Applications of Derivatives: Rates, Optimization, Tangents & Approximations

TAKEAWAYS:

• Rates of change: Derivatives describe how quantities (distance, cost, volume) change over time or other variables.
• Optimization: Use derivatives to find maximum profit, minimum cost, or optimal dimensions (e.g., maximizing area with fixed perimeter).
• Tangent lines: The derivative at a point gives the slope of the tangent line to a curve.
• Approximations: Linear approximations (differentials) estimate small changes in functions.
• Critical points: Solve $f^{\prime }(x)=0$ or $f^{\prime }(x)$ undefined to find potential maxima/minima.
• Second derivative test: $f^{\prime \prime }(x)>0$ → concave up (minimum); $f^{\prime \prime }(x)<0$ → concave down (maximum).

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### **1. Rates of Change (Related Rates)**
**What it is**:
Derivatives help find how fast one quantity changes with respect to another. For example, if a balloon’s radius increases, how fast does its volume change?

**Key formula**:
If \( y = f(x) \), then the rate of change of \( y \) with respect to \( x \) is \( \frac{dy}{dx} \).
For related rates, use the **chain rule**:
\[
\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}
\]

**Worked Example 1**:
*A spherical balloon is being inflated. If its radius increases at 2 cm/s, how fast is the volume increasing when the radius is 5 cm?*
**Solution**:
1. Volume of a sphere: \( V = \frac{4}{3}\pi r^3 \).
2. Differentiate with respect to time \( t \):
   \[
   \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}
   \]
3. Given \( \frac{dr}{dt} = 2 \) cm/s and \( r = 5 \) cm:
   \[
   \frac{dV}{dt} = 4\pi (5)^2 (2) = 200\pi \text{ cm}^3/\text{s}
   \]
**Answer**: The volume increases at \( 200\pi \) cm³/s.



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### **2. Optimization Problems (Maxima/Minima)**
**What it is**:
Find the maximum or minimum value of a function (e.g., profit, area, cost) using derivatives.

**Steps**:
1. Find the derivative \( f'(x) \).
2. Set \( f'(x) = 0 \) and solve for \( x \) (critical points).
3. Use the **second derivative test**:
   - If \( f''(x) > 0 \), it’s a **minimum**.
   - If \( f''(x) < 0 \), it’s a **maximum**.
4. Evaluate \( f(x) \) at critical points and endpoints.

**Worked Example 2**:
*A farmer has 100 meters of fencing to enclose a rectangular field. What dimensions maximize the area?*
**Solution**:
1. Let length = \( x \), width = \( y \). Perimeter constraint:
   \[
   2x + 2y = 100 \implies y = 50 - x
   \]
2. Area \( A = xy = x(50 - x) = 50x - x^2 \).
3. Find \( A'(x) = 50 - 2x \). Set \( A'(x) = 0 \):
   \[
   50 - 2x = 0 \implies x = 25
   \]
4. Second derivative: \( A''(x) = -2 < 0 \) → **maximum**.
5. Width \( y = 50 - 25 = 25 \) meters.
**Answer**: A square field (25 m × 25 m) maximizes the area.

```figure
{"type":"rectangle","length":"25 m","breadth":"25 m","caption":"Optimal dimensions for maximum area"}

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3. Tangent Lines and Normal Lines

What it is:

• Tangent line: A straight line that touches a curve at exactly one point (slope = derivative at that point).
• Normal line: Perpendicular to the tangent line (slope = $-\frac{1}{f^{\prime }(x)}$).

Worked Example 3: Find the equation of the tangent line to $y=x^2$ at $x=2$. Solution:

1. Find $f^{\prime }(x)=2x$. At $x=2$, slope $m=4$.
2. Point of tangency: $(2,f(2))=(2,4)$.
3. Equation of tangent line (point-slope form): $$y-4=4(x-2)⟹y=4x-4$$ Answer: $y=4x-4$.

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4. Linear Approximations (Differentials)

What it is: Estimate small changes in a function using its derivative. The approximation is: $$\Delta y\approx f^{\prime }(x)\Delta x$$ or $$f(x+\Delta x)\approx f(x)+f^{\prime }(x)\Delta x$$

Worked Example 4: Approximate $\sqrt{16.1}$ using $f(x)=\sqrt{x}$ near $x=16$. Solution:

1. $f(x)=\sqrt{x}$, $f^{\prime }(x)=\frac{1}{2\sqrt{x}}$.
2. At $x=16$, $f(16)=4$, $f^{\prime }(16)=\frac{1}{8}$.
3. $\Delta x=0.1$, so: $$\sqrt{16.1}\approx 4+\frac{1}{8}(0.1)=4.0125$$ Answer: $\sqrt{16.1}\approx 4.0125$.

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5. Comparison Table: Applications of Derivatives

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6. Common Mistakes to Avoid

• Forgetting the second derivative test: Always check $f^{\prime \prime }(x)$ to confirm maxima/minima.
• Ignoring endpoints: In optimization, evaluate the function at critical points and endpoints.
• Misapplying the chain rule: For related rates, ensure all variables are accounted for.
• Sign errors in slopes: The normal line’s slope is the negative reciprocal of the tangent’s slope.

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Exam Tip

1. NEB loves word problems: Always define variables clearly (e.g., "Let $x$ = length, $y$ = width").
2. Show all steps: Partial credit is given for correct reasoning, even if the final answer is wrong.
3. Units matter: Include units in your final answer (e.g., cm/s, m²).
4. Graphs help: Sketch the function and mark critical points to visualize maxima/minima.
5. Practice mixed questions: Combine rates of change with optimization (e.g., "A ladder slides down a wall...").

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NEB Board-Style Questions

Short Answer (5 marks each):

1. A ladder 10 m long leans against a wall. If the bottom slides away at 2 m/s, how fast is the top sliding down when it is 6 m above the ground?
2. Find the dimensions of a rectangular box with volume 1000 cm³ that minimize the surface area.
3. Find the equation of the normal line to $y=x^3$ at $x=1$.
4. Use linear approximation to estimate $\sin (31^{\circ })$ (use $f(x)=\sin x$ near $x=30^{\circ }$).

Long Answer (10 marks): 5. A company’s profit (in thousands of rupees) is given by $P(x)=-2x^3+60x^2-300x+1000$, where $x$ is the number of units sold.

• Find the number of units that maximizes profit.
• Determine the maximum profit.
• If the cost function is $C(x)=50x+100$, find the break-even points.

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Key Formula Sheet for Revision: $$\begin{array}{rl}\text{Rate of change:}& \frac{dy}{dt}=\frac{dy}{dx}\cdot \frac{dx}{dt}\\ \text{Optimization:}& f^{\prime }(x)=0,\text{ then }{f}^{\prime \prime }(x)\text{ test}\\ \text{Tangent line:}& y-f(a)=f^{\prime }(a)(x-a)\\ \text{Linear approx.:}& f(x+\Delta x)\approx f(x)+f^{\prime }(x)\Delta x\end{array}$$