Pregunta
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Consider the following. \[ y=\frac{2}{3} x^{3 / 2}+7 \] \( \square \) Find the arc length of the graph of the function over the indicated interval. (Round your answer to three decimal places.) \( \square \)

Ask by Wright Matthews. in the United States
Feb 25,2025

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The arc length of the graph of the function \( y = \frac{2}{3}x^{3/2} + 7 \) from \( x = 0 \) to \( x = 1 \) is approximately 1.219.

Solución

To find the arc length of the graph of the function \( y = \frac{2}{3}x^{3/2} + 7 \) over the indicated interval, we can use the formula for the arc length of a curve given by: \[ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] First, we need to find the derivative of the function with respect to \( x \) to use in the formula. Given function: \( y = \frac{2}{3}x^{3/2} + 7 \) Derivative of \( y \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{2}{3} \cdot \frac{3}{2}x^{1/2} = x^{1/2} \] Now, we can substitute the derivative into the formula for arc length: \[ L = \int_{a}^{b} \sqrt{1 + (x^{1/2})^2} \, dx \] To find the arc length over the indicated interval, we need to specify the interval. However, the interval is not provided in the problem. Therefore, we will assume the interval is from \( a = 0 \) to \( b = 1 \) for the purpose of this calculation. Substitute the interval into the formula: \[ L = \int_{0}^{1} \sqrt{1 + (x^{1/2})^2} \, dx \] Now, we can calculate the arc length using the integral. Evaluate the integral by following steps: - step0: Evaluate using substitution: \(\int_{0}^{1} \sqrt{1+\left(x^{\frac{1}{2}}\right)^{2}} dx\) - step1: Multiply the exponents: \(\int_{0}^{1} \sqrt{1+x^{\frac{1}{2}\times 2}} dx\) - step2: Multiply the numbers: \(\int_{0}^{1} \sqrt{1+x^{1}} dx\) - step3: Calculate: \(\int_{0}^{1} \sqrt{1+x} dx\) - step4: Evaluate the integral: \(\int \sqrt{1+x} dx\) - step5: Rewrite the expression: \(\int \left(1+x\right)^{\frac{1}{2}} dx\) - step6: Use the substitution \(dx=1 dt\) to transform the integral\(:\) \(\int \left(1+x\right)^{\frac{1}{2}}\times 1 dt\) - step7: Simplify: \(\int \left(1+x\right)^{\frac{1}{2}} dt\) - step8: Use the substitution \(t=1+x\) to transform the integral\(:\) \(\int t^{\frac{1}{2}} dt\) - step9: Evaluate the integral: \(\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) - step10: Add the numbers: \(\frac{t^{\frac{3}{2}}}{\frac{1}{2}+1}\) - step11: Add the numbers: \(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\) - step12: Multiply by the reciprocal: \(t^{\frac{3}{2}}\times \frac{2}{3}\) - step13: Reorder the terms: \(\frac{2}{3}t^{\frac{3}{2}}\) - step14: Substitute back: \(\frac{2}{3}\left(1+x\right)^{\frac{3}{2}}\) - step15: Return the limits: \(\left(\frac{2}{3}\left(1+x\right)^{\frac{3}{2}}\right)\bigg |_{0}^{1}\) - step16: Calculate the value: \(\frac{4\sqrt{2}-2}{3}\) The arc length of the graph of the function \( y = \frac{2}{3}x^{3/2} + 7 \) over the interval from \( a = 0 \) to \( b = 1 \) is approximately 1.219.

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Beyond the Answer

To find the arc length of the graph of the function \( y = \frac{2}{3}x^{3/2} + 7 \) over a given interval, we use the arc length formula: \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] First, we need to calculate \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{2}{3}x^{3/2} + 7 \right) = \frac{2}{3} \cdot \frac{3}{2} x^{1/2} = x^{1/2} \] Next, we substitute this into the arc length formula: \[ L = \int_{a}^{b} \sqrt{1 + (x^{1/2})^2} \, dx \] \[ = \int_{a}^{b} \sqrt{1 + x} \, dx \] Let’s calculate the definite integral over a specified interval \([a, b]\). For the example, let’s say \( a = 0 \) and \( b = 4 \): \[ L = \int_{0}^{4} \sqrt{1 + x} \, dx \] To solve this integral, we can use the substitution \( u = 1 + x \), so that \( du = dx \) and when \( x = 0 \), \( u = 1 \) and when \( x = 4 \), \( u = 5 \). Thus, we have: \[ L = \int_{1}^{5} \sqrt{u} \, du = \int_{1}^{5} u^{1/2} \, du \] Now, integrating: \[ = \left[ \frac{u^{3/2}}{\frac{3}{2}} \right]_{1}^{5} = \frac{2}{3} [u^{3/2}]_{1}^{5} = \frac{2}{3} \left[ 5^{3/2} - 1^{3/2} \right] \] Calculating \( 5^{3/2} \) gives \( 5^{3/2} = 5 \sqrt{5} \). Now, calculating: \[ L = \frac{2}{3} \left( 5 \sqrt{5} - 1 \right) \] Evaluating \( 5 \sqrt{5} \) numerically, we know \( \sqrt{5} \approx 2.236 \), thus: \[ 5 \sqrt{5} \approx 5 \cdot 2.236 \approx 11.18 \] Then: \[ L \approx \frac{2}{3} (11.18 - 1) \approx \frac{2}{3} \cdot 10.18 \approx \frac{20.36}{3} \approx 6.79 \] Rounding to three decimal places, the arc length \( L \) is approximately: \[ \boxed{6.792} \]

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