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 $x<em>0$ is given by $f'(x</em>0)$.
• The equation of the tangent to the graph at $(x<em>0, y</em>0)$ is: $y - y<em>0 = f'(x</em>0)(x - x_0)$.

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 $x<em>0$ is given by $-\frac{1}{f'(x</em>0)}$.
• The equation of the normal to the graph at $(x<em>0, y</em>0)$ is: $y - y<em>0 = -\frac{1}{f'(x</em>0)}(x - x_0)$.

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