Applications of Derivatives: Optimization, Rates, Tangents & Curve Sketching

TAKEAWAYS:

• Optimization: Use derivatives to maximize profit (e.g., Daraz’s pricing) or minimize cost (e.g., NTC’s cable routing).
• Rates of change: Model real-time scenarios like traffic flow (Kathmandu’s congestion) or loan interest (Nepal Bank’s EMI).
• Tangents & normals: Apply to design (e.g., Pathao’s delivery route slopes) or physics (e.g., Ncell’s signal propagation).
• Curve sketching: Visualize functions (e.g., NEPSE stock trends) using first/second derivatives.
• Error analysis: Estimate measurement errors (e.g., land surveying in Nepal) via differentials.
• Business decisions: Use derivatives to predict demand (e.g., eSewa’s transaction limits) or supply chains (e.g., Daraz’s warehouse locations).

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1. Optimization Problems: Maximizing Profit, Minimizing Cost

Definition: Optimization uses derivatives to find local maxima (highest points) or local minima (lowest points) of a function $f(x)$. Steps:

1. Find $f^{\prime }(x)$ and set $f^{\prime }(x)=0$ to locate critical points.
2. Use the Second Derivative Test:
   • If $f^{\prime \prime }(x)>0$: local minimum (concave up).
   • If $f^{\prime \prime }(x)<0$: local maximum (concave down).
3. Compare function values at critical points and endpoints to find the absolute maximum/minimum.

How it works:

Worked Example 1: Daraz’s Optimal Pricing Problem: Daraz wants to maximize profit $P(x)=-2x^3+60x^2+1000x-5000$ (where $x$ = price in ₹1000). Find the optimal price and maximum profit.

Solution:

1. Find $P^{\prime }(x)$: $$P^{\prime }(x)=\frac{d}{dx}(-2x^3+60x^2+1000x-5000)=-6x^2+120x+1000$$

2. Set $P^{\prime }(x)=0$: $$-6x^2+120x+1000=0⟹x^2-20x-\frac{1000}{6}=0$$ Solve using the quadratic formula: $$x=\frac{20\pm \sqrt{400+\frac{10000}{6}}}{2}=\frac{20\pm \sqrt{400+1666.67}}{2}=\frac{20\pm \sqrt{2066.67}}{2}$$ $$x\approx 25.8\text{ or }x\approx -5.8(\text{Discard negative price})$$ Critical point: $x=25.8$ (₹25,800).

3. Second Derivative Test: $$P^{\prime \prime }(x)=-12x+120$$ At $x=25.8$: $$P^{\prime \prime }(25.8)=-12(25.8)+120=-309.6+120=-189.6<0⟹\text{Local maximum}$$

4. Calculate Maximum Profit: $$P(25.8)=-2(25.8{)}^3+60(25.8{)}^2+1000(25.8)-5000\approx ₹1,650,000$$

Visual:

Real-World Tie-In: Daraz uses similar optimization to set dynamic pricing during sales (e.g., reducing prices if inventory is high). The derivative helps balance revenue and demand.

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2. Rates of Change: Related Rates Problems

Definition: Related rates problems involve functions of two or more variables where one variable changes with time $t$. Steps:

1. Differentiate both sides with respect to $t$ (using the chain rule).
2. Substitute known values to find the unknown rate.

How it works:

Worked Example 2: Kathmandu Traffic Congestion Problem: In Kathmandu, traffic flow on Ring Road is modeled by $V=50-0.2t^2$ (vehicles per minute at time $t$ hours). If the flow decreases at 3 vehicles/min² at $t=5$ hours, find the rate of change of congestion $C$, where $C=\frac{1}{V}$.

Solution:

1. Differentiate $V$ w.r.t. $t$: $$\frac{dV}{dt}=-0.4t$$ Given $\frac{dV}{dt}=-3$ at $t=5$: $$-0.4(5)=-2\ne -3\text{(Adjust model or assume typo; proceed with given rate)}$$ Assume $\frac{dV}{dt}=-3$ (as per problem).

2. Differentiate $C=V^{-1}$ w.r.t. $t$: $$\frac{dC}{dt}=-V^{-2}\cdot \frac{dV}{dt}$$

3. Substitute $V(5)=50-0.2(25)=45$ and $\frac{dV}{dt}=-3$: $$\frac{dC}{dt}=-\frac{1}{45^2}\cdot (-3)=\frac{3}{2025}\approx 0.0015\text{ min/vehicle}$$ Interpretation: Congestion increases at 0.0015 min/vehicle per hour.

Visual:

Real-World Tie-In: NTC uses related rates to predict network congestion (e.g., data speeds dropping during peak hours). The derivative helps preemptively reroute traffic or allocate bandwidth.

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3. Tangents and Normals

Definition:

• Tangent: A line touching the curve at a point with slope $f^{\prime }(x)$.
• Normal: A line perpendicular to the tangent at the point (slope $=-\frac{1}{f^{\prime }(x)}$).

Equation of Tangent: $$y-f(a)=f^{\prime }(a)(x-a)$$ Equation of Normal: $$y-f(a)=-\frac{1}{f^{\prime }(a)}(x-a)$$

Worked Example 3: Pathao’s Delivery Route Slope Problem: Pathao’s delivery cost $C(x)=0.5x^2+10x+100$ (₹) depends on distance $x$ km. Find the tangent and normal at $x=5$ km.

Solution:

1. Find $C^{\prime }(x)$: $$C^{\prime }(x)=x+10$$ At $x=5$: $$C^{\prime }(5)=15\text{(slope of tangent)}$$ $$C(5)=0.5(25)+50+100=137.5\text{ ₹}$$
2. Tangent Equation: $$y-137.5=15(x-5)⟹y=15x-75+137.5⟹y=15x+62.5$$
3. Normal Equation (slope $=-\frac{1}{15}$): $$y-137.5=-\frac{1}{15}(x-5)⟹y=-\frac{1}{15}x+\frac{5}{15}+137.5⟹y=-\frac{1}{15}x+137.83$$

Visual:

Real-World Tie-In: Pathao uses tangent lines to estimate delivery time changes (e.g., a small detour’s cost impact). Normals help in route optimization (e.g., avoiding steep inclines).

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4. Curve Sketching Using Derivatives

Steps:

1. Domain: Find where $f(x)$ is defined.
2. Intercepts:
   • $x$-intercepts: $f(x)=0$.
   • $y$-intercept: $f(0)$.
3. Symmetry: Check $f(-x)$ (even/odd).
4. Asymptotes:
   • Vertical: Where $f(x)\to \infty$ (denominator zero).
   • Horizontal/Oblique: Limits as $x\to \pm \infty$.
5. Critical Points: $f^{\prime }(x)=0$ or undefined.
6. Increasing/Decreasing: Sign of $f^{\prime }(x)$.
7. Concavity: Sign of $f^{\prime \prime }(x)$ (inflection points where $f^{\prime \prime }(x)=0$).
8. Sketch: Plot key points and draw the curve.

Worked Example 4: NEPSE Stock Trend Problem: Sketch $f(x)=\frac{x^2-4}{x^2-1}$.

Solution:

1. Domain: $x\ne \pm 1$.
2. Intercepts:
   • $x$-intercepts: $x^2-4=0⟹x=\pm 2$.
   • $y$-intercept: $f(0)=-4$.
3. Symmetry: $f(-x)=f(x)$ (even function).
4. Asymptotes:
   • Vertical: $x=\pm 1$.
   • Horizontal: ${\lim }_{x\to \pm \infty }f(x)=1$.
5. Critical Points: $$f^{\prime }(x)=\frac{(2x)(x^2-1)-(x^2-4)(2x)}{(x^2-1{)}^2}=\frac{2x^3-2x-2x^3+8x}{(x^2-1{)}^2}=\frac{6x}{(x^2-1{)}^2}$$ Set $f^{\prime }(x)=0⟹x=0$.
6. Increasing/Decreasing:
   • $f^{\prime }(x)>0$ for $x>0$ (increasing).
   • $f^{\prime }(x)<0$ for $x<0$ (decreasing).
7. Concavity: $$f^{\prime \prime }(x)=\frac{6(x^2-1{)}^2-6x\cdot 2(x^2-1)\cdot 2x}{(x^2-1{)}^4}=\frac{6(x^2-1)[(x^2-1)-8x^2]}{(x^2-1{)}^4}=\frac{6(-7x^2-1)}{(x^2-1{)}^3}$$ $f^{\prime \prime }(x)<0$ for all $x\ne \pm 1$ (concave down everywhere).

Visual:

Real-World Tie-In: NEPSE analysts use curve sketching to predict stock trends (e.g., identifying peaks/troughs in $f(x)$ for buy/sell signals).

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5. Error Analysis Using Differentials

Definition: For a function $y=f(x)$, the differential $dy=f^{\prime }(x)dx$ approximates the change in $y$ for small $\Delta x$.

Formula: $$\Delta y\approx dy=f^{\prime }(x)\Delta x$$

Worked Example 5: Land Surveying in Nepal Problem: A farmer measures a rectangular field’s side as 100 m with a 0.5 m error. If the area $A=x^2$, estimate the maximum error in area.

Solution:

1. $A=x^2⟹\frac{dA}{dx}=2x$.
2. At $x=100$ m, $\frac{dA}{dx}=200$.
3. $\Delta x=0.5$ m: $$\Delta A\approx dA=200\times 0.5=100{\text{ m}}^2$$ Actual error: $(100.5{)}^2-100^2=100.5{\text{ m}}^2$ (close approximation).

Visual:

Real-World Tie-In: Nepal’s land revenue department uses differentials to estimate tax errors from measurement inaccuracies.

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6. Comparison Table: Applications of Derivatives

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In the Real World

1. eSewa’s Transaction Limits: eSewa uses optimization (derivatives) to set daily transaction caps. For example, if profit $P(x)=-0.001x^2+100x-5000$ (where $x$ = transactions), the optimal cap is found by setting $P^{\prime }(x)=0$, maximizing revenue while minimizing fraud risk.

2. Ncell’s Network Congestion: Ncell monitors data speeds $S(t)=50-0.5t^2$ (Mbps at time $t$ hours). Using related rates, they predict when speeds drop below 10 Mbps: $$\frac{dS}{dt}=-t⟹\text{At }S=10,\frac{dS}{dt}=-4\text{ Mbps/hour}$$ This triggers automatic bandwidth reallocation.

3. Khalti’s Loan Interest: Khalti’s loan interest is modeled by $I(t)=1000(1.02{)}^t$. The derivative $I^{\prime }(t)=1000\ln (1.02)(1.02{)}^t$ shows the instantaneous interest rate, helping users compare loans.

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Exam Tip

1. Optimization Questions:

   • Always check if critical points are maxima/minima using $f^{\prime \prime }(x)$.
   • Compare values at critical points and endpoints for absolute extrema.
   • Common mistake: Forgetting to discard non-physical solutions (e.g., negative prices).

2. Related Rates:

   • Label all variables clearly (e.g., $\frac{dV}{dt}$, $\frac{dr}{dt}$).
   • Use the chain rule correctly: $\frac{dy}{dt}=\frac{dy}{dx}\cdot \frac{dx}{dt}$.
   • Common mistake: Mixing up $\frac{dx}{dt}$ and $\frac{dt}{dx}$.

3. Curve Sketching:

   • Plot all critical points, intercepts, and asymptotes.
   • Use a table to organize signs of $f^{\prime }(x)$ and $f^{\prime \prime }(x)$.
   • Common mistake: Skipping the second derivative test.

4. Tangents/Normals:

   • Write the equation in slope-intercept form ($y=mx+c$).
   • For normals, remember the slope is the negative reciprocal of $f^{\prime }(x)$.

5. Error Analysis:

   • Use $dy=f^{\prime }(x)dx$ for small errors.
   • Common mistake: Forgetting the approximation sign ($\approx$).

Past Exam Pitfall: In the question "Find all second order partial derivatives when $f(x,y)=x^4-5x^3{y}^2+7x^2{y}^3+4x{y}^5$", students often:

• Miss mixed partials (e.g., $\frac{{\partial }^{2}f}{\partial x\partial y}\ne \frac{{\partial }^{2}f}{\partial y\partial x}$ unless $f$ is smooth).
• Forget to simplify terms like $-30x^2{y}^2$ in $f_{xy}$. Solution:

f_x = 4x^3 - 15x^2y^2 + 14xy^3 + 4y^5
f_{xx} = 12x^2 - 30x^2y^2 + 14y^3
f_{xy} = -30x^2y + 42xy^2 + 20y^4
f_{yy} = -10x^3y + 21x^2y^2 + 20xy^4
f_{yx} = f_{xy} \quad \text{(Clairaut's theorem)}
f_{xxy} = -60xy + 42y^2 + 80y^3