Comprehensive Calculus Notes: Kinematics, Trigonometric Derivatives, and Chain Rule

Course Logistics & Test 1 Overview

• Test 1 Schedule and Logistics:

  • Date: Thursday, September 17.
  • Time: Administered during regular class time. The test is designed as a 50-minute test, but approximately one hour will be provided, accounting for initial setup.
  • Office Hours Adjustment: Office hours for test week are moved to Wednesday, September 16, from 10:00 AM to 11:00 AM (replacing the usual Friday office hours).
  • Homework Due Date: WebAssign homework covering Sections 2.4 to 2.5 is due next Wednesday, September 16.
  • Test Environment: Paper-based exam. No electronics or calculators of any kind are permitted. Bags must be placed at the front or sides of the room before testing begins.
  • Absence Policy: Documented excuse/university-sanctioned absences must be communicated as early as possible. Makeup exams or score refunds are handled at the end of the semester to prevent students from falling behind.

• Test Structure & Content Coverage:

  • Scope: Covers Section 2.4 through Section 2.5 (Parametric equations, limits, derivatives up through derivative rules, trig derivatives, and the chain rule).
  • Format: Free-response / written exam on paper (not multiple choice), consisting of approximately 6 written questions.
  • Grading: Partial credit is awarded; complete step-by-step work must be displayed.
  • Question Types: Similar in style and rigor to class worksheets and textbook homework problems found on WebAssign. Questions may require formal application of theorems such as the Intermediate Value Theorem (IVT) or the Squeeze Theorem.
  • Precalculus Prerequisites: Graphing techniques and fundamental trigonometric identities from precalculus are required background knowledge for solving limit and derivative problems, though precalculus will not be explicitly tested in isolation.

• Study Resources on Moodle:

  • Derivative flashcards.
  • Review sheet listing recommended textbook practice problems.
  • A mock practice worksheet structured like the exam, to be completed in class prior to test day.
  • Posted worked-out examples from class sessions.

• Student Review Topic Survey Results:

  • Key topics identified for targeted review: Parametric equations (converting to Cartesian equations and identifying graphs), continuity, formal applications of IVT and Squeeze Theorem, and basic derivative definitions/rules.

Kinematics: Position, Velocity, Acceleration, and Signs

• Relationship Between Velocity and Acceleration Signs:

  • Accelerating (Speeding Up): Velocity $v(t)$ and acceleration $a(t)$ have the same sign (both positive or both negative).
  • Decelerating (Slowing Down): Velocity $v(t)$ and acceleration $a(t)$ have opposite signs.
  • Physical Meaning of Acceleration: Acceleration is the instantaneous rate of change of velocity ($a(t) = v'(t)$).
  • Negative Velocity Case: If $v(t) < 0$ and $a(t) < 0$, velocity is decreasing (becoming more negative), meaning the magnitude of velocity ($\text{speed} = |v(t)|$) is increasing in the negative direction. Therefore, the object is accelerating.
  • If $v(t) < 0$ and $a(t) > 0$, positive acceleration counteracts negative velocity, causing the magnitude of velocity to decrease (decelerating).

• Detailed Kinematic Example 1:

  • Position function: $s(t) = 24t - 16t^2$
  • Velocity function: $v(t) = s'(t) = 24 - 32t$
  • Acceleration function: $a(t) = v'(t) = -32$
  • Evaluation at $t = 2$:
  • Velocity: $v(2) = 24 - 32(2) = -40\,m/s$
  • Acceleration: $a(2) = -32\,m/s^2$
  • Conclusion at $t = 2$: Since $v(2) = -40\,m/s$ and $a(2) = -32\,m/s^2$ share the same sign (both negative), the object is accelerating in the negative direction.

• Graphical Properties of Kinematic Functions:

  • Position graph $s(t)$: Parabola opening downward; $s(t) = 0$ represents the ground level.
  • Velocity graph $v(t)$: Line with negative slope. Intersects the $t$-axis ($v(t) = 0$) at $t = 0.75$. This intercept represents the instantaneous point of rest where direction changes from increasing position to decreasing position.
  • Acceleration graph $a(t)$: Horizontal constant line at $a = -32$.

• Detailed Kinematic Example 2 (Particle Motion on a 1D Line):

  • Position function: $s(t) = t^3 - 9t^2 + 24t + 4$ for $t \ge 0$
  • Velocity: $v(t) = s'(t) = 3t^2 - 18t + 24$
  • Rest Points: Set $v(t) = 0$
  • $3(t^2 - 6t + 8) = 0$
  • $3(t - 2)(t - 4) = 0$
  • Particle is at rest at $t = 2$ and $t = 4$
  • Moving to the Right ($v(t) > 0$):
  • Factored inequality: $3(t - 2)(t - 4) > 0$
  • Both factors positive: $t > 2$ and $t > 4 \implies (4, \infty)$
  • Both factors negative: $t < 2$ and $t < 4 \implies (-\infty, 2)$ (or $[0, 2)$ for $t \ge 0$)
  • Intervals of rightward motion: $(-\infty, 2)$ and $(4, \infty)$
  • Moving to the Left ($v(t) < 0$):
  • Occurs when factors have opposite signs: $(2, 4)$
  • Acceleration: $a(t) = v'(t) = 6t - 18$
  • Acceleration Intervals:
  • Accelerating in positive direction ($v(t) > 0$ and $a(t) > 0$):
    • $a(t) = 6t - 18 > 0 \implies t > 3$
    • Overlap with $v(t) > 0$ occurs on the interval $(4, \infty)$
  • Accelerating in negative direction ($v(t) < 0$ and $a(t) < 0$):
    • $v(t) < 0 \implies t \in (2, 4)$
    • $a(t) = 6t - 18 < 0 \implies t < 3$
    • Overlap occurs on the interval $(2, 3)$
  • Motion Breakdown: At $t = 2$, the particle stops, turns left, and speeds up in the negative direction until $t = 3$. From $t = 3$ to $t = 4$, $a(t) > 0$ while $v(t) < 0$, so it slows down until coming to rest at $t = 4$, after which it accelerates to the right.

Derivation of Trigonometric Derivatives

• Fundamental Reciprocal and Quotient Identities:

  • $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$
  • $\sec(\theta) = \frac{1}{\cos(\theta)}$
  • $\csc(\theta) = \frac{1}{\sin(\theta)}$
  • $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$

• Limit Definition Derivation of $\frac{d}{d\theta}[\sin(\theta)]$:

  • Limit setup: $\frac{d}{d\theta}[\sin(\theta)] = \lim_{h \to 0} \frac{\sin(\theta + h) - \sin(\theta)}{h}$
  • Apply sum-of-angles identity $\sin(\theta + h) = \sin(\theta)\cos(h) + \cos(\theta)\sin(h)$
  • Rearrange terms:     $\lim_{h \to 0} \frac{\sin(\theta)\cos(h) + \cos(\theta)\sin(h) - \sin(\theta)}{h} = \sin(\theta) \lim_{h \to 0} \frac{\cos(h) - 1}{h} + \cos(\theta) \lim_{h \to 0} \frac{\sin(h)}{h}$
  • Direct substitution yields indeterminate forms $\frac{0}{0}$, requiring geometric and algebraic evaluation of fundamental limits.

• Proof 1: Geometric Squeeze Theorem Proof for $\lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1$:

  • Consider a unit circle ($r = 1$) in the first quadrant ($0 < \theta < \frac{\pi}{2}$):
  • Inner triangle $COA$ with base $OA = \cos(\theta)$ and height $AC = \sin(\theta)$
  • Circular sector $COB$ with radius $r = 1$ and angle $\theta$
  • Outer triangle $DOB$ with base $OB = 1$ and height $BD = \tan(\theta)$
  • Area Inequality: $\text{Area}(\triangle COA) < \text{Area}(\text{Sector } COB) < \text{Area}(\triangle DOB)$
  • Substitute geometric area formulas:     $\frac{1}{2} \cos(\theta) \sin(\theta) < \frac{1}{2} (1)^2 \theta < \frac{1}{2} (1) \tan(\theta)$
  • Multiply by $2$ and divide through by $\sin(\theta)$ (since $\sin(\theta) > 0$ in Q1):     $\cos(\theta) < \frac{\theta}{\sin(\theta)} < \frac{1}{\cos(\theta)}$
  • Invert terms (reversing inequality signs):     $\cos(\theta) < \frac{\sin(\theta)}{\theta} < \frac{1}{\cos(\theta)}$
  • Take limits as $\theta \to 0$:     $\lim_{\theta \to 0} \cos(\theta) = 1 \quad \text{and} \quad \lim_{\theta \to 0} \frac{1}{\cos(\theta)} = 1$
  • By Squeeze Theorem: $\lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1$

• Proof 2: Algebraic Proof for $\lim_{\theta \to 0} \frac{\cos(\theta) - 1}{\theta} = 0$:

  • Multiply numerator and denominator by conjugate $(\cos(\theta) + 1)$$\lim_{\theta \to 0} \frac{\cos(\theta) - 1}{\theta} \cdot \frac{\cos(\theta) + 1}{\cos(\theta) + 1} = \lim_{\theta \to 0} \frac{\cos^2(\theta) - 1}{\theta(\cos(\theta) + 1)} = \lim_{\theta \to 0} \frac{-\sin^2(\theta)}{\theta(\cos(\theta) + 1)}$
  • Separate product limits:     $\left(\lim_{\theta \to 0} \frac{\sin(\theta)}{\theta}\right) \cdot \left(\lim_{\theta \to 0} \frac{-\sin(\theta)}{\cos(\theta) + 1}\right) = (1) \cdot \left(\frac{0}{1 + 1}\right) = 0$

• Final Derivative Conclusion for Sine:

  • Substitute fundamental limits back into definition:     $\frac{d}{d\theta}[\sin(\theta)] = \sin(\theta) \cdot (0) + \cos(\theta) \cdot (1) = \cos(\theta)$
  • Graphical validation: At $\theta = \frac{\pi}{2}$ and $\theta = \frac{3\pi}{2}$, $\sin(\theta)$ has horizontal tangent lines ($m = 0$), matching $\cos\left(\frac{\pi}{2}\right) = 0$ and $\cos\left(\frac{3\pi}{2}\right) = 0$

Summary of Trigonometric Derivatives and Computations

• Complete Set of Standard Trigonometric Derivatives:

  • $\frac{d}{dx}[\sin(x)] = \cos(x)$
  • $\frac{d}{dx}[\cos(x)] = -\sin(x)$
  • $\frac{d}{dx}[\tan(x)] = \sec^2(x)$ (derived via Quotient Rule on $\frac{\sin(x)}{\cos(x)}$)
  • $\frac{d}{dx}[\sec(x)] = \sec(x)\tan(x)$ (derived via Quotient Rule on $\frac{1}{\cos(x)}$)
  • $\frac{d}{dx}[\csc(x)] = -\csc(x)\cot(x)$ (derived via Quotient Rule on $\frac{1}{\sin(x)}$)
  • $\frac{d}{dx}[\cot(x)] = -\csc^2(x)$ (derived via Quotient Rule on $\frac{\cos(x)}{\sin(x)}$)

• Example 1: Tangent Line Equation:

  • Find tangent line equation for $y = \sin(x)\cos(x)$ at $x = \frac{\pi}{4}$
  • Derivative (Product Rule):     $y' = \cos(x)\cos(x) + \sin(x)(-\sin(x)) = \cos^2(x) - \sin^2(x)$
  • Evaluate slope $m$ at $x = \frac{\pi}{4}$:     $m = \cos^2\left(\frac{\pi}{4}\right) - \sin^2\left(\frac{\pi}{4}\right) = \left(\frac{\sqrt{2}}{2}\right)^2 - \left(\frac{\sqrt{2}}{2}\right)^2 = 0$
  • Evaluate $y$ point value at $x = \frac{\pi}{4}$:     $y\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{1}{2}$
  • Equation using Point-Slope Form $y - y_1 = m(x - x_1)$:     $y - \frac{1}{2} = 0\left(x - \frac{\pi}{4}\right) \implies y = \frac{1}{2}$

• Example 2: Quotient Rule and Identity Pre-simplification:

  • Function: $y = \frac{3\tan(x)}{5\sec(x)}$
  • Method 1 (Direct Quotient Rule):
  • Numerator $f(x) = 3\tan(x) \implies f'(x) = 3\sec^2(x)$
  • Denominator $g(x) = 5\sec(x) \implies g'(x) = 5\sec(x)\tan(x)$
  • $y' = \frac{3\sec^2(x) \cdot 5\sec(x) - 3\tan(x) \cdot 5\sec(x)\tan(x)}{(5\sec(x))^2}$
  • Method 2 (Trigonometric Simplification prior to differentiation):
  • $y = \frac{3 \cdot \frac{\sin(x)}{\cos(x)}}{5 \cdot \frac{1}{\cos(x)}} = \frac{3}{5}\sin(x)$
  • $y' = \frac{3}{5}\cos(x)$
  • Rule of Thumb: Simplifying trigonometric expressions algebraically before taking the derivative is often substantially easier than simplifying post-differentiation.

The Chain Rule

• Definition and Formula:

  • Used to differentiate composite functions $h(x) = f(g(x))$
  • Prime Notation: $h'(x) = f'(g(x)) \cdot g'(x)$
  • Leibniz Notation: If $y = f(u)$ where $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
  • Strategy: Identify inner function $u = g(x)$, identify outer function $f(u)$, differentiate outer function evaluated at unchanged inner function, and multiply by the derivative of the inner function.

• Chain Rule Example 1:

  • Differentiate $h(x) = \sin(x^2)$
  • Outer function: $f(u) = \sin(u) \implies f'(u) = \cos(u)$
  • Inner function: $u = x^2 \implies u' = 2x$
  • Result: $h'(x) = \cos(u) \cdot u' = \cos(x^2) \cdot 2x = 2x\cos(x^2)$

• Chain Rule Example 2 (Comparison of Methods):

  • Differentiate $f(x) = (3x + 2)^2$
  • Method 1 (Algebraic Expansion):
  • $f(x) = 9x^2 + 12x + 4$
  • $f'(x) = 18x + 12$
  • Method 2 (Chain Rule):
  • Outer function: $f(u) = u^2 \implies f'(u) = 2u$
  • Inner function: $u = 3x + 2 \implies u' = 3$
  • Result: $f'(x) = 2(3x + 2) \cdot 3 = 6(3x + 2) = 18x + 12$
  • Both methods yield identical results, validating the power rule composition.