Differential Curve Sketching, Extreme Values & Optimization
Comprehensive applications of derivatives: Fermat's interior extremum theorem, critical numbers, First and Second Derivative Tests, concavity and inflection points, universal 7-step analytical curve sketching algorithm, applied geometric and physical optimization problems, and the related rates framework.
Β§8.1 Critical Numbers, Fermat's Theorem on Stationary Points & Extrema
1. Local Extrema & Critical Numbers
2. Fermat's Interior Extremum Theorem
Proof for Local Maximum: Suppose $f$ has a local maximum at $c$. Then there exists $\delta > 0$ such that $f(x) \le f(c)$ for all $x \in (c - \delta, c + \delta)$.
- For $h \in (0, \delta)$ (approaching from right): $f(c + h) - f(c) \le 0 \implies \frac{f(c + h) - f(c)}{h} \le 0 \implies f'_+(c) \le 0$.
- For $h \in (-\delta, 0)$ (approaching from left): $f(c + h) - f(c) \le 0 \implies \frac{f(c + h) - f(c)}{h} \ge 0 \implies f'_-(c) \ge 0$.
Since $f$ is differentiable at $c$, $f'(c) = f'_+(c) = f'_-(c)$. The only real number satisfying both $f'(c) \le 0$ and $f'(c) \ge 0$ is $f'(c) = 0 \quad \blacksquare$
3. The Closed Interval Method for Absolute Extrema
To find the absolute maximum and minimum of a continuous function $f$ on a closed bounded interval $[a, b]$:
- Find all critical numbers $c_i \in (a, b)$.
- Evaluate $f(c_i)$ at every critical number.
- Evaluate $f(a)$ and $f(b)$ at the endpoints.
- The greatest evaluated value is the absolute maximum; the least is the absolute minimum.
Β§8.2 First & Second Derivative Tests, Concavity & Inflection Points
1. The First Derivative Test for Local Extrema
Let $c$ be a critical number of a continuous function $f$:
- If $f'(x)$ changes from positive to negative as $x$ increases through $c$, then $f(c)$ is a local maximum.
- If $f'(x)$ changes from negative to positive as $x$ increases through $c$, then $f(c)$ is a local minimum.
- If $f'(x)$ does not change sign across $c$ (e.g. $+ \to +$ or $- \to -$), then $f(c)$ is neither a local maximum nor a local minimum (e.g. $f(x) = x^3$ at $c = 0$).
2. Concavity & Inflection Points
Point of Inflection: A point $(c, f(c))$ on the curve where the function is continuous and the concavity changes (from upward to downward, or vice versa). A necessary condition for an inflection point is $f''(c) = 0$ or $f''(c)$ undefined.
3. The Second Derivative Test for Local Extrema
Suppose $f''(x)$ is continuous near a stationary point $c$ with $f'(c) = 0$:
- If $f''(c) > 0$, then the curve is concave upward at $c$, so $f(c)$ is a local minimum.
- If $f''(c) < 0$, then the curve is concave downward at $c$, so $f(c)$ is a local maximum.
- If $f''(c) = 0$, the test is inconclusive (e.g. $x^4$ has a minimum, $-x^4$ has a maximum, $x^3$ has an inflection point); one must revert to the First Derivative Test.
Β§8.3 Systematic 7-Step Protocol for Mathematical Curve Sketching
The Universal 7-Step Curve Sketching Protocol
To sketch the graph of an arbitrary function $y = f(x)$ with analytical precision without guesswork:
- Step 1: Domain & Symmetries β Identify $\operatorname{Dom}(f)$. Test for even symmetry $f(-x) = f(x)$, odd symmetry $f(-x) = -f(x)$, or periodicity $f(x+T) = f(x)$.
- Step 2: Intercepts β $y$-intercept at $(0, f(0))$. $x$-intercepts by solving $f(x) = 0$.
- Step 3: Asymptotes β Vertical asymptotes where $Q(x) = 0$ with $\lim |f(x)| = \infty$. Horizontal asymptotes $\lim_{x \to \pm\infty} f(x) = L$. Slant asymptotes $y = mx + b$ if $\lim [f(x) - (mx+b)] = 0$.
- Step 4: First Derivative $f'(x)$ β Critical numbers ($f'(x) = 0$ or undefined). Sign chart for $f'(x)$ to determine intervals of increase and decrease.
- Step 5: Local Extrema β Apply First or Second Derivative Test to classify each critical point.
- Step 6: Second Derivative $f''(x)$ β Determine $f''(x) = 0$ or undefined. Sign chart for $f''(x)$ to determine intervals of concavity upward/downward and exact inflection points.
- Step 7: Global Assembly & Plot β Plot intercepts, asymptotes, extrema, and inflection points; sketch the smooth curve following the concavity and monotonicity signatures.
Β§8.4 Applied Real-World Optimization & Related Rates Framework
1. Applied Mathematical Optimization Protocol
- Variable Identification & Sketch: Assign variables to all quantities. Draw a diagram.
- Objective Function: Write an explicit formula for the quantity $Q$ to be maximized or minimized (e.g. volume, area, cost, material).
- Constraint Equations: Relate secondary variables via geometric or physical constraints (e.g. surface area budget, perimeter, Pythagorean theorem) to express $Q = f(x)$ purely in terms of a single independent variable.
- Domain Specification: Establish the feasible physical domain $x \in [a, b]$ or $(a, b)$.
- Extremum Determination: Compute $f'(x) = 0$. Use the Closed Interval Method or Second Derivative Test to verify that the critical point is indeed the global optimum.
2. The Systematic Related Rates Protocol
When physical variables are related by an equation $F(x, y, z) = 0$ and change over time $t$:
- Identify the given rates (e.g. $\frac{dx}{dt}$) and the target rate (e.g. $\frac{dy}{dt}$) at a specific instant $t_0$.
- Formulate a geometric equation relating the static variables (never substitute numerical values that change with time before differentiating!).
- Differentiate implicitly with respect to time $t$ using the Chain Rule: $\frac{d}{dt}[y^2] = 2y \frac{dy}{dt}$.
- Substitute the instantaneous values and solve algebraically for the desired rate of change.
Step-by-Step Solved Examination Problems
Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.
For the polynomial function $f(x) = x^4 - 4x^3 + 10$: (a) Find all critical numbers. (b) Determine the intervals of increase and decrease. (c) Classify all local extrema. (d) Find all points of inflection and intervals of concavity.
Critical numbers occur at $x = 0$ and $x = 3$.
$f$ is strictly decreasing on $(-\infty, 3)$ and strictly increasing on $(3, \infty)$. Notice $f'$ does NOT change sign across $x = 0$.
At $x = 0$, $f'$ changes from negative to negative, so $(0, 10)$ is a stationary horizontal inflection point, NOT an extremum.
$f''(x) > 0$ on $(-\infty, 0) \cup (2, \infty)$ (Concave Up). $f''(x) < 0$ on $(0, 2)$ (Concave Down). Inflection points at $(0, 10)$ and $(2, -6)$.
\text{Local/Abs Min at } (3, -17); \quad \text{Inflection Points at } (0, 10) \text{ and } (2, -6); \quad \text{Concave Up on } (-\infty, 0) \cup (2, \infty)
A manufacturer wishes to design a closed cylindrical can of volume $V = 1000\pi\text{ cm}^3$ using the minimum possible surface area of metal. (a) Formulate the total surface area $A$ as a function of the radius $r$. (b) Determine the radius $r$ and height $h$ that minimize the surface area. (c) Prove that for any optimum cylinder, the height must equal the diameter ($h = 2r$).
Total surface area is two circular ends plus the cylindrical lateral mantle: $A(r) = 2\pi r^2 + 2\pi r h$.
This expresses the objective function purely in terms of radius $r$.
Setting $A'(r) = 0$ gives the unique positive critical radius.
Since $A''(r) > 0$, the critical radius guarantees a strict global minimum. Computing $h$: $h = \frac{V}{\pi r^2} = \frac{\pi r^2 h}{\pi r^2} \implies h = \frac{2000}{2r^2} = 2 \left(\frac{1000}{2r^2}\right) = 2r$. Thus $h = 2r = 10\sqrt[3]{4} ext{ cm}$.
r = 5\sqrt[3]{4} \approx 7.94\text{ cm}, \qquad h = 10\sqrt[3]{4} \approx 15.87\text{ cm}, \qquad h = 2r \text{ (Height equals Diameter)}
An inverted conical water tank has a height of $H = 6.0\text{ m}$ and a top radius of $R = 2.0\text{ m}$. Water leaks out of a hole in the bottom vertex at a constant rate of $k = 0.50\text{ m}^3/\text{min}$. (a) Find a formula relating the water volume $V$ directly to the instantaneous water depth $h$. (b) Derive the rate at which the water depth is falling $\frac{dh}{dt}$ as a function of depth $h$. (c) Compute $\frac{dh}{dt}$ when the water is $h = 3.0\text{ m}$ deep and when $h = 1.0\text{ m}$ deep, explaining why the water level drops with accelerating speed as it empties.
Because the cone has straight cross-sectional walls, the water radius $r$ scales strictly proportionally with depth $h$.
Substituting $r = h/3$ yields $V(h)$ purely in terms of depth.
Given that water is leaking out at $0.50 ext{ m}^3/ ext{min}$, we have $\frac{dV}{dt} = -0.50 = -1/2 ext{ m}^3/ ext{min}$.
As the tank empties, the cross-sectional surface area $\pi r^2 \propto h^2$ shrinks quadratically. To sustain the constant volumetric drain rate, the water depth recession rate $\left|\frac{dh}{dt}\right| \propto 1/h^2$ increases inversely with the square of depth, accelerating as $h o 0$.
\frac{dh}{dt} = -\frac{9}{2\pi h^2}; \quad \text{At } h = 3\text{ m}: -\frac{1}{2\pi} \approx -0.16\text{ m/min}; \quad \text{At } h = 1\text{ m}: -\frac{9}{2\pi} \approx -1.43\text{ m/min}