Application Of Derivatives: CBSE Class 12 Maths NCERT

Overview of Applications of Derivatives

Derivatives, fundamentally, measure how sensitive a function is to changes in its input. This simple concept branches out into several powerful applications that are integral to understanding various phenomena in mathematics, science, and engineering. The core idea is that the derivative $f'(x)$ of a function $f(x)$ at a point $x$ gives us the instantaneous rate of change of $f(x)$ with respect to $x$. Geometrically, it represents the slope of the tangent to the curve $y=f(x)$ at that point.

We will explore several key applications:

1. Rate of Change of Quantities: Determining how one quantity changes with respect to another. For example, finding how the volume of a sphere changes as its radius increases.
2. Increasing and Decreasing Functions: Identifying intervals where a function's value is either rising or falling. This helps in understanding the behaviour of graphs and optimizing processes.
3. Tangents and Normals: Finding the equations of lines that touch a curve at a single point (tangents) and lines perpendicular to these tangents (normals). This has applications in optics and curve analysis.
4. Approximations: Using derivatives to estimate the change in a quantity for small changes in its input, a powerful tool for complex calculations.
5. Maxima and Minima: Locating the maximum or minimum values of a function, which is critical for optimization problems in fields ranging from economics (maximizing profit) to physics (minimizing energy). This includes finding local extrema and absolute extrema.

Key Concepts and Definitions

Rate of Change

If $y = f(x)$, then the derivative $\frac{dy}{dx}$ represents the instantaneous rate of change of $y$ with respect to $x$. If $x$ is time ($t$), then $\frac{dy}{dt}$ denotes the rate of change of $y$ over time.

Increasing Function

A function $f(x)$ is said to be increasing on an interval $(a, b)$ if for any $x_1, x_2 \in (a, b)$ with $x_1 < x_2$, we have $f(x_1) \le f(x_2)$. If $f'(x) \ge 0$ for all $x$ in $(a, b)$, then $f(x)$ is increasing. If $f'(x) > 0$, it is strictly increasing.

Decreasing Function

A function $f(x)$ is said to be decreasing on an interval $(a, b)$ if for any $x_1, x_2 \in (a, b)$ with $x_1 < x_2$, we have $f(x_1) \ge f(x_2)$. If $f'(x) \le 0$ for all $x$ in $(a, b)$, then $f(x)$ is decreasing. If $f'(x) < 0$, it is strictly decreasing.

Critical Point

A critical point of a function $f(x)$ is a point $c$ in the domain of $f$ where $f'(c) = 0$ or $f'(c)$ is undefined. These points are candidates for local maxima or minima.

Local Maxima/Minima (Extrema)

A function $f(x)$ has a local maximum at $c$ if $f(c)$ is the largest value in some open interval containing $c$. Similarly, it has a local minimum if $f(c)$ is the smallest value. Together, they are called local extrema.

Process: Finding Local Maxima and Minima

1. Step 1: Find the First Derivative — Calculate $f'(x)$, the first derivative of the given function $f(x)$.
2. Step 2: Find Critical Points — Set $f'(x) = 0$ and solve for $x$. These values of $x$ are the critical points. Also, check for points where $f'(x)$ is undefined, if any.
3. Step 3 (Option A): First Derivative Test — Examine the sign of $f'(x)$ around each critical point. - If $f'(x)$ changes from positive to negative as $x$ increases through $c$, then $f(c)$ is a local maximum. - If $f'(x)$ changes from negative to positive as $x$ increases through $c$, then $f(c)$ is a local minimum. - If $f'(x)$ does not change sign, $c$ is a point of inflection (neither maximum nor minimum).
4. Step 3 (Option B): Second Derivative Test — Calculate $f''(x)$, the second derivative of the function. - Substitute each critical point $c$ into $f''(x)$. - If $f''(c) < 0$, then $f(c)$ is a local maximum. - If $f''(c) > 0$, then $f(c)$ is a local minimum. - If $f''(c) = 0$, the test fails, and you must use the first derivative test.
5. Step 4: Determine Local Extrema — Substitute the values of $x$ (for local maxima or minima) back into the original function $f(x)$ to find the maximum or minimum values.

Worked Examples

• Example 1: Rate of Change A spherical balloon is being inflated. Its volume is increasing at the rate of 10 cubic centimeters per second. At what rate is the radius of the balloon increasing when the radius is 15 cm? Solution: Step 1: Understand the given information and what needs to be found. Let $V$ be the volume of the sphere and $r$ be its radius. Given: $\frac{dV}{dt} = 10 \text{ cm}^3/s$. We need to find $\frac{dr}{dt}$ when $r = 15 \text{ cm}$. Step 2: Write down the formula for the volume of a sphere. $V = \frac{4}{3}\pi r^3$ Step 3: Differentiate the volume formula with respect to time $t$. Using the chain rule: $\frac{dV}{dt} = \frac{d}{dt} \left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi (3r^2) \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}$ Step 4: Substitute the given values and solve for the unknown. $10 = 4\pi (15)^2 \frac{dr}{dt}$ $10 = 4\pi (225) \frac{dr}{dt}$ $10 = 900\pi \frac{dr}{dt}$ $\frac{dr}{dt} = \frac{10}{900\pi} = \frac{1}{90\pi}$ cm/s Final Answer: The radius of the balloon is increasing at the rate of $\frac{1}{90\pi}$ cm/s when the radius is 15 cm.
• Example 2: Increasing/Decreasing Functions Find the intervals in which the function $f(x) = 2x^3 - 3x^2 - 36x + 7$ is strictly increasing or strictly decreasing. Solution: Step 1: Find the first derivative of the function. $f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 36x + 7) = 6x^2 - 6x - 36$ Step 2: Find the critical points by setting $f'(x) = 0$. $6x^2 - 6x - 36 = 0$ $x^2 - x - 6 = 0$ (Dividing by 6) $(x-3)(x+2) = 0$ So, $x = 3$ or $x = -2$. These are the critical points that divide the number line into intervals: $(-\infty, -2)$, $(-2, 3)$, and $(3, \infty)$. Step 3: Test the sign of $f'(x)$ in each interval. - Interval $(-\infty, -2)$: Pick a test value, e.g., $x=-3$. $f'(-3) = 6(-3)^2 - 6(-3) - 36 = 6(9) + 18 - 36 = 54 + 18 - 36 = 36 > 0$. Since $f'(x) > 0$, $f(x)$ is strictly increasing in $(-\infty, -2)$. - Interval $(-2, 3)$: Pick a test value, e.g., $x=0$. $f'(0) = 6(0)^2 - 6(0) - 36 = -36 < 0$. Since $f'(x) < 0$, $f(x)$ is strictly decreasing in $(-2, 3)$. - Interval $(3, \infty)$: Pick a test value, e.g., $x=4$. $f'(4) = 6(4)^2 - 6(4) - 36 = 6(16) - 24 - 36 = 96 - 24 - 36 = 36 > 0$. Since $f'(x) > 0$, $f(x)$ is strictly increasing in $(3, \infty)$. Final Answer: The function $f(x)$ is strictly increasing in $(-\infty, -2) \cup (3, \infty)$ and strictly decreasing in $(-2, 3)$.
• Example 3: Maxima and Minima (Second Derivative Test) Find the local maximum and minimum values of the function $f(x) = x^3 - 6x^2 + 9x + 15$. Solution: Step 1: Find the first derivative. $f'(x) = 3x^2 - 12x + 9$ Step 2: Find critical points by setting $f'(x) = 0$. $3x^2 - 12x + 9 = 0$ $x^2 - 4x + 3 = 0$ $(x-1)(x-3) = 0$ Critical points are $x=1$ and $x=3$. Step 3: Find the second derivative. $f''(x) = \frac{d}{dx}(3x^2 - 12x + 9) = 6x - 12$ Step 4: Apply the Second Derivative Test. - At $x=1$: $f''(1) = 6(1) - 12 = -6$. Since $f''(1) < 0$, $x=1$ is a point of local maximum. Local maximum value: $f(1) = (1)^3 - 6(1)^2 + 9(1) + 15 = 1 - 6 + 9 + 15 = 19$. - At $x=3$: $f''(3) = 6(3) - 12 = 18 - 12 = 6$. Since $f''(3) > 0$, $x=3$ is a point of local minimum. Local minimum value: $f(3) = (3)^3 - 6(3)^2 + 9(3) + 15 = 27 - 54 + 27 + 15 = 15$. Final Answer: The local maximum value is 19 at $x=1$, and the local minimum value is 15 at $x=3$.

Exam Tips for Application Of Derivatives

To excel in the Application of Derivatives chapter, consistency and conceptual clarity are key. Always read the question carefully to understand what is being asked – is it a rate of change, an interval for increasing/decreasing, an equation of tangent, or finding extrema?

1. Visualize the Problem: For rate of change or optimization problems, draw a diagram if possible. This helps in forming the correct mathematical model or relation between variables.
2. Chain Rule is Your Best Friend: Many problems involve implicit differentiation with respect to time ($t$). Remember to apply the chain rule correctly (e.g., $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$). Neglecting this is a common mistake.
3. Correct Test for Extrema: Understand when to use the First Derivative Test vs. the Second Derivative Test. The second derivative test is often quicker if $f''(x)$ is easy to compute and non-zero at critical points. If $f''(x)=0$, you must revert to the first derivative test.
4. Intervals for Increasing/Decreasing: When stating intervals, use open intervals for strictly increasing/decreasing functions. For just increasing/decreasing, closed intervals are used in some contexts, but stick to open intervals for strict cases as per NCERT.
5. Units: Always include appropriate units in your final answer, especially for rate of change problems (e.g., cm/s, m$^3$/min).

Practice Questions with Solutions

• Q: A particle moves along the curve $y = \frac{2}{3}x^3 + 1$. Find the points on the curve at which the $y$-coordinate is changing 8 times as fast as the $x$-coordinate. A: Step 1: Set up the given condition. Given that $\frac{dy}{dt} = 8 \frac{dx}{dt}$. Step 2: Differentiate the curve equation with respect to $t$. $y = \frac{2}{3}x^3 + 1 \implies \frac{dy}{dt} = \frac{d}{dt}(\frac{2}{3}x^3 + 1) = \frac{2}{3}(3x^2)\frac{dx}{dt} = 2x^2\frac{dx}{dt}$. Step 3: Substitute the condition from Step 1 into the differentiated equation. $8\frac{dx}{dt} = 2x^2\frac{dx}{dt}$. Step 4: Solve for $x$. Since $\frac{dx}{dt}$ cannot be zero (otherwise $y$ would not be changing), we can divide by $\frac{dx}{dt}$: $8 = 2x^2 \implies x^2 = 4 \implies x = \pm 2$. Step 5: Find the corresponding $y$-coordinates. If $x=2$, $y = \frac{2}{3}(2)^3 + 1 = \frac{16}{3} + 1 = \frac{19}{3}$. Point: $(2, \frac{19}{3})$. If $x=-2$, $y = \frac{2}{3}(-2)^3 + 1 = \frac{-16}{3} + 1 = \frac{-13}{3}$. Point: $(-2, \frac{-13}{3})$. Final answer: The points on the curve are $(2, \frac{19}{3})$ and $(-2, \frac{-13}{3})$.
• Q: Find the values of $x$ for which $f(x) = x^2 e^{-x}$ is strictly increasing. A: Step 1: Find the first derivative of $f(x)$. Using the product rule: $f'(x) = \frac{d}{dx}(x^2)e^{-x} + x^2\frac{d}{dx}(e^{-x}) = 2xe^{-x} + x^2(-e^{-x}) = e^{-x}(2x - x^2) = xe^{-x}(2 - x)$. Step 2: For $f(x)$ to be strictly increasing, $f'(x) > 0$. $xe^{-x}(2 - x) > 0$. Step 3: Analyze the sign of the factors. Since $e^{-x}$ is always positive for all real $x$, we only need to consider the sign of $x(2-x)$. We need $x(2-x) > 0$. This inequality holds when $x$ and $(2-x)$ have the same sign. Case 1: $x > 0$ and $2-x > 0 \implies x > 0$ and $x < 2$. So, $0 < x < 2$. Case 2: $x < 0$ and $2-x < 0 \implies x < 0$ and $x > 2$. This case is impossible. Final answer: The function $f(x)$ is strictly increasing for $x \in (0, 2)$.
• Q: Find the equation of the tangent and normal to the curve $y = x^4 - 6x^3 + 13x^2 - 10x + 5$ at $(0, 5)$. A: Step 1: Find the slope of the tangent by calculating the derivative. $y = x^4 - 6x^3 + 13x^2 - 10x + 5$ $\frac{dy}{dx} = 4x^3 - 18x^2 + 26x - 10$. Step 2: Evaluate the slope at the given point $(0, 5)$. Slope of tangent $m_t = \left(\frac{dy}{dx}\right)_{(0,5)} = 4(0)^3 - 18(0)^2 + 26(0) - 10 = -10$. Step 3: Find the equation of the tangent. Using point-slope form: $y - y_1 = m_t(x - x_1)$. $y - 5 = -10(x - 0) \implies y - 5 = -10x \implies 10x + y - 5 = 0$. Step 4: Find the slope of the normal. The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{-10} = \frac{1}{10}$. Step 5: Find the equation of the normal. Using point-slope form: $y - y_1 = m_n(x - x_1)$. $y - 5 = \frac{1}{10}(x - 0) \implies 10(y - 5) = x \implies 10y - 50 = x \implies x - 10y + 50 = 0$. Final answer: Equation of tangent: $10x + y - 5 = 0$. Equation of normal: $x - 10y + 50 = 0$.
• Q: Find the maximum value of $f(x) = (x-1)^2 + 10$ on the interval $[-2, 3]$. A: Step 1: Find the derivative of the function. $f(x) = (x-1)^2 + 10 \implies f'(x) = 2(x-1)$. Step 2: Find critical points by setting $f'(x) = 0$. $2(x-1) = 0 \implies x = 1$. This critical point lies within the interval $[-2, 3]$. Step 3: Evaluate the function at the critical point and at the endpoints of the interval. - At critical point $x=1$: $f(1) = (1-1)^2 + 10 = 0 + 10 = 10$. - At endpoint $x=-2$: $f(-2) = (-2-1)^2 + 10 = (-3)^2 + 10 = 9 + 10 = 19$. - At endpoint $x=3$: $f(3) = (3-1)^2 + 10 = (2)^2 + 10 = 4 + 10 = 14$. Step 4: Compare the values to find the maximum. The values are $10, 19, 14$. The maximum among these is 19. Final answer: The maximum value of the function on the interval $[-2, 3]$ is 19.

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