Module V — Trigonometry, Optimization, and Linearization

Course: MAT1CJ101 — Differential Calculus
Hours: 12 | Assessment: Internal only (not assessed in the external 70-mark exam)
Topics: Trigonometric functions, formal limits, derivatives of trig functions, power rule for rational exponents, optimization, linearization and differentials.

1. Trigonometric Functions

1.1 Review — Angles and Radian Measure

Angles are measured in radians in calculus. One radian is the angle subtended at the centre of a unit circle by an arc of length 1.

\[\pi \text{ radians} = 180°\]

To convert degrees to radians: multiply by $\pi/180$
To convert radians to degrees: multiply by $180/\pi$

Degrees	0°	30°	45°	60°	90°	180°	270°	360°
Radians	$0$	$\pi/6$	$\pi/4$	$\pi/3$	$\pi/2$	$\pi$	$3\pi/2$	$2\pi$

1.2 The Unit Circle and Standard Values

On the unit circle (radius 1), the point corresponding to angle $\theta$ is $(\cos\theta, \sin\theta)$.

$\theta$ (radians)	$\sin\theta$	$\cos\theta$	$\tan\theta$
$0$	$0$	$1$	$0$
$\pi/6$	$1/2$	$\sqrt3/2$	$1/\sqrt3$
$\pi/4$	$1/\sqrt2$	$1/\sqrt2$	$1$
$\pi/3$	$\sqrt3/2$	$1/2$	$\sqrt3$
$\pi/2$	$1$	$0$	undefined
$\pi$	$0$	$-1$	$0$

1.3 Key Identities

Pythagorean:

$\sin^2\theta + \cos^2\theta = 1$ $1 + \tan^2\theta = \sec^2\theta$ $1 + \cot^2\theta = \csc^2\theta$

Double angle:

$\sin 2\theta = 2\sin\theta\cos\theta$ $\cos 2\theta = \cos^2\theta - \sin^2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1$

Half-angle (from double angle):

$\sin^2\theta = \frac{1 - \cos 2\theta}{2}$ $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$

2. Tangent Values and Formal Definitions of Limits

2.1 The Formal $\varepsilon$–$\delta$ Definition

Definition: $\lim_{x\to c} f(x) = L$ means:

For every $\varepsilon > 0$, there exists $\delta > 0$ such that:

\[0 < |x - c| < \delta \implies |f(x) - L| < \varepsilon\]

Reading this: No matter how small a tolerance $\varepsilon$ we specify around $L$, we can find a “window” of width $2\delta$ around $c$ such that all $x$ within that window (except possibly $c$ itself) give $f(x)$ within the tolerance.

Why the formal definition: The informal “approaches” language is sufficient for computation, but the $\varepsilon$–$\delta$ definition is needed for proofs about continuity, differentiability, and uniform convergence.

Worked Example 1 ($\varepsilon$–$\delta$ proof): Prove $\lim_{x\to 2}(3x - 1) = 5$.

We need $

(3x-1) - 5

< \varepsilon$, i.e. $

3x - 6

< \varepsilon$, i.e. $

x - 2

< \varepsilon/3$.

Choose $\delta = \varepsilon/3$.

Then: if $0 <

x-2

< \delta = \varepsilon/3$, then $

3x-6

= 3

x-2

< 3\cdot(\varepsilon/3) = \varepsilon$.

Therefore $\lim_{x\to 2}(3x-1) = 5$. $\blacksquare$

2.2 Important Trigonometric Limits

Two fundamental limits (proved via the squeeze theorem and geometry):

\[\lim_{x\to 0} \frac{\sin x}{x} = 1\] \[\lim_{x\to 0} \frac{1 - \cos x}{x} = 0\]

Proof sketch of the first: For small $x > 0$, compare areas:

Area of triangle $OAP = \frac12\sin x$
Area of sector $OAP = x/2$
Area of triangle $OAT = \frac12\tan x$

So $\sin x < x < \tan x$, hence $\sin x / x$ is trapped between $\cos x$ and $1$. Both $\to 1$ as $x \to 0$.

Worked Example 2: Evaluate $\lim_{x\to 0} \dfrac{\sin 3x}{x}$.

\[\lim_{x\to0}\frac{\sin 3x}{x} = \lim_{x\to0} 3\cdot\frac{\sin 3x}{3x} = 3\cdot 1 = 3\]

Worked Example 3: Evaluate $\lim_{x\to 0} \dfrac{1 - \cos x}{x^2}$.

Multiply numerator and denominator by $(1 + \cos x)$:

\[\frac{(1-\cos x)(1+\cos x)}{x^2(1+\cos x)} = \frac{\sin^2 x}{x^2(1+\cos x)} = \left(\frac{\sin x}{x}\right)^2 \cdot \frac{1}{1+\cos x}\]

As $x \to 0$: $\left(\dfrac{\sin x}{x}\right)^2 \to 1$ and $\dfrac{1}{1+\cos x} \to \dfrac12$.

Limit $= 1 \cdot \dfrac12 = \dfrac12$.

3. Derivatives of Trigonometric Functions

3.1 Derivation of $\frac{d}{dx}[\sin x]$

\[\frac{d}{dx}[\sin x] = \lim_{h\to0}\frac{\sin(x+h) - \sin x}{h}\]

Using $\sin(x+h) = \sin x\cos h + \cos x\sin h$:

$= \lim_{h\to0}\frac{\sin x\cos h + \cos x\sin h - \sin x}{h} = \lim_{h\to0}\left[\sin x\cdot\frac{\cos h - 1}{h} + \cos x\cdot\frac{\sin h}{h}\right]$ $= \sin x \cdot 0 + \cos x \cdot 1 = \cos x \qquad \text{[using the two fundamental limits]}$

3.2 Table of Trig Derivatives

Function	Derivative
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\sec^2 x$
$\cot x$	$-\csc^2 x$
$\sec x$	$\sec x \tan x$
$\csc x$	$-\csc x \cot x$

Memory aid for the “co-“ functions: All derivatives of “co-“ functions (cos, cot, csc) have a negative sign.

3.3 Using the Chain Rule with Trig Functions

Worked Example 4: Differentiate $y = \sin(x^2)$.

Outer: $\sin(u)$, derivative $= \cos(u)$
Inner: $u = x^2$, derivative $= 2x$

\[y' = \cos(x^2) \cdot 2x = 2x\cos(x^2)\]

Worked Example 5: Differentiate $y = \sin^2(3x)$.

Write as $y = [\sin(3x)]^2$

Outer: $u^2$, derivative $= 2u$
Middle: $\sin(3x)$, derivative $= 3\cos(3x)$
Inner: $3x$, derivative $= 3$

\[y' = 2\sin(3x) \cdot \cos(3x) \cdot 3 = 6\sin(3x)\cos(3x) = 3\sin(6x)\]

Worked Example 6: Differentiate $y = \tan(\sqrt x)$.

\[y = \tan(x^{1/2})\] \[y' = \sec^2(x^{1/2}) \cdot \tfrac12 x^{-1/2} = \frac{\sec^2(\sqrt x)}{2\sqrt x}\]

3.4 Higher Derivatives of Sine and Cosine

$(\sin x)' = \cos x$ $(\sin x)'' = -\sin x$ $(\sin x)''' = -\cos x$ $(\sin x)^{(4)} = \sin x \quad \text{← period-4 cycle}$

This cyclic behaviour is important in differential equations and Taylor series.

4. Power Rule for Rational Exponents

In Module II, we stated that $(x^{m/n})’ = \frac{m}{n}x^{m/n - 1}$. Here is the derivation using implicit differentiation.

4.1 Derivation

Let $y = x^{m/n}$ where $m, n$ are integers and $n > 0$.

Then $y^n = x^m$.

Differentiate both sides with respect to $x$:

\[n \cdot y^{n-1} \cdot \frac{dy}{dx} = m \cdot x^{m-1}\] \[\frac{dy}{dx} = \frac{m \cdot x^{m-1}}{n \cdot y^{n-1}}\]

Substitute $y = x^{m/n}$:

\[\frac{dy}{dx} = \frac{m \cdot x^{m-1}}{n \cdot x^{(m/n)(n-1)}} = \frac{m}{n} \cdot x^{m - 1 - \frac{m(n-1)}{n}} = \frac{m}{n} \cdot x^{m - 1 - m + m/n} = \frac{m}{n} \cdot x^{m/n - 1}\]

Worked Example 7: Differentiate $y = x^{5/3}$.

\[\frac{dy}{dx} = \frac53 x^{5/3 - 1} = \frac53 x^{2/3}\]

Worked Example 8: Differentiate $y = x^{3/4} \cdot (x + 1)$.

Use product rule: let $f = x^{3/4}$, $g = x+1$.
$f’ = \frac34 x^{-1/4}$, $g’ = 1$.

\[y' = \frac34 x^{-1/4}(x+1) + x^{3/4} = \frac34\frac{x^{3/4}}{x} + x^{3/4} = x^{3/4}\left[\frac{3/4}{x} + 1\right] = \frac{x^{3/4}(x + 3/4)}{x} = \frac{(4x + 3)x^{-1/4}}{4}\]

5. Optimization Problems

Optimization is the process of finding the maximum or minimum value of a quantity in a real-world problem. The derivative is the key tool.

5.1 Strategy

Step 1. Read the problem. Identify what is to be maximized or minimized.
Step 2. Draw a diagram and label variables.
Step 3. Write the objective function (what you're optimizing) in terms of one variable
        (use any given constraints to eliminate extra variables).
Step 4. Find the domain of the objective function.
Step 5. Find critical points (set derivative = 0, check where undefined).
Step 6. Use the Closed Interval Method (if bounded domain) or 
        the First/Second Derivative Test to classify.
Step 7. State the answer with units and verify it makes physical sense.

5.2 Worked Examples

Worked Example 9: A farmer wants to fence a rectangular plot using 200 m of fencing along three sides (the fourth side is a wall). What dimensions maximize the area?

Step 2. Let width $= x$, length $= y$. Three sides: $2x + y = 200$.
Step 3. Area $A = xy$. Substitute $y = 200 - 2x$: $A(x) = x(200 - 2x) = 200x - 2x^2$.
Step 4. Domain: $x > 0$ and $200 - 2x > 0$, so $0 < x < 100$.
Step 5. $A’(x) = 200 - 4x = 0 \Rightarrow x = 50$.
Step 6. $A’‘(x) = -4 < 0 \Rightarrow$ local (and absolute) maximum at $x = 50$.
Step 7. Width $= 50$ m, Length $= 200 - 100 = 100$ m, Area $= 5000\text{ m}^2$.

Worked Example 10: Find the point on the parabola $y = x^2$ nearest to the point $(0, 1)$.

Distance from $(x, x^2)$ to $(0, 1)$: $D = \sqrt{x^2 + (x^2-1)^2}$.

Minimize $D^2 = x^2 + (x^2-1)^2$ (easier). Let $f = x^2 + x^4 - 2x^2 + 1 = x^4 - x^2 + 1$.

\[f' = 4x^3 - 2x = 2x(2x^2-1) = 0 \implies x = 0 \text{ or } x = \pm\frac{1}{\sqrt2}\]

$f(0) = 1$, $f(\pm 1/\sqrt2) = \frac14 - \frac12 + 1 = \frac34$.

Nearest points: $(\pm 1/\sqrt2,\ 1/2)$. Distance $= \sqrt{3/4} = \sqrt3/2$.

Worked Example 11: A cylindrical can must hold 1 litre ($1000\text{ cm}^3$) of soup. Find the dimensions that minimize surface area (material cost).

Let radius $= r$, height $= h$.

Volume: $\pi r^2 h = 1000$, so $h = 1000/(\pi r^2)$.
Surface area: $S = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r \cdot 1000/(\pi r^2) = 2\pi r^2 + 2000/r$.

\[\frac{dS}{dr} = 4\pi r - \frac{2000}{r^2} = 0 \implies 4\pi r^3 = 2000 \implies r^3 = \frac{500}{\pi} \implies r = \left(\frac{500}{\pi}\right)^{1/3} \approx 5.42 \text{ cm}\] \[h = \frac{1000}{\pi r^2} = 2r \approx 10.84 \text{ cm}\]

Optimal can: height = diameter. (A classic result.)

6. Linearization and Differentials

6.1 The Linear Approximation

When $f$ is differentiable at $a$, the tangent line at $(a, f(a))$ is:

\[L(x) = f(a) + f'(a)(x - a)\]

For $x$ near $a$, $f(x) \approx L(x)$. This is the linear approximation or linearization of $f$ at $a$.

Why it works: Near $a$, a smooth curve is nearly straight. The tangent line is the best straight-line approximation.

Worked Example 12: Approximate $\sqrt{4.02}$ using linearization.

Let $f(x) = \sqrt x$, $a = 4$.
$f(4) = 2$, $f’(x) = \dfrac{1}{2\sqrt x}$, $f’(4) = \dfrac14$.

$L(x) = 2 + \frac14(x - 4)$ $L(4.02) = 2 + \frac14(0.02) = 2 + 0.005 = 2.005$

Actual: $\sqrt{4.02} \approx 2.00499\ldots$ ✓ (error $< 0.00001$)

Worked Example 13: Approximate $(1.001)^{10}$ using linearization.

$f(x) = x^{10}$, $a = 1$.
$f(1) = 1$, $f’(x) = 10x^9$, $f’(1) = 10$.

$L(x) = 1 + 10(x-1)$ $L(1.001) = 1 + 10(0.001) = 1.01$

Actual: $(1.001)^{10} \approx 1.01004\ldots$ ✓

6.2 Differentials

The differential $dx$ is an independent variable representing a small change in $x$.

The differential of $y$ is:

\[dy = f'(x)\, dx\]

Geometric meaning: $dy$ is the change in the tangent line, while $\Delta y$ is the actual change in $f$. For small $dx$, $dy \approx \Delta y$.

Worked Example 14: If $y = x^3 - x$, find $dy$ and use it to approximate $\Delta y$ when $x = 2$ and $\Delta x = 0.01$.

$dy = (3x^2 - 1)\, dx$ $\text{At } x = 2: \quad dy = (12 - 1)(0.01) = 0.11$

Actual: $\Delta y = f(2.01) - f(2) = [(2.01)^3 - 2.01] - [8 - 2] = [8.120601 - 2.01] - 6 \approx 0.1106$

6.3 Error Estimation using Differentials

If $y = f(x)$ and $x$ has a measurement error of $\Delta x \approx dx$, then the propagated error in $y$ is approximately:

\[\Delta y \approx dy = f'(x)\, dx\]

Relative error: $\dfrac{dy}{y} = \dfrac{f’(x)\, dx}{f(x)}$.

Worked Example 15: A sphere’s radius is measured as $r = 10$ cm with possible error $0.1$ cm. Estimate the error in the volume.

$V = \frac43 \pi r^3$ $dV = 4\pi r^2\, dr = 4\pi(100)(0.1) = 40\pi \approx 125.66 \text{ cm}^3$

Relative error: $\dfrac{dV}{V} = \dfrac{40\pi}{\frac43\pi(1000)} = \dfrac{40}{4000/3} = \dfrac{40 \cdot 3}{4000} = 0.03 = 3\%$

7. Summary

Topic	Key Formula/Result
Radian conversion	radians $= $ degrees $\times \pi/180$
$\sin x / x$ limit	$\lim_{x\to0} \sin x/x = 1$
$\varepsilon$–$\delta$ definition	$\forall \varepsilon>0,\ \exists \delta>0: \lvert x-c\rvert<\delta \implies \lvert f(x)-L\rvert<\varepsilon$
Trig derivatives	$(\sin x)’ = \cos x$, $(\cos x)’ = -\sin x$, $(\tan x)’ = \sec^2 x$
Power rule (rational)	$(x^{m/n})’ = \frac{m}{n}x^{m/n-1}$
Linearization	$L(x) = f(a) + f’(a)(x-a)$
Differential	$dy = f’(x)\,dx$
Optimization	Find critical points; classify using derivative tests
Optimal cylinder	$h = 2r$ minimizes surface area for fixed volume

8. Practice Problems

(Internal assessment style — all question types)

Convert to radians: $135°, 210°, 315°$. Convert to degrees: $7\pi/6, 5\pi/4, 2\pi/3$.
Prove $\lim_{x\to0} \sin x / x = 1$ using the Squeeze Theorem (with geometric argument).
Evaluate: (a) $\lim_{x\to0} \sin(5x)/(3x)$, (b) $\lim_{x\to0} (\tan 2x)/x$, (c) $\lim_{x\to0} x\cdot\cot x$.
Prove the $\varepsilon$–$\delta$ definition: show $\lim_{x\to3} (2x+1) = 7$.
Differentiate using trig differentiation rules: (a) $y = x\sin x$, (b) $y = \cos(x^2)$, (c) $y = \tan(3x+1)$, (d) $y = \sin^2 x + \cos^2 x$.
Verify that $\dfrac{d}{dx}[\sin^2 x + \cos^2 x] = 0$. What does this confirm?
Derive the formula for $\dfrac{d}{dx}[\tan x]$ using the quotient rule and sin/cos derivatives.
Find all local maxima and minima of $f(x) = \sin x + \cos x$ on $[0, 2\pi]$.
A ladder 10 m long leans against a vertical wall. The bottom slides away at 1 m/s. How fast is the top sliding down when the bottom is 6 m from the wall? (This is a related rates problem — uses the chain rule.)
Use linearization to approximate: (a) $\sin(31°)$, (b) $\sqrt[3]{26}$.
A box with a square base and no top must have a volume of $32\text{ m}^3$. Find the dimensions that minimize the surface area.
A metal cylinder with volume $500\text{ cm}^3$ must be constructed. What radius and height minimize the total surface area (including both ends)?