Week 11 Notes: Implicit Differentiation and Curve Sketching

Week 11: Differentiation and Curve Sketching

Implicit Functions

• Notation: Consider an expression like x^2 + y^2 + 5x + 6y - 13xy = 8, where y is a function of x. The goal is to find \frac{dy}{dx}.
• Definition: An implicit function is defined by a relation F(x, y) = 0, where F(x, y) is an expression in terms of two variables x and y.

Example

• Find \frac{dy}{dx} from x^2 + y^2 + 5x + 6y - 13xy = 8.

Example

• Find \frac{dy}{dx} from \sin(y) + x^2y^3 - \cos(x) = 2y. The solution involves differentiating each term with respect to x and then solving for y'.
  • \cos(y) \cdot y' + 2xy^3 + 3x^2y^2y' + \sin(x) = 2y'
  • Rearranging terms: (\cos(y) + 3x^2y^2 - 2)y' = -2xy^3 - \sin(x)
  • y' = \frac{-2xy^3 - \sin(x)}{\cos(y) + 3x^2y^2 - 2}

Equation of Tangent to a Graph

• Consider a function y = f(x).
• The slope of the tangent at a point x0 is given by f'(x0).
• The equation of the tangent to the graph at (x0, y0) is: y - y0 = f'(x0)(x - x_0).

Example

• Find the equation of the tangent to y = x^3 + 3x^2 - 5x + 4 at x = 5.

Equation of Normal to a Curve

• Consider a function y = f(x).
• The slope of the normal at a point x0 is given by -\frac{1}{f'(x0)}.
• The equation of the normal to the graph at (x0, y0) is: y - y0 = -\frac{1}{f'(x0)}(x - x_0).

Example

• Find the equation of the normal to the curve y = x + \frac{1}{x} at x = 2.

Example

• Find the equation of the tangent and normal to the curve x^2 + y^2 + 5x + 6y - 5xy = 8 at (x, y) = (1, 1).

Curve Sketching

• Increasing Function: When \frac{dy}{dx} > 0, the function is increasing.
• Decreasing Function: When \frac{dy}{dx} < 0, the function is decreasing.
• Stationary Points: Points where the tangent to the graph is horizontal, i.e., \frac{dy}{dx} = 0.
• Turning Points: Points where the function changes direction from increasing to decreasing or vice versa. At these points, \frac{dy}{dx} = 0. Turning points are stationary points, but not all stationary points are turning points.

Maximum and Minimum Turning Points

• Local Minimum: A point where the function is lower than its nearby points within a certain neighborhood.
• Local Maximum: A point where the function is higher than its nearby points within a certain neighborhood.

Distinguishing Maximum and Minimum Points

• Minimum: The derivative changes from decreasing to increasing ($\searrow \nearrow$).
• Maximum: The derivative changes from increasing to decreasing ($\nearrow \searrow$).

First Derivative Test

• A stationary point is a:
  • Minimum: If the sign diagram of the first derivative is of the form ($\searrow \nearrow$).
  • Maximum: If the sign diagram of the first derivative is of the form ($\nearrow \searrow$).

Example

• Find stationary points of the function y = \frac{(x-1)^2}{x} and classify them.
  • y = \frac{x^2 - 2x + 1}{x} = x - 2 + \frac{1}{x}
  • y' = 1 - \frac{1}{x^2} = 0
  • Setting y' = 0: 1 - \frac{1}{x^2} = 0 \Rightarrow x = -1, 1
  • Sign analysis of y':
    • For x = -2, y' = 1 - \frac{1}{4} = \frac{3}{4} > 0
    • For x = \frac{1}{2}, y' = 1 - 4 = -3 < 0
    • For x = 2, y' = \frac{3}{4} > 0
  • Conclusion:
    • x = -1: local maximum
    • x = 1: local minimum

Concavity of a Curve

• At a given point x = c:
  • If \frac{d^2y}{dx^2} > 0, the curve is concave up.
  • If \frac{d^2y}{dx^2} < 0, the curve is concave down.

Second Derivative Test

• At a stationary point:
  • If \frac{d^2y}{dx^2} > 0 (concave up): minimum
  • If \frac{d^2y}{dx^2} < 0 (concave down): maximum

Example

• Find stationary points of the function y = x^3 - 3x + 2 and classify them. Use this information to sketch the graph of the function, hint: x^3 - 3x + 2 = (x - 1)^2(x + 2).

Point of Inflection (Inflexion)

• A point of inflection occurs at a point P if \frac{d^2y}{dx^2} = 0 at point P and \frac{d^2y}{dx^2} changes sign as x increases through point P.

Example

• Sketch the graph of 3x^4 - 4x^3 - 12x^2.
  • First derivative test information:
    • Sign analysis where y' possess the values + 0 - 0 + 0 for x values -1, 0, 2 respectively
  • Second derivative test:
    • y' = 12x^3 - 12x^2 - 24x
    • \frac{d^2y}{dx^2} = y'' = 36x^2 - 24x - 24
    • x = -1 \Rightarrow y'' > 0
    • x = 0 \Rightarrow y'' = -24 < 0
    • x = 2 \Rightarrow y'' > 0
  • Conclusion:
    • -1, 2: local minimum
    • 0: local maximum