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: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Under the Conceptual Framework section, what are the three key steps/elements required for evaluating trigonometric functions of any angle using the unit circle?
(b) Look at Worked Example A. Why is $\sin(3\pi/4)$ positive, and what is its reference angle?
(c) Read the linked Engineering Application under the Engineering Applications section. Why are right-triangle trigonometric ratios crucial in bridge truss force decomposition?

Solution Key:

(a) According to the Conceptual Framework section, unit circle evaluation is done by:
1. Finding which quadrant the angle belongs to.
2. Finding the reference angle $\theta_R$ .
3. Applying coordinate signs based on the quadrant (using the CAST rule).
(b) In Worked Example A, the angle $\theta = \frac{3\pi}{4}$ is in Quadrant II (where sine represents positive vertical coordinates). The reference angle is $\theta_R = \pi - \frac{3\pi}{4} = \frac{\pi}{4}$ . Thus, $\sin(\frac{3\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ .
(c) As outlined in the Engineering Applications section under "Bridge Truss Force Decomposition", right-triangle trigonometry allows civil and mechanical engineers to resolve forces acting along angled structural components into orthogonal (vertical and horizontal) force vectors for static analysis.

Show your steps and calculations here.

Practice Problem 2

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 3

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 4

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 5

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 6: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Under the Conceptual Framework section, how do we ensure a piecewise function is continuous and differentiable at its boundary?
(b) Look at Worked Example B. Explain why we set the derivatives of the two pieces equal to each other at $x = -3$ to find the value of $a$ .
(c) Trace Worked Example C. How is the perimeter equation used to express the window area in terms of a single variable $y$ before maximizing?

Solution Key:

(a) As explained in the Conceptual Framework section, a piecewise function is continuous at a boundary point $x=c$ if the left limit, right limit, and function value are all equal ( $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$ ). It is differentiable if the derivative limits match at the boundary ( $\lim_{x \to c^-} f'(x) = \lim_{x \to c^+} f'(x)$ ).
(b) In Worked Example B, the differentiability condition at the boundary $x = -3$ is applied. The derivative of the left piece is $3$ , and the derivative of the right piece is $4x + a$ . Setting them equal at $x = -3$ gives $3 = 4(-3) + a$ , which immediately isolates $a = 15$ .
(c) Tracing Worked Example C, the window area depends on both $x$ (height) and $y$ (width). Using the perimeter constraint $2x + y + \pi(y/2) = 26$ , we isolate $2x = 26 - y(1 + \pi/2)$ . Substituting this $x$ into the area equation $A = xy + \frac{\pi y^2}{8}$ yields a single-variable quadratic function in terms of $y$ , allowing us to find the maximum using single-variable calculus.

Show your steps and calculations here.

Practice Problem 7

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 8

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 9

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 10

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 11

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 12

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 13: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Under the Conceptual Framework section, what is the difference between net signed area and total area when evaluating definite integrals?
(b) According to the Key Equations & Structural Rules section, how is the accumulated quantity $F(b)$ computed from the initial value $F(a)$ and the rate of change $F'(t)$ ?
(c) Trace Worked Example A. Describe the geometric shape represented by the integrand $y = \sqrt{-(x+5)^2 + 4}$ , including its center and radius.

Solution Key:

(a) As outlined in the Conceptual Framework section, when evaluating an integral, net signed area subtracts regions below the $x$ -axis from regions above, whereas total area takes the absolute value of all regions so they are all added positively.
(b) The Net Change Theorem formula in the Key Equations & Structural Rules section states that the final value $F(b)$ equals the initial value $F(a)$ plus the net accumulated change, which is the definite integral of the rate function $F'(t)$ from $a$ to $b$ .
(c) Tracing the solution to Worked Example A, the equation of the curve is $y = \sqrt{4 - (x+5)^2}$ . Squaring both sides and rearranging yields $(x+5)^2 + y^2 = 2^2$ (with $y \ge 0$ ), which represents a semicircle sitting above the $x$ -axis, centered at $(-5,0)$ with a radius of $2$ .

Show your steps and calculations here.

Practice Problem 14

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 15

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 16

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.