Calculus, a branch of mathematics that deals with the study of continuous change, has been a cornerstone of modern mathematics and science. It consists of two main branches: Differential Calculus, which concerns rates of change and slopes of curves, and Integral Calculus, which deals with the accumulation of quantities. The application of calculus is vast, ranging from physics and engineering to economics and computer science. Here, we delve into seven calculus problems, exploring both differential and integral calculus, and provide step-by-step solutions to understand the concepts better.
Problem 1: Finding the Derivative of a Function
Find the derivative of the function (f(x) = 3x^2 + 2x - 5).
Solution: To find the derivative, we apply the power rule of differentiation, which states that if (f(x) = x^n), then (f’(x) = nx^{n-1}). - For (3x^2), the derivative is (3 \cdot 2x^{2-1} = 6x). - For (2x), the derivative is (2 \cdot 1x^{1-1} = 2). - The derivative of a constant is 0, so the derivative of (-5) is (0). Therefore, (f’(x) = 6x + 2).
Problem 2: Optimization Problem
A farmer wants to fence a rectangular area of 100 square meters. What are the dimensions of the rectangle that will minimize the amount of fencing used?
Solution: Let (x) and (y) be the length and width of the rectangle, respectively. The perimeter (P = 2x + 2y), which we want to minimize, and the area (A = xy = 100). From (A = xy = 100), we get (y = \frac{100}{x}). Substituting (y) in the perimeter equation gives (P = 2x + 2\left(\frac{100}{x}\right)). To find the minimum, we take the derivative of (P) with respect to (x) and set it equal to 0: [P’ = 2 - \frac{200}{x^2} = 0] [2 = \frac{200}{x^2}] [x^2 = 100] [x = 10] Since (y = \frac{100}{x} = \frac{100}{10} = 10), the dimensions that minimize the fencing are (x = y = 10) meters, resulting in a square.
Problem 3: Evaluating a Definite Integral
Evaluate the definite integral (\int_{0}^{2} (x^2 + 1) dx).
Solution: The integral of (x^2 + 1) with respect to (x) is (\frac{x^3}{3} + x + C). Evaluating this from 0 to 2: [\left[\frac{(2)^3}{3} + 2\right] - \left[\frac{(0)^3}{3} + 0\right]] [= \left[\frac{8}{3} + 2\right] - [0]] [= \frac{8}{3} + 2] [= \frac{8}{3} + \frac{6}{3}] [= \frac{14}{3}]
Problem 4: Finding the Area Between Two Curves
Find the area between the curves (y = x^2) and (y = x + 1) from (x = 0) to (x = 1).
Solution: First, we need to determine which curve is on top within the given interval. Let’s evaluate both functions at (x = 0) and (x = 1): - At (x = 0), (y = 0^2 = 0) and (y = 0 + 1 = 1). So, (y = x + 1) is on top. - At (x = 1), (y = 1^2 = 1) and (y = 1 + 1 = 2). So, (y = x + 1) is still on top. The area (A) between the two curves from (x = 0) to (x = 1) is given by the integral: [A = \int{0}^{1} [(x + 1) - x^2] dx] [= \int{0}^{1} (x + 1 - x^2) dx] [= \left[\frac{x^2}{2} + x - \frac{x^3}{3}\right]_{0}^{1}] [= \left(\frac{1}{2} + 1 - \frac{1}{3}\right) - 0] [= \frac{1}{2} + 1 - \frac{1}{3}] [= \frac{3}{6} + \frac{6}{6} - \frac{2}{6}] [= \frac{7}{6}]
Problem 5: Volume of a Solid of Revolution
Find the volume of the solid formed by revolving the region bounded by (y = x^2), (x = 1), and (x = 2) about the x-axis.
Solution: The volume (V) of the solid formed by revolving the region about the x-axis is given by: [V = \pi \int{a}^{b} [f(x)]^2 dx] Here, (f(x) = x^2), (a = 1), and (b = 2). [V = \pi \int{1}^{2} (x^2)^2 dx] [= \pi \int{1}^{2} x^4 dx] [= \pi \left[\frac{x^5}{5}\right]{1}^{2}] [= \pi \left(\frac{2^5}{5} - \frac{1^5}{5}\right)] [= \pi \left(\frac{32}{5} - \frac{1}{5}\right)] [= \pi \left(\frac{31}{5}\right)] [= \frac{31\pi}{5}]
Problem 6: Finding the Equation of a Curve
Given that (y = f(x)) and (\frac{dy}{dx} = 3x^2 - 2x), find the equation of the curve if it passes through the point ((1, 2)).
Solution: To find (y), we integrate (\frac{dy}{dx}) with respect to (x): [y = \int (3x^2 - 2x) dx] [= x^3 - x^2 + C] Given the curve passes through ((1, 2)), we substitute (x = 1) and (y = 2): [2 = (1)^3 - (1)^2 + C] [2 = 1 - 1 + C] [2 = C] So, the equation of the curve is (y = x^3 - x^2 + 2).
Problem 7: Applications in Physics - Motion Along a Line
A particle moves along a line with its position given by (s(t) = t^3 - 2t^2 + t + 1), where (s) is in meters and (t) is in seconds. Find the velocity and acceleration of the particle at (t = 2) seconds.
Solution: The velocity (v(t)) is the derivative of (s(t)) with respect to time: [v(t) = \frac{ds}{dt} = 3t^2 - 4t + 1] At (t = 2), the velocity is: [v(2) = 3(2)^2 - 4(2) + 1] [= 12 - 8 + 1] [= 5 \, \text{m/s}] The acceleration (a(t)) is the derivative of (v(t)): [a(t) = \frac{dv}{dt} = 6t - 4] At (t = 2), the acceleration is: [a(2) = 6(2) - 4] [= 12 - 4] [= 8 \, \text{m/s}^2]
These problems illustrate the diversity and depth of calculus, from finding derivatives and integrals to applying these concepts to real-world problems in physics and optimization. Each problem requires a deep understanding of the underlying principles of calculus and the ability to apply these principles in a logical and systematic way.
