Step 1: Identifying the Mathematical Relationship

The area $A$ of a circle as a function of its radius $r$ is given by the geometric formula: $$A = \pi r^2$$ [Per the definition of the area of a circle in Euclidean geometry]

Step 2: Differentiating with respect to $r$

To find the rate of change of the area with respect to the radius, we calculate the derivative $\frac{dA}{dr}$. Using the Power Rule for differentiation, $\frac{d}{dx}(x^n) = nx^{n-1}$:

Step 3: Evaluation at the specified radius

We are required to evaluate the rate of change at $r = 6$ cm. Substituting the value into the derived expression:

Visual Representation of the Rate of Change

Unit Analysis

Since the area $A$ is measured in $\text{cm}^2$ and the radius $r$ is measured in $\text{cm}$, the units of the rate of change $\frac{dA}{dr}$ are:

Thus, the final value is $12\pi \text{ cm}^2/\text{cm}$, which simplifies to $12\pi \text{ cm}$.

Final Solution: The rate of change of the area of the circle with respect to its radius at $r = 6$ cm is $12\pi \text{ cm}$.

Step 1: Identifying the Mathematical Relationship

The area $A$ of a circle as a function of its radius $r$ is given by the formula: $A = \pi r^2$ [Per the geometric definition of the area of a circle].

Step 2: Defining the Rate of Change

We are required to find the rate of change of the area $A$ with respect to the radius $r$. Mathematically, this is expressed as the derivative $\frac{dA}{dr}$. To find this, we differentiate $A$ with respect to $r$: $\frac{dA}{dr} = \frac{d}{dr}(\pi r^2)$

Step 3: Executing the Differentiation

Applying the Power Rule for differentiation, $\frac{d}{dx}(x^n) = nx^{n-1}$ [Fundamental Theorem of Calculus/Differentiation Rules]: $\frac{dA}{dr} = \pi \cdot (2r^{2-1})$ $\frac{dA}{dr} = 2\pi r$

Step 4: Substitution of Given Conditions

We must evaluate the rate of change when $r = 3$ cm. Substituting $r = 3$ into the expression derived in Step 3: $\left( \frac{dA}{dr} \right)_{r=3} = 2\pi(3)$ $\left( \frac{dA}{dr} \right)_{r=3} = 6\pi$

Visual Representation of the Rate Function

Unit Analysis

Since the area is measured in $\text{cm}^2$ and the radius is measured in $\text{cm}$, the rate of change $\frac{dA}{dr}$ is measured in units of: $\frac{[\text{Area}]}{[\text{Radius}]} = \frac{\text{cm}^2}{\text{cm}} = \text{cm}$

Final Solution: The rate of change of the area of the circle with respect to its radius when $r = 3$ cm is $6\pi \text{ cm}^2/\text{cm}$, or simply $6\pi \text{ cm}$.

Step 1: Given Variables and Rates of Change

Based on the problem statement, we define the variables and their time-dependent derivatives as follows:

• Length of the rectangle: $x = 8 \text{ cm}$
• Width of the rectangle: $y = 6 \text{ cm}$
• Rate of change of length: $\frac{dx}{dt} = -5 \text{ cm/min}$ [Negative sign indicates decrease]
• Rate of change of width: $\frac{dy}{dt} = 4 \text{ cm/min}$ [Positive sign indicates increase]

Step 2: Defining the Perimeter Function

The perimeter $P$ of a rectangle is given by the formula:

$P = 2(x + y)$

[By the geometric definition of a rectangle perimeter]

Step 3: Differentiation with respect to time ($t$)

To find the rate of change of the perimeter, we differentiate both sides of the equation with respect to $t$ using the Chain Rule [Per the derivative operator properties]:

$\frac{dP}{dt} = \frac{d}{dt} [2(x + y)]$

$\frac{dP}{dt} = 2 \left( \frac{dx}{dt} + \frac{dy}{dt} \right)$

Step 4: Substituting Values and Calculation

Substitute the given rates $\frac{dx}{dt} = -5$ and $\frac{dy}{dt} = 4$ into the derived expression:

$\frac{dP}{dt} = 2 (-5 + 4)$

$\frac{dP}{dt} = 2 (-1)$

$\frac{dP}{dt} = -2 \text{ cm/min}$

Visual Representation of Rates

Step 5: Interpretation

The negative result signifies that the perimeter is decreasing at the specified moment. [Per the interpretation of first-order derivatives in physical applications].

Final Solution: The rate of change of the perimeter is $-2 \text{ cm/min}$, which indicates the perimeter is decreasing at a rate of $2 \text{ cm/min}$.

Step 1: Theoretical Foundation of Marginal Revenue

In economic mathematics, the total revenue function $R(x)$ represents the total income from selling $x$ units. Marginal Revenue ($MR$) is defined as the instantaneous rate of change of the total revenue with respect to the number of units sold. Mathematically, this is the first derivative of the revenue function: $MR = \frac{dR}{dx}$ [Per the definition of Marginal Analysis in calculus].

Step 2: Defining the Given Revenue Function

The total revenue function is provided as: $R(x) = 3x^2 + 36x + 5$

Step 3: Differentiation of the Revenue Function

To find the marginal revenue function $MR(x)$, we differentiate $R(x)$ with respect to $x$ using the Power Rule of differentiation, which states $\frac{d}{dx}(x^n) = nx^{n-1}$ [By the Power Rule of Calculus]:

$\frac{dR}{dx} = \frac{d}{dx}(3x^2) + \frac{d}{dx}(36x) + \frac{d}{dx}(5)$

$\frac{dR}{dx} = 3(2x^{2-1}) + 36(1x^{1-1}) + 0$

$MR(x) = 6x + 36$

Step 4: Evaluation at the Specified Quantity

The question requires the calculation of marginal revenue when $x = 15$. We substitute $x = 15$ into the derived expression for $MR(x)$:

$MR(15) = 6(15) + 36$

$MR(15) = 90 + 36$

$MR(15) = 126$

Visual Representation of the Rate of Change

Step 5: Analysis of Results

The marginal revenue represents the revenue generated by selling one additional unit when 15 units have already been produced. The value $126$ indicates the slope of the revenue curve at that specific point.

Final Solution: The Marginal Revenue at $x = 15$ is ₹ 126.

Step 1: Identifying Given Variables and Geometric Relationships

Let $V$ be the volume of the conical pile of sand, $r$ be the radius of its base, and $h$ be its height at any time $t$.

• The rate of change of volume is given as $\frac{dV}{dt} = 12 \text{ cm}^3/\text{s}$.
• The geometric relationship provided is $h = \frac{1}{6}r$, which implies $r = 6h$.
• We are required to find the rate of change of height $\frac{dh}{dt}$ when $h = 4 \text{ cm}$.

Step 2: Establishing the Volume Formula

The volume $V$ of a cone is given by the formula [Geometric Formula for Volume of a Right Circular Cone]:

$V = \frac{1}{3}\pi r^2 h$

Substituting $r = 6h$ into the volume equation to express $V$ as a function of $h$ solely:

$V = \frac{1}{3}\pi (6h)^2 h$

$V = \frac{1}{3}\pi (36h^2) h$

$V = 12\pi h^3$

Step 3: Differentiation with respect to time $t$

Differentiating both sides with respect to $t$ using the [Chain Rule of Calculus]:

$\frac{dV}{dt} = \frac{d}{dt} (12\pi h^3)$

$\frac{dV}{dt} = 12\pi \cdot 3h^2 \cdot \frac{dh}{dt}$

$\frac{dV}{dt} = 36\pi h^2 \frac{dh}{dt}$

Step 4: Substitution and Solving for $\frac{dh}{dt}$

Substitute the known values $\frac{dV}{dt} = 12$ and $h = 4$ into the differentiated equation:

$12 = 36\pi (4)^2 \frac{dh}{dt}$

$12 = 36\pi (16) \frac{dh}{dt}$

$12 = 576\pi \frac{dh}{dt}$

Isolating $\frac{dh}{dt}$:

$\frac{dh}{dt} = \frac{12}{576\pi}$

Reducing the fraction by $12$:

$\frac{dh}{dt} = \frac{1}{48\pi} \text{ cm/s}$

Final Solution: The height of the sand cone is increasing at the rate of $\frac{1}{48\pi} \text{ cm/s}$ when the height is $4 \text{ cm}$.

Step 1: Given Variables and Rates of Change

Based on the problem statement, we define the following variables and their temporal derivatives (rates of change with respect to time $t$):

• Length of the rectangle: $x = 8\text{ cm}$
• Width of the rectangle: $y = 6\text{ cm}$
• Rate of change of length: $\frac{dx}{dt} = -5\text{ cm/min}$ [Negative sign denotes decreasing rate]
• Rate of change of width: $\frac{dy}{dt} = 4\text{ cm/min}$ [Positive sign denotes increasing rate]

Step 2: Defining the Area Function

The area $A$ of a rectangle is defined as the product of its length and width:

$A = x \cdot y$

Step 3: Differentiation with Respect to Time

To find the rate of change of the area with respect to time $\frac{dA}{dt}$, we apply the Product Rule of differentiation, which states that for a product of two functions $u$ and $v$, $\frac{d}{dt}(uv) = u\frac{dv}{dt} + v\frac{du}{dt}$ [Per the Leibniz Product Rule]:

$\frac{dA}{dt} = \frac{d}{dt}(x \cdot y)$

$\frac{dA}{dt} = x\left(\frac{dy}{dt}\right) + y\left(\frac{dx}{dt}\right)$

Step 4: Substitution and Calculation

Substitute the given values into the derived differential equation:

$\frac{dA}{dt} = (8\text{ cm}) \cdot (4\text{ cm/min}) + (6\text{ cm}) \cdot (-5\text{ cm/min})$

$\frac{dA}{dt} = 32\text{ cm}^2\text{/min} - 30\text{ cm}^2\text{/min}$

$\frac{dA}{dt} = 2\text{ cm}^2\text{/min}$

Visual Representation of the Rate Vectors

Final Solution: The area of the rectangle is increasing at the rate of $2\text{ cm}^2\text{/min}$.

Step 1: Establishing the Mathematical Model

An air bubble is assumed to be spherical in shape. The volume $V$ of a sphere with radius $r$ is given by the formula:

$V = \frac{4}{3} \pi r^3$ [Geometric formula for the volume of a sphere]

Step 2: Identifying Given Variables and Rates

From the problem statement, we have the following:

• Rate of change of radius with respect to time: $\frac{dr}{dt} = \frac{1}{2} \, \text{cm/s}$
• Instantaneous radius at which the rate of volume change is required: $r = 1 \, \text{cm}$
• We are required to find the rate of change of volume with respect to time: $\frac{dV}{dt}$

Step 3: Differentiation with Respect to Time

To find the rate of change of volume, we differentiate $V$ with respect to time $t$ using the Chain Rule [Per the Chain Rule for differentiation, $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$]:

$\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right)$

$\frac{dV}{dt} = \frac{4}{3} \pi \cdot (3r^2) \cdot \frac{dr}{dt}$ [By the Power Rule $\frac{d}{dx}x^n = nx^{n-1}$]

$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$

Step 4: Substitution and Calculation

Substitute the known values ($r = 1 \, \text{cm}$ and $\frac{dr}{dt} = \frac{1}{2} \, \text{cm/s}$) into the derived expression:

$\frac{dV}{dt} = 4 \pi (1)^2 \cdot \left( \frac{1}{2} \right)$

$\frac{dV}{dt} = 4 \pi \cdot 1 \cdot \frac{1}{2}$

$\frac{dV}{dt} = 2 \pi \, \text{cm}^3/\text{s}$

Step 5: Visual Representation

Final Solution: The volume of the air bubble is increasing at the rate of $2\pi \, \text{cm}^3/\text{s}$.

Given Variables & Initial Setup

Let $r$ represent the radius of the circle at any time $t$, and $A$ represent the area of the circle. We are provided with the following information:

• Rate of change of the radius with respect to time: $\frac{dr}{dt} = 3 \text{ cm/s}$
• Specific radius at which the rate of change of area is required: $r = 10 \text{ cm}$

Step 1: Establishing the Geometric Relationship

The area $A$ of a circle in terms of its radius $r$ is given by the formula [Per the standard formula for the area of a circle]:

$A = \pi r^2$

Step 2: Differentiating with respect to Time ($t$)

To find the rate of change of the area with respect to time, we apply the Chain Rule of calculus [By the Chain Rule: $\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}$]:

$\frac{d}{dt}(A) = \frac{d}{dt}(\pi r^2)$

$\frac{dA}{dt} = \pi \cdot \frac{d}{dr}(r^2) \cdot \frac{dr}{dt}$

$\frac{dA}{dt} = \pi \cdot (2r) \cdot \frac{dr}{dt}$

$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$

Step 3: Calculating the Result at $r = 10 \text{ cm}$

Substitute the given values $r = 10 \text{ cm}$ and $\frac{dr}{dt} = 3 \text{ cm/s}$ into the derived equation:

$\frac{dA}{dt} = 2 \cdot \pi \cdot (10 \text{ cm}) \cdot (3 \text{ cm/s})$

$\frac{dA}{dt} = 60\pi \text{ cm}^2/\text{s}$

Visual Representation of the Expansion

Summary of Calculation

Final Solution: The area of the circle is increasing at the rate of $60\pi \text{ cm}^2/\text{s}$.

Step 1: Establishing Variables and Constants

Let $x$ represent the length of the edge of the cube in cm. Let $V$ represent the volume of the cube and $S$ represent the surface area of the cube.

• Edge length at the given instant: $x = 12 \text{ cm}$
• Rate of change of volume: $\frac{dV}{dt} = 8 \text{ cm}^3/\text{s}$
• Objective: Find the rate of change of surface area, $\frac{dS}{dt}$, when $x = 12 \text{ cm}$

Step 2: Defining Geometric Formulas

The geometric formulas for a cube with edge $x$ are:

• Volume: $V = x^3$ [Geometric property of a cube]
• Surface Area: $S = 6x^2$ [Geometric property of a cube]

Step 3: Differentiating Volume with respect to Time

Differentiating $V = x^3$ with respect to time $t$ using the chain rule [Per the Chain Rule of Differentiation]:

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

Substitute the given values into the derivative expression:

$8 = 3(12)^2 \frac{dx}{dt}$

$8 = 3(144) \frac{dx}{dt}$

$8 = 432 \frac{dx}{dt}$

$\frac{dx}{dt} = \frac{8}{432} = \frac{1}{54} \text{ cm/s}$

Step 4: Differentiating Surface Area with respect to Time

Differentiating $S = 6x^2$ with respect to time $t$ using the chain rule:

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

Step 5: Final Computation

Substitute $x = 12$ and $\frac{dx}{dt} = \frac{1}{54}$ into the equation derived in Step 4:

$\frac{dS}{dt} = 12(12) \left( \frac{1}{54} \right)$

$\frac{dS}{dt} = \frac{144}{54}$

Simplifying the fraction by dividing both numerator and denominator by their greatest common divisor ($18$):

$\frac{dS}{dt} = \frac{144 \div 18}{54 \div 18} = \frac{8}{3} \text{ cm}^2/\text{s}$

Final Solution: The surface area is increasing at a rate of $\frac{8}{3} \text{ cm}^2/\text{s}$.

Step 1: Establishing the Geometric Variables

The volume $V$ of a sphere with radius $r$ is given by the formula: $V = \frac{4}{3}\pi r^3$ [Standard formula for volume of a sphere]

The problem states the diameter $D$ is a function of $x$: $D(x) = \frac{3}{2}(2x + 1)$

Since the radius $r$ is half of the diameter, we define $r$ as: $r = \frac{D}{2} = \frac{1}{2} \cdot \left[ \frac{3}{2}(2x + 1) \right] = \frac{3}{4}(2x + 1)$

Step 2: Formulating the Volume Function

Substitute the expression for $r$ into the volume formula to express $V$ as a function of $x$: $V(x) = \frac{4}{3}\pi \left[ \frac{3}{4}(2x + 1) \right]^3$

Expanding the cube: $V(x) = \frac{4}{3}\pi \cdot \left( \frac{3}{4} \right)^3 \cdot (2x + 1)^3$ $V(x) = \frac{4}{3}\pi \cdot \frac{27}{64} \cdot (2x + 1)^3$

Simplifying the coefficients: $V(x) = \left( \frac{4}{64} \right) \cdot \left( \frac{27}{3} \right) \cdot \pi \cdot (2x + 1)^3$ $V(x) = \frac{1}{16} \cdot 9 \cdot \pi \cdot (2x + 1)^3$ $V(x) = \frac{9\pi}{16}(2x + 1)^3$

Step 3: Differentiation with Respect to $x$

To find the rate of change of volume with respect to $x$, we compute $\frac{dV}{dx}$ using the Chain Rule [Per the Chain Rule of differentiation, $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$]:

$\frac{dV}{dx} = \frac{d}{dx} \left[ \frac{9\pi}{16}(2x + 1)^3 \right]$ $\frac{dV}{dx} = \frac{9\pi}{16} \cdot \frac{d}{dx} \left[ (2x + 1)^3 \right]$

Applying the Power Rule $\frac{d}{dx}u^n = nu^{n-1} \frac{du}{dx}$: $\frac{dV}{dx} = \frac{9\pi}{16} \cdot 3(2x + 1)^2 \cdot \frac{d}{dx}(2x + 1)$ $\frac{dV}{dx} = \frac{27\pi}{16} \cdot (2x + 1)^2 \cdot (2)$

Step 4: Final Simplification

$\frac{dV}{dx} = \frac{54\pi}{16} (2x + 1)^2$ Reducing the fraction $\frac{54}{16}$ by a factor of 2: $\frac{dV}{dx} = \frac{27\pi}{8} (2x + 1)^2$

Final Solution: The rate of change of the volume with respect to $x$ is $\frac{27\pi}{8}(2x + 1)^2$.

Step 1: Establishing the Geometric Model

The object is a sphere with a variable radius $r$. The volume $V$ of a sphere is defined by the geometric formula:

$V = \frac{4}{3} \pi r^3$

[By the geometric definition of the volume of a sphere]

Step 2: Defining the Rate of Change

We are required to find the rate of change of volume $V$ with respect to the radius $r$. This is expressed mathematically as the derivative $\frac{dV}{dr}$.

Applying the power rule for differentiation, $\frac{d}{dx}(x^n) = nx^{n-1}$, where $n=3$:

$\frac{dV}{dr} = \frac{d}{dr} \left( \frac{4}{3} \pi r^3 \right)$

$\frac{dV}{dr} = \frac{4}{3} \pi \cdot (3r^2)$

$\frac{dV}{dr} = 4 \pi r^2$

[By the Power Rule of Differentiation and the linearity of the derivative operator]

Step 3: Evaluating at the Specific Point

Given that the radius $r = 10 \text{ cm}$, we substitute this value into the derived expression for the rate of change:

$\left. \frac{dV}{dr} \right|_{r=10} = 4 \pi (10)^2$

$\left. \frac{dV}{dr} \right|_{r=10} = 4 \pi (100)$

$\left. \frac{dV}{dr} \right|_{r=10} = 400 \pi$

Step 4: Dimensional Analysis and Verification

The units for volume are $\text{cm}^3$ and for radius are $\text{cm}$. Therefore, the rate of change $\frac{dV}{dr}$ is expressed in $\text{cm}^3/\text{cm}$, which simplifies to $\text{cm}^2$.

Final Solution: The rate at which the volume of the balloon is increasing with respect to the radius is $400 \pi \text{ cm}^2$.

Step 1: Variables Definition and Geometric Setup

Let the wall lie along the $y$-axis and the ground along the $x$-axis. Let the position of the ladder be represented by the hypotenuse of a right-angled triangle, where the base $x$ represents the distance of the foot of the ladder from the wall, and the height $y$ represents the height of the ladder on the wall.

• Length of the ladder, $L = 5\text{ m}$.
• Distance of the foot of the ladder from the wall: $x$.
• Height of the ladder on the wall: $y$.
• Rate of change of the foot of the ladder: $\frac{dx}{dt} = 2\text{ cm/s} = 0.02\text{ m/s}$.
• Goal: Find the rate of decrease of height, $-\frac{dy}{dt}$, when $x = 4\text{ m}$.

Step 2: Establishing the Mathematical Relationship

By the [Pythagorean Theorem], in a right-angled triangle with hypotenuse $L$:

$x^2 + y^2 = L^2$

$x^2 + y^2 = 5^2 = 25$ --- (Equation 1)

Step 3: Differentiation with Respect to Time ($t$)

Differentiating Equation 1 with respect to $t$ using the [Chain Rule]:

$\frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(25)$

$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$

$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$ --- (Equation 2)

Step 4: Determining Values at the Specific Instant

When $x = 4\text{ m}$, we substitute into Equation 1 to find $y$:

$4^2 + y^2 = 25$

$16 + y^2 = 25 \implies y^2 = 9 \implies y = 3\text{ m}$

Step 5: Solving for $\frac{dy}{dt}$

Substitute $x = 4\text{ m}$, $y = 3\text{ m}$, and $\frac{dx}{dt} = 0.02\text{ m/s}$ into Equation 2:

$4(0.02) + 3 \left( \frac{dy}{dt} \right) = 0$

$0.08 + 3 \frac{dy}{dt} = 0$

$3 \frac{dy}{dt} = -0.08$

$\frac{dy}{dt} = -\frac{0.08}{3} \text{ m/s} \approx -0.0267 \text{ m/s}$

Converting back to cm/s: $-\frac{0.08}{3} \times 100 = -\frac{8}{3} \text{ cm/s} \approx -2.67 \text{ cm/s}$.

Final Solution: The height of the ladder on the wall is decreasing at the rate of $\frac{8}{3}$ cm/s (or approximately $2.67$ cm/s).

Given Variables & Initial Setup

• Volume of a spherical balloon: $V = \frac{4}{3}\pi r^3$
• Rate of change of volume: $\frac{dV}{dt} = 900 \text{ cm}^3/\text{s}$
• Specific radius at the moment of interest: $r = 15 \text{ cm}$
• Target variable: Rate of change of radius, $\frac{dr}{dt}$

Step 1: Establishing the Mathematical Relationship

To find the relationship between the rate of change of volume and the rate of change of radius, we differentiate the volume formula with respect to time $t$ using the Chain Rule [Per the Chain Rule in Calculus]:

$\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3}\pi r^3 \right)$

$\frac{dV}{dt} = \frac{4}{3}\pi \cdot (3r^2) \cdot \frac{dr}{dt}$

$\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}$

Step 2: Substitution of Known Values

Substitute $\frac{dV}{dt} = 900$ and $r = 15$ into the derived equation:

$900 = 4\pi (15)^2 \cdot \frac{dr}{dt}$

Step 3: Solving for $\frac{dr}{dt}$

Calculate the square of the radius:

$900 = 4\pi (225) \cdot \frac{dr}{dt}$

$900 = 900\pi \cdot \frac{dr}{dt}$

Divide both sides by $900\pi$:

$\frac{dr}{dt} = \frac{900}{900\pi}$

$\frac{dr}{dt} = \frac{1}{\pi}$

Visual Representation of Spherical Expansion

Unit Analysis

Since the volume is in $\text{cm}^3/\text{s}$ and the radius is in $\text{cm}$, the rate of change of the radius must be in units of $\text{cm/s}$.

Final Solution: The radius of the balloon is increasing at a rate of $\frac{1}{\pi} \text{ cm/s}$ when the radius is $15 \text{ cm}$.

Step 1: Theoretical Definition of Marginal Cost

The marginal cost ($MC$) represents the instantaneous rate of change of the total cost function $C(x)$ with respect to the number of units $x$ produced. Mathematically, it is defined as the first derivative of the total cost function [Per the definition of derivative as the rate of change]:

$MC = \frac{d}{dx}[C(x)]$

Step 2: Differentiating the Cost Function

Given the total cost function: $C(x) = 0.007x^3 - 0.003x^2 + 15x + 4000$.

Applying the Power Rule $\frac{d}{dx}(ax^n) = anx^{n-1}$ to each term individually [Per the linearity property of differentiation]:

$\frac{d}{dx}(0.007x^3) = 0.007 \cdot 3x^2 = 0.021x^2$

$\frac{d}{dx}(-0.003x^2) = -0.003 \cdot 2x = -0.006x$

$\frac{d}{dx}(15x) = 15 \cdot 1 = 15$

$\frac{d}{dx}(4000) = 0$ [Since the derivative of a constant is zero]

Thus, $MC(x) = 0.021x^2 - 0.006x + 15$.

Step 3: Calculating Marginal Cost for $x = 17$

Substitute $x = 17$ into the expression for $MC(x)$:

$MC(17) = 0.021(17)^2 - 0.006(17) + 15$

$MC(17) = 0.021(289) - 0.102 + 15$

$MC(17) = 6.069 - 0.102 + 15$

$MC(17) = 5.967 + 15$

$MC(17) = 20.967$

Step 4: Visualizing the Marginal Cost Trend

Final Solution: The marginal cost when 17 units are produced is Rupees 20.967.

Step 1: Identification of Variables and Rates

Let $r$ represent the radius of the circular wave at any time $t$, and let $A$ represent the area of the circular wave at any time $t$.

• The rate at which the radius is increasing is given as: $\frac{dr}{dt} = 5 \text{ cm/s}$.
• We are required to find the rate of change of the area $\frac{dA}{dt}$ at the specific instant when $r = 8 \text{ cm}$.

Step 2: Establishing the Mathematical Relationship

The area $A$ of a circle with radius $r$ is defined by the geometric formula:

$A = \pi r^2$ [Definition of Area of a Circle]

Step 3: Differentiation with respect to Time ($t$)

To determine the rate of change of the area, we differentiate both sides of the equation with respect to $t$ using the Chain Rule of calculus:

$\frac{dA}{dt} = \frac{d}{dt}(\pi r^2)$

$\frac{dA}{dt} = \pi \cdot \frac{d}{dr}(r^2) \cdot \frac{dr}{dt}$ [By Chain Rule of Differentiation]

$\frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt}$

Step 4: Substitution and Calculation

Substitute the given values $r = 8 \text{ cm}$ and $\frac{dr}{dt} = 5 \text{ cm/s}$ into the derived expression:

$\frac{dA}{dt} = 2 \cdot \pi \cdot (8 \text{ cm}) \cdot (5 \text{ cm/s})$

$\frac{dA}{dt} = 2 \cdot \pi \cdot 40 \text{ cm}^2/\text{s}$

$\frac{dA}{dt} = 80\pi \text{ cm}^2/\text{s}$

Visual Representation

Step 5: Dimensional Analysis

The units are verified as follows: $[L] \times [L/T] = [L^2/T]$. Since the area is measured in $\text{cm}^2$ and time in $\text{s}$, the rate of change is in $\text{cm}^2/\text{s}$, which is consistent with the physics of the problem.

Final Solution: The enclosed area is increasing at a rate of $80\pi \text{ cm}^2/\text{s}$.

Step 1: Identifying Given Variables and Constraints

Let $r$ represent the radius of the circle and $C$ represent its circumference. The rate of change of the radius with respect to time $t$ is given as a constant value.

• Rate of change of radius: $\frac{dr}{dt} = 0.7 \, \text{cm/s}$
• Function of circumference: $C = 2\pi r$

Step 2: Mathematical Formulation of the Rate of Change

To find the rate of increase of the circumference, we must differentiate the circumference equation with respect to time $t$ using the Chain Rule of calculus [Per the Chain Rule of Differentiation: $\frac{d}{dt}f(r) = \frac{df}{dr} \cdot \frac{dr}{dt}$].

Taking the derivative of both sides of $C = 2\pi r$ with respect to $t$:

$$\frac{dC}{dt} = \frac{d}{dt}(2\pi r)$$

Since $2$ and $\pi$ are constants, we extract them from the derivative operator:

$$\frac{dC}{dt} = 2\pi \cdot \frac{dr}{dt}$$

Step 3: Geometric Visualization

Step 4: Substitution and Numerical Calculation

Substitute the given value $\frac{dr}{dt} = 0.7 \, \text{cm/s}$ into the derived equation:

$$\frac{dC}{dt} = 2 \cdot \pi \cdot (0.7)$$

$$\frac{dC}{dt} = 1.4\pi \, \text{cm/s}$$

Using the standard approximation $\pi \approx \frac{22}{7}$ for a numerical decimal representation:

$$\frac{dC}{dt} = 1.4 \cdot \frac{22}{7} = 0.2 \cdot 22 = 4.4 \, \text{cm/s}$$

Final Solution: The rate of increase of the circumference is $1.4\pi \, \text{cm/s}$ or approximately $4.4 \, \text{cm/s}$.

Step 1: Identification of Variables and Given Rates

Let the edge length of the variable cube be denoted by $x$ (in cm) and its volume be denoted by $V$ (in $\text{cm}^3$). We are given the rate of change of the edge with respect to time $t$:

$\frac{dx}{dt} = 3 \text{ cm/s}$

We are required to determine the rate of change of the volume, $\frac{dV}{dt}$, at the specific instant when:

$x = 10 \text{ cm}$

Step 2: Establishing the Mathematical Relationship

The volume $V$ of a cube with edge length $x$ is defined by the geometric formula:

$V = x^3$ [Geometric definition of cubic volume]

Step 3: Differentiation with Respect to Time

To find the rate of change of the volume, we differentiate the volume formula with respect to time $t$ using the Chain Rule of calculus [Per the Chain Rule of Differentiation]:

$\frac{d}{dt}(V) = \frac{d}{dt}(x^3)$

$\frac{dV}{dt} = 3x^2 \cdot \frac{dx}{dt}$ [Applying the Power Rule $n \cdot x^{n-1}$ followed by the Chain Rule]

Step 4: Substitution and Calculation

Substitute the known values ($x = 10 \text{ cm}$ and $\frac{dx}{dt} = 3 \text{ cm/s}$) into the derived derivative equation:

$\frac{dV}{dt} = 3(10 \text{ cm})^2 \cdot (3 \text{ cm/s})$

$\frac{dV}{dt} = 3(100 \text{ cm}^2) \cdot (3 \text{ cm/s})$

$\frac{dV}{dt} = 900 \text{ cm}^3\text{/s}$

Step 5: Visual Representation

Final Solution: The volume of the cube is increasing at a rate of 900 cm³/s.

Step 1: Theoretical Foundation of Marginal Revenue

The total revenue function $R(x)$ represents the total income generated from selling $x$ units. In economic mathematics, the Marginal Revenue (MR) is defined as the instantaneous rate of change of the total revenue with respect to the number of units sold. Mathematically, this is expressed as the derivative of the revenue function with respect to $x$:

$MR = \frac{dR}{dx}$ [Per the definition of marginal analysis in calculus]

Step 2: Differentiation of the Revenue Function

Given the revenue function:

$R(x) = 13x^2 + 26x + 15$

To find the marginal revenue, we apply the power rule of differentiation, $\frac{d}{dx}(ax^n) = anx^{n-1}$, to each term in the polynomial:

$\frac{dR}{dx} = \frac{d}{dx}(13x^2) + \frac{d}{dx}(26x) + \frac{d}{dx}(15)$

$\frac{dR}{dx} = 13(2x^{2-1}) + 26(1x^{1-1}) + 0$ [By the constant rule of differentiation, $\frac{d}{dx}(C) = 0$]

$MR(x) = 26x + 26$

Step 3: Calculating Marginal Revenue at $x = 7$

To find the specific marginal revenue when $7$ units are sold, we substitute $x = 7$ into the derived function $MR(x)$:

$MR(7) = 26(7) + 26$

$MR(7) = 182 + 26$

$MR(7) = 208$

Visual Representation of the Rate of Change

Final Solution: The marginal revenue when $x = 7$ is Rs $208$.

Step 1: Establishing the Mathematical Model

Let the area of a circle be denoted by $A$ and its radius by $r$. The relationship between the area and the radius is defined by the fundamental geometric formula:

$A = \pi r^2$ [Per the definition of the area of a circle in Euclidean geometry]

Step 2: Defining the Rate of Change

To find the rate of change of the area with respect to the radius, we must calculate the derivative of $A$ with respect to $r$. This is denoted as $\frac{dA}{dr}$.

$\frac{dA}{dr} = \frac{d}{dr}(\pi r^2)$

Applying the Power Rule of differentiation, where $\frac{d}{dr}(r^n) = nr^{n-1}$ [By the Power Rule of Calculus]:

$\frac{dA}{dr} = \pi \cdot (2r^{2-1})$

$\frac{dA}{dr} = 2\pi r$

Step 3: Calculating at the Given Condition

The problem requests the rate of change when the radius $r = 4$ cm. Substituting $r = 4$ into our derivative function:

$\left. \frac{dA}{dr} \right|_{r=4} = 2 \cdot \pi \cdot 4$

$\left. \frac{dA}{dr} \right|_{r=4} = 8\pi$

Step 4: Dimensional Analysis and Visual Representation

The units for $\frac{dA}{dr}$ are calculated as $\frac{\text{units of Area}}{\text{units of Length}} = \frac{\text{cm}^2}{\text{cm}} = \text{cm}$.

Final Solution: The rate of change of the area of the circle with respect to its radius when $r = 4$ cm is $8\pi$ $\text{cm}^2/\text{cm}$ (or $8\pi$ $\text{cm}$).

Step 1: Establishing the Mathematical Relationship

The movement of the particle is constrained by the curve equation:

$6y = x^3 + 2$

Differentiating both sides of the equation with respect to time $t$ using the Chain Rule [Per the rules of Implicit Differentiation]:

$\frac{d}{dt}(6y) = \frac{d}{dt}(x^3 + 2)$

$6 \frac{dy}{dt} = 3x^2 \frac{dx}{dt} + 0$

$6 \frac{dy}{dt} = 3x^2 \frac{dx}{dt}$

Step 2: Incorporating the Given Rate of Change

The problem states that the $y$-coordinate is changing $8$ times as fast as the $x$-coordinate. This is expressed mathematically as:

$\frac{dy}{dt} = 8 \frac{dx}{dt}$

Substitute this relation into the differentiated equation obtained in Step 1:

$6(8 \frac{dx}{dt}) = 3x^2 \frac{dx}{dt}$

$48 \frac{dx}{dt} = 3x^2 \frac{dx}{dt}$

Step 3: Solving for the $x$-coordinate

Assuming the particle is in motion ($\frac{dx}{dt} \neq 0$), we divide both sides by $\frac{dx}{dt}$:

$48 = 3x^2$

$x^2 = \frac{48}{3}$

$x^2 = 16$

$x = \pm 4$

Step 4: Calculating Corresponding $y$-coordinates

We substitute the values of $x$ back into the original curve equation $6y = x^3 + 2$ to find the corresponding points.

Case 1: When $x = 4$

$6y = (4)^3 + 2$

$6y = 64 + 2$

$6y = 66 \implies y = 11$

Point: $(4, 11)$

Case 2: When $x = -4$

$6y = (-4)^3 + 2$

$6y = -64 + 2$

$6y = -62 \implies y = -\frac{62}{6} = -\frac{31}{3}$

Point: $(-4, -\frac{31}{3})$

Visual Representation of the Curve

Final Solution: The required points on the curve are $(4, 11)$ and $(-4, -\frac{31}{3})$.