Problem 11 - Calculus: Early Transcendentals

Question

Find the arc length of the function on the given interval. $$ f(x)=\sqrt{8} x \text { on }[-1,1] $$

Step-by-Step Solution

Answer

The arc length is 6.

Step 1: Understand the Arc Length Formula

The formula to find the arc length of a function \( f(x) \) over an interval \([a, b]\) is given by \( L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx \). Here, \[a = -1, \ b = 1\].We need to first find the derivative \( f'(x) \) of \( f(x) = \sqrt{8} x \).

Step 2: Find the Derivative

The derivative \( f'(x) \) of \( f(x) = \sqrt{8} x \) is simply the constant coefficient of \( x \), which is \( \sqrt{8} \).Thus, \( f'(x) = \sqrt{8} \).

Step 3: Plug the Derivative into the Formula

Now that we have \( f'(x) = \sqrt{8} \), substitute it into the arc length formula.\[ L = \int_{-1}^{1} \sqrt{1 + (\sqrt{8})^2} \, dx \]This simplifies to \[ L = \int_{-1}^{1} \sqrt{1 + 8} \, dx = \int_{-1}^{1} \sqrt{9} \, dx = \int_{-1}^{1} 3 \, dx \].

Step 4: Integrate to Find Arc Length

Calculate the integral of the constant function \( 3 \) over the interval \[-1, 1\]:\[ L = 3 \int_{-1}^{1} 1 \, dx = 3 \left[ x \right]_{-1}^{1} \].Evaluate this to find the arc length:\[ L = 3([1 - (-1)]) = 3 \times 2 = 6. \]

Step 5: Final Step: Conclusion

Thus, the arc length of the function \( f(x) = \sqrt{8} x \) on the interval \([-1,1]\) is 6.

Key Concepts

Calculus

Calculus is a branch of mathematics that focuses on rates of change and the accumulation of quantities. It is built on two fundamental concepts: differentiation and integration. Understanding calculus is essential for solving problems that involve motion, area, volume, and other complex ideas. One of the key applications of calculus is finding the arc length of a curve, as shown in the exercise. The arc length formula in calculus helps to determine the length of a segment of a curve. It involves the integral of the square root of the sum of one plus the square of the derivative of the function. This highlights how both the rate of change (derivative) and accumulation (integral) are intertwined in calculus. It is an important skill to learn how to apply these principles to various mathematical problems.

Derivatives

Derivatives, a core concept within calculus, measure how a function changes as its input changes. In simpler terms, a derivative gives us the rate at which a function is changing at any point, often referred to as the slope. When dealing with arc length, we use derivatives to understand how steep or flat a function is at particular points along its curve. In the exercise, the derivative of the function \( f(x) = \sqrt{8} x \) is constant, \( f'(x) = \sqrt{8} \), since it’s a linear function. This means the rate of change is uniform across the interval considered. Knowing how to compute derivatives allows us to better analyze and understand the behavior of functions, ultimately making it an indispensable tool for finding solutions to real-world problems.

Integration

Integration is the process of finding the integral of a function. It is essentially the reverse of differentiation and is used to calculate areas, volumes, and other quantities under a curve. In the context of finding the arc length, integration helps us accumulate the small segments of the curve's length to find the total arc length over a specific interval. The step-by-step solution showed us how to integrate the constant derived from the arc length formula. Integrating \( \sqrt{9} \) over the interval \([-1, 1]\) results in a simple multiplication due to constant integration, giving us the arc length of 6. Understanding integration is crucial for solving complex problems, making it a powerful component of calculus that offers endless possibilities in mathematical analysis.