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)).

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.

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.

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{h\to 0} \frac{(2 + 7h + 5h^2) - 2}{h} m = \lim{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 (x1, y1) = (1, 2) and the slope is m = 7.
y - y1 = m(x - x1)
y - 2 = 7(x - 1)
y - 2 = 7x - 7
y = 7x - 5
The equation of the tangent line is y = 7x - 5.

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{h\to 0} \frac{\frac{4 \cdot 3 - 4(3 + h)}{3(3 + h)}}{h} m = \lim{h\to 0} \frac{\frac{12 - 12 - 4h}{3(3 + h)}}{h}
  m = \lim{h\to 0} \frac{-4h}{3h(3 + h)} m = \lim{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 (x1, y1) = (5, \frac{4}{3}) and the slope is m = -\frac{4}{9}.
y - y1 = m(x - x1)
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}.

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.

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(t1) at time t1 and f(t2) at time t2, its average velocity over the interval [t1, t2] is:
\text{Average Velocity} = \frac{\text{Change in Displacement}}{\text{Change in Time}} = \frac{f(t2) - f(t1)}{t2 - t1}

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}

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

Given: t1 = 4, t2 = 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(t2) - f(t1)}{t2 - t1} = \frac{f(12) - f(4)}{12 - 4}
\text{Average Velocity} = \frac{5 - 3}{8} = \frac{2}{8} = \frac{1}{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{h\to 0} \frac{\sqrt{9 + 2h} - 3}{h} \cdot \frac{\sqrt{9 + 2h} + 3}{\sqrt{9 + 2h} + 3} = \lim{h\to 0} \frac{(\sqrt{9 + 2h})^2 - 3^2}{h(\sqrt{9 + 2h} + 3)} = \lim{h\to 0} \frac{9 + 2h - 9}{h(\sqrt{9 + 2h} + 3)} = \lim{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.