Unit 3: Calculus
Differentiation and its Applications
Differentiation is the process of finding the derivative of a function. For a function \( y = f(x) \), its derivative \( f’(x) \) or \( \frac{dy}{dx} \) is defined as:
\[ f’(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \]
Maxima and Minima
A common application of derivatives is finding the local maximum and minimum values of a function.
- Find the first derivative \( f’(x) \).
- Set \( f’(x) = 0 \) to find the critical points.
- Use the second derivative test \( f’‘(x) \) to determine if it is a maximum (\( f’‘(x) < 0 \)) or minimum (\( f’’(x) > 0 \)).
Integration and its Applications
Integration is the reverse process of differentiation. The indefinite integral of a function represents a family of curves, while the definite integral gives the area under a curve.
\[ \int f(x) dx = F(x) + C \] where \( F’(x) = f(x) \).
Consumer and Producer Surplus
- Consumer Surplus: Area representing the difference between what consumers are willing to pay and what they actually pay.
- Producer Surplus: Area representing the difference between the actual price and the minimum price at which producers are willing to supply the good.
\[ CS = \int_{0}^{x_0} f(x) dx - p_0 \cdot x_0 \]
Differential Equations
A differential equation contains the derivative of an unknown function. We solve simple differential equations using the Variable Separable Method.
Competency-Based Questions
Q1. A company calculates that its marginal cost for producing \( x \) units is \( MC(x) = 3x^2 - 10x + 5 \). Find the total cost function \( C(x) \) if the fixed cost is ₹ 500. Formulate the differential equation and solve it.
Q2. The equilibrium price and quantity for a commodity are ₹ 15 and 200 units, respectively. The demand function is given by \( p = 35 - 0.1x \). Calculate the consumer surplus at the point of market equilibrium.
Q3. A population model predicts that the rate of change of the population \( P \) with respect to time \( t \) is proportional to \( P \). Write down the differential equation governing this growth. If initially the population is 10,000 and it doubles in 5 years, determine the population after 10 years using integration.
Q4. A firm’s revenue function is \( R(x) = -2x^2 + 400x \). Evaluate the quantity \( x \) that maximizes the revenue. Plot the graph and show the maxima point.
Answers
Ans 1. \( C(x) = \int (3x^2 - 10x + 5) dx = x^3 - 5x^2 + 5x + C_1 \). Since fixed cost \( C(0) = 500 \), \( C_1 = 500 \). So \( C(x) = x^3 - 5x^2 + 5x + 500 \).
Ans 2. Consumer Surplus \( = \int_0^{200} (35 - 0.1x) dx - (15 \times 200) = [35x - 0.05x^2]_0^{200} - 3000 = (7000 - 2000) - 3000 = 2000 \). CS is ₹ 2,000.
Ans 3. Diff. Equation: \( \frac{dP}{dt} = kP \). Solving gives \( P(t) = P_0 e^{kt} \). If it doubles in 5 years, in 10 years it doubles twice, meaning \( 10000 \times 4 = 40,000 \).
Ans 4. Find derivative: \( R’(x) = -4x + 400 \). Setting to 0 gives \( x = 100 \). Second derivative \( R’’(x) = -4 < 0 \), confirming it’s a maxima.