Understanding the instantaneous rate of change via limits is essential for analyzing movements or changes that occur precisely at a given moment. This concept forms the backbone of calculus, offering a mathematical lens to study changes that happen instantaneously. By utilizing limits, calculus transcends the average rate of change, providing insights into the exact rate at which a variable changes at any given point.

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The Concept of Instantaneous Rate of Change

The instantaneous rate of change at a point on a function is akin to the concept of speed at an instant for a moving object. It is the rate at which a function's output value changes with respect to a change in the input value, precisely at a specific point.

Definition: The instantaneous rate of change of a function $f(x)$ at a point $c$ is defined as:

$\lim_{h \to 0} \dfrac{f(c+h) - f(c)}{h}$

This limit, when exists, represents the slope of the tangent line to the curve $y=f(x)$ at the point $(c, f(c))$.

Understanding Through Examples

Let's delve into examples to thoroughly grasp the concept and computational methods of determining the instantaneous rate of change.

Example 1: Linear Function

Consider the linear function $f(x) = 2x + 3$. Find the instantaneous rate of change at $x = 1$.

• Solution:

$\begin{aligned} \lim{h \to 0} \frac{f(1+h) - f(1)}{h} &= \lim{h \to 0} \frac{(2(1+h) + 3) - (2\cdot1 + 3)}{h} \\ &= \lim{h \to 0} \frac{2 + 2h + 3 - 5}{h} \\ &= \lim{h \to 0} \frac{2h}{h} \\ &= \lim_{h \to 0} 2 \\ &= 2 \end{aligned}$

The instantaneous rate of change of $f(x)$ at $x = 1$ is 2, which is consistent across all points for a linear function, indicating a constant slope.

Example 2: Quadratic Function

Consider the function $f(x) = x^2 - 4x + 4$. Calculate the instantaneous rate of change at $x = 2$.

• Solution:

$\begin{aligned} \lim{h \to 0} \frac{f(2+h) - f(2)}{h} &= \lim{h \to 0} \frac{((2+h)^2 - 4(2+h) + 4) - (2^2 - 4\cdot2 + 4)}{h} \\ &= \lim{h \to 0} \frac{(4 + 4h + h^2 - 8 - 4h + 4) - 0}{h} \\ &= \lim{h \to 0} \frac{4h + h^2}{h} \\ &= \lim_{h \to 0} (4 + h) \\ &= 4 \end{aligned}$

Here, the instantaneous rate of change at $x = 2$ is 4, illustrating how the slope of the tangent line to the curve at that point can be determined using limits.

Example 3: Cubic Function

Find the instantaneous rate of change of $f(x) = x^3$ at $x = 1$.

• Solution:

$\begin{aligned} \lim{h \to 0} \frac{f(1+h) - f(1)}{h} &= \lim{h \to 0} \frac{(1+h)^3 - 1^3}{h} \\ &= \lim{h \to 0} \frac{1 + 3h + 3h^2 + h^3 - 1}{h} \\ &= \lim{h \to 0} \frac{3h + 3h^2 + h^3}{h} \\ &= \lim_{h \to 0} (3 + 3h + h^2) \\ &= 3 \end{aligned}$

For $f(x) = x^3$ at $x = 1$, the instantaneous rate of change is 3, which varies at different points on the curve, reflecting the changing slope of the tangent as we move along the function.

Applying the Concept to Real-World Problems

Consider a ball thrown upwards with its height above the ground given by the function $s(t) = -16t^2 + 64t$, where $s$ is the height in feet, and $t$ is the time in seconds. Calculate the instantaneous velocity of the ball at $t = 2$ seconds.

Real-World Example: Instantaneous Velocity

The instantaneous velocity of an object at a given time is the instantaneous rate of change of its position with respect to time.

• Given: $s(t) = -16t^2 + 64t$

• Find: Instantaneous velocity at $t = 2$ seconds.

• Solution:

$\begin{aligned} v(t) &= \lim{h \to 0} \frac{s(2+h) - s(2)}{h} \\ &= \lim{h \to 0} \frac{-16(2+h)^2 + 64(2+h) - (-16\cdot2^2 + 64\cdot2)}{h} \\ &= \lim{h \to 0} \frac{-16(4 + 4h + h^2) + 64 + 64h - (-64 + 128)}{h} \\ &= \lim{h \to 0} \frac{-64 - 64h - 16h^2 + 64 + 64h - 64 + 128}{h} \\ &= \lim{h \to 0} \frac{-16h^2}{h} \\ &= \lim{h \to 0} -16h \\ &= 0 \end{aligned}$

At $t = 2$ seconds, the instantaneous velocity of the ball is 0. This means that at this exact moment, the ball has reached its maximum height and its velocity is momentarily zero before it starts descending.

Understanding the Result

The calculation shows how calculus enables us to capture the behavior of dynamic systems at specific instances. The result of 0 for the instantaneous velocity at $t = 2$ seconds illustrates a key point in the motion of the ball: its peak height. This demonstrates how the concept of limits and instantaneous rates of change is not only theoretical but also has practical applications in understanding real-world phenomena.

Practice Questions

Question 1

A particle moves along a straight line with its position at time

given by the function $s(t) = t^3 - 6t^2 + 9t - 4$, where $s$ is measured in meters and $t$ in seconds. Calculate the instantaneous velocity of the particle at $t = 2$ seconds.

Question 2

The temperature, $T$, in degrees Celsius, of a substance at time $t$ hours is given by $T(t) = 4t^2 - 16t + 25$. Find the rate of temperature change with respect to time at $t = 3$ hours.

Question 3

Given the function $f(x) = \sqrt{x}$, determine the instantaneous rate of change at $x = 4$.

Solutions to Practice Questions

Solution to Question 1

Given: $s(t) = t^3 - 6t^2 + 9t - 4$

Find: Instantaneous velocity at $t = 2$ seconds.

• Solution:

$v(t) = \lim_{h \to 0} \frac{s(2+h) - s(2)}{h}$

Substituting $s(t)$:

$\begin{aligned} &= \lim{h \to 0} \frac{(2+h)^3 - 6(2+h)^2 + 9(2+h) - 4 - (8 - 24 + 18 - 4)}{h} \\ &= \lim{h \to 0} \frac{8 + 12h + 6h^2 + h^3 - 24 - 24h - 6h^2 + 18 + 9h - 4 + 8}{h} \\ &= \lim{h \to 0} \frac{12h + h^3 + 9h}{h} \\ &= \lim{h \to 0} (12 + h^2 + 9) \\ &= 21 \end{aligned}$

The instantaneous velocity of the particle at $t = 2$ seconds is 21 meters per second.

Solution to Question 2

Given: $T(t) = 4t^2 - 16t + 25$

Find: Rate of temperature change at $t = 3$ hours.

• Solution:

$\begin{aligned} T'(t) &= \lim{h \to 0} \frac{T(3+h) - T(3)}{h} \\ &= \lim{h \to 0} \frac{4(3+h)^2 - 16(3+h) + 25 - (36 - 48 + 25)}{h} \\ &= \lim{h \to 0} \frac{36 + 24h + 4h^2 - 48 - 16h + 25 + 1}{h} \\ &= \lim{h \to 0} \frac{24h + 4h^2 - 16h}{h} \\ &= \lim_{h \to 0} (8 + 4h) \\ &= 8 \end{aligned}$

The rate of temperature change at $t = 3$ hours is 8 degrees Celsius per hour.

Solution to Question 3

Given: $f(x) = \sqrt{x}$

Find: Instantaneous rate of change at $x = 4$.

• Solution:

$f'(x) = \lim{h \to 0} \frac{f(4+h) - f(4)}{h} = \lim{h \to 0} \frac{\sqrt{4+h} - \sqrt{4}}{h}$

Applying the conjugate multiplication technique to rationalize the numerator:

$\begin{aligned} &= \lim{h \to 0} \frac{\sqrt{4+h} - 2}{h} \cdot \frac{\sqrt{4+h} + 2}{\sqrt{4+h} + 2} \\ &= \lim{h \to 0} \frac{(4+h) - 4}{h(\sqrt{4+h} + 2)} \\ &= \lim{h \to 0} \frac{h}{h(\sqrt{4+h} + 2)} \\ &= \lim{h \to 0} \frac{1}{\sqrt{4+h} + 2} \\ &= \frac{1}{\sqrt{4} + 2} \\ &= \frac{1}{2 + 2} \\ &= \frac{1}{4} \end{aligned}$

The instantaneous rate of change of the function $f(x) = \sqrt{x}$ at (x = 4) is $\frac{1}{4}$. This result represents the slope of the tangent line to the curve $y = \sqrt{x}$ at the point $x = 4$, indicating how rapidly the function is changing at that specific point.

Other:

# Simple simulation of the Fetch-Execute Cycle

# Example program stored in memory (a list of instructions)
memory = [
    "LOAD 5",     # Load the number 5 into the accumulator
    "ADD 3",      # Add 3 to the accumulator
    "STORE 10",   # Store the result at memory location 10
]

# Simple registers
program_counter = 0    # Holds the address of the next instruction
accumulator = 0        # Holds the result of calculations
memory_store = [0] * 20  # Create a simple memory store with 20 locations

while program_counter < len(memory):
    # FETCH
    instruction = memory[program_counter]

    # DECODE
    parts = instruction.split()
    operation = parts[0]
    value = int(parts[1])

    # EXECUTE
    if operation == "LOAD":
        accumulator = value
    elif operation == "ADD":
        accumulator += value
    elif operation == "STORE":
        memory_store[value] = accumulator

    # Move to the next instruction
    program_counter += 1

# Output final result stored in memory
print("Final stored value at location 10:", memory_store[10])