Chapter 6: Application of Derivatives - Board Exam Notes

Exam Weightage & Blueprint

Total: ~8-10 Marks

This chapter applies differentiation to real-world problems. Maxima and Minima word problems are a favorite for 5-mark questions (or Case Studies).

Question Type	Marks	Frequency	Focus Topic
MCQ	1	High	Rate of Change, Increasing/Decreasing Intervals
Short Answer (2M)	2	Medium	Rate of Change, Slope of Tangent (if in syllabus)
Short Answer (3M)	3	High	Finding Intervals for Inc/Dec, Local Maxima/Minima
Long Answer / Case Study	4/5	Very High	Optimization Word Problems (Maxima/Minima)

Last 24-Hour Checklist

Rate of Change: $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$
Strictly Increasing: $f'(x) > 0$
Strictly Decreasing: $f'(x) < 0$
Critical Points: Solve $f'(x) = 0$

First Derivative Test: Sign change of $f'(x)$
Second Derivative Test: $f''(c) < 0 \Rightarrow$ Maxima
Mensuration Formulas: Cone, Cylinder, Sphere (Memorize!)

Rate of Change of Quantities

If a quantity $y$ varies with another quantity $x$, then $\frac{dy}{dx}$ represents the rate of change of $y$ with respect to $x$.
If both vary with time $t$: $$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$$

Example: Area of circle $A = \pi r^2$. Rate of change w.r.t radius is $\frac{dA}{dr} = 2\pi r$. Rate of change w.r.t time is $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.

Increasing and Decreasing Functions

Strictly Increasing

$x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$

Condition: $f'(x) > 0$ for all $x \in (a, b)$.

Strictly Decreasing

$x_1 < x_2 \Rightarrow f(x_1) > f(x_2)$

Condition: $f'(x) < 0$ for all $x \in (a, b)$.

Steps to find Intervals:
1. Find $f'(x)$.
2. Put $f'(x) = 0$ to find critical points.
3. Divide the domain into disjoint intervals.
4. Check the sign of $f'(x)$ in each interval.

Maxima and Minima

Second Derivative Test (Most Used)

Let $f$ be a function defined on an interval $I$. Let $c \in I$ be a critical point ($f'(c)=0$).
1. If $f''(c) < 0$, then $c$ is a point of Local Maxima.
2. If $f''(c) > 0$, then $c$ is a point of Local Minima.
3. If $f''(c) = 0$, the test fails (use First Derivative Test).

Absolute Maxima/Minima (Closed Interval)

To find absolute max/min in $[a, b]$:
1. Find all critical points $c_1, c_2, \dots$ in $(a, b)$.
2. Evaluate $f(x)$ at all critical points AND endpoints $a$ and $b$.
3. The largest value is Absolute Max, smallest is Absolute Min.

Solved Examples (Board Marking Scheme)

Q1. The volume of a cube is increasing at the rate of 8 cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm? (3 Marks)

Step 1: Given Data 0.5 Mark

Let side $= x$, Volume $V = x^3$, Surface Area $S = 6x^2$.

Given $\frac{dV}{dt} = 8$ cm³/s.

Step 2: Differentiate Volume 1 Mark

$\frac{dV}{dt} = \frac{d}{dt}(x^3) = 3x^2 \frac{dx}{dt}$.

$8 = 3x^2 \frac{dx}{dt} \Rightarrow \frac{dx}{dt} = \frac{8}{3x^2}$.

Step 3: Differentiate Surface Area 1 Mark

$\frac{dS}{dt} = \frac{d}{dt}(6x^2) = 12x \frac{dx}{dt}$.

Substitute $\frac{dx}{dt}$: $\frac{dS}{dt} = 12x \left(\frac{8}{3x^2}\right) = \frac{32}{x}$.

Step 4: Final Calculation 0.5 Mark

At $x = 12$: $\frac{dS}{dt} = \frac{32}{12} = \frac{8}{3}$ cm²/s.

Q2. Find the intervals in which $f(x) = 2x^3 - 3x^2 - 36x + 7$ is strictly increasing. (3 Marks)

Step 1: Find f'(x) 1 Mark

$f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x-3)(x+2)$.

Step 2: Critical Points 0.5 Mark

$f'(x) = 0 \Rightarrow x = 3, x = -2$.

Step 3: Test Intervals 1.5 Marks

Intervals: $(-\infty, -2), (-2, 3), (3, \infty)$.

For $(-\infty, -2)$: Test $x=-3 \Rightarrow f'(-3) = 6(-6)(-1) > 0$. (Increasing)

For $(-2, 3)$: Test $x=0 \Rightarrow f'(0) = -36 < 0$. (Decreasing)

For $(3, \infty)$: Test $x=4 \Rightarrow f'(4) = 6(1)(6) > 0$. (Increasing)

Ans: Strictly increasing in $(-\infty, -2) \cup (3, \infty)$.

Q3. Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area. (5 Marks)

Step 1: Setup 1 Mark

Let circle radius $= a$ (constant). Let rectangle sides be $x$ and $y$.

$x^2 + y^2 = (2a)^2 = 4a^2 \Rightarrow y = \sqrt{4a^2 - x^2}$.

Area $A = xy = x\sqrt{4a^2 - x^2}$.

Step 2: Maximize $A^2$ (Easier) 1 Mark

Let $Z = A^2 = x^2(4a^2 - x^2) = 4a^2x^2 - x^4$.

$\frac{dZ}{dx} = 8a^2x - 4x^3$.

Step 3: Critical Point 1 Mark

$\frac{dZ}{dx} = 0 \Rightarrow 4x(2a^2 - x^2) = 0$. Since $x \neq 0$, $x^2 = 2a^2 \Rightarrow x = \sqrt{2}a$.

Step 4: Second Derivative Test 1 Mark

$\frac{d^2Z}{dx^2} = 8a^2 - 12x^2$. At $x^2 = 2a^2$: $8a^2 - 24a^2 = -16a^2 < 0$.

Hence, Area is Maximum.

Step 5: Conclusion 1 Mark

If $x^2 = 2a^2$, then $y^2 = 4a^2 - 2a^2 = 2a^2$. So $x = y$.

Since sides are equal, the rectangle is a square.

Previous Year Questions (PYQs)

2023 (1 Mark MCQ): The function $f(x) = x^3 - 3x^2 + 12x - 18$ is:
(A) Strictly increasing on R (B) Strictly decreasing on R (C) Neither (D) None
Ans: (A). $f'(x) = 3x^2 - 6x + 12 = 3(x^2 - 2x + 4) = 3[(x-1)^2 + 3]$. Since $(x-1)^2 \ge 0$, $f'(x) \ge 9 > 0$ for all real $x$.

2022 (2 Marks): Find the rate of change of the area of a circle with respect to its radius $r$ when $r = 5$ cm.
Ans: $A = \pi r^2 \Rightarrow \frac{dA}{dr} = 2\pi r$. At $r=5$, Rate $= 10\pi$ cm²/cm.

2020 (5 Marks): A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?
Hint: Let pieces be $x$ and $28-x$. Perimeter of square $= x \Rightarrow$ side $= x/4$. Circumference $= 28-x \Rightarrow r = (28-x)/2\pi$. Minimize $A = (x/4)^2 + \pi r^2$.

Exam Strategy & Mistake Bank

Common Mistakes

Mistake 1: In Rate of Change, forgetting the chain rule term (e.g., $\frac{dr}{dt}$). Always check if differentiation is w.r.t time or radius.

Mistake 2: In Maxima/Minima word problems, forgetting to verify the second derivative test. Marks are deducted for this!

Mistake 3: Confusing "Increasing" ($f'(x) \ge 0$) with "Strictly Increasing" ($f'(x) > 0$).

Scoring Tips

Tip 1: For complex functions (like roots), try maximizing the square of the function ($Z = A^2$) to avoid square root differentiation.

Tip 2: Draw a diagram for word problems (Cone, Cylinder, etc.) and label variables clearly.

Tip 3: Memorize mensuration formulas (Volume of Cone, Sphere, Cylinder). You cannot solve 5-mark questions without them.

Practice Problems (Self-Assessment)

Level 1: Basic (1 Mark Each)

Q1. Find the slope of the tangent to the curve $y = 3x^4 - 4x$ at $x = 4$.

Answer: $\frac{dy}{dx} = 12x^3 - 4$. At $x=4$, Slope $= 12(64) - 4 = 768 - 4 = 764$.

Q2. The total revenue in Rupees received from the sale of $x$ units is given by $R(x) = 3x^2 + 36x + 5$. Find the marginal revenue when $x = 15$.

Answer: $MR = \frac{dR}{dx} = 6x + 36$. At $x=15$, $MR = 90 + 36 = 126$.

Level 2: Intermediate (2-3 Marks Each)

Q3. Find the local maxima and local minima of $f(x) = x^3 - 6x^2 + 9x + 15$.

Hint: $f'(x) = 3(x-1)(x-3)$. Critical points $1, 3$. $f''(x) = 6x - 12$.
At $x=1, f'' < 0$ (Max). At $x=3, f'' > 0$ (Min).

Q4. Find the intervals in which $f(x) = \sin x + \cos x, 0 \le x \le 2\pi$ is strictly increasing.

Answer: $f'(x) = \cos x - \sin x$. $f'(x) > 0 \Rightarrow \cos x > \sin x$. In $[0, 2\pi]$, this holds for $[0, \pi/4) \cup (5\pi/4, 2\pi]$.

Level 3: Advanced (5 Marks Each)

Q5. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $R$ is $\frac{2R}{\sqrt{3}}$.

Hint: Let cylinder height $h$, radius $r$. $r^2 + (h/2)^2 = R^2 \Rightarrow r^2 = R^2 - h^2/4$.
Maximize $V = \pi r^2 h = \pi (R^2h - h^3/4)$. Differentiate w.r.t $h$.