Applications of Derivatives: Rate of Change and Marginals

Derivatives as a Rate Measure (Rate of Change of Quantities)

The derivative of a function provides a powerful tool for analyzing how quantities change. One of the most direct applications of the derivative is to represent the instantaneous rate of change of one variable with respect to another. This concept is widely used in various fields, including physics, engineering, economics, and biology.

Fundamental Interpretation

Recall the definition of the derivative. If $y$ is a function of $x$, denoted by $y = f(x)$, the derivative $\frac{dy}{dx} = f'(x)$ captures the instantaneous rate at which $y$ changes for a unit change in $x$, exactly at a specific value of $x$.

• The average rate of change of $y$ with respect to $x$ over an interval $[x_1, x_2]$ is the change in $y$ divided by the change in $x$:

  Average Rate of Change $= \frac{\Delta y}{\Delta x} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$

  If we consider a small interval starting at $x$ with length $\Delta x = h$, the interval is $[x, x+h]$, and the average rate of change is $\frac{f(x+h) - f(x)}{h}$.

• The instantaneous rate of change of $y$ with respect to $x$ at a specific value $x=a$ is the limit of the average rate of change as the interval shrinks to zero ($\Delta x \to 0$ or $h \to 0$). This is the definition of the derivative at $x=a$.

  Instantaneous Rate of Change at $x=a = f'(a) = \lim\limits_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a}$

  For a general value of $x$, the instantaneous rate of change is $f'(x)$.

The units of the instantaneous rate of change $\frac{dy}{dx}$ are the units of the dependent variable $y$ divided by the units of the independent variable $x$. For example, if $y$ is measured in meters (m) and $x$ in seconds (s), $\frac{dy}{dx}$ would be in m/s.

Examples in Different Contexts

The interpretation of the derivative as a rate of change appears naturally in many different fields:

1. Motion (Physics): If the position $s$ of an object is a function of time $t$, $s = f(t)$, then:

   • Velocity: $v(t) = \frac{ds}{dt} = f'(t)$. This is the instantaneous rate of change of position with respect to time. Units are distance/time (e.g., meters per second, km/hour).
   • Acceleration: $a(t) = \frac{dv}{dt} = v'(t) = s''(t)$. This is the instantaneous rate of change of velocity with respect to time (the second derivative of position). Units are velocity/time (e.g., m/s², km/hour²).

2. Geometry: Consider how geometric quantities change as their dimensions change.

   • Area of a circle: $A(r) = \pi r^2$, where $r$ is the radius. The rate of change of area with respect to radius is $\frac{dA}{dr} = 2\pi r$. This is the circumference. Imagine increasing the radius by a small amount $\Delta r$; the added area is approximately a thin ring with area $2\pi r \Delta r$.
   • Volume of a sphere: $V(r) = \frac{4}{3}\pi r^3$. The rate of change of volume with respect to radius is $\frac{dV}{dr} = 4\pi r^2$. This is the surface area. Increasing the radius by $\Delta r$ adds approximately a thin spherical shell with volume $4\pi r^2 \Delta r$.

3. Biology / Population Studies: If $P(t)$ represents the population size of a species at time $t$, then $\frac{dP}{dt} = P'(t)$ is the instantaneous rate of population growth (or decline) at time $t$. Units could be individuals per year, bacteria per hour, etc.

4. Chemistry: If $C$ is the concentration of a reactant in a chemical reaction at time $t$, $C = C(t)$, then $\frac{dC}{dt} = C'(t)$ is the instantaneous rate of reaction. Units are concentration per unit time (e.g., moles per litre per second). The rate might be negative if the substance is being consumed.

5. Economics: As discussed in a later section, rates of change of cost, revenue, and profit are called marginals.

Answer:

Related Rates Problems

Related rates problems are a common application of differentiation where we are given the rate of change of one or more quantities and are asked to find the rate of change of another quantity. The key is that these quantities are related to each other by some equation. Differentiation with respect to time ($t$), along with the Chain Rule, is the tool used to solve these problems.

In essence, if quantities are related by an equation, their rates of change are also related by the derivative of that equation with respect to time.

Problem-Solving Strategy for Related Rates

Solving related rates problems systematically is crucial. Here is a step-by-step approach:

1. Understand the Problem and Identify Variables:
   • Read the problem carefully to understand the physical situation and identify all quantities that are changing. Assign variable names to these quantities.
   • Identify any quantities that remain constant throughout the process.
   • Determine what rate(s) are given (e.g., $\frac{dx}{dt}$, $\frac{dV}{dt}$) and what rate you need to find (e.g., $\frac{dy}{dt}$, $\frac{dr}{dt}$).
   • Note any specific moment in time or specific values of the variables at which you need to find the unknown rate.
2. Draw a Diagram:
   • Sketch a diagram of the situation if possible. This helps visualize the relationships between the variables.
   • Label the diagram with the assigned variables. If a quantity is constant, label it with its constant value. If a quantity is changing, label it with its variable name.
3. Write Down the Given Information and What You Need to Find:
   • Translate the given rates of change into mathematical notation using derivatives with respect to time $t$ (e.g., "volume increases at 100 cm³/s" means $\frac{dV}{dt} = 100$). Pay attention to units and signs (e.g., "decreasing" means a negative rate).
   • Write down the rate you need to find in the same notation (e.g., find $\frac{dr}{dt}$).
   • List the values of the variables at the specific moment in question. These values should **not** be substituted into the relating equation until *after* differentiation.
4. Find a Relating Equation:
   • Establish an equation that relates the variables involved in the problem. This equation describes how the quantities are connected at *any* moment in time during the process.
   • This equation often comes from geometric formulas (e.g., area, volume, Pythagorean theorem, similar triangles), physical laws, or other given relationships. Ensure the equation involves the variables whose rates are known and the variable whose rate you need to find.
5. Differentiate the Relating Equation Implicitly with Respect to Time ($t$):
   • Differentiate both sides of the relating equation found in Step 4 with respect to the time variable $t$.
   • This is the crucial step where you apply implicit differentiation. Remember that all variables representing quantities that change with time are functions of $t$. When you differentiate a term involving a variable, say $V$, with respect to $t$, you differentiate $V$ with respect to $V$ (using the appropriate rule) and then multiply by $\frac{dV}{dt}$ (by the Chain Rule). For example, $\frac{d}{dt}(V) = \frac{dV}{dt}$, and $\frac{d}{dt}(r^3) = 3r^2 \frac{dr}{dt}$.
   • Apply all standard differentiation rules (Power Rule, Product Rule, Quotient Rule, Chain Rule) as needed during this differentiation step. Remember that constants differentiate to zero, and variables that are constant throughout the process (not just at a specific moment) are treated as constants during differentiation.
6. Substitute Known Values and Solve for the Unknown Rate:
   • After differentiating, substitute all the known values for the variables *and* the known rates (from Step 3) into the differentiated equation from Step 5.
   • Solve the resulting algebraic equation for the unknown rate.
7. Check and Interpret the Answer:
   • Review your answer. Does the sign of the rate make sense in the context of the problem (positive for increasing quantities, negative for decreasing quantities)?
   • Make sure the units are correct (units of the quantity divided by units of time).
   • Write a concluding sentence that answers the original question in plain language, including the calculated rate and its units.

Answer:

Step 1 & 2: Understand and Diagram

The quantities changing are the volume ($V$) and the radius ($r$) of the spherical balloon. Both depend on time ($t$).

We want to find the rate of change of the radius, $\frac{dr}{dt}$. We are given the rate of change of the volume, $\frac{dV}{dt}$.

Draw a diagram of a sphere and label its radius $r$.

Step 3: Write Down Knowns and Unknowns

Given rate: The volume increases at 100 cm³/s. So, $\frac{dV}{dt} = 100$ cm³/s.

Rate to find: How fast is the radius increasing? Find $\frac{dr}{dt}$.

Specific moment: When the diameter is 50 cm. The radius at this moment is $r = \frac{\text{Diameter}}{2} = \frac{50}{2} = 25$ cm.

Step 4: Find a Relating Equation

The quantities $V$ and $r$ are related by the formula for the volume of a sphere:

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

[Volume of a sphere]

Step 5: Differentiate Implicitly w.r.t. Time ($t$)

Differentiate both sides of the equation $V = \frac{4}{3}\pi r^3$ with respect to $t$. Remember that $V$ and $r$ are functions of $t$.

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

[Differentiate both sides w.r.t. $t$]

Left side: $\frac{d}{dt}(V) = \frac{dV}{dt}$.

Right side: Use the Constant Multiple Rule and the Chain Rule. $\frac{4}{3}\pi$ is a constant. Differentiate $r^3$ with respect to $t$. The outer function is $u^3$, the inner is $r(t)$. $\frac{d}{du}(u^3) = 3u^2$. So, $\frac{d}{dt}(r^3) = 3r^2 \cdot \frac{dr}{dt}$.

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

[Applying Chain Rule to $r^3$]

Simplify the equation:

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

... (i)

Step 6: Substitute Known Values and Solve

Now, substitute the values known at the specific moment in question into the differentiated equation (i):

Given $\frac{dV}{dt} = 100$ cm³/s.

At the moment of interest, $r = 25$ cm.

"$100 = 4\pi (25)^2 \frac{dr}{dt}$"

[Substitute values into (i)]

Calculate $(25)^2 = 625$:

"$100 = 4\pi (625) \frac{dr}{dt}$"

"$100 = 2500\pi \frac{dr}{dt}$"

Solve for $\frac{dr}{dt}$ by dividing both sides by $2500\pi$:

"$\frac{dr}{dt} = \frac{100}{2500\pi}$"

[Solve for $\frac{dr}{dt}$]

Simplify the fraction:

"$\frac{dr}{dt} = \frac{1}{25\pi}$"

Step 7: Check and Interpret

The calculated rate $\frac{dr}{dt} = \frac{1}{25\pi}$ is positive, which is consistent with the balloon's volume increasing (meaning the radius must also be increasing). The units are cm/s, which are appropriate units for a rate of change of radius with respect to time.

Conclusion: The radius of the balloon is increasing at a rate of $\frac{1}{25\pi}$ cm/s when the diameter is 50 cm.

Marginal Cost and Marginal Revenue using Derivatives (Economic Applications)

In economics, differentiation is a powerful tool for analyzing the impact of small changes in production or sales levels on total cost, total revenue, and total profit. The concept of "marginal" in economics specifically refers to the rate of change of a total quantity with respect to a change in the number of units produced or sold. In calculus terms, marginal quantities are represented by the derivatives of the corresponding total functions.

Cost Function and Marginal Cost

Let $x$ represent the number of units of a product produced or sold. The Total Cost Function, denoted by $C(x)$, gives the total cost incurred in producing $x$ units.

The Marginal Cost (MC) is defined as the rate of change of the total cost with respect to the number of units produced. Using differentiation, the marginal cost at a production level of $x$ units is the derivative of the total cost function:

Marginal Cost $= MC(x) = C'(x) = \frac{dC}{dx}$

Interpretation of Marginal Cost:

The derivative $C'(x)$ gives the instantaneous rate of change of total cost at the production level $x$. In practical economic terms, $C'(x)$ represents the approximate cost of producing one additional unit when the current production level is $x$.

That is, $C'(x) \approx C(x+1) - C(x)$, which is the cost added by producing the $(x+1)$th unit.

For instance, if $C'(100) = \textsf{₹} 50$, it means that when 100 units are being produced, the estimated cost to produce the 101st unit is approximately $\textsf{₹} 50$.

Revenue Function and Marginal Revenue

Let $x$ again represent the number of units of a product sold. The Total Revenue Function, denoted by $R(x)$, gives the total revenue earned from selling $x$ units.

The revenue function is often related to the price per unit. If $p(x)$ is the price per unit when $x$ units are demanded (this is the demand function, price depends on quantity sold), then the total revenue is the price per unit multiplied by the number of units sold:

$R(x) = p(x) \cdot x$

The Marginal Revenue (MR) is defined as the rate of change of the total revenue with respect to the number of units sold. Using differentiation, the marginal revenue at a sales level of $x$ units is the derivative of the total revenue function:

Marginal Revenue $= MR(x) = R'(x) = \frac{dR}{dx}$

Interpretation of Marginal Revenue:

The derivative $R'(x)$ gives the instantaneous rate of change of total revenue at the sales level $x$. In economic terms, $R'(x)$ represents the approximate additional revenue generated by selling one more unit when the current sales level is $x$.

That is, $R'(x) \approx R(x+1) - R(x)$, which is the revenue added by selling the $(x+1)$th unit.

For instance, if $R'(50) = \textsf{₹} 80$, it means that when 50 units are being sold, the estimated additional revenue from selling the 51st unit is approximately $\textsf{₹} 80$.

Profit Function and Marginal Profit

The Total Profit Function, denoted by $P(x)$, is the difference between the total revenue and the total cost for producing and selling $x$ units:

$P(x) = R(x) - C(x)$

The Marginal Profit (MP) is defined as the rate of change of the total profit with respect to the number of units produced and sold. It is the derivative of the total profit function:

Marginal Profit $= MP(x) = P'(x) = \frac{dP}{dx}$

Using the properties of derivatives (Difference Rule), the marginal profit is also the difference between marginal revenue and marginal cost:

$P'(x) = \frac{d}{dx}(R(x) - C(x)) = \frac{d}{dx}(R(x)) - \frac{d}{dx}(C(x)) = R'(x) - C'(x)$

Marginal Profit $= MP(x) = MR(x) - MC(x)$

Interpretation of Marginal Profit:

$P'(x)$ represents the approximate additional profit obtained from producing and selling one more unit when the current level is $x$. It is the net gain (or loss) from the next unit.

In economic theory, profit is often maximized at the production/sales level $x$ where Marginal Revenue equals Marginal Cost ($MR(x) = MC(x)$), provided that the second derivative of the profit function is negative at that point (indicating a maximum).

Answer: