Rate of Change of Quantities

Meaning of Derivative as Rate of Change

Recall that the derivative $\frac{ds}{dt}$ represents the rate of change of distance $s$ with respect to time $t$. Similarly, whenever a quantity $y$ varies with another quantity $x$ satisfying some rule $y = f(x)$, then $\frac{dy}{dx}$ (or $f'(x)$) represents the rate of change of $y$ with respect to $x$.

• $\left. \frac{dy}{dx} \right|_{x=x_0}$ or $f'(x_0)$ represents the rate of change of $y$ with respect to $x$ precisely at the instant when $x = x_0$.
• Positive Rate: If $\frac{dy}{dx}$ is positive, it means $y$ is increasing as $x$ increases.
• Negative Rate: If $\frac{dy}{dx}$ is negative, it means $y$ is decreasing as $x$ increases.
• Zero Rate: If $\frac{dy}{dx}=0$ at a point, then at that instant the quantity $y$ is not changing with respect to $x$.

Related Rates (Parametric Rate of Change)

Often, two variables $x$ and $y$ are both varying with respect to a third variable, usually time $t$. If $x$ and $y$ are linked by a given equation (for example, $y = f(x)$ or an implicit relation involving $x$ and $y$), we can find how their rates of change with respect to time are related by differentiating both sides with respect to $t$ using the Chain Rule:

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

This concept is extensively used in physics and geometry, such as finding the rate at which the volume of a balloon increases when the radius is expanding at a known rate.

Important idea

In related-rates problems, we usually:

1. Write the relation between the variables.
2. Differentiate with respect to time $t$.
3. Substitute the numerical values at the required instant.
4. Interpret the sign of the answer.

Marginal Cost and Marginal Revenue

Derivatives have direct applications in business and economics to find marginal values.

Marginal Cost (MC)

It is the instantaneous rate of change of the total cost $C(x)$ with respect to the number of items $x$ produced. $$MC = \frac{dC}{dx}$$

Marginal Revenue (MR)

It is the instantaneous rate of change of the total revenue $R(x)$ with respect to the number of items $x$ sold. $$MR = \frac{dR}{dx}$$

In practical terms, the marginal value tells us approximately how much the total cost or total revenue changes when production or sales increase by one additional unit.

Example 1: Rate of Change of Area of a Circle

Find the rate of change of the area of a circle with respect to its radius $r$ when $r = 5 \text{ cm}$.

Solution: Step 1: Write the formula for the area of a circle. The area $A$ of a circle with radius $r$ is $A = \pi r^2$.

Step 2: Differentiate the area with respect to the radius $r$. $$\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = \pi \cdot 2r = 2\pi r.$$

Step 3: Evaluate the derivative at the specific point $r = 5 \text{ cm}$. $$\left. \frac{dA}{dr} \right|_{r=5} = 2\pi (5) = 10\pi.$$

Answer: $10\pi \text{ cm}^2/\text{cm}$. This means the area is changing at a rate of $10\pi$ square centimeters for every 1 cm change in radius.

Example 2: Volume of a Cube

The volume of a cube is increasing at a rate of $8 \text{ cm}^3/\text{s}$. How fast is the surface area increasing when the length of an edge is $12 \text{ cm}$?

Solution: Step 1: Let $x$ be the length of an edge, $V$ be the volume, and $S$ be the surface area of the cube at any time $t$. Given: $$\frac{dV}{dt} = 8 \text{ cm}^3/\text{s}.$$ We know: $$V = x^3, \qquad S = 6x^2.$$

Step 2: Differentiate $V$ with respect to time $t$ using the chain rule. $$\frac{dV}{dt} = \frac{d}{dt}(x^3) = 3x^2 \frac{dx}{dt}.$$

Step 3: Substitute the known rate $\frac{dV}{dt} = 8$ to find $\frac{dx}{dt}$. $$8 = 3x^2 \frac{dx}{dt} \implies \frac{dx}{dt} = \frac{8}{3x^2}.$$

Step 4: Differentiate $S$ with respect to time $t$. $$\frac{dS}{dt} = \frac{d}{dt}(6x^2) = 12x \frac{dx}{dt}.$$

Step 5: Substitute $\frac{dx}{dt}$ into the $\frac{dS}{dt}$ equation. $$\frac{dS}{dt} = 12x \left( \frac{8}{3x^2} \right) = \frac{32}{x}.$$

Step 6: Evaluate when $x = 12 \text{ cm}$. $$\left. \frac{dS}{dt} \right|_{x=12} = \frac{32}{12} = \frac{8}{3}.$$

Answer: $\frac{8}{3} \text{ cm}^2/\text{s}$.

Example 3: Expanding Ripples in a Lake

A stone is dropped into a quiet lake and waves move in circles at a speed of $4 \text{ cm/s}$. At the instant when the radius of the circular wave is $10 \text{ cm}$, how fast is the enclosed area increasing?

Solution: Step 1: Let $r$ be the radius and $A$ be the area of the circular wave. Given: $$\frac{dr}{dt} = 4 \text{ cm/s}.$$ The area is: $$A = \pi r^2.$$

Step 2: Differentiate $A$ with respect to time $t$. $$\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \cdot \frac{dr}{dt}.$$

Step 3: Substitute the known values $r = 10$ and $\frac{dr}{dt} = 4$. $$\frac{dA}{dt} = 2\pi (10)(4) = 80\pi.$$

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

Example 4: Changing Dimensions of a Rectangle

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.

Solution: Step 1: Identify the given rates. Since length $x$ is decreasing, $$\frac{dx}{dt} = -3 \text{ cm/min}.$$ Since width $y$ is increasing, $$\frac{dy}{dt} = 2 \text{ cm/min}.$$

Step 2: Part (a) - Rate of change of perimeter $P$. The perimeter is: $$P = 2(x+y).$$ Differentiate with respect to time: $$\frac{dP}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right).$$ Substitute the values: $$\frac{dP}{dt} = 2(-3+2) = 2(-1) = -2 \text{ cm/min}.$$ The negative sign means the perimeter is decreasing.

Step 3: Part (b) - Rate of change of area $A$. The area is: $$A = xy.$$ Differentiate using the product rule: $$\frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt}.$$ Substitute $x=10$, $y=6$, $\frac{dx}{dt}=-3$, $\frac{dy}{dt}=2$: $$\frac{dA}{dt} = 10(2) + 6(-3) = 20 - 18 = 2 \text{ cm}^2/\text{min}.$$ The positive sign means the area is increasing.

Answer: (a) Perimeter is decreasing at $2 \text{ cm/min}$. (b) Area is increasing at $2 \text{ cm}^2/\text{min}$.

Example 5: Marginal Cost

The total cost $C(x)$ in Rupees associated with the production of $x$ units of an item is given by $C(x) = 0.005x^3 - 0.02x^2 + 30x + 5000$. Find the marginal cost when $3$ units are produced.

Solution: Step 1: Recall the definition of marginal cost. $$MC = \frac{dC}{dx}.$$

Step 2: Differentiate the cost function. $$MC = \frac{d}{dx}(0.005x^3 - 0.02x^2 + 30x + 5000)$$ $$MC = 3(0.005)x^2 - 2(0.02)x + 30$$ $$MC = 0.015x^2 - 0.04x + 30.$$

Step 3: Evaluate at $x=3$. $$MC = 0.015(3)^2 - 0.04(3) + 30$$ $$MC = 0.015(9) - 0.12 + 30 = 0.135 - 0.12 + 30 = 30.015.$$

Answer: The marginal cost when 3 units are produced is ₹ 30.015.

Example 6: Marginal Revenue

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

Solution: Step 1: Recall the definition of marginal revenue. $$MR = \frac{dR}{dx}.$$

Step 2: Differentiate the revenue function. $$MR = \frac{d}{dx}(3x^2 + 36x + 5) = 6x + 36.$$

Step 3: Evaluate at $x=5$. $$MR = 6(5) + 36 = 30 + 36 = 66.$$

Answer: The marginal revenue when 5 units are sold is ₹ 66.

Example 7: Volume of a Sphere

A balloon, which always remains spherical, has a variable radius. Find the rate at which its volume is increasing with respect to its radius when the radius is $10 \text{ cm}$.

Solution: Step 1: Write the formula for the volume of a sphere. $$V = \frac{4}{3}\pi r^3.$$

Step 2: Differentiate with respect to radius $r$. $$\frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi \cdot 3r^2 = 4\pi r^2.$$

Step 3: Evaluate at $r=10 \text{ cm}$. $$\left. \frac{dV}{dr} \right|_{r=10} = 4\pi (10)^2 = 400\pi.$$

Answer: $400\pi \text{ cm}^3/\text{cm}$.

Example 8: The Sliding Ladder Problem

A ladder $5 \text{ m}$ long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of $2 \text{ cm/s}$. How fast is its height on the wall decreasing when the foot of the ladder is $4 \text{ m}$ away from the wall?

Solution: Step 1: Let $x$ be the distance of the bottom of the ladder from the wall and $y$ be the height of the top of the ladder on the wall. The ladder length is constant, so $$x^2 + y^2 = 5^2 = 25.$$ Use consistent units. Since the given rate is $2 \text{ cm/s}$, convert it to meters per second: $$\frac{dx}{dt} = 0.02 \text{ m/s}.$$

Step 2: When $x=4 \text{ m}$, find $y$. $$4^2 + y^2 = 25 \implies 16 + y^2 = 25 \implies y^2 = 9 \implies y = 3 \text{ m}.$$

Step 3: Differentiate the equation $x^2+y^2=25$ with respect to time $t$. $$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0.$$ Dividing by 2, $$x\frac{dx}{dt} + y\frac{dy}{dt} = 0.$$

Step 4: Substitute the values $x=4$, $y=3$, $\frac{dx}{dt}=0.02$. $$4(0.02) + 3\frac{dy}{dt} = 0$$ $$0.08 + 3\frac{dy}{dt} = 0$$ $$3\frac{dy}{dt} = -0.08$$ $$\frac{dy}{dt} = -\frac{0.08}{3} = -\frac{2}{75} \text{ m/s}.$$

Step 5: Convert to cm/s if desired. $$-\frac{2}{75} \text{ m/s} = -\frac{200}{75} \text{ cm/s} = -\frac{8}{3} \text{ cm/s}.$$ The negative sign shows the height is decreasing.

Answer: The height is decreasing at the rate of $\frac{8}{3} \text{ cm/s}$.

Example 9: Particle Moving on a Curve

A particle moves along the curve $6y = x^3 + 2$. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

Solution: Step 1: Translate the statement about rates. "The y-coordinate is changing 8 times as fast as the x-coordinate" means $$\frac{dy}{dt} = 8\frac{dx}{dt}.$$

Step 2: Differentiate the curve equation with respect to time $t$. Given: $$6y = x^3 + 2.$$ Differentiating, $$6\frac{dy}{dt} = 3x^2\frac{dx}{dt}.$$

Step 3: Substitute $\frac{dy}{dt} = 8\frac{dx}{dt}$. $$6\left(8\frac{dx}{dt}\right) = 3x^2\frac{dx}{dt}.$$ Assuming the particle is actually moving so that $\frac{dx}{dt} \neq 0$, cancel $\frac{dx}{dt}$: $$48 = 3x^2 \implies x^2 = 16 \implies x = \pm 4.$$

Step 4: Find the corresponding $y$-coordinates. If $x=4$, $$6y = 4^3 + 2 = 64 + 2 = 66 \implies y = 11.$$ If $x=-4$, $$6y = (-4)^3 + 2 = -64 + 2 = -62 \implies y = -\frac{31}{3}.$$

Answer: The points are $(4, 11)$ and $(-4, -31/3)$.

Example 10: Sand Pouring to Form a Cone

Sand is pouring from a pipe at the rate of $12 \text{ cm}^3/\text{s}$. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is $4 \text{ cm}$?

Solution: Step 1: Let $V$ be the volume, $r$ be the base radius, and $h$ be the height of the cone. Given: $$\frac{dV}{dt} = 12 \text{ cm}^3/\text{s}.$$ Also, the height is always one-sixth of the radius: $$h = \frac{1}{6}r \implies r = 6h.$$ We need to find $\frac{dh}{dt}$ when $h=4 \text{ cm}$.

Step 2: Use the volume formula for a cone. $$V = \frac{1}{3}\pi r^2 h.$$ Substitute $r=6h$ so that the formula contains only $h$: $$V = \frac{1}{3}\pi (6h)^2 h = \frac{1}{3}\pi (36h^2)h = 12\pi h^3.$$

Step 3: Differentiate with respect to time. $$\frac{dV}{dt} = \frac{d}{dt}(12\pi h^3) = 36\pi h^2 \frac{dh}{dt}.$$

Step 4: Substitute the known values $\frac{dV}{dt}=12$ and $h=4$. $$12 = 36\pi (4)^2 \frac{dh}{dt} = 36\pi (16) \frac{dh}{dt} = 576\pi \frac{dh}{dt}.$$

Step 5: Solve for $\frac{dh}{dt}$. $$\frac{dh}{dt} = \frac{12}{576\pi} = \frac{1}{48\pi} \text{ cm/s}.$$

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