Calculus Notes: Derivatives and Rates of Change

Derivatives and Rates of Change

We begin by considering a curve represented by the function $y = f(x)$. On this curve, we identify two points: $P(a, f(a))$ and $Q(x, f(x))$.

Slope of the Secant Line (Average Rate of Change)

• For any two distinct points $P(a, f(a))$ and $Q(x, f(x))$ on the curve, where $x \neq a$, the slope of the secant line passing through these points can be calculated as:
  $m_{\text{secant}} = \frac{f(x) - f(a)}{x - a}$
• This formula represents the average rate of change of the function $f(x)$ over the interval from $x = a$ to $x$. It indicates how much the function's output changes on average with respect to the change in its input over that interval.

Slope of the Tangent Line (Instantaneous Rate of Change / Derivative)

• As the point $Q(x, f(x))$ gets progressively closer to the point $P(a, f(a))$ (i.e., as $x$ approaches $a$), the secant line $PQ$ gradually transforms and approaches the tangent line to the curve at point $P$.
• The slope of this tangent line is defined by the limit of the secant line's slope:
  $m = \lim_{x\to a} \frac{f(x) - f(a)}{x - a}$
• Alternative Definition (using $h$):
  • To express this limit in a different, often more convenient, form, let $h$ be the difference between $x$ and $a$, so $h = x - a$.
  • As $x$ approaches $a$, it follows that $h$ approaches $0$.
  • Furthermore, we can write $x$ in terms of $a$ and $h$ as $x = a + h$. Consequently, $f(x)$ becomes $f(a + h)$.
  • Substituting these into the limit definition, the slope of the tangent line to the curve $y = f(x)$ at point $P$ can be redefined as:
    $m = \lim_{h\to 0} \frac{f(a + h) - f(a)}{h}$
• This slope is also known as the instantaneous rate of change of the function at $x = a$
  ahd, most importantly, it is the derivative of the function at $x = a$.

Example 1: Finding the Equation of a Tangent Line

• Problem: Find the equation of the tangent line to the graph of $f(x) = 5x^2 - 3x$ at the point $(1, 2)$.
• Given: $a = 1$, $f(a) = f(1) = 2$.
• Step 1: Calculate the slope ($m$) of the tangent line. We use the definition: $m = \lim_{h\to 0} \frac{f(a + h) - f(a)}{h}$
  • Substitute $a = 1$:
    $m = \lim_{h\to 0} \frac{f(1 + h) - f(1)}{h}$
  • Calculate $f(1 + h)$:
    $f(1 + h) = 5(1 + h)^2 - 3(1 + h)$
    $= 5(1 + 2h + h^2) - 3 - 3h$
    $= 5 + 10h + 5h^2 - 3 - 3h$
    $= 2 + 7h + 5h^2$
  • Calculate $f(1)$:
    $f(1) = 5(1)^2 - 3(1) = 5 - 3 = 2$
  • Substitute these into the limit expression:
    $m = \lim<em>{h\to 0} \frac{(2 + 7h + 5h^2) - 2}{h}$$m = \lim</em>{h\to 0} \frac{7h + 5h^2}{h}$
    $m = \lim_{h\to 0} (7 + 5h)$
  • Evaluate the limit:
    $m = 7 + 5(0) = 7$
• Step 2: Use the point-slope form to find the equation of the line.
  The point is $(x<em>1, y</em>1) = (1, 2)$ and the slope is $m = 7$.
  $y - y<em>1 = m(x - x</em>1)$
  $y - 2 = 7(x - 1)$
  $y - 2 = 7x - 7$
  $y = 7x - 5$
  The equation of the tangent line is $y = 7x - 5$.

Example 2: Finding the Equation of a Tangent Line for a Rational Function

• Problem: Find the equation of the tangent line to the graph of $f(x) = \frac{4}{x - 2}$ at $x = 5$.
• Given: $a = 5$.
• Step 1: Find the y-coordinate of the point of tangency.
  $f(5) = \frac{4}{5 - 2} = \frac{4}{3}$
  So, the point of tangency is $(5, \frac{4}{3})$.
• Step 2: Calculate the slope ($m$) of the tangent line.$m = \lim_{h\to 0} \frac{f(a + h) - f(a)}{h}$
  • Substitute $a = 5$:
    $m = \lim_{h\to 0} \frac{f(5 + h) - f(5)}{h}$
  • Calculate $f(5 + h)$:
    $f(5 + h) = \frac{4}{(5 + h) - 2} = \frac{4}{3 + h}$
  • Calculate $f(5)$:
    $f(5) = \frac{4}{3}$
  • Substitute these into the limit expression:
    $m = \lim_{h\to 0} \frac{\frac{4}{3 + h} - \frac{4}{3}}{h}$
  • Combine the fractions in the numerator:
    $m = \lim<em>{h\to 0} \frac{\frac{4 \cdot 3 - 4(3 + h)}{3(3 + h)}}{h}$$m = \lim</em>{h\to 0} \frac{\frac{12 - 12 - 4h}{3(3 + h)}}{h}$
    $m = \lim<em>{h\to 0} \frac{-4h}{3h(3 + h)}$$m = \lim</em>{h\to 0} \frac{-4}{3(3 + h)}$
  • Evaluate the limit:
    $m = \frac{-4}{3(3 + 0)} = \frac{-4}{9}$
• Step 3: Use the point-slope form to find the equation of the line.
  The point is $(x<em>1, y</em>1) = (5, \frac{4}{3})$ and the slope is $m = -\frac{4}{9}$.
  $y - y<em>1 = m(x - x</em>1)$
  $y - \frac{4}{3} = -\frac{4}{9}(x - 5)$
  $y = -\frac{4}{9}x + \frac{20}{9} + \frac{4}{3}$
  $y = -\frac{4}{9}x + \frac{20}{9} + \frac{12}{9}$
  $y = -\frac{4}{9}x + \frac{32}{9}$
  The equation of the tangent line is $y = -\frac{4}{9}x + \frac{32}{9}$.

Applications to Motion: Position and Velocity

• When an object moves along a straight line, its motion can be described by an equation of motion given by $s = f(t)$. Here, $s$ represents the displacement (which is the directed distance) of the object from its origin at a specific time $t$.
• The function $f$ in this context is known as the position function of the object, as it defines the object's position at any given time.

Average Velocity

• The concept of the average velocity of an object is directly analogous to the average rate of change. It describes the overall speed and direction of the object over a specific time interval.
• If an object has a position $f(t<em>1)$ at time $t</em>1$ and $f(t<em>2)$ at time $t</em>2$, its average velocity over the interval $[t<em>1, t</em>2]$ is:
  $\text{Average Velocity} = \frac{\text{Change in Displacement}}{\text{Change in Time}} = \frac{f(t<em>2) - f(t</em>1)}{t<em>2 - t</em>1}$

Instantaneous Velocity

• The instantaneous velocity represents the velocity (speed and direction) of the object at a precise moment in time, $t = a$. It is an instantaneous measure, unlike average velocity which is over an interval.
• It corresponds to the instantaneous rate of change and is calculated as the derivative of the position function at that specific time $t = a$:
  $\text{Instantaneous Velocity at } t = a = \lim_{h\to 0} \frac{f(a + h) - f(a)}{h}$

Example 3: Calculating Average and Instantaneous Velocity

• Problem: An object is moving in a straight line at time $t$ with a position function $f(t) = \sqrt{2t + 1}$.

(a) What is the average velocity of the object over the time interval from $t = 4$ to $t = 12$?

• Given: $t<em>1 = 4$, $t</em>2 = 12$.
• Step 1: Calculate the position at each time.
  • At $t_1 = 4$: $f(4) = \sqrt{2(4) + 1} = \sqrt{8 + 1} = \sqrt{9} = 3$
  • At $t_2 = 12$: $f(12) = \sqrt{2(12) + 1} = \sqrt{24 + 1} = \sqrt{25} = 5$
• Step 2: Calculate the average velocity.
  $\text{Average Velocity} = \frac{f(t<em>2) - f(t</em>1)}{t<em>2 - t</em>1} = \frac{f(12) - f(4)}{12 - 4}$
  $\text{Average Velocity} = \frac{5 - 3}{8} = \frac{2}{8} = \frac{1}{4}$

(b) What is the instantaneous velocity of the object at time $t = 4$?

• Given: $a = 4$.
• Step 1: Use the limit definition for instantaneous velocity.$\text{Instantaneous Velocity} = \lim_{h\to 0} \frac{f(a + h) - f(a)}{h}$
  • Substitute $a = 4$:
    $\lim_{h\to 0} \frac{f(4 + h) - f(4)}{h}$
  • Calculate $f(4 + h)$:
    $f(4 + h) = \sqrt{2(4 + h) + 1} = \sqrt{8 + 2h + 1} = \sqrt{9 + 2h}$
  • We know $f(4) = 3$ from part (a).
  • Substitute these into the limit expression:
    $\lim_{h\to 0} \frac{\sqrt{9 + 2h} - 3}{h}$
• Step 2: Evaluate the limit using conjugation. To resolve the indeterminate form $\frac{0}{0}$ (when $h=0$), multiply the numerator and denominator by the conjugate of the numerator: $\lim<em>{h\to 0} \frac{\sqrt{9 + 2h} - 3}{h} \cdot \frac{\sqrt{9 + 2h} + 3}{\sqrt{9 + 2h} + 3}$$= \lim</em>{h\to 0} \frac{(\sqrt{9 + 2h})^2 - 3^2}{h(\sqrt{9 + 2h} + 3)}$$= \lim<em>{h\to 0} \frac{9 + 2h - 9}{h(\sqrt{9 + 2h} + 3)}$$= \lim</em>{h\to 0} \frac{2h}{h(\sqrt{9 + 2h} + 3)}$
  • Cancel out $h$ (since $h \neq 0$ as $h \to 0$):
    $= \lim_{h\to 0} \frac{2}{\sqrt{9 + 2h} + 3}$
  • Evaluate the limit by direct substitution of $h = 0$:
    $= \frac{2}{\sqrt{9 + 2(0)} + 3} = \frac{2}{\sqrt{9} + 3} = \frac{2}{3 + 3} = \frac{2}{6} = \frac{1}{3}$
    The instantaneous velocity of the object at time $t = 4$ is $\frac{1}{3}$ units per unit time.