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Chapter 6: Problem 34

Find a curve that passes through the point (1,5) and has an arc length on the interval [2,6] given by \(\int_{2}^{6} \sqrt{1+16 x^{-6}} d x\)

Short Answer

Expert verified
Answer: The equation of the curve is \(f(x) = -2x^{-2} + 7\).

Step by step solution

01

Set up the arc length formula and find \(f'(x)\)

Given the arc length formula from the analysis section: $$L = \int_a^b \sqrt{1 + f'(x)^2} dx = \int_{2}^{6} \sqrt{1+16 x^{-6}} d x$$ The integrand is already in the required form, so we can match the terms inside the square root: $$1 + f'(x)^2 = 1 + 16 x^{-6}$$ Now, we can find \(f'(x)\) by isolating the term: $$f'(x) = \sqrt{16 x^{-6}} = 4x^{-3}$$
02

Integrate \(f'(x)\) to find the curve equation

Now, let's integrate \(f'(x)\) to find the curve equation: $$f(x) = \int 4x^{-3} dx$$ Using the power rule for integration: $$f(x) = -\frac{4}{2}x^{-2} + C$$ Simplify the equation: $$f(x) = -2x^{-2} + C$$
03

Determine the constant of integration C

We are given that the curve passes through point (1,5). Let's use this information to find the constant of integration. Plug in the coordinates (1, 5) into the equation: $$5 = -2(1)^{-2} + C$$ Solve for C: $$C = 2 + 5 =7$$
04

Write the final curve equation

Now that we have the constant of integration, we can write the final curve equation: $$f(x) = -2x^{-2} + 7$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Calculus
Calculus is a powerful tool in mathematics that deals with the study of change. It's particularly important for understanding how quantities vary, providing a framework with which we can explore concepts of motion and change.

In the context of this exercise, calculus helps us determine the arc length of a curve, which is essentially the distance along a curved path. Calculus enables us to precisely calculate this by integrating the rate of change along the curve.

Arc length problems are common in calculus, showcasing the practical application of the subject in various fields. For our exercise, we use the arc length formula:
  • The formula for arc length from point \(a\) to \(b\) is: \(L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx\).
  • This uses both differentiation to find \(f'(x)\) and integration to sum up the changes over the interval.
The problem provides us with this integral representation, illustrating the immediate application of calculus principles.
Exploring Integration
Integration is one of the core operations in calculus that allows us to find quantities like areas and lengths by summing infinitesimal changes. In this problem, we integrate to find both the arc length and the original function from its derivative.

When we integrate the derivative \(f'(x)\), we're essentially "undoing" differentiation. This process leads us to find the original function for the curve.
  • The function here is obtained by integrating \(f'(x) = 4x^{-3}\): \(f(x) = \int 4x^{-3} \, dx\).
  • Using the power rule of integration, the antiderivative of \(4x^{-3}\) is \(-2x^{-2} + C\), where \(C\) is a constant.
This constant \(C\) is crucial as it personalizes the generic antiderivative to fit the specific conditions of the problem, such as a known point on the curve.
Demystifying Differentiation
Differentiation is a fundamental concept in calculus, defining how a function's value changes as its input changes. It's represented by \(f'(x)\), the derivative of \(f(x)\), and here, it helps describe the curve's slope.

In this problem, differentiation tells us how steep the curve is at any point along its length. By working backwards from the arc length formula, we determine \(f'(x)\):
  • We start with \(1 + (f'(x))^2 = 1 + 16x^{-6}\), leading to \(f'(x) = 4x^{-3}\).
  • This result shows us not just the change, but provides exactly what we need to find the original function via integration.
Differentiation thus bridges the gap between the geometric and the algebraic, informing the calculations necessary for understanding the curve's true path.

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Most popular questions from this chapter

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