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 definite integral (area under a curve).

The Two Main Problems of Calculus

Calculus addresses two classical problems:

• The Tangent Line Problem: Finding the instantaneous rate of change of a function at a point, which is the slope of the tangent line to the curve at that point.

• The Area Problem: Determining the area of a region bounded by a curve, which leads to the concept of integration.

Additional info: The tangent line problem is the focus of differentiation (Chapters 3 and 4), while the area problem 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 (slope) as the curve at that point.

• Secant Line: Used to approximate the slope between two points on a curve.

• Tangent Line: Represents the instantaneous rate of change at a single point.

Example 1: Slope of the Secant Line

Given the function $f(x) = x^2$, find the slope of the secant line between the points $(\frac{1}{2}, \frac{1}{4})$ and $(1, 1)$.

• Formula for slope between $(x_1, y_1)$ and $(x_2, y_2)$:

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

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

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

Example 1 (continued): Slope for Closer Points

Find the slope of the secant line between $(\frac{1}{2}, \frac{1}{4})$ and $(\frac{3}{4}, \frac{9}{16})$:

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

Estimating the Slope of the Tangent Line

As the two points get closer, the slope of the secant line approaches the slope of the tangent line at $(\frac{1}{2}, \frac{1}{4})$.

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

As the interval narrows, the slope approaches 1. The tangent line's slope at $x = 0.5$ is estimated to be 1.

Average and Instantaneous Velocity

Velocity and Secant/Tangent Lines

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

• 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 given by $y = 40t - 16t^2$. Find the average velocity for the time period beginning when $t = 2$ and lasting:

• a) 0.5 seconds

• b) 0.1 seconds

• c) 0.01 seconds

Then, estimate the instantaneous velocity when $t = 2$.

Interval	Average Velocity
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

As the interval shortens, the average velocity approaches -24 ft/s. Thus, the instantaneous velocity at $t = 2$ is approximately -24 ft/s.

Finding Points with Zero Slope

Example 3: Tangent Line with Zero Slope

Consider the function $f(x) = x^2 + 4x - 1$. To find the point where the tangent line has a slope of zero, set the derivative equal to zero:

• Compute the derivative: $f'(x) = 2x + 4$

• Set $f'(x) = 0$:

$2x + 4 = 0$

• Solve for $x$:

$x = -2$

• Find the corresponding $y$ value:

$f(-2) = (-2)^2 + 4(-2) - 1 = 4 - 8 - 1 = -5$

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