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Magazine Chapter 5: Problem 18
For the curve \(y=\sqrt{x},\) between \(x=0\) and \(x=2,\) find: The arc length.
Short Answer
Expert verified
The arc length is 4.
Step by step solution
01
Understand the arc length formula
For a curve described by the function y = f(x) the arc length from x = a to x = b is given by the integral: \[ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]
02
Find the derivative of the function
For the curve y = \sqrt{x} the derivative is: \[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \]
03
Substitute the derivative into the arc length formula
Substitute \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \) into the arc length formula: \[ L = \int_0^2 \sqrt{1 + \left( \frac{1}{2\sqrt{x}} \right)^2} \, dx \] This simplifies to: \[ L = \int_0^2 \sqrt{1 + \frac{1}{4x}} \, dx \]
04
Simplify the integrand expression
Combine the terms inside the square root: \[ L = \int_0^2 \sqrt{\frac{4x + 1}{4x}} \, dx \] Which can be simplified to: \[ L = \int_0^2 \sqrt{\frac{4x + 1}{4x}} = \int_0^2 \frac{\sqrt{4x + 1}}{2\sqrt{x}} \, dx \]
05
Perform a u-substitution
Let \( u = 4x + 1 \). Then \( du = 4 \, dx \), and \( dx = \frac{du}{4} \). Also, when \( x = 0 \), \( u = 1 \) and when \( x = 2 \), \( u = 9 \). Substituting these values into the integral gives: \[ L = \int_1^9 \frac{\sqrt{u}}{2\sqrt{\frac{u-1}{4}}} \cdot \frac{du}{4} = \int_1^9 \frac{\sqrt{u}}{\frac{2\sqrt{u-1}}{2}} \cdot \frac{du}{4} = \int_1^9 \frac{\sqrt{u}}{\sqrt{u-1}} \cdot \frac{du}{4} = \int_1^9 \frac{du}{2} \]
06
Integrate and evaluate the definite integral
Integrate and evaluate: \[ L = \int_1^9 \frac{du}{2} = \frac{1}{2} \, (u)\bigg|_1^9 = \frac{1}{2}(9-1) = 4 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
arc length formula
To find the length of a curve, we use the arc length formula. The arc length \(L\) of a function \(y = f(x)\) from \(x = a\) to \(x = b\) is given by:
\[ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] This formula provides the total length of the curve by integrating the square root of the sum of 1 and the square of the derivative of the function.
It's a way to measure the distance along the curve rather than a straight line. First, you need to compute the derivative \(\frac{dy}{dx}\) and then square it to substitute back into the formula.
\[ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] This formula provides the total length of the curve by integrating the square root of the sum of 1 and the square of the derivative of the function.
It's a way to measure the distance along the curve rather than a straight line. First, you need to compute the derivative \(\frac{dy}{dx}\) and then square it to substitute back into the formula.
- Arc length calculates the actual 'distance' along the curve.
- Derivatives play a crucial role in this formula.
- Understanding how to manipulate and simplify the integrand is essential.
u-substitution
U-substitution is a method for simplifying integrals by substituting a part of the integrand with a new variable, usually denoted as u. This technique is very handy in integrals that involve complicated expressions.
Let's take an example from the arc length problem:
We had an integral \int_0^2 \frac{\sqrt{4x + 1}}{2\sqrt{x}} \, dx\. To simplify, we set \( u = 4x + 1 \). Then, our differential \( dx \) changes as well: \( du = 4 \, dx \) or \( dx = \frac{du}{4} \).
When \( x = 0 \), \( u = 1\), and when \( x = 2, u = 9\). This transforms the integral into:
Let's take an example from the arc length problem:
We had an integral \int_0^2 \frac{\sqrt{4x + 1}}{2\sqrt{x}} \, dx\. To simplify, we set \( u = 4x + 1 \). Then, our differential \( dx \) changes as well: \( du = 4 \, dx \) or \( dx = \frac{du}{4} \).
When \( x = 0 \), \( u = 1\), and when \( x = 2, u = 9\). This transforms the integral into:
- The variable limits change according to the new substitution.
- The integral becomes simpler and often more manageable to solve.
- Finally, involve back-substitution to finish the integration.
definite integral evaluation
Definite integrals are used to find the total value of the function over a specific interval. These integrals have upper and lower limits and are noted as follows: \int_{a}^{b} f(x) \, dx\. They give the net area between the function graph and the x-axis.
Using our example problem, after performing u-substitution, we got the integral: \[ L = \int_1^9 \frac{du}{2} \] This is much simpler to evaluate. The antiderivative of \(\frac{1}{2} u \) is straightforward:
\[ \frac{1}{2} \, (u)\bigg|_{1}^{9} \] Finally, substitute the limits of integration into the antiderivative:
\[ \frac{1}{2}(9) - \frac{1}{2}(1) = 4 \] The definite integral evaluates to 4, which gives us the arc length.
Using our example problem, after performing u-substitution, we got the integral: \[ L = \int_1^9 \frac{du}{2} \] This is much simpler to evaluate. The antiderivative of \(\frac{1}{2} u \) is straightforward:
\[ \frac{1}{2} \, (u)\bigg|_{1}^{9} \] Finally, substitute the limits of integration into the antiderivative:
\[ \frac{1}{2}(9) - \frac{1}{2}(1) = 4 \] The definite integral evaluates to 4, which gives us the arc length.
- Evaluation turns the integral function value into tangible results.
- Consistent simplification and back-substitution are key.
