Problem 33 - Calculus Early Transcendentals

Question

What differentiable functions have an arc length on the interval \([a, b]\) given by the following integrals? Note that the answers are not unique. Give a family of functions that satisfy the conditions. a. \(\int_{a}^{b} \sqrt{1+16 x^{4}} d x \quad\) b. \(\int_{a}^{b} \sqrt{1+36 \cos ^{2} 2 x} d x\)

Step-by-Step Solution

Answer

Question: Given the following arc length integrals of differentiable functions over the interval [a, b], find a family of functions that satisfy the conditions: a. ∫(a to b) √(1 + 16x^4) dx b. ∫(a to b) √(1 + 36cos²(2x)) dx Answer: The family of differentiable functions satisfying the conditions are: a. f(x) = (4/3)x^3 + C b. f(x) = 3sin(2x) + C

Step 1: Identify the derivative of the function from the given arc length integral

For part (a), the given arc length integral is \(\int_{a}^{b} \sqrt{1+16 x^{4}} d x\). Comparing it with the arc length formula, we have \(1+(f'(x))^2 = 1+16x^4\). Solve for \(f'(x)\): \(f'(x)^2 = 16x^4\) Take the square root and get \(f'(x) = 4x^2\).

Step 2: Calculate the function

Now, we need to find the function \(f(x)\) by integrating the derivative \(f'(x)\): \(f(x) = \int f'(x)dx = \int 4x^2 dx\) This integral gives us: \(f(x) = \frac{4}{3}x^3 + C\), where \(C\) is the constant of integration. Thus, the family of functions satisfying the given condition is \(f(x) = \frac{4}{3}x^3 + C\). Now, let's move on to part b.

Step 3: Identify the derivative of the function from the given arc length integral

For part (b), the given arc length integral is \(\int_{a}^{b} \sqrt{1 + 36\cos^2(2x)} dx\). Comparing it with the arc length formula, we have \(1+(f'(x))^2 = 1+36\cos^2(2x)\). Solve for \(f'(x)\): \(f'(x)^2 = 36\cos^2(2x)\) Take the square root and get \(f'(x) = 6\cos(2x)\).

Step 4: Calculate the function

Now, we need to find the function \(f(x)\) by integrating the derivative \(f'(x)\): \(f(x) = \int f'(x) dx = \int 6\cos(2x) dx\) This integral gives us: \(f(x) = 3\sin(2x) + C\), where \(C\) is the constant of integration. Thus, the family of functions satisfying the given condition is \(f(x) = 3\sin(2x) + C\). In conclusion, the family of differentiable functions satisfying the conditions are: a. \(f(x) = \frac{4}{3}x^3 + C\) b. \(f(x) = 3\sin(2x) + C\)

Key Concepts

Differentiable Functions

Differentiable functions play a crucial role in understanding and solving problems in calculus, including the calculation of arc lengths. A differentiable function is one that has a derivative at each point within its domain. This means that the function's graph has a tangent line at every point, and the function is smooth with no sharp corners or discontinuities. Differentiability implies continuity, but not vice versa. It's important because it allows us to apply calculus principles like differentiation and integration to find function properties effectively. In solving for arc lengths, knowing if a function is differentiable helps ensure reliable calculations when deriving or integrating.

Integral Calculus

Integral calculus is a branch of calculus focusing on finding functions and quantities related to areas and accumulation. Its primary operation is integration, which is essentially the reverse process of differentiation. In the context of arc lengths, integral calculus helps to calculate the length of a curve by integrating functions. To find the arc length on a specific interval, we rely on solving an integral that often contains a square root for calculating the small changes along the arc. Mastering integral calculus allows us to analyze and understand complex geometric shapes and curves effectively.

Arc Length Formula

The arc length formula is a foundational tool in calculus used to determine the length of a curve between two points on a plane. The general formula is \(\int_{a}^{b} \sqrt{1+(f'(x))^2} \, dx\), where \(f'(x)\) is the derivative of the function defining the curve. This formula essentially sums up the slanted line elements of a curve, taking into account both horizontal and vertical changes. In practical terms, it requires determining the derivative of the function first, then applying it within the integral. Calculating an arc length involves substituting this derivative into the formula, integrating over the given interval, and interpreting the result geometrically.

Integrating Functions

Integrating functions is the process of finding the antiderivative, or integral, of a given function. It is a key skill in calculus that allows us to reverse the process of differentiation. In practice, when dealing with the arc length, you will often integrate to find the original function from its derivative, as seen in the example problems. This is done by setting up the integral with the known derivative, calculating the integral to find the function, and including a constant of integration to represent a family of functions. Mastery of integration techniques, such as substitution or partial fraction decomposition, is essential for solving a broad range of problems in calculus, including the computation of arc lengths.