Mathematics: Class 12 (Book 1) — Chapter 6: Application of Derivatives

If a variable quantity $y$ varies with another quantity $x$ according to a rule $y = f(x)$, then the derivative $\frac{dy}{dx} = f'(x)$ represents the instantaneous rate of change of $y$ with respect to $x$.

The rate of change of $y$ with respect to $x$ at a specific point $x = x_0$ is denoted by:

If two variables $x$ and $y$ are both varying with respect to another variable $t$ (such as time), i.e., $x = f(t)$ and $y = g(t)$, then by the Chain Rule, the rate of change of $y$ with respect to $x$ is calculated as:

Mathematical Problem Statement: The volume of a cube is increasing at a rate of $9\text{ cm}^3/\text{s}$. How fast is the surface area increasing when the length of an edge is $10\text{ cm}$?

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Step-by-Step Algorithmic Solution Trace:

1. Let $x$ be the edge length, $V$ be the volume, and $S$ be the surface area of the cube. Thus, $V = x^3$ and $S = 6x^2$.
2. We are given the rate of change of volume: $\frac{dV}{dt} = 9\text{ cm}^3/\text{s}$.
3. By the Chain Rule, differentiate $V$ with respect to $t$:$$\frac{dV}{dt} = 3x^2 \frac{dx}{dt} \implies 9 = 3x^2 \frac{dx}{dt} \implies \frac{dx}{dt} = \frac{3}{x^2}$$
4. Differentiate the surface area $S = 6x^2$ with respect to $t$:$$\frac{dS}{dt} = 12x \frac{dx}{dt}$$
5. Substitute $\frac{dx}{dt} = \frac{3}{x^2}$ into the expression for $\frac{dS}{dt}$:$$\frac{dS}{dt} = 12x \left(\frac{3}{x^2}\right) = \frac{36}{x}$$
6. Evaluate at $x = 10\text{ cm}$:$$\frac{dS}{dt} = \frac{36}{10} = 3.6\text{ cm}^2/\text{s}$$

Final Evaluation Output: The surface area is increasing at a rate of $3.6\text{ cm}^2/\text{s}$.

Mathematical Problem Statement: The length $x$ of a rectangle is decreasing at the rate of $3\text{ cm/minute}$ and the width $y$ is increasing at the rate of $2\text{ cm/minute}$. When $x = 10\text{ cm}$ and $y = 6\text{ cm}$, find the rates of change of (a) the perimeter and (b) the area of the rectangle.

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Step-by-Step Algorithmic Solution Trace:

1. Identify the given rates: $\frac{dx}{dt} = -3\text{ cm/min}$ (negative because length is decreasing) and $\frac{dy}{dt} = 2\text{ cm/min}$.
2. (a) Perimeter $P = 2(x + y)$:$$\frac{dP}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right) = 2(-3 + 2) = -2\text{ cm/min}$$
3. (b) Area $A = x \cdot y$: Using the product rule:$$\frac{dA}{dt} = \frac{dx}{dt} \cdot y + x \cdot \frac{dy}{dt}$$Substitute the known parameters:$$\frac{dA}{dt} = -3(6) + 10(2) = -18 + 20 = 2\text{ cm}^2/\text{min}$$

Final Evaluation Output: (a) The perimeter is decreasing at a rate of $2\text{ cm/min}$. (b) The area is increasing at a rate of $2\text{ cm}^2/\text{min}$.

Let $I$ be an interval contained in the domain of a real-valued function $f$. Then $f$ is said to be:

• increasing on $I$ if $x_1 < x_2 \text{ in } I \implies f(x_1) \le f(x_2)$ for all $x_1, x_2 \in I$.
• strictly increasing on $I$ if $x_1 < x_2 \text{ in } I \implies f(x_1) < f(x_2)$ for all $x_1, x_2 \in I$.
• decreasing on $I$ if $x_1 < x_2 \text{ in } I \implies f(x_1) \ge f(x_2)$ for all $x_1, x_2 \in I$.
• strictly decreasing on $I$ if $x_1 < x_2 \text{ in } I \implies f(x_1) > f(x_2)$ for all $x_1, x_2 \in I$.
• constant on $I$ if $f(x) = c$ for all $x \in I$, where $c$ is a constant.

Let $f$ be continuous on $[a, b]$ and differentiable on the open interval $(a, b)$. Then:

1. $f$ is increasing in $[a, b]$ if $f'(x) \ge 0$ for each $x \in (a, b)$.
2. $f$ is decreasing in $[a, b]$ if $f'(x) \le 0$ for each $x \in (a, b)$.
3. $f$ is a constant function in $[a, b]$ if $f'(x) = 0$ for each $x \in (a, b)$.

Mathematical Problem Statement: Find the intervals in which the function $f$ given by $f(x) = x^2 - 4x + 6$ is (a) strictly increasing, (b) strictly decreasing.

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Step-by-Step Algorithmic Solution Trace:

1. Differentiate the function: $f'(x) = 2x - 4$.
2. Set $f'(x) = 0$ to find the critical point: $2x - 4 = 0 \implies x = 2$.
3. The point $x = 2$ divides the real line into two disjoint intervals: $(-\infty, 2)$ and $(2, \infty)$.
4. Test the sign of $f'(x)$ in each interval:
   • In $(-\infty, 2)$, choose $x = 0 \implies f'(0) = -4 < 0$. Thus, $f$ is strictly decreasing.
   • In $(2, \infty)$, choose $x = 3 \implies f'(3) = 2 > 0$. Thus, $f$ is strictly increasing.

Final Evaluation Output: $f$ is strictly decreasing in $(-\infty, 2]$ and strictly increasing in $[2, \infty)$.

Let $f$ be a real-valued function and let $c$ be an interior point in the domain of $f$.

• $c$ is a point of local maxima if there exists an $h > 0$ such that $f(c) \ge f(x)$ for all $x \in (c - h, c + h)$. The value $f(c)$ is called the local maximum value of $f$.
• $c$ is a point of local minima if there exists an $h > 0$ such that $f(c) \le f(x)$ for all $x \in (c - h, c + h)$. The value $f(c)$ is called the local minimum value of $f$.
• A point $c$ in the domain of a function $f$ at which either $f'(c) = 0$ or $f$ is not differentiable is called a critical point of $f$.

Let $f$ be continuous at a critical point $c$ in an open interval $I$.

1. If $f'(x)$ changes sign from positive to negative as $x$ increases through $c$, then $c$ is a point of local maxima.
2. If $f'(x)$ changes sign from negative to positive as $x$ increases through $c$, then $c$ is a point of local minima.
3. If $f'(x)$ does not change sign as $x$ increases through $c$, then $c$ is defined as a point of inflection.

Let $f$ be a function defined on an interval $I$ and $c \in I$. Let $f$ be twice differentiable at $c$.

1. $x = c$ is a point of local maxima if $f'(c) = 0$ and $f''(c) < 0$.
2. $x = c$ is a point of local minima if $f'(c) = 0$ and $f''(c) > 0$.
3. The test fails if $f'(c) = 0$ and $f''(c) = 0$. In this case, we revert to the First Derivative Test.

Mathematical Problem Statement: Find the local maximum and local minimum values of the function $f(x) = 3x^4 + 4x^3 - 12x^2 + 12$.

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Step-by-Step Algorithmic Solution Trace:

1. Find the first derivative of $f(x)$:$$f'(x) = 12x^3 + 12x^2 - 24x = 12x(x - 1)(x + 2)$$
2. Set $f'(x) = 0$ to find the critical points: $x = 0, x = 1, x = -2$.
3. Find the second derivative of $f(x)$:$$f''(x) = 36x^2 + 24x - 24 = 12(3x^2 + 2x - 2)$$
4. Apply the Second Derivative Test at each critical point:
   • At $x = 0$: $f''(0) = -24 < 0 \implies x = 0$ is a point of local maxima.
   • At $x = 1$: $f''(1) = 36 > 0 \implies x = 1$ is a point of local minima.
   • At $x = -2$: $f''(-2) = 72 > 0 \implies x = -2$ is a point of local minima.
5. Calculate the local maximum and minimum values:
   • Local maximum value at $x = 0$: $f(0) = 12$.
   • Local minimum value at $x = 1$: $f(1) = 3(1)^4 + 4(1)^3 - 12(1)^2 + 12 = 7$.
   • Local minimum value at $x = -2$: $f(-2) = -20$.

Final Evaluation Output: Local maximum value is $12$ (at $x = 0$), and local minimum values are $7$ (at $x = 1$) and $-20$ (at $x = -2$).

1. Find all critical points of $f$ in the interval $(a, b)$ where either $f'(x) = 0$ or $f$ is not differentiable.
2. Take the endpoints $a$ and $b$ of the interval.
3. Calculate the values of $f(x)$ at all points identified in Steps 1 and 2.
4. Identify the maximum and minimum values from this set. The largest value is the absolute maximum value, and the smallest is the absolute minimum value.

Mathematical Problem Statement: Find the absolute maximum and minimum values of the function $f(x) = 2x^3 - 15x^2 + 36x + 1$ on the interval $[1, 5]$.

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Step-by-Step Algorithmic Solution Trace:

1. Differentiate the function:$$f'(x) = 6x^2 - 30x + 36 = 6(x - 3)(x - 2)$$
2. Find critical points by setting $f'(x) = 0$: $x = 2, x = 3$. Both lie within $[1, 5]$.
3. Evaluate $f(x)$ at the critical points and endpoints ($x = 1, 2, 3, 5$):
   • $f(1) = 2(1)^3 - 15(1)^2 + 36(1) + 1 = 24$
   • $f(2) = 2(2)^3 - 15(2)^2 + 36(2) + 1 = 29$
   • $f(3) = 2(3)^3 - 15(3)^2 + 36(3) + 1 = 28$
   • $f(5) = 2(5)^3 - 15(5)^2 + 36(5) + 1 = 56$
4. Compare the values: the maximum value is $56$ and the minimum value is $24$.

Final Evaluation Output: The absolute maximum value of $f$ on $[1, 5]$ is $56$ (occurring at $x = 5$), and the absolute minimum value is $24$ (occurring at $x = 1$).

Mathematical Problem Statement: Find two positive numbers whose sum is $15$ and the sum of whose squares is minimum.

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Step-by-Step Algorithmic Solution Trace:

1. Let the first number be $x$. Then the second number is $(15 - x)$.
2. Define the function $S(x)$ representing the sum of their squares:$$S(x) = x^2 + (15 - x)^2 = 2x^2 - 30x + 225$$
3. Find the first derivative of $S(x)$:$$S'(x) = 4x - 30$$
4. Set $S'(x) = 0$ to find the critical point: $4x - 30 = 0 \implies x = \frac{15}{2}$.
5. Find the second derivative to verify minimality:$$S''(x) = 4 > 0$$Since $S''(x) > 0$, $x = \frac{15}{2}$ is a point of local minima.
6. Calculate the two numbers: First number $= \frac{15}{2}$, Second number $= 15 - \frac{15}{2} = \frac{15}{2}$.

Final Evaluation Output: The two positive numbers are $\frac{15}{2}$ and $\frac{15}{2}$.