MATH-283 (Calculus III) Curriculum

Session 01 - Course Overview + Prerequisite Review

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

By the end of this session, you will be able to:

Apply right-triangle trigonometric ratios, angle conversions, and unit circle values.
Solve limits, evaluate piecewise continuity/differentiability, and optimize geometric functions using derivatives.
Calculate net signed area, total area, and net change using definite integrals and geometric formulas.

Micro-Lecture

1. Trigonometric Unit Circle & Right Triangle Geometry

Mathematical Foundation: A right triangle with acute angle $\theta$ , adjacent side $x$ , opposite side $y$ , and hypotenuse $r = \sqrt{x^2 + y^2}$ defines the standard trigonometric ratios: $\sin\theta = \frac{y}{r}, \quad \cos\theta = \frac{x}{r}, \quad \tan\theta = \frac{y}{x}$ Angles are converted between radians and degrees using the ratio $\pi \text{ rad} = 180^\circ$ . The unit circle maps angles to coordinates $(x,y) = (\cos\theta, \sin\theta)$ on $x^2 + y^2 = 1$ , where sign conventions are determined by the quadrant (CAST rule).
Engineering Context: Crucial for static force decompositions in structural trusses, coordinate system rotations, robotics kinematics, and analyzing alternating current (AC) circuits.

2. Limits, Continuity, and Differentiability

Mathematical Foundation: Limits describe function behavior as the input $x$ approaches a value $c$ . Indeterminate forms like $\frac{0}{0}$ are resolved algebraically via factoring or rationalizing. A function $f(x)$ is continuous at $c$ if $\lim_{x \to c} f(x) = f(c)$ , and differentiable if the derivative limit $\lim_{x \to c} f'(x)$ exists and is continuous at $c$ .
Engineering Context: Modeling state transitions in physical processes (such as thermodynamics or fluid mechanics) requires continuous and differentiable functions to avoid non-physical shocks or discontinuities in stress, pressure, or temperature.

3. Derivatives and Optimization

Mathematical Foundation: The derivative $f'(x)$ measures the instantaneous rate of change. Critical points occur where $f'(x) = 0$ or is undefined. The second derivative $f''(x)$ dictates concavity (concave up if $f''(x) > 0$ , concave down if $f''(x) < 0$ ), and inflection points occur where concavity changes sign. Optimization uses these derivatives to locate maximum or minimum values under specific constraints.
Engineering Context: Used to minimize material cost (e.g., structural optimization), maximize structural load capacities, and determine peak operating efficiencies.

4. Definite Integrals and the Net Change Theorem

Mathematical Foundation: A definite integral $\int_a^b f(x) dx$ represents the net signed area bounded by the curve and the $x$ -axis. Total area treats all bounded regions as positive. The Net Change Theorem states that the integral of a rate of change function $F'(t)$ yields the net accumulated change: $F(b) - F(a) = \int_a^b F'(t) dt$
Engineering Context: Enables calculating total physical accumulation (e.g., total volume of water discharged, total charge accumulated, or total work done) from a time-varying rate.

Skill Block 1

Worked Example A: Unit Circle Evaluation

Evaluate the following, giving your answer in exact form:

\sin\left(\frac{3\pi}{4}\right)

Worked Solution:

Identify the Quadrant: The angle $\theta = \frac{3\pi}{4}$ lies in Quadrant II, where the sine function is positive.
Find the Reference Angle ( $\theta_R$ ): $\theta_R = \pi - \frac{3\pi}{4} = \frac{\pi}{4}$
Evaluate: $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$ Since sine is positive in Quadrant II, the exact value is: $\sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}$

Worked Example B: Right Triangle Side Lengths

Consider triangle $ABC$ , a right triangle with a right angle at $C$ . Given the hypotenuse $c = 12$ and adjacent side $b = 9$ , find the exact length of the third side $a$ .

Worked Solution:

Apply the Pythagorean Theorem: $a^2 + b^2 = c^2 \implies a^2 + 9^2 = 12^2$
Isolate $a^2$ : $a^2 + 81 = 144 \implies a^2 = 63$
Solve for $a$ (exact form): $a = \sqrt{63} = \sqrt{9 \cdot 7} = 3\sqrt{7}$

Worked Example C: Right Triangle Trigonometric Ratios

Consider triangle $ABC$ , a right triangle with a right angle at $C$ . Given adjacent side $b = 6$ and opposite side $a = 6$ , find the exact values of $\sin A$ and $\cos A$ .

Worked Solution:

Find the Hypotenuse ( $c$ ): $c^2 = a^2 + b^2 = 6^2 + 6^2 = 36 + 36 = 72 \implies c = \sqrt{72} = 6\sqrt{2}$
Evaluate $\sin A$ : $\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$
Evaluate $\cos A$ : $\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

Practice Problem 1

Consider triangle $ABC$ , a right triangle with a right angle at $C$ . Given the hypotenuse $c = 18$ and angle $A = 63^\circ$ , find the length $a$ . Round your answer to 2 decimal places.

Solution Key:

Identify the Given Information:
- Angle $A = 63^\circ$
- Hypotenuse $c = 18$
- Side $a$ is opposite to angle $A$ .
Apply the Sine Ratio: $\sin(A) = \frac{\text{Opposite}}{\text{Hypotenuse}} \implies \sin(63^\circ) = \frac{a}{18}$
Solve for $a$ : $a = 18 \cdot \sin(63^\circ) \approx 18 \cdot 0.8910065 \approx 16.038 \implies a \approx 16.04$

Show your steps and calculations here.

Practice Problem 2

Express $\frac{7\pi}{12}$ rad in degrees. (Enter exact values.)

Solution Key:

\text{Degrees} = \left(\frac{7\pi}{12}\right) \times \frac{180^\circ}{\pi} = \frac{7}{12} \times 180^\circ = 7 \times 15^\circ = 105^\circ

Show your steps and calculations here.

Practice Problem 3

Express $23^\circ$ in radians. (Enter exact values.)

Solution Key:

\text{Radians} = 23^\circ \times \frac{\pi}{180^\circ} = \frac{23\pi}{180} \text{ rad}

Show your steps and calculations here.

Practice Problem 4

A wooden ramp is to be built with one end on the ground and the other end at the top of a short staircase. If the top of the staircase is $6\text{ ft}$ from the ground and the angle between the ground and the ramp is to be $26^\circ$ , how long does the ramp need to be? Round your answer to four decimal places.

Solution Key:

Apply the Sine Ratio: $\sin(26^\circ) = \frac{6}{x} \implies x = \frac{6}{\sin(26^\circ)}$
Calculate: $x \approx \frac{6}{0.438371} \approx 13.68703 \implies x \approx 13.6870\text{ ft}$

Show your steps and calculations here.

Regroup 1

Review unit circle quadrant signs and reference angles.
Review radian-degree conversion factors.

Skill Block 2

Worked Example A: Algebraic Limit Evaluation

Evaluate the following limit:

\lim_{x \to -2} \frac{-x^2 + 5x + 14}{-5x^2 - 2x + 16}

Worked Solution:

Check for Indeterminate Form: Substituting $x = -2$ gives $\frac{0}{0}$ .
Factor Numerator and Denominator:
- Numerator: $-x^2 + 5x + 14 = -(x - 7)(x + 2)$
- Denominator: $-5x^2 - 2x + 16 = (x + 2)(-5x + 8)$
Cancel Common Factor $(x + 2)$ : $\lim_{x \to -2} \frac{-(x - 7)}{-5x + 8}$
Evaluate by Substitution: $\frac{-(-2 - 7)}{-5(-2) + 8} = \frac{9}{18} = \frac{1}{2}$

Worked Example B: Piecewise Continuity and Differentiability

Find the values of $a$ and $b$ that make the following piecewise defined function both continuous and differentiable everywhere:

f(x) = \begin{cases} 3x + 4, & x < -3 \ 2x^2 + ax + b, & x \ge -3 \end{cases}

Worked Solution:

Apply Continuity Condition at $x = -3$ : $3(-3) + 4 = 2(-3)^2 + a(-3) + b \implies -5 = 18 - 3a + b \implies -23 = -3a + b \quad \text{--- (Eq. 1)}$
Apply Differentiability Condition at $x = -3$ : $f'(x) = \begin{cases} 3, & x < -3 \ 4x + a, & x > -3 \end{cases}$ $3 = 4(-3) + a \implies 3 = -12 + a \implies a = 15$
Solve for $b$ : $-23 = -3(15) + b \implies -23 = -45 + b \implies b = 22$

Worked Example C: Norman Window Optimization

A Norman window is constructed by adjoining a semicircle to the top of a rectangular window. What is the maximum possible area, in square meters, of a Norman window with a perimeter of $26$ meters?

Worked Solution:

Set Up Perimeter and Area Equations: Let the rectangle have width $y$ and height $x$ . The semicircle has radius $r = y/2$ .
- Perimeter: $P = 2x + y + \pi(y/2) = 26 \implies 2x = 26 - y(1 + \pi/2)$
- Area: $A = xy + \frac{1}{2}\pi(y/2)^2 = xy + \frac{\pi y^2}{8}$
Substitute $x$ to express $A$ in terms of $y$ : $A = y\left(13 - \frac{y}{2} - \frac{\pi y}{4}\right) + \frac{\pi y^2}{8} = 13y - \frac{y^2}{2} - \frac{\pi y^2}{8} = 13y - y^2\left(\frac{4 + \pi}{8}\right)$
Maximize Area by Taking the Derivative: $\frac{dA}{dy} = 13 - y\left(\frac{4 + \pi}{4}\right) = 0 \implies y = \frac{52}{4 + \pi}$
Calculate Maximum Area: $A = 13\left(\frac{52}{4 + \pi}\right) - \left(\frac{52}{4 + \pi}\right)^2\left(\frac{4 + \pi}{8}\right) = \frac{338}{4 + \pi} \text{ m}^2$

Practice Problem 5

Evaluate:

\lim_{x \to -5} \frac{\sqrt{-x-2} - \sqrt{3}}{-x-5}

Solution Key:

Multiply by the Conjugate: $\lim_{x \to -5} \frac{(\sqrt{-x-2} - \sqrt{3})(\sqrt{-x-2} + \sqrt{3})}{(-x-5)(\sqrt{-x-2} + \sqrt{3})} = \lim_{x \to -5} \frac{-x - 5}{(-x - 5)(\sqrt{-x-2} + \sqrt{3})}$
Cancel Common Factor $(-x - 5)$ and Evaluate: $\lim_{x \to -5} \frac{1}{\sqrt{-x-2} + \sqrt{3}} = \frac{1}{\sqrt{3} + \sqrt{3}} = \frac{1}{2\sqrt{3}} = \frac{\sqrt{3}}{6}$

Show your steps and calculations here.

Practice Problem 6

Consider a function $f(x)$ with critical points at $-2$ , $-\frac{3}{2}$ , $-\frac{3}{5}$ , and $-\frac{1}{2}$ . Based on the sign table for $f'(x)$ below, on which of the intervals is $f(x)$ increasing?

Interval	$(-\infty, -2)$	$\left(-2, -\frac{3}{2}\right)$	$\left(-\frac{3}{2}, -\frac{3}{5}\right)$	$\left(-\frac{3}{5}, -\frac{1}{2}\right)$	$\left(-\frac{1}{2}, \infty\right)$
Sign of $f'(x)$	$+$	$-$	$+$	$+$	$+$

Solution Key:

Read the table and select the intervals with a positive ( $+$ ) derivative sign. All intervals except $\left(-2, -\frac{3}{2}\right)$ have $f'(x) > 0$ , indicating the function is increasing.

Show your steps and calculations here.

Practice Problem 7

The graph of the derivative $f'(x)$ is given below. On what interval(s) is the function $f(x)$ concave up?

Solution Key:

Identify the intervals where the graph of $f'(x)$ has a positive slope (is rising):

Rising from $x = -6$ to $x = -4$
Rising from $x = 0$ to $x = \infty$ Thus, the function $f(x)$ is concave up on $(-6, -4) \cup (0, \infty)$ .

Show your steps and calculations here.

Practice Problem 8

Consider the function $f(x) = 4x^5 - 7x^3 - 5x + 1$ . Find the $x$ -values of all inflection points.

Solution Key:

Find $f'(x)$ and $f''(x)$ : $f'(x) = 20x^4 - 21x^2 - 5 \implies f''(x) = 80x^3 - 42x$
Solve $f''(x) = 0$ : $2x(40x^2 - 21) = 0 \implies x = 0 \quad \text{or} \quad x^2 = \frac{21}{40} \implies x = \pm \sqrt{\frac{21}{40}} = \pm \frac{\sqrt{210}}{20}$ Since $f''(x)$ changes signs at each of these roots, they are all inflection points.

Show your steps and calculations here.

Practice Problem 9

Consider the function $f(x) = \frac{x^4}{4} + \frac{22x^3}{3} + \frac{153x^2}{2} + 324x + 3$ . Over what open interval(s) is the function increasing and concave up?

Solution Key:

Find Increasing Intervals ( $f'(x) > 0$ ): $f'(x) = x^3 + 22x^2 + 153x + 324 = (x + 4)(x + 9)^2 = 0 \implies \text{Increasing on } (-4, \infty)$
Find Concave Up Intervals ( $f''(x) > 0$ ): $f''(x) = 3x^2 + 44x + 153 = (3x + 17)(x + 9) = 0 \implies \text{Concave up on } (-\infty, -9) \cup \left(-\frac{17}{3}, \infty\right)$
Find the Intersection: The intersection of $(-4, \infty)$ and $(-\infty, -9) \cup \left(-\frac{17}{3}, \infty\right)$ is $(-4, \infty)$ .

Show your steps and calculations here.

Practice Problem 10

A cell phone plan charges $30 per month for unlimited calls and texts, and each gigabyte (GB) of data is charged at $10 per GB (prorated). If a user uses more than $15\text{ GB}$ of data, the cost is capped at the cost for $15\text{ GB}$ . Write a piecewise-defined function for the cost $C$ as a function of data $d$ in GB.

Solution Key:

For $0 \le d \le 15$ : $C(d) = 30 + 10d$
For $d > 15$ : Cost is capped at $C(15) = 30 + 10(15) = 180$ . $C(d) = \begin{cases} 30 + 10d, & 0 \le d \le 15 \\ 180, & d > 15 \end{cases}$

Show your steps and calculations here.

Regroup 2

Review the definition of derivative, concavity, and critical points.
Discuss standard optimization setup (objective function and constraint).

Skill Block 3

Worked Example A: Geometric Definite Integral (Semicircle)

Evaluate the definite integral using geometric formulas:

\int_{-7}^{-3} \left(\sqrt{-(x+5)^2 + 4}\right) dx

Worked Solution:

Identify the Curve: The integrand $y = \sqrt{4 - (x+5)^2}$ represents a semicircle centered at $(-5, 0)$ with radius $R = \sqrt{4} = 2$ , sitting above the $x$ -axis.
Check the Bounds: The bounds $[-7, -3]$ represent the full domain of the semicircle (from $-5 - 2$ to $-5 + 2$ ).
Calculate Semicircle Area: $\text{Area} = \frac{1}{2}\pi R^2 = \frac{1}{2}\pi (2)^2 = 2\pi$

Worked Example B: Net Signed Area

Find the net signed area between the graph of $f(x) = 4 - \frac{5x}{4}$ and the $x$ -axis over the interval $\left[\frac{6}{5}, \frac{46}{5}\right]$ .

Worked Solution:

Set Up the Definite Integral: $\text{Net Signed Area} = \int_{6/5}^{46/5} \left(4 - \frac{5x}{4} \right) dx$
Find the Antiderivative: $F(x) = 4x - \frac{5x^2}{8}$
Evaluate at the Bounds:
- Upper bound: $F(46/5) = 4(46/5) - \frac{5(46/5)^2}{8} = \frac{184}{5} - \frac{529}{10} = -\frac{161}{10}$
- Lower bound: $F(6/5) = 4(6/5) - \frac{5(6/5)^2}{8} = \frac{24}{5} - \frac{9}{10} = \frac{39}{10}$
Subtract: $\text{Net Signed Area} = -\frac{161}{10} - \frac{39}{10} = -\frac{200}{10} = -20$

Worked Example C: Net Change from a Graph

A function $F(x)$ has an initial value $F(0) = -1$ . The graph of its derivative $F'(x)$ is a line passing through $(0,0)$ with a constant slope of $-2$ . What is the value of $F(3)$ ?

Worked Solution:

Apply the Net Change Theorem: $F(3) = F(0) + \int_{0}^{3} F'(x) dx$
Calculate the Area under $F'(x)$ Geometrically: From $x = 0$ to $x = 3$ , the graph of $F'(x) = -2x$ forms a right triangle below the $x$ -axis.
- Base = 3
- Height = $F'(3) = -6 \implies$ vertical length is 6.
- Area (below $x$ -axis) = $-\frac{1}{2} \times 3 \times 6 = -9$ .
Solve for $F(3)$ : $F(3) = -1 + (-9) = -10$

Practice Problem 11

Evaluate the definite integral using geometric formulas:

\int_{1}^{7} (3 - |-4 + x|) dx

Solution Key:

Find Triangle Dimensions:
- Base: from $x = 1$ to $x = 7 \implies b = 6$ .
- Height: peak is at $x = 4 \implies h = f(4) = 3 - 0 = 3$ .
Calculate Area: $\text{Area} = \frac{1}{2} \times 6 \times 3 = 9$

Show your steps and calculations here.

Practice Problem 12

The piecewise function $f(x)$ is graphed below. Use geometric formulas to find the total area between the graph of the function and the $x$ -axis.

Solution Key:

Calculate Area of Triangle 1 (above axis):
- Base $b_1 = 4$ , Height $h_1 = 1 \implies A_1 = \frac{1}{2} \times 4 \times 1 = 2$
Calculate Area of Triangle 2 (below axis):
- Base $b_2 = 4$ , Height $h_2 = 3 \implies A_2 = \frac{1}{2} \times 4 \times 3 = 6$
Sum the Areas: $\text{Total Area} = A_1 + A_2 = 2 + 6 = 8$

Show your steps and calculations here.

Practice Problem 13

Water leaks out of a tank at a rate of $r(t) = 18 - \frac{t^2}{2}$ for $t \ge 0$ , measured in gallons per minute. Initially the tank has $65$ gallons of water in the tank. How much water is left in the tank after $3$ minutes?

Solution Key:

Set Up the Equation: $\text{Water Left} = 65 - \int_{0}^{3} \left(18 - \frac{t^2}{2}\right) dt$
Integrate: $\int_{0}^{3} \left(18 - \frac{t^2}{2}\right) dt = \left[18t - \frac{t^3}{6}\right]_0^3 = 18(3) - \frac{27}{6} = 54 - 4.5 = 49.5\text{ gallons lost}$
Subtract: $\text{Remaining Water} = 65 - 49.5 = 15.5\text{ gallons}$

Show your steps and calculations here.

Regroup 3

Discuss the difference between net signed area and total area.
Review the Net Change Theorem.

Synthesis Wrap-up

Re-emphasize key calculus prerequisites.
Discuss how limits, derivatives, optimization, and integration are generalized to multi-dimensional surfaces in Calculus 3.