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IB Mathematics AA AHL Implicit differentiation Study Notes

IB Mathematics AA AHL Implicit differentiation Study Notes - New Syllabus

IB Mathematics AA AHL Implicit differentiation Study Notes

LEARNING OBJECTIVE

  • Implicit differentiation

Key Concepts: 

  • Implicit differentiation
  • Rates of change. 
  • Optimisation problems

MAA HL and SL Notes – All topics

Implicit Differentiation

Implicit Differentiation

In many equations, \( y \) is not explicitly written as a function of \( x \). When an equation involves both \( x \) and \( y \) intertwined (i.e., not solved for \( y \)), we use implicit differentiation.

Key Concept:

Differentiate both sides of the equation with respect to \( x \), treating \( y \) as a function of \( x \). When differentiating terms with \( y \), apply the chain rule and multiply by \( \frac{dy}{dx} \).

Why Use It?

  • When it is difficult or impossible to isolate \( y \) in terms of \( x \)
  • To find \( \frac{dy}{dx} \) in relations such as circles, ellipses, or curves like \( x^2 + y^2 = 25 \)

Method:

  1. Differentiate both sides of the equation with respect to \( x \)
  2. Apply chain rule to any term involving \( y \): e.g., \( \frac{d}{dx}(y^n) = n y^{n-1} \cdot \frac{dy}{dx} \)
  3. Solve the resulting equation for \( \frac{dy}{dx} \)

Example

Differentiate implicitly: \( x^2 + y^2 = 25 \)

▶️ Answer/Explanation

Differentiate both sides with respect to \( x \):

$ \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(25) $ $ 2x + 2y \frac{dy}{dx} = 0 $

Solve for \( \frac{dy}{dx} \):
$ 2y \frac{dy}{dx} = -2x \Rightarrow \frac{dy}{dx} = -\frac{x}{y} $
Result:
\( \frac{dy}{dx} = -\frac{x}{y} \)

Related Rates of Change

Related Rates of Change

Related rates involve finding the rate of change of one quantity in terms of another quantity that is changing over time. These problems typically arise in dynamic scenarios where multiple variables depend on time t.

Key Concept:

If a variable \( x \) changes with time, then \( \dfrac{dx}{dt} \) is its rate of change. When two or more variables are related by an equation, we can differentiate both sides with respect to \( t \), using the chain rule as needed.

Steps to Solve Related Rates Problems:

  1. Identify the given rates (e.g., \( \dfrac{dx}{dt} \)) and the required rate (e.g., \( \dfrac{dy}{dt} \))
  2. Write an equation relating all the relevant variables
  3. Differentiate implicitly with respect to time \( t \)
  4. Substitute known values of variables and rates
  5. Solve for the required rate

Common Scenarios:

  • Expanding/contracting circles or spheres
  • Water pouring into or draining from a container
  • Moving objects (cars, planes, shadows)
  • Right triangles involving distance, height, and length

Always use the chain rule correctly and maintain the units of the rates throughout the process.

Example

A balloon is rising vertically at a rate of 3 m/s. A boy is cycling along a straight road at 4 m/s. At the moment the boy is 5 meters from the point on the ground directly beneath the balloon At this moment, assume the balloon has risen to 12 m , how fast is the distance between the boy and the balloon increasing?

▶️ Answer/Explanation

Let \( x \) be the horizontal distance from the boy to the balloon base, and \( y \) be the balloon height. Let \( z \) be the distance between the boy and the balloon.

 \[ \frac{dx}{dt} = 4 \text{ m/s}, \quad \frac{dy}{dt} = 3 \text{ m/s}, \quad x = 5, \quad y = \text{unknown} \]

Use Pythagoras: \[ z^2 = x^2 + y^2 \]
Differentiate both sides: \[ 2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt} \]
Divide both sides by 2: \[ z\frac{dz}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt} \]
\[ z = \sqrt{x^2 + y^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \]
 \[ 13\frac{dz}{dt} = 5 \cdot 4 + 12 \cdot 3 = 20 + 36 = 56 \Rightarrow \frac{dz}{dt} = \frac{56}{13} \approx 4.31 \text{ m/s} \]
Final Answer: The distance is increasing at approximately 4.31 m/s.

Optimisation Problems

Optimisation Problems

Optimisation problems involve finding maximum or minimum values of a quantity subject to certain constraints. These problems are common in geometry, economics, physics, and engineering.

 Key Concept:

To solve an optimisation problem, express the quantity to be maximised or minimised as a function of a single variable, then use differentiation to find local extrema (maximum or minimum points).

General Strategy:

  1. Understand the scenario: Identify what is to be maximised or minimised.
  2. Draw a diagram (if appropriate) and label all variables.
  3. Write an equation for the quantity to be optimised (objective function).
  4. Use constraints to reduce the function to a single variable.
  5. Differentiate the function and find critical points.
  6. Evaluate endpoints if the variable is within a closed interval.
  7. Justify whether each critical point is a maximum or minimum.

Techniques:

  • Use the chain rule or implicit differentiation if the function is defined implicitly.
  • Check endpoints when the domain is restricted.
  • Use the second derivative test or a sign chart to confirm maxima or minima.

Optimisation is closely linked to calculus applications such as related rates, marginal analysis, and geometric modelling.

Example

A farmer wants to build a rectangular enclosure against a barn wall (so one side of the rectangle is already provided). He has 40 meters of fencing and wants to maximise the enclosed area. What dimensions should he use?

▶️ Answer/Explanation
Let the width perpendicular to the barn be \( x \), and the length parallel to the barn be \( y \).
Since the barn provides one side, the fencing is used for 2 widths and 1 length:
$ 2x + y = 40 \quad \text{(Constraint equation)} $
$ A = x \cdot y $ Use the constraint to write \( y = 40 – 2x \):
$ A = x(40 – 2x) = 40x – 2x^2 $
$ \frac{dA}{dx} = 40 – 4x $
Set derivative to 0: $ 40 – 4x = 0 \Rightarrow x = 10 $ Then \( y = 40 – 2(10) = 20 \)
$ \frac{d^2A}{dx^2} = -4 < 0 \quad \Rightarrow \text{Maximum} $
Final Answer:
Maximum area is achieved when the width is 10 m and the length is 20 m.
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