CELE Differential Calculus — Study Notes

Super Tutor's review for GELE (as “Mathematics”) uses these same chapters, so they live here in the CELE section.

Differential Calculus - Study Notes

Differential calculus is the mathematical foundation for analyzing rates of change, optimizing structural designs, and solving motion problems essential to civil engineering practice. This study module covers limits, continuity, derivatives, extrema, and related rates—topics that recur annually on the MSTE (Mathematics, Science and Technology) section of the PRC Licensure Examination. Mastery of these concepts enables engineers to solve real-world problems such as minimizing material cost in beam design, finding critical buckling loads, and analyzing traffic flow dynamics. This material is pitched at the professional review level, emphasizing board-style worked examples in SI units and practical applications aligned with NSCP 2015, AISC 360, and ACI 318 standards.

1. Limits and Continuity — Foundation of the Derivative

A limit describes the value that a function approaches as the independent variable moves toward a specified point. Formally, $\lim_{x\to a}f(x) = L$ means that for any small positive number $\epsilon$ , there exists a small positive number $\delta$ such that whenever $0 < |x - a| < \delta$ , we have $|f(x) - L| < \epsilon$ .

Continuity at a point: A function $f(x)$ is continuous at $x = a$ if and only if three conditions hold:

$f(a)$ is defined
$\lim_{x\to a}f(x)$ exists
$\lim_{x\to a}f(x) = f(a)$

Indeterminate forms arise when direct substitution yields $\frac{0}{0}$ , $\frac{\infty}{\infty}$ , $0 \cdot \infty$ , $\infty - \infty$ , $0^0$ , $1^{\infty}$ , or $\infty^0$ . For these cases, L'Hôpital's Rule provides a systematic solution: if $\lim_{x\to a}\frac{f(x)}{g(x)}$ yields an indeterminate form and both $f'(x)$ and $g'(x)$ exist near $a$ , then

\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}

provisioning the latter limit exists. You may apply L'Hôpital's Rule repeatedly if necessary.

Limit laws (for limits that exist):

Sum rule: $\lim[f(x) + g(x)] = \lim f(x) + \lim g(x)$
Product rule: $\lim[f(x) \cdot g(x)] = [\lim f(x)] \cdot [\lim g(x)]$
Quotient rule: $\lim\frac{f(x)}{g(x)} = \frac{\lim f(x)}{\lim g(x)}$ (provided $\lim g(x) \neq 0$ )
Power rule: $\lim[f(x)]^n = [\lim f(x)]^n$

Important standard limits:

\lim_{x\to 0}\frac{\sin x}{x} = 1, \quad \lim_{x\to 0}\frac{1 - \cos x}{x^2} = \frac{1}{2}, \quad \lim_{x\to \infty}\left(1 + \frac{1}{x}\right)^x = e

These limits appear frequently in engineering applications—for example, when analyzing small-angle approximations in structural dynamics or wave propagation.

Examples

Evaluate $\lim_{x\to 2}\frac{x^2 - 4}{x - 2}$ .

Solution

Direct substitution gives $\frac{0}{0}$ , an indeterminate form. Factor the numerator:

\lim_{x\to 2}\frac{x^2 - 4}{x - 2} = \lim_{x\to 2}\frac{(x-2)(x+2)}{x-2} = \lim_{x\to 2}(x+2) = 4

Alternatively, apply L'Hôpital's Rule:

\lim_{x\to 2}\frac{x^2 - 4}{x - 2} = \lim_{x\to 2}\frac{2x}{1} = 4

Use L'Hôpital's Rule to evaluate $\lim_{x\to 0}\frac{e^x - 1 - x}{x^2}$ .

Solution

Direct substitution: $\frac{e^0 - 1 - 0}{0^2} = \frac{0}{0}$ , indeterminate. Apply L'Hôpital's:

\lim_{x\to 0}\frac{e^x - 1 - x}{x^2} = \lim_{x\to 0}\frac{e^x - 1}{2x}

This is still $\frac{0}{0}$ , so apply L'Hôpital's again:

\lim_{x\to 0}\frac{e^x - 1}{2x} = \lim_{x\to 0}\frac{e^x}{2} = \frac{1}{2}

Note: Each application of L'Hôpital's Rule requires differentiation of the numerator and denominator separately.

Determine where $f(x) = \frac{x^2 - 1}{x - 1}$ is continuous.

Solution

The function has a removable discontinuity at $x = 1$ because the denominator vanishes there. For $x \neq 1$ :

f(x) = \frac{(x-1)(x+1)}{x-1} = x + 1

Thus $\lim_{x\to 1}f(x) = 2$ , but $f(1)$ is undefined. If we redefine $f(1) = 2$ , the function becomes continuous everywhere. For all other points, the function is continuous by the limit laws.

Key points

A limit is the value a function approaches, not necessarily the value it attains at a point
Continuity requires the limit to exist and equal the function value; check all three conditions
Indeterminate forms must be resolved algebraically or via L'Hôpital's Rule before evaluating
L'Hôpital's Rule applies only to indeterminate forms and requires differentiation of numerator and denominator separately
Standard limits like $\sin x / x \to 1$ as $x \to 0$ are essential for engineering calculations

2. The Derivative — Definition and Geometric Interpretation

The derivative of $f(x)$ at point $x$ measures the instantaneous rate of change of $f$ with respect to $x$ . Geometrically, it is the slope of the tangent line to the curve $y = f(x)$ at that point.

Definition (limit form):

f'(x) = \lim_{h\to 0}\frac{f(x+h) - f(x)}{h}

Alternatively, at a specific point $x = a$ :

f'(a) = \lim_{x\to a}\frac{f(x) - f(a)}{x - a}

Notation: The derivative of $y = f(x)$ is denoted as $f'(x)$, $\frac{dy}{dx}$ , $\frac{df}{dx}$ , or $y'$.

Equation of the tangent line at point $(x_0, f(x_0))$ :

y - f(x_0) = f'(x_0)(x - x_0)

Differentiability: A function is differentiable at $x = a$ if $f'(a)$ exists. Differentiability implies continuity, but continuity does not guarantee differentiability (e.g., $y = |x|$ is continuous at $x = 0$ but not differentiable there due to a sharp corner).

Physical interpretation in civil engineering: If $y$ represents displacement and $x$ represents time, then $\frac{dy}{dx}$ is velocity. If stress $\sigma$ varies with strain $\epsilon$ , then $\frac{d\sigma}{d\epsilon}$ is the elastic modulus. Derivatives quantify how quickly structural responses change under loading—critical for analyzing beam deflections, column buckling, and dynamic behavior.

Examples

Using the limit definition, find the derivative of $f(x) = x^2$ at $x = 3$ .

Solution

Apply the definition:

f'(3) = \lim_{h\to 0}\frac{f(3+h) - f(3)}{h} = \lim_{h\to 0}\frac{(3+h)^2 - 9}{h}

Expand $(3+h)^2 = 9 + 6h + h^2$ :

f'(3) = \lim_{h\to 0}\frac{9 + 6h + h^2 - 9}{h} = \lim_{h\to 0}\frac{6h + h^2}{h} = \lim_{h\to 0}(6 + h) = 6

Therefore, $f'(3) = 6$ .

Find the equation of the tangent line to $y = x^2 - 2x$ at the point where $x = 1$ .

Solution

First, find the point on the curve: $y(1) = 1 - 2 = -1$ , so the point is $(1, -1)$.
Next, find the derivative: $y' = 2x - 2$ , so $y'(1) = 2(1) - 2 = 0$ .
The tangent line has slope $m = 0$ and passes through $(1, -1)$:

y - (-1) = 0(x - 1) \Rightarrow y = -1

The tangent line is horizontal, which makes sense because $x = 1$ is the vertex of the parabola $y = (x-1)^2 - 1$ .

Analyze the differentiability of $f(x) = |x|$ at $x = 0$ .

Solution

Check the left and right derivatives:

Left derivative: $\lim_{h\to 0^-}\frac{f(0+h) - f(0)}{h} = \lim_{h\to 0^-}\frac{|h| - 0}{h} = \lim_{h\to 0^-}\frac{-h}{h} = -1$
Right derivative: $\lim_{h\to 0^+}\frac{f(0+h) - f(0)}{h} = \lim_{h\to 0^+}\frac{|h| - 0}{h} = \lim_{h\to 0^+}\frac{h}{h} = 1$

Since the left and right derivatives differ ( $-1 \neq 1$ ), the derivative does not exist at $x = 0$ . The function $f(x) = |x|$ is continuous but not differentiable at $x = 0$ due to the sharp corner (cusp). This illustrates that a function can be continuous everywhere but differentiable only where it is smooth.

Key points

The derivative is the limit of the difference quotient; it measures instantaneous rate of change
Geometrically, $f'(x_0)$ is the slope of the tangent line at $(x_0, f(x_0))$
Differentiability implies continuity, but not vice versa; check for sharp corners or discontinuities
The tangent line equation is $y - y_0 = f'(x_0)(x - x_0)$
In engineering: derivatives represent rates of change (velocity, strain-stress relationships, reaction rates)

3. Derivative Rules — Power, Product, Quotient, and Chain Rules

Rather than use the limit definition repeatedly, we apply systematic rules to find derivatives efficiently.

Power Rule: For any real constant $n$ ,

\frac{d}{dx}x^n = nx^{n-1}

This applies even when $n$ is negative or fractional (e.g., $\frac{d}{dx}x^{-2} = -2x^{-3}$ , $\frac{d}{dx}\sqrt{x} = \frac{d}{dx}x^{1/2} = \frac{1}{2}x^{-1/2}$ ).

Constant Multiple Rule:

\frac{d}{dx}[cf(x)] = c\frac{df}{dx}

Sum and Difference Rule:

\frac{d}{dx}[f(x) \pm g(x)] = \frac{df}{dx} \pm \frac{dg}{dx}

Product Rule: For the product of two functions $u(x)$ and $v(x)$,

(uv)' = u'v + uv'

Mnemonic: "first times derivative of second, plus second times derivative of first."

Quotient Rule: For the quotient $\frac{u}{v}$ where $v \neq 0$ ,

\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}

Mnemonic: "low $d$ -high minus high $d$ -low, over low low" (denominator-derivative high minus high-derivative low, divided by low squared).

Chain Rule: If $y = f(u)$ and $u = g(x)$ , then

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = f'(g(x)) \cdot g'(x)

Alternatively, if $y = f(g(x))$ , then $y' = f'(g(x)) \cdot g'(x)$ . The chain rule is crucial for composite functions and appears constantly in engineering calculations.

Derivatives of trigonometric functions:

\frac{d}{dx}\sin x = \cos x, \quad \frac{d}{dx}\cos x = -\sin x, \quad \frac{d}{dx}\tan x = \sec^2 x

\frac{d}{dx}\cot x = -\csc^2 x, \quad \frac{d}{dx}\sec x = \sec x \tan x, \quad \frac{d}{dx}\csc x = -\csc x \cot x

Derivatives of exponential and logarithmic functions:

\frac{d}{dx}e^x = e^x, \quad \frac{d}{dx}a^x = a^x \ln a \quad (a > 0, a \neq 1)

\frac{d}{dx}\ln x = \frac{1}{x}, \quad \frac{d}{dx}\log_a x = \frac{1}{x \ln a}

Inverse trigonometric function derivatives:

\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}, \quad \frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1-x^2}}, \quad \frac{d}{dx}\arctan x = \frac{1}{1+x^2}

These rules form the toolkit for all derivative calculations. Mastering them enables rapid solution of board-exam problems.

Examples

Differentiate $y = (3x^2 + 2x - 1)(x^3 - 4)$ .

Solution

Use the product rule with $u = 3x^2 + 2x - 1$ and $v = x^3 - 4$ .
First, find the derivatives:

$u' = 6x + 2$
$v' = 3x^2$

Apply the product rule:

y' = u'v + uv' = (6x + 2)(x^3 - 4) + (3x^2 + 2x - 1)(3x^2)

Expand the first term: $(6x + 2)(x^3 - 4) = 6x^4 - 24x + 2x^3 - 8$
Expand the second term: $(3x^2 + 2x - 1)(3x^2) = 9x^4 + 6x^3 - 3x^2$
Combine:

y' = 6x^4 - 24x + 2x^3 - 8 + 9x^4 + 6x^3 - 3x^2 = 15x^4 + 8x^3 - 3x^2 - 24x - 8

Differentiate $y = \frac{2x^2 + 3}{x - 1}$ .

Solution

Use the quotient rule with $u = 2x^2 + 3$ and $v = x - 1$ .
Find the derivatives:

$u' = 4x$
$v' = 1$

Apply the quotient rule:

y' = \frac{u'v - uv'}{v^2} = \frac{4x(x-1) - (2x^2 + 3)(1)}{(x-1)^2} = \frac{4x^2 - 4x - 2x^2 - 3}{(x-1)^2} = \frac{2x^2 - 4x - 3}{(x-1)^2}

Differentiate $y = (5x^3 - 2)^7$ .

Solution

Use the chain rule. Let $u = 5x^3 - 2$ , so $y = u^7$ .
Apply the chain rule:

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 7u^6 \cdot 15x^2 = 7(5x^3 - 2)^6 \cdot 15x^2 = 105x^2(5x^3 - 2)^6

Differentiate $y = e^{\sin x}$ .

Solution

Use the chain rule. Let $u = \sin x$ , so $y = e^u$ .
Apply the chain rule:

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = e^u \cdot \cos x = e^{\sin x} \cos x

Differentiate $y = \ln(x^2 + 3x)$ .

Solution

Use the chain rule. Let $u = x^2 + 3x$ , so $y = \ln u$ .
Apply the chain rule:

\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} = \frac{1}{x^2 + 3x} \cdot (2x + 3) = \frac{2x + 3}{x^2 + 3x}

Key points

Power rule: $\frac{d}{dx}x^n = nx^{n-1}$ applies to all real exponents, including fractions and negatives
Product rule and quotient rule have specific, easy-to-confuse forms; practice to avoid sign errors
Chain rule is essential for nested functions; apply from outside to inside
Memorize standard derivatives of trig, exponential, and log functions—they recur constantly
Combine rules systematically: often a single problem requires power, product, and chain rules together

4. Higher-Order Derivatives and Concavity

The second derivative $f''(x)$ (or $\frac{d^2y}{dx^2}$ ) is the derivative of the first derivative. It measures how quickly the rate of change is changing—i.e., the curvature or concavity of the graph.

Concavity interpretation:

Concave up (convex): $f''(x) > 0$. The graph curves upward; tangent lines lie below the curve. Geometrically, the function is "holding water."
Concave down: $f''(x) < 0$. The graph curves downward; tangent lines lie above the curve.
Inflection point: A point where the concavity changes, where $f''(x) = 0$ (and the sign of $f''$ changes).

Second Derivative Test for Extrema: At a critical point $x = c$ where $f'(c) = 0$ :

If $f''(c) > 0$, then $f$ has a local minimum at $c$ .
If $f''(c) < 0$, then $f$ has a local maximum at $c$ .
If $f''(c) = 0$ , the test is inconclusive; use the first derivative test instead.

Third and higher derivatives: Successive differentiation yields $f'''(x)$, $f^{(4)}(x)$ , etc. In engineering applications, the third derivative (often called "jerk" in motion analysis) represents the rate of change of acceleration and is important in vehicle dynamics and seismic analysis. Higher derivatives are rare in basic optimization but essential in Taylor series approximations.

Application in civil engineering: When analyzing beam deflection under load, the deflection function $y(x)$ has a physical meaning:

$y'(x)$ = slope of the beam (angle of rotation)
$y''(x)$ = curvature (proportional to bending moment $M$ ; the beam equation is $M = EI\frac{d^2y}{dx^2}$ where $E$ is elastic modulus and $I$ is second moment of area)
$y'''(x)$ = shear force (related to internal shear $V$ ; $V = EI\frac{d^3y}{dx^3}$ )
$y^{(4)}(x)$ = distributed load $q$ (the load equation is $q = EI\frac{d^4y}{dx^4}$ )

This direct relationship between derivatives and physical quantities makes differential calculus indispensable in structural mechanics.

Examples

For $f(x) = x^3 - 3x^2 + 2$ , find the second derivative and identify concavity regions.

Solution

Find the first derivative: $f'(x) = 3x^2 - 6x$ .
Find the second derivative: $f''(x) = 6x - 6 = 6(x - 1)$ .

Set $f''(x) = 0$ : $x = 1$ is the inflection point.

For $x < 1$ : $f''(x) < 0$ → concave down
For $x > 1$ : $f''(x) > 0$ → concave up

At $x = 1$ , the concavity changes from down to up, confirming the inflection point.

A cantilever beam of length $L = 4$ m has deflection $y(x) = -\frac{w_0}{24EI}x^2(x^2 - 4Lx + 6L^2)$ , where $w_0$ is the distributed load. Using beam theory, find the bending moment at $x = 2$ m. (Given: $E = 200$ GPa, $I = 8 \times 10^{-5}$ m⁴, $w_0 = 10$ kN/m.)

Solution

The bending moment is $M(x) = EI\frac{d^2y}{dx^2}$ . Rather than expand the full deflection formula, recognize that the deflection equation encodes the load distribution. Differentiate twice:

First derivative: $\frac{dy}{dx} = -\frac{w_0}{24EI}\frac{d}{dx}[x^2(x^2 - 4Lx + 6L^2)]$

Using the product rule: $\frac{dy}{dx} = -\frac{w_0}{24EI}[2x(x^2 - 4Lx + 6L^2) + x^2(2x - 4L)]$

Simplifying: $\frac{dy}{dx} = -\frac{w_0}{24EI}[2x^3 - 8Lx^2 + 12L^2x + 2x^3 - 4Lx^2] = -\frac{w_0}{24EI}[4x^3 - 12Lx^2 + 12L^2x]$

Second derivative: $\frac{d^2y}{dx^2} = -\frac{w_0}{24EI}[12x^2 - 24Lx + 12L^2] = -\frac{w_0}{2EI}[x^2 - 2Lx + L^2] = -\frac{w_0}{2EI}(x - L)^2$

Therefore, $M(x) = EI \cdot \left(-\frac{w_0}{2EI}(x - L)^2\right) = -\frac{w_0}{2}(x - L)^2$

At $x = 2$ m with $L = 4$ m:

M(2) = -\frac{10 \times 10^3}{2}(2 - 4)^2 = -\frac{10000}{2} \times 4 = -20,000\text{ N}\cdot\text{m} = -20\text{ kN}\cdot\text{m}

The negative sign indicates the beam is in hogging (negative moment) at $x = 2$ m from the cantilever support.

Key points

Second derivative $f''(x)$ measures concavity; $f'' > 0$ is concave up, $f'' < 0$ is concave down
Inflection points occur where $f''(x) = 0$ and the sign of $f''$ changes
Second derivative test: at critical point, $f'' > 0$ → minimum, $f'' < 0$ → maximum
In beam theory, $M = EI\frac{d^2y}{dx^2}$ relates bending moment to deflection curvature
Higher derivatives encode physical information: slope, curvature, shear, and load in structural systems

5. Maxima, Minima, and Optimization — Critical Points and Extrema

Finding extrema (maximum and minimum values of a function) is among the most important applications of calculus in engineering. Optimal structural designs minimize cost or weight while meeting safety constraints—a fundamental principle in NSCP 2015 and modern engineering practice.

Critical points: A critical point occurs at $x = c$ if:

$f'(c) = 0$ (stationary point), or
$f'(c)$ does not exist (e.g., sharp corner, cusp).

Every local extremum in the interior of a domain must occur at a critical point.

First Derivative Test (always reliable):

Find all critical points by solving $f'(x) = 0$ (or locating where $f'$ is undefined).
Examine the sign of $f'(x)$ on intervals around each critical point:
- If $f'$ changes from positive to negative as $x$ increases through $c$ , then $c$ is a local maximum.
- If $f'$ changes from negative to positive as $x$ increases through $c$ , then $c$ is a local minimum.
- If $f'$ does not change sign, then $c$ is neither a maximum nor a minimum (saddle point or inflection).

Second Derivative Test (convenient, sometimes inconclusive):
At a critical point $c$ where $f'(c) = 0$ :

If $f''(c) > 0$, then $c$ is a local minimum.
If $f''(c) < 0$, then $c$ is a local maximum.
If $f''(c) = 0$ , the test is inconclusive; use the first derivative test.

Global extrema on a closed interval $[a,b]$:
Evaluate $f$ at:

All critical points in $(a, b)$
The endpoints $f(a)$ and $f(b)$

The largest value is the global maximum, and the smallest is the global minimum. This is essential for constrained optimization in engineering design.

Optimization strategy:

Identify the quantity to optimize and express it as a function of a single variable using constraint equations.
Find the domain (physical bounds on the variable).
Differentiate and set equal to zero to find critical points.
Check endpoints if the domain is closed.
Verify using the second derivative test or by examining sign changes in the first derivative.
Interpret the result physically—does it make sense given the context?

Example context in civil engineering: Minimizing the cost of concrete in a rectangular foundation of fixed volume, or maximizing the load capacity of a beam of given cross-section—these are prototypical optimization problems solved via calculus.

Examples

Find the absolute maximum and minimum of $f(x) = x^3 - 3x^2 - 9x + 5$ on the closed interval $[-2, 4]$.

Solution

Step 1: Find critical points.

f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1)

Setting $f'(x) = 0$ : $x = 3$ or $x = -1$ . Both lie in $[-2, 4]$.

Step 2: Evaluate $f$ at critical points and endpoints.

f(-2) = (-2)^3 - 3(-2)^2 - 9(-2) + 5 = -8 - 12 + 18 + 5 = 3

f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10

f(3) = (3)^3 - 3(3)^2 - 9(3) + 5 = 27 - 27 - 27 + 5 = -22

f(4) = (4)^3 - 3(4)^2 - 9(4) + 5 = 64 - 48 - 36 + 5 = -15

Step 3: Compare values.

Absolute maximum: $f(-1) = 10$
Absolute minimum: $f(3) = -22$

A rectangular garden is to be enclosed by 80 m of fencing, with one side bounded by an existing wall (no fence needed). Maximize the enclosed area.

Solution

Step 1: Set up variables. Let $x$ = width (perpendicular to wall), $y$ = length (parallel to wall).

Step 2: Constraint (fence equation): The three free sides require $2x + y = 80$ , so $y = 80 - 2x$ .

Step 3: Express area in one variable: $A(x) = xy = x(80 - 2x) = 80x - 2x^2$ .

Step 4: Domain: Since $x > 0$ and $y > 0$ , we need $80 - 2x > 0$ , so $0 < x < 40$ .

Step 5: Find critical points.

A'(x) = 80 - 4x = 0 \Rightarrow x = 20

Step 6: Verify using second derivative test.

A''(x) = -4 < 0 \Rightarrow \text{local maximum at } x = 20

Step 7: Find corresponding dimensions and area.

x = 20\text{ m}, \quad y = 80 - 2(20) = 40\text{ m}

A_{\max} = 20 \times 40 = 800\text{ m}^2

Interpretation: To maximize the garden area with 80 m of fencing and one side against a wall, make the width perpendicular to the wall 20 m and the length parallel to the wall 40 m.

Design a cylindrical storage tank (open top) with volume $V = 1000$ m³. Minimize the surface area of the metal used (base plus lateral surface). Assume radius $r$ and height $h$ .

Solution

Step 1: Volume constraint: $V = \pi r^2 h = 1000$ , so $h = \frac{1000}{\pi r^2}$ .

Step 2: Surface area (base + lateral, no top): $S = \pi r^2 + 2\pi r h$ (base area + lateral area).

Step 3: Substitute $h$ :

S(r) = \pi r^2 + 2\pi r \cdot \frac{1000}{\pi r^2} = \pi r^2 + \frac{2000}{r}

Step 4: Find critical points.

S'(r) = 2\pi r - \frac{2000}{r^2} = 0

2\pi r = \frac{2000}{r^2} \Rightarrow 2\pi r^3 = 2000 \Rightarrow r^3 = \frac{1000}{\pi} \Rightarrow r = \sqrt[3]{\frac{1000}{\pi}} \approx 6.83\text{ m}

Step 5: Verify with second derivative test.

S''(r) = 2\pi + \frac{4000}{r^3} > 0 \quad \text{(always positive)} \Rightarrow \text{minimum}

Step 6: Find height.

h = \frac{1000}{\pi r^2} = \frac{1000}{\pi (6.83)^2} \approx 6.83\text{ m}

Note that the optimal design has $h = r$ —the height equals the radius. This is a classic result: a cylinder of fixed volume has minimum lateral + base surface area when height equals diameter.

Minimum surface area: $S = \pi(6.83)^2 + \frac{2000}{6.83} \approx 146.4 + 292.8 = 439.2\text{ m}^2$ .

Key points

Critical points occur where $f'(x) = 0$ or $f'(x)$ does not exist; all interior extrema occur at critical points
First derivative test (sign change) is always valid; second derivative test is convenient but sometimes inconclusive
Always check endpoints and domain boundaries—the global maximum/minimum may not occur at a critical point
Optimization strategy: express the quantity in one variable via constraint, differentiate, solve $f' = 0$ , verify
Physical interpretation: verify that the critical point makes physical sense in the engineering context

6. Related Rates — Time-Dependent Optimization

Related rates problems involve quantities that change with respect to time and are connected by constraint equations. The goal is to find the rate of change of one quantity given the rates of others.

Strategy:

Draw and label a diagram with all quantities and given rates.
Identify the variables and constants. Variables usually change with time; constants do not.
Write the constraint equation relating all variables (geometric formula, volume, etc.).
Differentiate both sides with respect to time $t$ , treating all variables as functions of $t$ (use chain rule systematically).
Substitute the known values (rates, positions) to solve for the unknown rate.
Verify units and sign—positive rates mean increasing, negative means decreasing.

Key principle: Differentiate before substituting numerical values. Substitution after differentiation prevents errors from missing rate terms.

Common constraint equations in related rates:

Pythagorean theorem: $x^2 + y^2 = c^2$ (distances)
Circle area: $A = \pi r^2$ ; circumference: $C = 2\pi r$
Sphere volume: $V = \frac{4}{3}\pi r^3$
Cylinder volume: $V = \pi r^2 h$
Cone volume: $V = \frac{1}{3}\pi r^2 h$
Similar triangles: $\frac{x_1}{y_1} = \frac{x_2}{y_2}$ (proportional dimensions)

Application in civil engineering: In traffic flow analysis, if the number of vehicles $N(t)$ passing a point increases at a known rate $\frac{dN}{dt}$ , and the total distance traveled $D$ is related to $N$ by $D = L \cdot N$ (where $L$ is the route length), then the rate of distance increase is $\frac{dD}{dt} = L \frac{dN}{dt}$ . Monitoring such rates helps engineers optimize traffic signal timing and estimate congestion dynamics.

Examples

A spherical balloon is inflated at a constant rate of $\frac{dV}{dt} = 50$ cm³/s. At what rate is the radius increasing when $r = 10$ cm?

Solution

Step 1: Identify variables and rates.

$V(t)$ = volume (cm³)
$r(t)$ = radius (cm)
Given: $\frac{dV}{dt} = 50$ cm³/s
Find: $\frac{dr}{dt}$ when $r = 10$ cm

Step 2: Write the volume constraint.

V = \frac{4}{3}\pi r^3

Step 3: Differentiate with respect to $t$ .

\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}

Step 4: Solve for $\frac{dr}{dt}$ .

\frac{dr}{dt} = \frac{1}{4\pi r^2}\frac{dV}{dt}

Step 5: Substitute $r = 10$ cm and $\frac{dV}{dt} = 50$ cm³/s.

\frac{dr}{dt} = \frac{1}{4\pi (10)^2} \times 50 = \frac{50}{400\pi} = \frac{1}{8\pi} \approx 0.0398\text{ cm/s}

Interpretation: When the radius is 10 cm, it is increasing at approximately 0.04 cm/s (or about 0.4 mm/s). As the balloon inflates, the radius grows more slowly as $r$ increases (because the surface area $4\pi r^2$ increases with $r$ ).

A 5 m ladder leans against a vertical wall. The base slides away from the wall at $1$ m/s. When the base is 3 m from the wall, how fast is the top of the ladder sliding down the wall?

Solution

Step 1: Draw a diagram. Let $x$ = distance from base to wall, $y$ = height of ladder on wall. The ladder has fixed length $\ell = 5$ m.

Step 2: Identify rates.

Given: $\frac{dx}{dt} = 1$ m/s (base moving away)
Find: $\frac{dy}{dt}$ when $x = 3$ m (rate at top moving down)

Step 3: Write the Pythagorean constraint.

x^2 + y^2 = 25

Step 4: Differentiate with respect to $t$ .

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0

x\frac{dx}{dt} + y\frac{dy}{dt} = 0

Step 5: Solve for $\frac{dy}{dt}$ .

\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}

Step 6: Find $y$ when $x = 3$ m.

3^2 + y^2 = 25 \Rightarrow y^2 = 16 \Rightarrow y = 4\text{ m}

Step 7: Substitute.

\frac{dy}{dt} = -\frac{3}{4} \times 1 = -0.75\text{ m/s}

Interpretation: The top of the ladder is sliding down at 0.75 m/s (negative sign indicates downward motion). Note that as the base gets farther from the wall, $x$ increases and $y$ decreases, so $\frac{dy}{dt}$ is negative. The rate increases in magnitude as $x$ approaches 5 m (the top approaches the ground faster near the end).

Water flows into a conical tank (vertex down) at $2$ m³/min. The tank has a fixed height of $4$ m and base radius of $2$ m. How fast is the water level rising when the water depth is $1$ m?

Solution

Step 1: Identify variables. Let $h(t)$ = depth of water, $r(t)$ = radius of water surface, $V(t)$ = volume of water.

Given: $\frac{dV}{dt} = 2$ m³/min
Find: $\frac{dh}{dt}$ when $h = 1$ m

Step 2: Set up the cone. The tank has fixed height 4 m and base radius 2 m. By similar triangles, the water radius at depth $h$ is:

\frac{r}{h} = \frac{2}{4} = \frac{1}{2} \Rightarrow r = \frac{h}{2}

Step 3: Write the volume equation (cone).

V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{12}

Step 4: Differentiate with respect to $t$ .

\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt}

Step 5: Solve for $\frac{dh}{dt}$ .

\frac{dh}{dt} = \frac{4}{\pi h^2}\frac{dV}{dt}

Step 6: Substitute $h = 1$ m and $\frac{dV}{dt} = 2$ m³/min.

\frac{dh}{dt} = \frac{4}{\pi (1)^2} \times 2 = \frac{8}{\pi} \approx 2.546\text{ m/min}

Interpretation: When the water is 1 m deep, the water level rises at approximately 2.55 m/min. As the tank fills, $h$ increases, and $h^2$ increases, so $\frac{dh}{dt}$ decreases—the water level rises more slowly as the tank fills (the cone widens).

Key points

Draw a clear diagram and label all quantities, including rates of change
Write the constraint equation before differentiating
Differentiate with respect to time $t$ , applying the chain rule to all variables
Substitute numerical values after differentiating—never before
Check that units match and the sign makes physical sense (positive = increasing, negative = decreasing)

7. Tangent Lines and Linear Approximation

The tangent line to a curve $y = f(x)$ at a point $(x_0, f(x_0))$ is a straight line that touches the curve at that point and has the same slope as the curve there.

Equation of the tangent line:

y - f(x_0) = f'(x_0)(x - x_0)

Alternatively, in slope-intercept form:

y = f(x_0) + f'(x_0)(x - x_0)

The tangent line is the best linear approximation to $f$ near $x_0$ . For values of $x$ close to $x_0$ :

f(x) \approx f(x_0) + f'(x_0)(x - x_0)

This is called linear approximation or differential approximation. The error $E = f(x) - [f(x_0) + f'(x_0)(x - x_0)]$ is small when $x$ is near $x_0$ and becomes negligible for very small intervals (e.g., small measurement uncertainties).

Differential notation: The differential $dy$ represents the change in the tangent line over an interval $dx$ :

dy = f'(x)\,dx

This is useful for estimating small changes in $y$ due to small changes in $x$ . For example, if a structural parameter changes slightly, engineers can estimate the resulting change in response without full recalculation.

Normal line: The line perpendicular to the tangent at $(x_0, f(x_0))$ has slope $-\frac{1}{f'(x_0)}$ (negative reciprocal) and equation:

y - f(x_0) = -\frac{1}{f'(x_0)}(x - x_0)

Application in engineering: In structural analysis, if a load parameter varies slightly from its design value, the engineer can use linear approximation to quickly estimate the change in deflection or stress without re-solving the full problem. This enables sensitivity analysis and uncertainty quantification.

Examples

Find the equation of the tangent line to $y = x^2 - 3x$ at the point $(2, -2)$.

Solution

Step 1: Verify the point is on the curve.

y(2) = (2)^2 - 3(2) = 4 - 6 = -2 \quad \checkmark

Step 2: Find the derivative.

y' = 2x - 3

Step 3: Evaluate the slope at $x = 2$ .

y'(2) = 2(2) - 3 = 1

Step 4: Write the tangent line.

y - (-2) = 1(x - 2)

y + 2 = x - 2

y = x - 4

Use linear approximation to estimate $\sqrt{25.1}$ based on $f(x) = \sqrt{x}$ at $x_0 = 25$ .

Solution

Step 1: Identify the function and point.

f(x) = \sqrt{x}, \quad x_0 = 25, \quad f(25) = 5

Step 2: Find the derivative.

f'(x) = \frac{1}{2\sqrt{x}}, \quad f'(25) = \frac{1}{2\sqrt{25}} = \frac{1}{10}

Step 3: Use linear approximation with $x = 25.1$ , so $\Delta x = 0.1$ .

f(x) \approx f(x_0) + f'(x_0)\Delta x = 5 + \frac{1}{10}(0.1) = 5 + 0.01 = 5.01

Step 4: Check with a calculator.

\sqrt{25.1} \approx 5.00999 \approx 5.01

The linear approximation is accurate to two decimal places. The small error arises because the second derivative (curvature) slightly bends the curve.

A cylindrical pipe has radius $r = 10$ cm. Due to corrosion, the radius decreases by $dr = 0.1$ cm. Estimate the decrease in cross-sectional area using a differential.

Solution

Step 1: Set up the area function.

A(r) = \pi r^2

Step 2: Find the differential.

dA = \frac{dA}{dr}\,dr = 2\pi r \, dr

Step 3: Substitute $r = 10$ cm and $dr = -0.1$ cm (decrease).

dA = 2\pi (10)(-0.1) = -2\pi \approx -6.28\text{ cm}^2

Step 4: Interpret.
The cross-sectional area decreases by approximately 6.28 cm² when the radius decreases by 0.1 cm.

Verification (exact): $A(10) = 100\pi$ , $A(9.9) = (9.9)^2\pi = 98.01\pi$ . The exact decrease is $100\pi - 98.01\pi = 1.99\pi \approx 6.25$ cm². The differential approximation gives 6.28 cm², a very close estimate (error < 0.5%).

Key points

The tangent line is the best linear approximation near a point; use it to estimate function values close to $x_0$
Tangent line equation: $y = f(x_0) + f'(x_0)(x - x_0)$
Differential $dy = f'(x)\,dx$ estimates the change in $y$ for a small change $dx$
Linear approximation error is small for small $|x - x_0|$ and shrinks as $x \to x_0$
Normal line: perpendicular to tangent; slope is $-\frac{1}{f'(x_0)}$

8. Curve Sketching and Radius of Curvature

Complete curve sketching synthesizes derivatives, critical points, concavity, and asymptotes to create an accurate graph. This skill is valuable for visualizing structural responses, force distributions, and optimization problems.

Curve-sketching procedure:

Domain: Identify where the function is defined.
Intercepts: Find $x$ -intercepts (set $y = 0$ ) and $y$ -intercepts (set $x = 0$ ).
Asymptotes:
- Vertical: Denominator = 0 (check left and right limits)
- Horizontal: $\lim_{x \to \pm\infty} f(x)$
- Oblique (slant): Polynomial long division when degree of numerator exceeds denominator by 1
First derivative: Find critical points ( $f'(x) = 0$ ), determine intervals of increase/decrease.
Second derivative: Find inflection points ( $f''(x) = 0$ ), determine concavity.
Sketch: Plot intercepts, critical points, inflection points, and asymptotes; connect with smooth curves respecting concavity and monotonicity.

Radius of curvature: The curvature $\kappa$ measures how sharply a curve bends. The radius of curvature $R$ is the reciprocal:

\kappa = \frac{|f''(x)|}{[1 + (f'(x))^2]^{3/2}}, \quad R = \frac{1}{\kappa} = \frac{[1 + (f'(x))^2]^{3/2}}{|f''(x)|}

Geometrically, $R$ is the radius of the osculating circle (best-fitting circle) at a point on the curve.

Physical meaning:

Large $R$ → gentle curve (small curvature)
Small $R$ → sharp curve (high curvature)
At inflection points where $f''(x) = 0$ , the radius becomes infinite (curvature is zero).

Application in civil engineering: Road and railway alignment design use curvature formulas. For a highway curve, the minimum radius of curvature is constrained by speed and friction to prevent skidding: $R_{\min} = \frac{v^2}{g(e + f)}$ , where $v$ is speed, $e$ is banking angle, $f$ is friction coefficient, and $g$ is gravity. The radius of curvature of the designed centerline must exceed this minimum everywhere.

Examples

Sketch the curve $y = \frac{x}{x^2 - 1}$ .

Solution

Step 1: Domain. The denominator $x^2 - 1 = 0$ at $x = \pm 1$ , so domain is $(-\infty, -1) \cup (-1, 1) \cup (1, \infty)$ .

Step 2: Intercepts.

$y$ -intercept: $x = 0 \Rightarrow y = 0$ . Point $(0, 0)$.
$x$ -intercept: $y = 0 \Rightarrow x = 0$ . Same point.

Step 3: Asymptotes.

Vertical: $x = -1$ and $x = 1$ (denominator zero)
- As $x \to -1^-$ : $y \to -\infty$ (negative numerator, small positive denom)
- As $x \to -1^+$ : $y \to +\infty$ (negative numerator, small negative denom)
- As $x \to 1^-$ : $y \to +\infty$ (positive numerator, small negative denom)
- As $x \to 1^+$ : $y \to -\infty$ (positive numerator, small positive denom)
Horizontal: $\lim_{x \to \pm\infty}\frac{x}{x^2-1} = \lim_{x \to \pm\infty}\frac{1/x}{1 - 1/x^2} = 0$ . So $y = 0$ is a horizontal asymptote.

Step 4: First derivative.

y' = \frac{(x^2-1) - x(2x)}{(x^2-1)^2} = \frac{-x^2 - 1}{(x^2-1)^2} = \frac{-(x^2+1)}{(x^2-1)^2}

The numerator $-(x^2+1) < 0$ always, and the denominator $(x^2-1)^2 > 0$ always. Thus $y' < 0$ everywhere the function is defined: the function is decreasing on each connected interval.

Step 5: Second derivative (for concavity). Differentiate $y' = \frac{-(x^2+1)}{(x^2-1)^2}$ . Using the quotient rule:

y'' = \frac{-2x(x^2-1)^2 - (-(x^2+1)) \cdot 2(x^2-1) \cdot 2x}{(x^2-1)^4}

Simplify (factor $(x^2-1)$ from numerator):

y'' = \frac{-2x(x^2-1) + 4x(x^2+1)}{(x^2-1)^3} = \frac{-2x^3 + 2x + 4x^3 + 4x}{(x^2-1)^3} = \frac{2x^3 + 6x}{(x^2-1)^3} = \frac{2x(x^2+3)}{(x^2-1)^3}

Sign analysis:

$x^2 + 3 > 0$ always
Sign of $y''$ depends on $x$ and $(x^2-1)^3$ :
- $x < -1$: $x < 0$ , $(x^2-1)^3 < 0$ → $y'' > 0$ (concave up)
- $-1 < x < 0$: $x < 0$ , $(x^2-1)^3 > 0$ → $y'' < 0$ (concave down)
- $0 < x < 1$ : $x > 0$ , $(x^2-1)^3 < 0$ → $y'' < 0$ (concave down)
- $x > 1$ : $x > 0$ , $(x^2-1)^3 > 0$ → $y'' > 0$ (concave up)

Inflection point at $x = 0$ (where the concavity changes).

Step 6: Sketch.

Plot the point $(0, 0)$.
Draw vertical asymptotes at $x = -1$ and $x = 1$ .
Draw horizontal asymptote at $y = 0$ .
On $(-\infty, -1)$ : decreasing, concave up, approaches $y = 0$ as $x \to -\infty$ , approaches $-\infty$ as $x \to -1^-$ .
On $(-1, 1)$: passes through $(0, 0)$, decreasing, concave down for $x < 0$ , concave down for $0 < x < 1$ . Approaches $+\infty$ as $x \to -1^+$ and $+\infty$ as $x \to 1^-$ .
On $(1, \infty)$ : decreasing, concave up, approaches $-\infty$ as $x \to 1^+$ , approaches $0$ from below as $x \to \infty$ .

Find the radius of curvature of $y = x^2$ at $x = 0$ and at $x = 1$ .

Solution

Step 1: Find derivatives.

y' = 2x, \quad y'' = 2

Step 2: Apply the radius of curvature formula.

R = \frac{[1 + (y')^2]^{3/2}}{|y''|} = \frac{[1 + (2x)^2]^{3/2}}{2} = \frac{[1 + 4x^2]^{3/2}}{2}

At $x = 0$ :

R(0) = \frac{[1 + 0]^{3/2}}{2} = \frac{1}{2} = 0.5

The osculating circle at the origin has radius 0.5 (the parabola is sharply curved at the vertex).

At $x = 1$ :

R(1) = \frac{[1 + 4]^{3/2}}{2} = \frac{(5)^{3/2}}{2} = \frac{5\sqrt{5}}{2} \approx \frac{11.18}{2} \approx 5.59

At $(1, 1)$, the curve is gentler; the osculating circle has radius ≈ 5.59. This makes intuitive sense: as we move away from the vertex of the parabola, it becomes less sharply curved.

Key points

Follow a systematic curve-sketching procedure: domain, intercepts, asymptotes, derivatives, concavity, then sketch
Critical points (where $f'(x) = 0$ ) and inflection points (where $f''(x) = 0$ ) are key features
Radius of curvature $R = \frac{[1+(y')^2]^{3/2}}{|y''|}$ quantifies the tightness of a curve
High curvature (small $R$ ) indicates sharp bending; low curvature (large $R$ ) indicates gentle bending
In road design, minimum radius of curvature is determined by vehicle dynamics and friction

9. Partial Derivatives and Multivariable Optimization

Many engineering problems involve functions of multiple independent variables. For example, the deflection of a beam depends on load, span length, material properties, and geometry—all varying independently.

Partial derivative notation: The partial derivative of $f(x, y)$ with respect to $x$ , denoted $\frac{\partial f}{\partial x}$ or $f_x$ , is the derivative of $f$ with respect to $x$ alone, treating all other variables as constants.

\frac{\partial f}{\partial x} = \lim_{h \to 0}\frac{f(x+h, y) - f(x, y)}{h}

Similarly, $\frac{\partial f}{\partial y} = f_y$ is the derivative with respect to $y$ , treating $x$ as a constant.

Second-order partial derivatives:

$\frac{\partial^2 f}{\partial x^2} = f_{xx}$ : differentiate twice with respect to $x$
$\frac{\partial^2 f}{\partial y^2} = f_{yy}$ : differentiate twice with respect to $y$
$\frac{\partial^2 f}{\partial x \partial y} = f_{xy}$ : differentiate first with respect to $x$ , then $y$
$\frac{\partial^2 f}{\partial y \partial x} = f_{yx}$ : differentiate first with respect to $y$ , then $x$

Schwarz's theorem: If $f$ and its partial derivatives are continuous, then $f_{xy} = f_{yx}$ (mixed partials are equal).

Critical points for extrema: A point $(x_0, y_0)$ is a critical point if $f_x(x_0, y_0) = 0$ and $f_y(x_0, y_0) = 0$ simultaneously.

Second Derivative Test (2D Hessian): At a critical point $(x_0, y_0)$ , define

D = f_{xx}(x_0, y_0) \cdot f_{yy}(x_0, y_0) - [f_{xy}(x_0, y_0)]^2

Then:

If $D > 0$ and $f_{xx} > 0$ : local minimum
If $D > 0$ and $f_{xx} < 0$ : local maximum
If $D < 0$ : saddle point (neither maximum nor minimum)
If $D = 0$ : test is inconclusive

Application in civil engineering: In foundation design optimization, the settlement $S$ depends on soil cohesion $c$ and angle of internal friction $\phi$ . Minimizing cost $C(c, \phi)$ subject to settlement limits requires finding critical points via partial derivatives. This approach scales to problems with dozens of design variables.

Examples

Find the critical points and classify them for $f(x, y) = x^2 + y^2 - 4x + 6y + 5$ .

Solution

Step 1: Find first partial derivatives.

f_x = 2x - 4, \quad f_y = 2y + 6

Step 2: Solve $f_x = 0$ and $f_y = 0$ .

2x - 4 = 0 \Rightarrow x = 2

2y + 6 = 0 \Rightarrow y = -3

The critical point is $(2, -3)$.

Step 3: Find second partial derivatives.

f_{xx} = 2, \quad f_{yy} = 2, \quad f_{xy} = 0

Step 4: Apply the second derivative test.

D = f_{xx} f_{yy} - (f_{xy})^2 = (2)(2) - 0^2 = 4 > 0

f_{xx} = 2 > 0

Since $D > 0$ and $f_{xx} > 0$ , the point $(2, -3)$ is a local minimum.

Step 5: Find the minimum value.

f(2, -3) = 4 + 9 - 8 - 18 + 5 = -8

Interpretation: The paraboloid $f(x, y) = (x-2)^2 + (y+3)^2 - 8$ has its vertex (global minimum) at $(2, -3, -8)$.

A rectangular box is to be designed with volume $V = 100$ m³. The bottom and top cost $5/m², the four sides cost $3/m². Find the dimensions that minimize the total cost.

Solution

Step 1: Set up variables. Let $x$ = length, $y$ = width, $h$ = height.

Step 2: Volume constraint: $V = xyh = 100$ , so $h = \frac{100}{xy}$ .

Step 3: Cost function. Bottom and top: $2xy$ at $5/m² = $10xy$. Four sides: $2(xh + yh) = 2h(x+y)$ at $3/m² = $6h(x+y)$.

C = 10xy + 6h(x+y) = 10xy + 6\cdot\frac{100}{xy}\cdot(x+y) = 10xy + \frac{600(x+y)}{xy}

Simplify:

C(x,y) = 10xy + \frac{600}{y} + \frac{600}{x}

Step 4: Find critical points.

\frac{\partial C}{\partial x} = 10y - \frac{600}{x^2} = 0 \Rightarrow 10y = \frac{600}{x^2} \Rightarrow y = \frac{60}{x^2}

\frac{\partial C}{\partial y} = 10x - \frac{600}{y^2} = 0 \Rightarrow 10x = \frac{600}{y^2} \Rightarrow x = \frac{60}{y^2}

Step 5: Solve simultaneously. Substitute the first into the second:

x = \frac{60}{(60/x^2)^2} = \frac{60 \cdot x^4}{3600} = \frac{x^4}{60}

60x = x^4 \Rightarrow x^3 = 60 \Rightarrow x = \sqrt[3]{60} \approx 3.91\text{ m}

From $y = \frac{60}{x^2}$ :

y = \frac{60}{(\sqrt[3]{60})^2} = \frac{60}{60^{2/3}} = 60^{1-2/3} = 60^{1/3} \approx 3.91\text{ m}

From $h = \frac{100}{xy}$ :

h = \frac{100}{(\sqrt[3]{60})^2} = \frac{100}{60^{2/3}} = \frac{100}{\sqrt[3]{3600}} \approx 6.57\text{ m}

Step 6: Verify with second derivative test (or note that cost is convex for positive dimensions). The critical point is a minimum by physical reasoning (cost increases as dimensions approach 0 or ∞).

Optimal dimensions: $x \approx y \approx 3.91$ m, $h \approx 6.57$ m. Note that the optimal base is square; this is a common feature in many optimization problems due to symmetry.

Key points

Partial derivatives treat one variable as changing; hold other variables constant during differentiation
Find critical points by solving $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$ simultaneously
Second Derivative Test (Hessian): $D = f_{xx} f_{yy} - f_{xy}^2$ classifies critical points (local min, max, or saddle)
Schwarz's theorem: mixed partial derivatives are equal if the function is sufficiently smooth
Multivariable optimization: reduce constraints and solve systems of partial derivative equations

Diagrams

Differential Calculus Concept Map

This mind map organizes the entire differential calculus curriculum into six major branches: foundational concepts (limits and the derivative), computational rules, practical applications (optimization and related rates), curve sketching, advanced derivatives, and multivariable extension. The hierarchical structure helps reviewees see how topics interconnect and prioritize study time.

Optimization Problem-Solving Workflow

This flowchart provides a step-by-step decision tree for solving optimization problems. It emphasizes the key decision points (finding critical points, classifying extrema, comparing candidates) and the importance of verifying that the mathematical solution makes physical sense. This workflow is applicable to virtually all board-exam optimization questions.

CELE Engineering Mathematics — Differential CalculusStudy Notes

Differential Calculus - Study Notes

1. Limits and Continuity — Foundation of the Derivative

2. The Derivative — Definition and Geometric Interpretation

3. Derivative Rules — Power, Product, Quotient, and Chain Rules

4. Higher-Order Derivatives and Concavity

5. Maxima, Minima, and Optimization — Critical Points and Extrema

6. Related Rates — Time-Dependent Optimization

7. Tangent Lines and Linear Approximation

8. Curve Sketching and Radius of Curvature

9. Partial Derivatives and Multivariable Optimization

Diagrams

Study Differential Calculus in the Super Tutor app