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Use a linear approximation to estimate v10001.

Use a linear approximation to estimate v10001.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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**Problem Statement:** Use a linear approximation to estimate \(\sqrt{10001}\). **Explanation:** This problem asks for the use of linear approximation, often associated with calculus, to estimate the square root of 10001. Linear approximation involves using the tangent line at a known point on a curve to approximate values near that point. ### Steps to Solve: 1. **Identify the Function:** - Consider the function \( f(x) = \sqrt{x} \). 2. **Choose a Point Near 10001:** - Choose \( x = 10000 \), since \(\sqrt{10000} = 100\) is a convenient value. 3. **Calculate the Derivative:** - The derivative \( f'(x) = \frac{1}{2\sqrt{x}} \). 4. **Evaluate the Derivative at \( x = 10000 \):** - \( f'(10000) = \frac{1}{200} = 0.005 \). 5. **Use the Linear Approximation Formula:** - Linear approximation \( L(x) = f(a) + f'(a)(x - a) \). - Here, \( a = 10000 \), \( f(a) = 100 \), and \( f'(a) = 0.005 \). - So, \( L(10001) = 100 + 0.005(10001 - 10000) \). 6. **Calculate the Approximation:** - \( L(10001) = 100 + 0.005 \times 1 = 100.005 \). Therefore, using a linear approximation, \(\sqrt{10001} \approx 100.005\).
Transcribed Image Text:**Problem Statement:** Use a linear approximation to estimate \(\sqrt{10001}\). **Explanation:** This problem asks for the use of linear approximation, often associated with calculus, to estimate the square root of 10001. Linear approximation involves using the tangent line at a known point on a curve to approximate values near that point. ### Steps to Solve: 1. **Identify the Function:** - Consider the function \( f(x) = \sqrt{x} \). 2. **Choose a Point Near 10001:** - Choose \( x = 10000 \), since \(\sqrt{10000} = 100\) is a convenient value. 3. **Calculate the Derivative:** - The derivative \( f'(x) = \frac{1}{2\sqrt{x}} \). 4. **Evaluate the Derivative at \( x = 10000 \):** - \( f'(10000) = \frac{1}{200} = 0.005 \). 5. **Use the Linear Approximation Formula:** - Linear approximation \( L(x) = f(a) + f'(a)(x - a) \). - Here, \( a = 10000 \), \( f(a) = 100 \), and \( f'(a) = 0.005 \). - So, \( L(10001) = 100 + 0.005(10001 - 10000) \). 6. **Calculate the Approximation:** - \( L(10001) = 100 + 0.005 \times 1 = 100.005 \). Therefore, using a linear approximation, \(\sqrt{10001} \approx 100.005\).
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