Applications of Derivative Class 12 NCERT Solutions & Notes

Core Concepts & Geometrical Significance

Understanding Applications of Derivatives begins with recognizing the geometric meaning of a derivative. If $y = f(x)$ represents a continuous curve, the derivative $\frac{dy}{dx}$ at any point represents the slope of the tangent to the curve at that specific point. Mathematically, this serves three main purposes in the rationalized Class 12 CBSE syllabus:

1. Rate of Change of Quantities: We use $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$ via the Chain Rule to calculate how one quantity changes with respect to another (e.g., volume change with respect to radius or time).
2. Increasing and Decreasing Functions: A function $f(x)$ is strictly increasing in an interval $(a, b)$ if $f'(x) > 0$ for all $x \in (a, b)$. Conversely, it is strictly decreasing if $f'(x) < 0$ for all $x \in (a, b)$.
3. Maxima and Minima: Points where the derivative changes sign or becomes zero ($f'(x) = 0$) are critical points. By applying the First or Second Derivative Test at these points, we can locate local and absolute extremum values which form the base of optimization problems.

The Second Derivative Test for Local Extrema

1. Step 1: Compute the First Derivative — Find the first-order derivative of the given function, $f'(x)$.
2. Step 2: Find the Critical Points — Set $f'(x) = 0$ and solve for $x$. Let the real roots (critical points) be $x = c_1, c_2, \dots$
3. Step 3: Compute the Second Derivative — Differentiate $f'(x)$ again to find $f''(x)$.
4. Step 4: Evaluate at Critical Points — Substitute each critical point $c$ into $f''(x)$: - If $f''(c) < 0$, then $x = c$ is a point of local maxima. - If $f''(c) > 0$, then $x = c$ is a point of local minima. - If $f''(c) = 0$, the test fails. You must resort to the First Derivative Test.

CBSE Board Exam Tips & Common Traps

• Absolute vs. Local Extrema: For absolute maximum/minimum on a closed interval $[a, b]$, always evaluate the function's values at critical points and boundary points ($x = a$ and $x = b$). Do not use the second derivative test for closed interval absolute extrema.
• Rate of Change Units: Never forget to write the units in your final answer (e.g., $\text{cm}^3/\text{s}$ for volume change, $\text{cm}^2/\text{s}$ for surface area change). Losing half a mark for missing units is a common trap!
• Strictly vs. Non-Strict Intervals: Note that 'strictly increasing' uses open intervals where $f'(x) > 0$. Make sure to exclude the critical boundary values where $f'(x) = 0$ unless specified otherwise.

Practice Questions with Solutions

• Q: The volume of a spherical balloon is increasing at a constant rate of $900 \text{ cm}^3/\text{s}$. Find the rate at which its radius is increasing when the radius is $15 \text{ cm}$. A: Step 1: Write down the volume formula for a sphere: $V = \frac{4}{3}\pi r^3$. Step 2: Differentiate both sides with respect to time $t$: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$. Step 3: Substitute the given values $\frac{dV}{dt} = 900 \text{ cm}^3/\text{s}$ and $r = 15 \text{ cm}$ into the equation: $900 = 4\pi (15)^2 \frac{dr}{dt} \implies 900 = 900\pi \frac{dr}{dt}$. Step 4: Solve for $\frac{dr}{dt}$: $\frac{dr}{dt} = \frac{1}{\pi} \text{ cm/s}$. Final answer: The rate of change of the radius is $\frac{1}{\pi} \text{ cm/s}$.
• Q: Find the intervals in which the function $f(x) = 2x^3 - 3x^2 - 36x + 7$ is strictly increasing or strictly decreasing. A: Step 1: Compute the derivative of the function: $f'(x) = 6x^2 - 6x - 36$. Step 2: Find critical points by setting $f'(x) = 0$: $6(x^2 - x - 6) = 0 \implies 6(x-3)(x+2) = 0$. So, the critical points are $x = -2$ and $x = 3$. These points divide the real line into three intervals: $(-\infty, -2)$, $(-2, 3)$, and $(3, \infty)$. Step 3: Test the sign of $f'(x)$ in each interval: - For $(-\infty, -2)$: Choose $x = -3 \implies f'(-3) = 6(-3-3)(-3+2) = 36 > 0$ (Strictly Increasing). - For $(-2, 3)$: Choose $x = 0 \implies f'(0) = -36 < 0$ (Strictly Decreasing). - For $(3, \infty)$: Choose $x = 4 \implies f'(4) = 6(4-3)(4+2) = 36 > 0$ (Strictly Increasing). Final answer: Strictly increasing on $(-\infty, -2) \cup (3, \infty)$ and strictly decreasing on $(-2, 3)$.
• Q: Find the absolute maximum and minimum values of the function $f(x) = 2x^3 - 15x^2 + 36x + 1$ on the closed interval $[1, 5]$. A: Step 1: Find the first derivative: $f'(x) = 6x^2 - 30x + 36$. Step 2: Equate $f'(x) = 0$ to find critical points: $6(x^2 - 5x + 6) = 0 \implies 6(x-2)(x-3) = 0$. The critical points are $x = 2$ and $x = 3$, both of which lie in the interval $[1, 5]$. Step 3: Evaluate $f(x)$ at the critical points and the endpoints of the interval ($x = 1, 2, 3, 5$): - $f(1) = 2(1)^3 - 15(1)^2 + 36(1) + 1 = 24$ - $f(2) = 2(8) - 15(4) + 36(2) + 1 = 29$ - $f(3) = 2(27) - 15(9) + 36(3) + 1 = 28$ - $f(5) = 2(125) - 15(25) + 36(5) + 1 = 56$ Step 4: Compare all values to find the absolute extreme values. Final answer: The absolute maximum value is $56$ at $x=5$, and the absolute minimum value is $24$ at $x=1$.
• Q: Show that of all the rectangles of a given perimeter, the square has the maximum area. A: Step 1: Let the sides of the rectangle be $x$ and $y$. The given constant perimeter is $P = 2(x + y)$, which means $y = \frac{P}{2} - x$. Step 2: Formulate the area function $A(x) = x \cdot y = x \left(\frac{P}{2} - x\right) = \frac{Px}{2} - x^2$. Step 3: Differentiate with respect to $x$ to find the critical point: $A'(x) = \frac{P}{2} - 2x$. Setting $A'(x) = 0 \implies 2x = \frac{P}{2} \implies x = \frac{P}{4}$. Step 4: Use the second derivative test to check for maximum area: $A''(x) = -2 < 0$. Since the second derivative is negative, $x = \frac{P}{4}$ maximizes the area. Step 5: Substitute $x = \frac{P}{4}$ back into the equation for $y$: $y = \frac{P}{2} - \frac{P}{4} = \frac{P}{4}$. Since $x = y$, the rectangle is a square. Final answer: Proven that the rectangle of maximum area for a given perimeter is a square.

Frequently Asked Questions