatives and Curve Tracing

Table of Contents

Partial Derivatives

This is the first step into multivariable calculus. 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).

Example: Let f(x, y) = x³ + x²y⁵ - 2y⁴

Higher-Order Partial Derivatives

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

Clairaut's Theorem (Equality of Mixed Partials)

For most "well-behaved" functions (if f, fₓ, fᵧ, fₓᵧ, and fᵧₓ are all continuous), the order of differentiation does not matter.

∂²f/∂y∂x = ∂²f/∂x∂y (or fₓᵧ = fᵧₓ)
Exam Tip: You may be asked to "verify" Clairaut's Theorem for a given function. This means you must calculate both fₓᵧ and fᵧₓ separately and show that they are equal.

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)

Quick Test:

If every term in a polynomial function has the same total degree, the function is homogeneous. The total degree of a term like xᵃyᵇ is a + b.

Example: f(x, y) = 3x⁵ + 2x²y³ - y⁵

Since all terms have degree 5, this is a homogeneous function of degree 5.

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

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.

Euler's Theorem on Homogeneous Functions

This theorem provides a beautiful relationship between a homogeneous function 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

Example: Verify Euler's Theorem for f(x, y) = 3x⁵ + 2x²y³ - y⁵ (n = 5)

  1. Find partial derivatives:
    • ∂f/∂x = 15x⁴ + 4xy³
    • ∂f/∂y = 6x²y² - 5y⁴
  2. Calculate x(∂f/∂x) + y(∂f/∂y): x(15x⁴ + 4xy³) + y(6x²y² - 5y⁴) = (15x⁵ + 4x²y³) + (6x²y³ - 5y⁵) = 15x⁵ + 10x²y³ - 5y⁵
  3. This doesn't look right. Let's re-check the example from before.
    f(x, y) = 3x⁵ + 2x²y³ - y⁵
    ∂f/∂x = 15x⁴ + (2*2)x¹y³ = 15x⁴ + 4xy³
    ∂f/∂y = 2x²(3y²) - 5y⁴ = 6x²y² - 5y⁴
    x(∂f/∂x) + y(∂f/∂y) = x(15x⁴ + 4xy³) + y(6x²y² - 5y⁴)
    = 15x⁵ + 4x²y³ + 6x²y³ - 5y⁵ = 15x⁵ + 10x²y³ - 5y⁵
    Hmm, my partial derivatives are correct. Let me re-check the function. f(x, y) = 3x⁵ + 2x²y³ - y⁵. All terms are degree 5. n=5.
    n * f = 5 * (3x⁵ + 2x²y³ - y⁵) = 15x⁵ + 10x²y³ - 5y⁵
    Ah, they match! My mental math was just shaky.
    Result: x(∂f/∂x) + y(∂f/∂y) = 15x⁵ + 10x²y³ - 5y⁵
    And: n * f = 5 * f = 5(3x⁵ + 2x²y³ - y⁵) = 15x⁵ + 10x²y³ - 5y⁵
    They are equal. The theorem is verified.

Euler's Theorem for Second-Order Derivatives

If f(x, y) is a homogeneous function of degree n, then: x²(∂²f/∂x²) + 2xy(∂²f/∂x∂y) + y²(∂²f/∂y²) = n(n-1) * f
Exam Tip: Euler's Theorem is a powerful shortcut. If a problem gives you a complex homogeneous function and asks to prove x(∂f/∂x) + y(∂f/∂y) = n*f, you do NOT need to calculate the derivatives. You just need to:
  1. Show the function is homogeneous (using the f(tx, ty) method).
  2. Identify the degree 'n'.
  3. State that by Euler's Theorem, the equality must hold.

Asymptotes

An asymptote is a line that a curve approaches (gets infinitely close to) as the x or y coordinates approach infinity.

1. Vertical Asymptotes

These occur where the function "blows up" to ±∞. For a rational function f(x) = P(x) / Q(x), vertical asymptotes are found at the x-values 'a' that make the denominator zero (and the numerator non-zero).

Example: y = 1 / (x - 2). The denominator is zero at x = 2. So, x = 2 is a vertical asymptote.

2. Horizontal Asymptotes

These describe the behavior of f(x) as x → ∞ and x → -∞. y = L is a horizontal asymptote if lim (x → ∞) f(x) = L or lim (x → -∞) f(x) = L.

For a rational function f(x) = (axⁿ + ...) / (bxᵐ + ...):

3. Oblique (Slant) Asymptotes

These occur in rational functions when the degree of the numerator is exactly one more than the degree of the denominator.

The equation of the slant asymptote (y = mx + c) is the quotient obtained by performing polynomial long division.

Example: f(x) = (x² + 1) / (x - 1)

  1. Degree of top (2) is one more than degree of bottom (1).
  2. Perform long division: (x² + 1) ÷ (x - 1) gives a quotient of (x + 1) and a remainder of 2.
  3. So, f(x) = (x + 1) + 2/(x - 1)
  4. As x → ∞, the 2/(x-1) term goes to 0. The function approaches y = x + 1.
  5. The oblique asymptote is y = x + 1.

Concavity and Points of Inflection

This topic uses the second derivative, f''(x), to describe the "bend" of a curve.

Concavity

Points of Inflection (or Inflexion)

A point of inflection is a point on the curve where the concavity changes (from CU to CD, or from CD to CU).

How to Find Points of Inflection:

  1. Find the second derivative, f''(x).
  2. Find "potential" inflection points by solving f''(x) = 0 or finding where f''(x) is undefined.
  3. Test the intervals on either side of these potential points.
  4. If the sign of f''(x) changes (from + to - or - to +) at a point, then it is a point of inflection.

Example: f(x) = x³

  1. f'(x) = 3x²
  2. f''(x) = 6x
  3. Set f''(x) = 0 => 6x = 0 => x = 0. This is a potential inflection point.
  4. Test intervals:
    • For x < 0 (e.g., x = -1), f''(-1) = -6 (Negative, Concave Down)
    • For x > 0 (e.g., x = 1), f''(1) = 6 (Positive, Concave Up)
  5. Since concavity changes at x = 0, the point (0, f(0)) = (0, 0) is a point of inflection.
Common Mistake: Assuming that f''(c) = 0 automatically means there is an inflection point. This is not true.
Counterexample: f(x) = x⁴.
f'(x) = 4x³, f''(x) = 12x².
f''(x) = 0 at x = 0.
But for x < 0, f''(x) is positive (CU). For x > 0, f''(x) is also positive (CU).
Since the concavity does not change, x = 0 is not an inflection point (it's a local minimum).

Tracing Graphs of Simple Algebraic and Rational Functions

This is a capstone topic that combines all the concepts from the course. To trace a graph of y = f(x), you should follow a systematic procedure.

Checklist for Curve Tracing:

  1. Domain and Range

    Find the set of all possible x-values (domain) and y-values (range). Look for:
    - Division by zero (for rational functions).
    - Square roots of negative numbers.

  2. Intercepts

    Y-intercept: Set x = 0 and solve for y. (Point is (0, y)).
    X-intercept(s): Set y = 0 and solve for x. (Points are (x, 0)).

  3. Symmetry

    Y-axis Symmetry (Even Function): Is f(-x) = f(x)? (e.g., y = x², y = cos x)
    Origin Symmetry (Odd Function): Is f(-x) = -f(x)? (e.g., y = x³, y = sin x)
    If f(x,y)=0 is the equation, check if replacing x with -x (y-axis) or y with -y (x-axis) or both (origin) leaves the equation unchanged.

  4. Asymptotes

    Find all Vertical, Horizontal, and Oblique asymptotes as described in the section above.

  5. First Derivative (f') - Increasing/Decreasing and Local Extrema

    Find f'(x).
    Find critical points by solving f'(x) = 0 or where f'(x) is undefined.
    Create a sign chart for f'(x) to find intervals of increasing (f' > 0) and decreasing (f' < 0).
    Use the First Derivative Test to classify critical points as local maxima or local minima.

  6. Second Derivative (f'') - Concavity and Inflection Points

    Find f''(x).
    Find potential inflection points by solving f''(x) = 0 or where f''(x) is undefined.
    Create a sign chart for f''(x) to find intervals of concave up (f'' > 0) and concave down (f'' < 0).
    Identify points of inflection (where concavity changes).

  7. Plot and Sketch

    Draw the axes and mark the asymptotes (as dashed lines).
    Plot all key points: intercepts, local max/mins, and inflection points.
    Connect the dots, making sure your curve follows the information from your sign charts (increasing/decreasing and concavity) and approaches the asymptotes correctly.