Applications of derivatives (related rates, MVT, optimization, L'Hôpital's rule) Knowledge-Based Study

Study notes

1. Understanding Related Rates

   In related rates problems, we study two or more continuously varying quantities that are linked through an underlying geometric or physical equation. When these quantities change over time, their rates of change are related by differentiating the governing equation with respect to time $t$ using the Chain Rule.

   General Solution Strategy

   1. Identify Given and Desired Quantities: Assign variables to all quantities changing with time and express known rates and target rates as derivatives with respect to $t$.
   2. Establish a Relation: Write an equation connecting the variables based on geometric formulas (such as Pythagorean theorem, volume, or similar triangles) or physical laws.
   3. Differentiate with Respect to Time: Apply implicit differentiation with respect to $t$, remembering to attach factors such as $\frac{dx}{dt}$ or $\frac{dy}{dt}$ to each differentiated variable.
   4. Substitute and Solve: Substitute the known values and instantaneous rates only after differentiating, then isolate the unknown rate.

   Illustrative Example

   A spherical balloon is inflated such that its volume increases at a constant rate of $100\text{ cm}^3/\text{s}$. How fast is the radius increasing when the diameter is $50\text{ cm}$?

   The volume of a sphere of radius $r$ is given by:

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

   Differentiating both sides with respect to $t$:

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

   When the diameter is $50\text{ cm}$, the radius is $r = 25\text{ cm}$. Given $\frac{dV}{dt} = 100\text{ cm}^3/\text{s}$:

   $$100 = 4\pi (25)^2 \frac{dr}{dt} = 2500\pi \frac{dr}{dt}$$

   $$\frac{dr}{dt} = \frac{100}{2500\pi} = \frac{1}{25\pi}\text{ cm/s}$$

   Rolle's Theorem

   Let $f$ be a function that satisfies three hypotheses:

   • $f$ is continuous on the closed interval $[a, b]$,
   • $f$ is differentiable on the open interval $(a, b)$,
   • $f(a) = f(b)$.

   Then there exists at least one number $c \in (a, b)$ such that $f'(c) = 0$.

   The Mean Value Theorem (MVT)

   The Mean Value Theorem generalizes Rolle's Theorem by removing the requirement that $f(a) = f(b)$.

   Theorem: If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists a number $c \in (a, b)$ such that:

   $$f'(c) = \frac{f(b) - f(a)}{b - a}$$

   Geometrically, this guarantees that there is at least one point on the curve where the instantaneous rate of change (tangent slope) equals the average rate of change (secant slope) across the interval.

   Illustrative Example

   Consider $f(x) = x^3 - x$ on $[0, 2]$. Because $f$ is a polynomial, it is continuous and differentiable everywhere. The average rate of change is:

   $$\frac{f(2) - f(0)}{2 - 0} = \frac{(8 - 2) - 0}{2} = 3$$

   Setting $f'(c) = 3$:

   $$3c^2 - 1 = 3 \implies 3c^2 = 4 \implies c = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$$

   Since $c = \frac{2}{\sqrt{3}} \approx 1.155 \in (0, 2)$, the theorem is confirmed.

   Mathematical Optimization

   Optimization involves finding the absolute maximum or minimum value of an objective function subject to practical constraints. According to the Extreme Value Theorem, any continuous function on a closed, bounded interval achieves both an absolute maximum and an absolute minimum.

   Systematic Optimization Procedure

   1. Define Variables and Objective: Identify the quantity to maximize or minimize and write an equation representing it.
   2. Formulate Constraints: Identify equations that relate the independent variables and express the objective function in terms of a single variable.
   3. Determine the Feasible Domain: Establish the physical or geometric boundaries of the single variable.
   4. Locate Critical Points: Find where the derivative of the objective function is zero or undefined within the feasible domain.
   5. Classify Extrema: Use the First Derivative Test, Second Derivative Test, or Closed Interval Method (evaluating the objective function at critical points and endpoints) to confirm the global optimum.

   Illustrative Example

   A farmer has $120\text{ m}$ of fencing to construct a rectangular enclosure divided into two equal pens by a fence parallel to one of the sides. Find the dimensions that maximize the total enclosed area.

   Let $x$ be the length of the dividing fence and the two parallel exterior sides (three sections of length $x$), and let $y$ be the length of the remaining two exterior sides.

   The perimeter constraint is:

   $$3x + 2y = 120 \implies y = 60 - \frac{3}{2}x$$

   The objective function is total area $A(x) = x \cdot y$:

   $$A(x) = x\left(60 - \frac{3}{2}x\right) = 60x - \frac{3}{2}x^2$$

   Taking the derivative and setting it to zero:

   $$A'(x) = 60 - 3x = 0 \implies x = 20$$

   Because $A''(x) = -3 < 0$, the function is concave down everywhere, ensuring that $x = 20\text{ m}$ gives a global maximum. Substituting into the constraint gives $y = 60 - \frac{3}{2}(20) = 30\text{ m}$. The maximum enclosed area is $20 \times 30 = 600\text{ m}^2$.

   Indeterminate Quotients

   When evaluating $\lim_{x \to a} \frac{f(x)}{g(x)}$, direct substitution may yield the indeterminate forms $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$. In such cases, the limit cannot be determined by substitution alone.

   Statement of L'Hôpital's Rule

   Theorem: Suppose $f$ and $g$ are differentiable on an open interval containing $a$ (except possibly at $a$), and $g'(x) \neq 0$ on this interval (except possibly at $a$). If:

   $$\lim_{x \to a} f(x) = 0 \quad \text{and} \quad \lim_{x \to a} g(x) = 0$$

   or

   $$\lim_{x \to a} f(x) = \pm\infty \quad \text{and} \quad \lim_{x \to a} g(x) = \pm\infty$$

   then:

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

   provided the limit on the right-hand side exists or equals $\pm\infty$. Here $x \to a$ may represent a two-sided limit, one-sided limit, or a limit at infinity.

   Handling Other Indeterminate Forms

   • Products ($0 \cdot \infty$): Rewrite as a quotient: $f(x) \cdot g(x) = \frac{f(x)}{1/g(x)}$ to obtain $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
   • Differences ($\infty - \infty$): Combine fractions via a common denominator or factor out dominant terms.
   • Powers ($0^0$, $\infty^0$, $1^\infty$): Let $y = [f(x)]^{g(x)}$, compute $\ln y = g(x) \ln(f(x))$, evaluate $L = \lim \ln y$, and recover $\lim y = e^L$.

   Illustrative Example

   Evaluate $\lim_{x \to 0} \frac{e^{2x} - 1 - 2x}{x^2}$.

   Direct substitution yields $\frac{1 - 1 - 0}{0} = \frac{0}{0}$. Applying L'Hôpital's Rule by differentiating numerator and denominator:

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

   Substituting $x = 0$ gives $\frac{2 - 2}{0} = \frac{0}{0}$ again. Applying the rule a second time:

   $$\lim_{x \to 0} \frac{4e^{2x}}{2} = \frac{4(1)}{2} = 2$$