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IB Mathematics AI SL Derivative interpreted as gradient function MAI Study Notes - New Syllabus

IB Mathematics AI SL Derivative interpreted as gradient function MAI Study Notes

LEARNING OBJECTIVE

  • Introduction to the concept of a limit.

Key Concepts: 

  • Concept of a limit.
  • Derivative as gradient function and as rate of change.

MAI HL and SL Notes – All topics

CONCEPTS OF LIMITS

◆ LIMITS

Limits describe the behavior of a function as the input approaches a certain value.

 

ESTIMATING LIMITS FROM TABLES AND GRAPHS

  • Use values close to the target \(x\) to estimate \(\lim_{x \to c} f(x)\).
  • Graphically, observe the \(y\)-value as the curve approaches \(x = c\) from both left and right.
  • A limit exists only if both sides approach the same value.

Example (Using graph)

Given the function $f(x) = \frac{x – 3}{x – 2}$, identify the vertical and horizontal asymptotes. Also, provide the formal limit-based explanation for each asymptote.

▶️Answer/Explanation

Solution:

Vertical Asymptote:
$x = 2$ is a vertical asymptote.

Formal Explanation:

$
\lim_{x \to 2^+} f(x) = +\infty \quad \text{and} \quad \lim_{x \to 2^-} f(x) = -\infty
$

Horizontal Asymptote:
$y = 1$ is a horizontal asymptote.

Formal Explanation:

$
\lim_{x \to \infty} f(x) = 1 \quad \text{and} \quad \lim_{x \to -\infty} f(x) = 1
$

INFORMAL LIMIT NOTATION

\(\lim_{x \to c} f(x) = L\)

  • This means that as \(x\) approaches \(c\), \(f(x)\) gets arbitrarily close to \(L\).

Example (Using table)

Investigate the limit $\lim_{x \to \infty} \left(1 + \frac{1}{x} \right)^x$ informally using a calculator. What value does this limit approach?

▶️Answer/Explanation

Solution:

Using values of $x$ approaching infinity:

The resulting limit is in fact the number:

$
e = 2.7182818\ldots
$

So we conclude:

$
\lim_{x \to \infty} \left(1 + \frac{1}{x} \right)^x = e
$

Example:

Function: \( f(x) = \dfrac{x^2 – 4}{x – 2} \)

Estimate: \( \displaystyle \lim_{x \to 2} f(x) \) using values of \(x\) close to 2.

xf(x)
1.93.9
1.993.99
2.014.01
2.14.1
▶️ Answer/Explanation

Solution:

\( f(x) = \frac{x^2 – 4}{x – 2} = \frac{(x – 2)(x + 2)}{x – 2} = x + 2 \), for \( x \ne 2 \).

Conclusion: As \( x \to 2 \), the function values approach 4.

Therefore,

 \( \displaystyle \lim_{x \to 2} \frac{x^2 – 4}{x – 2} = 4 \)

Note: The limit exists even though the function is undefined at \( x = 2 \).

DERIVATIVE AS A GRADIENT FUNCTION / RATE OF CHANGE

INTERPRETING DERIVATIVES AS SLOPES

The derivative at a point gives the slope of the tangent line to the curve at that point.

It represents the instantaneous rate of change of the function.

\( f'(x) = \lim_{h \to 0} \frac{f(x + h) – f(x)}{h} \)

A graph showing a curve with a tangent line at a specific point, illustrating the derivative as the slope of this line

Example : (GRADIENT) ΙΝ Α LINE

Consider the line \( f(x) = 2x + 3 \). What is the rate of change (gradient) of this line, and how can it be confirmed using different points on the line?

▶️ Answer/Explanation

Solution:

  • From \( x = 1 \) to \( x = 2 \):
    \( f(1) = 5 \), \( f(2) = 7 \)
    \( \dfrac{\Delta y}{\Delta x} = \dfrac{f(2) – f(1)}{2 – 1} = \dfrac{7 – 5}{2 – 1} = 2 \)
  • From \( x = 0 \) to \( x = 2 \):
    \( f(0) = 3 \), \( f(2) = 7 \)
    \( \dfrac{\Delta y}{\Delta x} = \dfrac{f(2) – f(0)}{2 – 0} = \dfrac{7 – 3}{2 – 0} = 2 \)
  • The rate of change is consistent and equal to 2. This common value is the gradient of the line.

Example : (GRADIENT) ΙΝ Α CURVE

In a curve which is not a straight line, the rate of change between any two points is not always the same. For example, consider the function \( f(x) = x^2 \).

▶️ Answer/Explanation

Solution:

  • Rate of change from \( x = 1 \) to \( x = 2 \):
    $ \frac{\Delta y}{\Delta x} = \frac{f(2) – f(1)}{2 – 1} = \frac{4 – 1}{1} = 3 $
  • Rate of change from \( x = 1 \) to \( x = 3 \):
    $ \frac{\Delta y}{\Delta x} = \frac{f(3) – f(1)}{3 – 1} = \frac{9 – 1}{2} = 4 $
  • This shows that for a curve like \( f(x) = x^2 \), the rate of change varies depending on the interval chosen.

However, we can also measure the instantaneous rate of change at a single point — this is the gradient (derivative) at that point.

 

  • From the diagram, the gradient values at some points on the curve \( f(x) = x^2 \) are:
    • Gradient at \( x = -1 \): \( m = -2 \)
    • Gradient at \( x = 0 \): \( m = 0 \)
    • Gradient at \( x = 1 \): \( m = 2 \)
    • Gradient at \( x = 2 \): \( m = 4 \)
    • Gradient at \( x = 3 \): \( m = 6 \)

BASIC NOTATIONS FOR DERIVATIVES

  • \(\frac{dy}{dx}\): Rate of change of \(y\) with respect to \(x\).
  • \(f'(x)\): Derivative of the function \(f\) at \(x\).
  • \(\frac{ds}{dt}\): Rate of change of \(s\) with respect to \(t\).

Example

Question:

If \(s(t) = 5t^3\), find \(\frac{ds}{dt}\).

▶️Answer/Explanation

\(\frac{ds}{dt} = 15t^2\)

NOTATIONS FOR DERIVATIVES

LEIBNIZ NOTATION

  • Emphasizes the derivative as the ratio of infinitesimal changes.
  • Written as: \(\frac{dy}{dx}, \frac{ds}{dt}\)

LAGRANGE NOTATION

  • Often used for simplicity when the function is expressed explicitly in terms of \(x\).
  • Written as: \(f'(x), V'(r)\)

Example

Question:

For \(y = x^2\), write the derivative in both Leibniz and Lagrange notations. Also, for \(V = \pi r^2 h\), find \(\frac{dV}{dr}\).

▶️Answer/Explanation

\(\frac{dy}{dx} = 2x\) (Leibniz notation)

\(f'(x) = 2x\) (Lagrange notation)

\(\frac{dV}{dr} = 2\pi rh\)

\(V'(r) = 2\pi rh\)

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