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Mobile App 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 approximately 4.33 units.
Step by step solution
01
Identify the Arc Length Formula
The formula to find the arc length of a curve given by a function \(y=f(x)\) from \(x=a\) to \(x=b\) is: \[\text{Arc Length} = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \,dx\]
02
Calculate \(\frac{dy}{dx}\)
For the function \(y=\sqrt{x}\), find the derivative \(y'\):\[y = x^{1/2}\]\[\frac{dy}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\]
03
Substitute and Simplify
Substitute \(\frac{dy}{dx}\) into the arc length formula:\[\int_{0}^{2} \sqrt{1 + \left(\frac{1}{2\sqrt{x}}\right)^2} \,dx = \int_{0}^{2} \sqrt{1 + \frac{1}{4x}} \,dx\]
04
Further Simplify the Integrand
Simplify the integrand further:\[\sqrt{1 + \frac{1}{4x}} = \sqrt{\frac{4x + 1}{4x}} = \frac{\sqrt{4x + 1}}{2\sqrt{x}}\].Thus, the integral becomes:\[\int_{0}^{2} \frac{\sqrt{4x + 1}}{2\sqrt{x}} \,dx\]
05
Use Substitution
Use substitution to solve the integral. Let \(u = 4x + 1\):\[du = 4 dx\]Notice when \(x = 0\), \(u = 1\)When \(x = 2\), \(u = 9\).The integral becomes:\[\int_{1}^{9} \frac{\sqrt{u}}{4\sqrt{x}} \cdot\frac{1}{4} du\]
06
Solve the Integral
To solve the integral, rewrite the integral in terms of u:\[= \int_{1}^{9} \sqrt{u} du = \frac{1}{8} \int_{1}^{9} \sqrt{u} du\].\[ = \frac{1}{4} \,\left[\frac{2}{3}u^{3/2} \right]_{1}^{9} = \frac{1}{6}\left[u^{3/2} \right]_{1}^{9} \]\[=\frac{1}{6}\left[(9^{3/2})-(1^{3/2})\right]=\frac{1}{6}(27-1)\]\[ = \frac{26}{6} = \frac{13}{3} = 4.33\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
arc length formula
To calculate the arc length of a curve given by a function, we use the arc length formula. This formula helps us find the length of the curve between two points on the x-axis. The general formula is:
\foremost{Arc Length} = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \; dx You should first know the function of the curve and the limits between which you need to find the arc length. In our case, the function is \( y = \sqrt{x} \) and the limits are \( x = 0 \) and \( x = 2 \). Knowing the function allows you to find the derivative necessary for this formula.
\foremost{Arc Length} = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \; dx You should first know the function of the curve and the limits between which you need to find the arc length. In our case, the function is \( y = \sqrt{x} \) and the limits are \( x = 0 \) and \( x = 2 \). Knowing the function allows you to find the derivative necessary for this formula.
integration by substitution
Integration by substitution is a useful technique for simplifying complex integrals. This method involves substituting a part of the integrand with a variable to make the integral easier to solve.
To implement this technique, we often perform variable changes. In the arc length problem, we used the substitution \( u = 4x + 1 \), simplifying the integral function. Note, integrating \( du \) involves transforming the entire integral and adjusting the limits of integration accordingly. For instance, in our problem, when \( x=0 \), \( u=1 \) and when \( x=2 \), \( u=9 \). This substitution resulted in the integral:
\fore\frac{1}{8} \int_{1}^{9} \sqrt{u} \; du Here, we have transformed the problem into a familiar integral which can be solved easily.
To implement this technique, we often perform variable changes. In the arc length problem, we used the substitution \( u = 4x + 1 \), simplifying the integral function. Note, integrating \( du \) involves transforming the entire integral and adjusting the limits of integration accordingly. For instance, in our problem, when \( x=0 \), \( u=1 \) and when \( x=2 \), \( u=9 \). This substitution resulted in the integral:
\fore\frac{1}{8} \int_{1}^{9} \sqrt{u} \; du Here, we have transformed the problem into a familiar integral which can be solved easily.
derivative calculation
The derivative calculation is a crucial step in using the arc length formula. The derivative \( \frac{dy}{dx} \) represents the slope of the curve at any given point. For the function \( y = \sqrt{x} \), we need to find \( y \' \), which involves using basic differentiation rules.
The function \( y = \sqrt{x} \) can be rewritten as \( y = x^{1/2} \). Using the power rule for differentiation, we get:
\foremost\frac{dy}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}. This derivative is then squared and substituted into the arc length formula.
The resulting integrand becomes more manageable after substitution.
The function \( y = \sqrt{x} \) can be rewritten as \( y = x^{1/2} \). Using the power rule for differentiation, we get:
\foremost\frac{dy}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}. This derivative is then squared and substituted into the arc length formula.
The resulting integrand becomes more manageable after substitution.
simplifying integrals
Simplifying integrals is an essential step in solving problems involving integrals. After finding the derivative and squaring it, we plug it into the arc length formula to make the integration process easier.
For our problem, the formula becomes:
\foremost\frac{dy}{dx} = \frac{1}{2} \sqrt{x}
and now simplify the integrand:
\foremost}style\frac{2x + 1}{\sqrt{x}} . Further transformation using substitution \( u= 4x + 1 \), yields: \int_{0}^{2} \frac{\sqrt{4x + 1}}{2\sqrt{x}}; making it easier. Through these transformations and substitutions, the integral becomes more straightforward to solve using known integration techniques.
For our problem, the formula becomes:
\foremost\frac{dy}{dx} = \frac{1}{2} \sqrt{x}
and now simplify the integrand:
\foremost}style\frac{2x + 1}{\sqrt{x}} . Further transformation using substitution \( u= 4x + 1 \), yields: \int_{0}^{2} \frac{\sqrt{4x + 1}}{2\sqrt{x}}; making it easier. Through these transformations and substitutions, the integral becomes more straightforward to solve using known integration techniques.
