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Chapter 7: Problem 15

In Exercises \(11-18\) , find the exact length of the curve analytically by antidifferentiation. You will need to simplify the integrand algebraically before finding an antiderivative. $$x=\left(y^{3 / 6}\right)+1 /(2 y) \quad \text { from } y=1 \text { to } y=2$$ $$\left[\text {Hint} : 1+(d x / d y)^{2} \text { is a perfect square. }\right]$$

Short Answer

Expert verified
The exact length of the curve analytically by antidifferentiation is \(3/2(2^{2/3} - 1) - 1\).

Step by step solution

01

Expression for dx/dy

Differentiate function \(x = y^{3/6} + 1/(2y)\) with respect to \(y\) to get \(dx/dy = (3/10)y^{-4/6} - 1/(2y^{2})\). Simplify this to get \(dx/dy = (1/2) y^{-1/3} - 1/(2y^{2})\).
02

Calculation of 1 + (dx/dy)^2

Calculate \(1 + (dx/dy)^{2}\) to obtain perfect square. After squaring dx/dy and adding 1 we get \(1 + (dx/dy)^{2} = 1 + [(1/2) y^{-1/3} - 1/(2y^{2})]^{2} = 1 + [1/4y^{-2/3} - 1/2 + 1/4y^{-4}] = 1/4y^{-2/3} + 1/4y^{-4}\). Simplify this to get perfect square, \(1 + (dx/dy)^{2} = (\sqrt {y^{-2/3} + y^{-2}})^2\).
03

Calculation of curve length

The length of the curve is given by the integral \(L = \int_{a}^{b} \sqrt{1 + (dx/dy)^{2}} dy\). Here, \(a = 1\) and \(b = 2\). Substituting the perfect square obtained above, we calculate the integral as \(L = \int_{1}^{2} y^{-1/3} + y{-1} dy = [3/2y^{2/3} - y]_{1}^{2} = 3/2 * 2^{2/3} - 2 - 3/2 + 1 = 3/2(2^{2/3} - 1) + 1 - 2 = 3/2(2^{2/3} - 1) - 1\).

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Key Concepts

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

Antidifferentiation
Antidifferentiation, also known as integration, is the inverse operation of differentiation in calculus. It involves finding a function whose derivative is the given function. This process is essential in solving problems that require finding the original function when the rate of change is known. For example, in determining the arc length of a curve, we might first need to find the derivative of the curve's function and then antidifferentiate to reverse the process.

In the given exercise, antidifferentiation plays a key role as it is required to find the exact length of the curve by integrating a function derived from the curve's equation. Simplifying the integrand algebraically before integrating is crucial for easing the calculation process.
Calculus
Calculus is a branch of mathematics focused on the study of change and motion. Its two fundamental operations are differentiation (finding the rate of change) and integration (antidifferentiation). Calculus is essential for solving problems in physics, engineering, economics, statistics, and many other fields.

When working with the arc length of a curve, calculus allows us to calculate the length by integrating the square root of the sum of 1 and the square of the derivative of the function defining the curve. The problem provided is a calculus problem because it uses differentiation to find the derivative of the curve's equation and then requires integration to find the arc length.
Definite Integrals
Definite integrals are integrals with set limits of integration. They give the net area under a curve within a given interval. In the context of arc length, the definite integral sums up infinitely many infinitesimally small lengths along the curve between two points, giving the total length of the curve in that interval.

For the curve in the exercise, the definite integral from y=1 to y=2 of the square root of '1 plus the square of dx/dy' essentially adds up all the tiny segments of the curve's length to find the exact total length. The process of evaluating the definite integral requires finding the antiderivative and then applying the limits of the integration to this antiderivative.
Curve Analysis
Curve analysis in calculus refers to examining the properties of a curve determined by a specific function. It includes studying the rate of change of the function (through its derivative), concavity, inflection points, and even arc length. Analyzing a curve thoroughly involves taking derivatives and also knowing how to handle the outcomes of these derivatives through antidifferentiation.

In the provided problem, the analysis of the curve defined by the equation x = y^(3/6) + 1/(2y) requires not only the differentiation step mentioned but also manipulation of the derivative to form a perfect square. This manipulation simplifies the integration process when evaluating the arc length, showcasing the interconnected nature of curve analysis and the calculus operations of differentiation and integration.

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Most popular questions from this chapter

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