Introduction to Calculus: Limits, Tangent Lines, and Velocity

Section 2.1: An Introduction to Calculus

The Idea of Limits

Calculus is fundamentally concerned with understanding how quantities change and accumulate. The concept of a limit is central to calculus, providing the foundation for both derivatives and integrals. Limits allow us to rigorously define instantaneous rates of change and the area under curves.

• Limit: The value that a function approaches as the input approaches a certain point.

• Limits are used to define both the derivative (instantaneous rate of change) and the integral (area under a curve).

The Two Main Problems of Calculus

• The Tangent Line Problem: Finding the instantaneous rate of change of a function at a point. This is the focus of differentiation (Chapters 3 and 4).

• The Area Problem: Finding the area of a region bounded by a function. This is the focus of integration (Chapters 5 and 6).

The Tangent Line Problem

Secant and Tangent Lines

A secant line joins two points on a curve, while a tangent line touches the curve at a single point and has the same direction as the curve at that point. The slope of the tangent line represents the instantaneous rate of change of the function at that point.

• Secant Line: Slope between two points, $(x_1, f(x_1))$ and $(x_2, f(x_2))$, is given by:

$m_{secant} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$

• Tangent Line: Slope at a single point, found as the limit of the secant slope as the two points approach each other.

$m_{tangent} = \lim_{x_2 \to x_1} \frac{f(x_2) - f(x_1)}{x_2 - x_1}$

Example 1: Slope of the Secant Line for $f(x) = x^2$

• Between $(\frac{1}{2}, \frac{1}{4})$ and $(1, 1)$:

$m = \frac{1 - \frac{1}{4}}{1 - \frac{1}{2}} = \frac{\frac{3}{4}}{\frac{1}{2}} = 1.5$

• Between $(\frac{1}{2}, \frac{1}{4})$ and $(\frac{3}{4}, \frac{9}{16})$:

$m = \frac{\frac{9}{16} - \frac{1}{4}}{\frac{3}{4} - \frac{1}{2}} = \frac{\frac{9}{16} - \frac{4}{16}}{\frac{3}{4} - \frac{2}{4}} = \frac{\frac{5}{16}}{\frac{1}{4}} = 1.25$

• As the second point approaches $x = \frac{1}{2}$, the slope approaches 1.

Interval	Slope of Secant Line
0.5 ≤ x ≤ 1	1.5
0.5 ≤ x ≤ 0.75	1.25
0.5 ≤ x ≤ 0.625	1.125
0.5 ≤ x ≤ 0.5625	1.0625
0.5 ≤ x ≤ 0.53125	1.03125
0.5 ≤ x ≤ 0.515625	1.015625

Conclusion: The slope of the tangent line to $f(x) = x^2$ at $x = \frac{1}{2}$ is approximately 1.

Average and Instantaneous Velocity

Velocity as a Rate of Change

If $f(t)$ describes the position of an object at time $t$, then:

• Average velocity over an interval is the slope of the secant line over that interval.

• Instantaneous velocity at a given time is the slope of the tangent line at the corresponding point.

Example 2: Ball Thrown into the Air

The height of a ball thrown with velocity 40 ft/s is $y = 40t - 16t^2$. Find the average velocity for the time period beginning when $t = 2$ and lasting:

• a) 0.5 seconds: $2 \leq t \leq 2.5$

• b) 0.1 seconds: $2 \leq t \leq 2.1$

• c) 0.01 seconds: $2 \leq t \leq 2.01$

Interval	Average Velocity (ft/s)
2 ≤ t ≤ 2.5	-32
2 ≤ t ≤ 2.1	-25.6
2 ≤ t ≤ 2.01	-24.16
2 ≤ t ≤ 2.001	-24.016
2 ≤ t ≤ 2.0001	-24.0016

Estimate: The instantaneous velocity at $t = 2$ is approximately $-24$ ft/s.

Finding Points with Zero Slope

Tangent Line with Slope Zero

For the function $f(x) = x^2 + 4x - 1$, the tangent line has a slope of zero where the derivative is zero.

• Find $f'(x)$:

$f'(x) = 2x + 4$

• Set $f'(x) = 0$:

$2x + 4 = 0 \implies x = -2$

• At $x = -2$, $f(-2) = (-2)^2 + 4(-2) - 1 = 4 - 8 - 1 = -5$

Conclusion: The point $(-2, -5)$ is where the tangent line to $f(x)$ has a slope of zero.