Calc 1: Sections 2.7 & 2.8 - Derivatives and Tangent Lines

Welcome back to Professor Baker's Math Class! In this post, we'll be covering the key concepts from sections 2.7 and 2.8 of Calculus 1. We'll explore the idea of a derivative and how it relates to the tangent line of a curve. Let's get started!

Understanding the Tangent Line

The tangent line to a curve $y = f(x)$ at a point $P(a, f(a))$ is the line that "just touches" the curve at that point. The slope of this tangent line is a fundamental concept in calculus. We can find it using a limit:

The slope, $m$, of the tangent line is given by:

$$m = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$$

Alternatively, we can express the slope as:

$$m = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$

The Derivative: A Formal Definition

The derivative of a function $f$ at a number $a$, denoted by $f'(a)$, is defined as:

$$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$

This limit, if it exists, gives us the instantaneous rate of change of the function at the point $a$. The derivative $f'(x)$ is itself a function, giving the slope of the tangent line at any point $x$ where the derivative exists.

There are several notations for derivatives, including:

  • Newton's Notation: $f'(x) = y'$
  • Leibniz's Notation: $\frac{dy}{dx} = \frac{df}{dx} = \frac{d}{dx}f(x)$
  • Operator Notation: $Df(x) = D_x f(x)$

Examples and Applications

Let's consider some examples to solidify our understanding:

Example 1: Finding the Tangent Line

Find an equation of the tangent line to the hyperbola $y = \frac{3}{x}$ at the point $(3, 1)$.

We can use the limit definition to find the slope:

$$m = \lim_{h \to 0} \frac{f(3 + h) - f(3)}{h} = \lim_{h \to 0} \frac{\frac{3}{3+h} - 1}{h}$$

After simplifying, we find:

$$m = -\frac{1}{3}$$

Example 2: Velocity as a Derivative

Suppose a ball is dropped from the CN Tower, 450 m above the ground. The position function is given by $f(x) = 4.9t^2$. What is the velocity of the ball after 5 seconds?

We need to find the derivative of the position function and evaluate it at $t = 5$:

$$v(t) = f'(t) = \lim_{h \to 0} \frac{4.9(t+h)^2 - 4.9t^2}{h}$$.

Which simplifies to $v(t) = 9.8t$. Thus, $v(5) = 9.8 * 5 = 49$ m/s.

Differentiability

A function is differentiable at a point if its derivative exists at that point. However, there are cases where a function is not differentiable. These include:

  • Corners: Sharp turns in the graph
  • Discontinuities: Breaks in the graph
  • Vertical Tangents: Where the tangent line is vertical (infinite slope)

Higher Order Derivatives

We can continue to take derivatives of derivatives! The second derivative, denoted as $f''(x)$, represents the rate of change of the first derivative. In physics, if $f(x)$ represents position, then:

  • $f'(x)$ is velocity
  • $f''(x)$ is acceleration
  • $f'''(x)$ is jerk

Keep practicing, and you'll master these concepts in no time! Good luck, and see you in the next section!