| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the curvature of the function \(f(x) = \sqrt{4x - x^2}\) at \(x = 2\), we need to use the formula for curvature, which is given by: \[ \kappa(x) = \frac{|f''(x)|}{(1 + [f'(x)]^2)^{3/2}} \] First, we need to find the first and second derivatives of \(f(x)\). Given: \[ f(x) = \sqrt{4x - x^2} = (4x - x^2)^{1/2} \] Find the first derivative, \(f'(x)\): \[ f'(x) = \frac{1}{2}(4x - x^2)^{-1/2}(4 - 2x) = \frac{2 - x}{\sqrt{4x - x^2}} \] Find the second derivative, \(f''(x)\): \[ f''(x) = \frac{d}{dx}\left( \frac{2 - x}{\sqrt{4x - x^2}} \right) = \frac{-\sqrt{4x - x^2} - (2 - x)\left( \frac{1}{2}(4x - x^2)^{-1/2}(-2x) \right)}{4x - x^2} \] Simplifying, \[ f''(x) = \frac{-\sqrt{4x - x^2} + (2 - x)\left( \frac{x}{\sqrt{4x - x^2}} \right)}{4x - x^2} = \frac{-\sqrt{4x - x^2} + \frac{x(2 - x)}{\sqrt{4x - x^2}}}{4x - x^2} \] \[ f''(x) = \frac{-\sqrt{4x - x^2} + \frac{2x - x^2}{\sqrt{4x - x^2}}}{4x - x^2} = \frac{-\sqrt{4x - x^2} + \frac{2x - x^2}{\sqrt{4x - x^2}}}{4x - x^2} \cdot \frac{\sqrt{4x - x^2}}{\sqrt{4x - x^2}} \] \[ f''(x) = \frac{-4x + x^2 + 2x - x^2}{(4x - x^2)^{3/2}} = \frac{-2x}{(4x - x^2)^{3/2}} \] Now, we evaluate \(f'(x)\) and \(f''(x)\) at \(x = 2\): \[ f'(2) = \frac{2 - 2}{\sqrt{4(2) - (2)^2}} = 0 \] \[ f''(2) = \frac{-2(2)}{(4(2) - (2)^2)^{3/2}} = \frac{-4}{(8 - 4)^{3/2}} = \frac{-4}{4^{3/2}} = \frac{-4}{8} = -\frac{1}{2} \] Then, substitute these values into the curvature formula: \[ \kappa(2) = \frac{|-\frac{1}{2}|}{(1 + 0^2)^{3/2}} = \frac{1/2}{1} = \frac{1}{2} \] Therefore, the curvature of \(f(x) = \sqrt{4x - x^2}\) at \(x = 2\) is \(\boxed{\frac{1}{2}}\). |
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| deepseekr1 | To find the curvature of the function \( f(x) = \sqrt{4x - x^2} \) at \( x = 2 \): 1. **First Derivative**: \[ f'(x) = \frac{2 - x}{\sqrt{4x - x^2}} \] Evaluating at \( x = 2 \): \[ f'(2) = \frac{2 - 2}{\sqrt{4 \cdot 2 - 2^2}} = 0 \] 2. **Second Derivative**: Using the quotient rule and simplifying: \[ f''(x) = \frac{-4}{(4x - x^2)^{3/2}} \] Evaluating at \( x = 2 \): \[ f''(2) = \frac{-4}{(4 \cdot 2 - 2^2)^{3/2}} = \frac{-4}{4^{3/2}} = \frac{-4}{8} = -\frac{1}{2} \] 3. **Curvature Formula**: \[ \kappa = \frac{|f''(x)|}{\left(1 + (f'(x))^2\right)^{3/2}} \] Substituting \( f'(2) = 0 \) and \( |f''(2)| = \frac{1}{2} \): \[ \kappa = \frac{\frac{1}{2}}{\left(1 + 0^2\right)^{3/2}} = \frac{1}{2} \] The curvature at \( x = 2 \) is \(\boxed{\dfrac{1}{2}}\). |
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| gpto1 | To find the curvature \(\kappa\) of the function \(f(x) = \sqrt{4x - x^2}\) at \(x = 2\), we'll follow these steps: 1. **Compute the first derivative \(f'(x)\).** 2. **Compute the second derivative \(f''(x)\).** 3. **Use the curvature formula to find \(\kappa\) at \(x = 2\).** --- **Step 1: Compute the first derivative \(f'(x)\).** Start by expressing \(f(x)\) in exponential form: \[ f(x) = [4x - x^2]^{1/2} \] Using the chain rule: \[ f'(x) = \dfrac{1}{2} [4x - x^2]^{-1/2} \cdot (4 - 2x) \] Simplify: \[ f'(x) = \dfrac{4 - 2x}{2\sqrt{4x - x^2}} \] **At \(x = 2\):** \[ f'(2) = \dfrac{4 - 2 \cdot 2}{2\sqrt{4 \cdot 2 - 2^2}} = \dfrac{0}{2\sqrt{8 - 4}} = \dfrac{0}{2 \cdot 2} = 0 \] --- **Step 2: Compute the second derivative \(f''(x)\).** Using \(f'(x)\): \[ f'(x) = \dfrac{4 - 2x}{2[4x - x^2]^{1/2}} \] Compute \(f''(x)\) by differentiating \(f'(x)\): First, denote \(n(x) = 4 - 2x\) and \(d(x) = 2[4x - x^2]^{1/2}\). Then: \[ f''(x) = \dfrac{n'(x)d(x) - n(x)d'(x)}{[d(x)]^2} \] Compute derivatives: \[ n'(x) = -2 \] \[ d'(x) = 2 \cdot \dfrac{1}{2}[4x - x^2]^{-1/2}(4 - 2x) = [4x - x^2]^{-1/2}(4 - 2x) \] Simplify numerator: \[ n'(x)d(x) - n(x)d'(x) = -2 \cdot 2[4x - x^2]^{1/2} - (4 - 2x) \cdot [4 - 2x][4x - x^2]^{-1/2} \] Simplify further: \[ n'(x)d(x) - n(x)d'(x) = -4[4x - x^2]^{1/2} - (4 - 2x)^2[4x - x^2]^{-1/2} \] Factor out \([4x - x^2]^{-1/2}\): \[ = \left(-4[4x - x^2] - (4 - 2x)^2\right)[4x - x^2]^{-1/2} \] Simplify expressions: \[ -4(4x - x^2) - (4 - 2x)^2 = -16x + 4x^2 - (16 - 16x + 4x^2) = -16 \] Therefore: \[ f''(x) = \dfrac{-16[4x - x^2]^{-1/2}}{4[4x - x^2]} = \dfrac{-4}{[4x - x^2]^{3/2}} \] **At \(x = 2\):** \[ f''(2) = \dfrac{-4}{[4 \cdot 2 - 2^2]^{3/2}} = \dfrac{-4}{[8 - 4]^{3/2}} = \dfrac{-4}{[4]^{3/2}} = \dfrac{-4}{8} = -\dfrac{1}{2} \] --- **Step 3: Use the curvature formula to find \(\kappa\) at \(x = 2\).** The curvature \(\kappa\) of a function \(y = f(x)\) is given by: \[ \kappa(x) = \dfrac{|f''(x)|}{\left[1 + (f'(x))^2\right]^{3/2}} \] **At \(x = 2\):** Since \(f'(2) = 0\) and \(f''(2) = -\dfrac{1}{2}\): \[ \kappa(2) = \dfrac{\left|-\dfrac{1}{2}\right|}{\left[1 + (0)^2\right]^{3/2}} = \dfrac{\dfrac{1}{2}}{1^{3/2}} = \dfrac{1}{2} \] --- **Answer:** The curvature of \( f(x) = \sqrt{4x - x^2} \) at \( x = 2 \) is: \[ \kappa(2) = \dfrac{1}{2} \] **Therefore, the curvature at \( x = 2 \) is \( \dfrac{1}{2} \).** |
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| GPT-4o | To find the curvature \(\kappa\) of the function \(f(x) = \sqrt{4x - x^2}\) at \(x = 2\), we can follow these steps: 1. **Find the first derivative \(f'(x)\):** Given \(f(x) = \sqrt{4x - x^2}\), let's rewrite it as: \[ f(x) = (4x - x^2)^{1/2} \] Using the chain rule, we find the first derivative: \[ f'(x) = \frac{d}{dx} \left( (4x - x^2)^{1/2} \right) = \frac{1}{2}(4x - x^2)^{-1/2} \cdot (4 - 2x) \] Simplifying, we get: \[ f'(x) = \frac{4 - 2x}{2 \sqrt{4x - x^2}} = \frac{4 - 2x}{2 \sqrt{4x - x^2}} = \frac{4 - 2x}{2(4x - x^2)^{1/2}} \] 2. **Find the second derivative \(f''(x)\):** To find the second derivative, we differentiate \(f'(x)\) with respect to \(x\): \[ f''(x) = \frac{d}{dx} \left( \frac{4 - 2x}{2 \sqrt{4x - x^2}} \right) \] This is a quotient, so we use the quotient rule: \[ f''(x) = \frac{(2 \sqrt{4x - x^2})(-2) - (4 - 2x) \frac{d}{dx} (2 \sqrt{4x - x^2})}{(2 \sqrt{4x - x^2})^2} \] Simplifying the numerator first: \[ \frac{d}{dx} (2 \sqrt{4x - x^2}) = 2 \cdot \frac{1}{2} (4x - x^2)^{-1/2} \cdot (4 - 2x) = \frac{4 - 2x}{\sqrt{4x - x^2}} \] Plugging this back into the quotient rule: \[ f''(x) = \frac{-4 \sqrt{4x - x^2} - (4 - 2x) \cdot \frac{4 - 2x}{\sqrt{4x - x^2}}}{4x - x^2} = \frac{-4 \sqrt{4x - x^2} - \frac{(4 - 2x)^2}{\sqrt{4x - x^2}}}{4x - x^2} \] Simplify: \[ f''(x) = \frac{-4 \sqrt{4x - x^2} - \frac{16 - 16x + 4x^2}{\sqrt{4x - x^2}}}{4x - x^2} = \frac{-4(4x - x^2) - (16 - 16x + 4x^2)}{(4x - x^2)^{3/2}} \] \[ f''(x) = \frac{-4(4x - x^2) - 16 + 16x - 4x^2}{(4x - x^2)^{3/2}} = \frac{-4(4x - x^2) - 16 + 16x - 4x^2}{(4x - x^2)^{3/2}} \] \[ f''(x) = \frac{-16x + 4x^2 - 16 + 16x - 4x^2}{(4x - x^2)^{3/2}} = \frac{-16}{(4x - x^2)^{3/2}} \] 3. **Evaluate \(f'(x)\) and \(f''(x)\) at \(x = 2\):** \[ f'(2) = \frac{4 - 2 \cdot 2}{2 \sqrt{4 \cdot 2 - 2^2}} = \frac{4 - 4}{2 \sqrt{8 - 4}} = \frac{0}{2 \cdot 2} = 0 \] \[ f |
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| llama3.1-70B-Instruct | To find the curvature of the function f(x) = √(4x - x^2) at x = 2, we need to follow these steps: 1. Find the first and second derivatives of the function: f'(x) = d(√(4x - x^2))/dx = (2 - x) / √(4x - x^2) f''(x) = d²(√(4x - x^2))/dx² = (-1) / √(4x - x^2) + (2 - x)² / (4x - x^2)³/² 2. Evaluate the first and second derivatives at x = 2: f'(2) = (2 - 2) / √(4(2) - 2²) = 0 / √(8 - 4) = 0 / √4 = 0 f''(2) = (-1) / √(4(2) - 2²) + (2 - 2)² / (4(2) - 2²)³/² = (-1) / √4 + 0 / 4³/² = -1/2 + 0 = -1/2 3. Calculate the curvature using the formula: κ(x) = |f''(x)| / (1 + f'(x)²)³/² κ(2) = |-1/2| / (1 + 0²)³/² = 1/2 / 1 = 1/2 Therefore, the curvature of the function f(x) = √(4x - x^2) at x = 2 is 1/2. |
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| llama3.1-405B-Instruct | To find the curvature of the function f(x) = √(4x - x^2) at x = 2, we need to follow these steps: 1. Find the first and second derivatives of the function: f'(x) = d(√(4x - x^2))/dx = (4 - 2x) / (2√(4x - x^2)) f''(x) = d((4 - 2x) / (2√(4x - x^2)))/dx = (-2) / (2√(4x - x^2)) - (4 - 2x)^2 / (4(4x - x^2)^(3/2)) 2. Evaluate the first and second derivatives at x = 2: f'(2) = (4 - 2(2)) / (2√(4(2) - 2^2)) = 0 / (2√(8 - 4)) = 0 f''(2) = (-2) / (2√(4(2) - 2^2)) - (4 - 2(2))^2 / (4(4(2) - 2^2)^(3/2)) = -2 / (2√(8 - 4)) - 0^2 / (4(8 - 4)^(3/2)) = -2 / (2√4) = -1 / 2 3. Calculate the curvature using the formula: K = |f''(x)| / (1 + f'(x)^2)^(3/2) K = |-1/2| / (1 + 0^2)^(3/2) = 1/2 / 1^(3/2) = 1/2 Therefore, the curvature of the function f(x) = √(4x - x^2) at x = 2 is 1/2. |
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| DeepSeek-Coder-V2 | To find the curvature for the function \( f(x) = \sqrt{4x - x^2} \) at \( x = 2 \), we will follow these steps: 1. **Find the first derivative \( f'(x) \)**: \[ f(x) = \sqrt{4x - x^2} \] Let \( u = 4x - x^2 \). Then \( f(x) = \sqrt{u} \). Using the chain rule: \[ f'(x) = \frac{d}{dx} \left( \sqrt{u} \right) = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} \] \[ \frac{du}{dx} = 4 - 2x \] Therefore: \[ f'(x) = \frac{1}{2\sqrt{4x - x^2}} \cdot (4 - 2x) = \frac{4 - 2x}{2\sqrt{4x - x^2}} = \frac{2 - x}{\sqrt{4x - x^2}} \] 2. **Find the second derivative \( f''(x) \)**: Using the quotient rule: \[ f'(x) = \frac{2 - x}{\sqrt{4x - x^2}} \] Let \( v = 2 - x \) and \( w = \sqrt{4x - x^2} \). Then: \[ f'(x) = \frac{v}{w} \] \[ f''(x) = \frac{w \cdot v' - v \cdot w'}{w^2} \] \[ v' = -1 \] \[ w' = \frac{d}{dx} \left( \sqrt{4x - x^2} \right) = \frac{4 - 2x}{2\sqrt{4x - x^2}} = \frac{2 - x}{\sqrt{4x - x^2}} \] Therefore: \[ f''(x) = \frac{\sqrt{4x - x^2} \cdot (-1) - (2 - x) \cdot \frac{2 - x}{\sqrt{4x - x^2}}}{(4x - x^2)} \] Simplifying: \[ f''(x) = \frac{-\sqrt{4x - x^2} - \frac{(2 - x)^2}{\sqrt{4x - x^2}}}{4x - x^2} \] \[ f''(x) = \frac{-(4x - x^2) - (2 - x)^2}{(4x - x^2)\sqrt{4x - x^2}} \] \[ f''(x) = \frac{-(4x - x^2) - (4 - 4x + x^2)}{(4x - x^2)\sqrt{4x - x^2}} \] \[ f''(x) = \frac{-4x + x^2 - 4 + 4x - x^2}{(4x - x^2)\sqrt{4x - x^2}} \] \[ f''(x) = \frac{-4}{(4x - x^2)\sqrt{4x - x^2}} \] 3. **Evaluate \( f'(2) \) and \( f''(2) \)**: \[ f'(2) = \frac{2 - 2}{\sqrt{4 \cdot 2 - 2^2}} = 0 \] \[ f''(2) = \frac{-4}{(4 \cdot 2 - 2^2)\sqrt{4 \cdot 2 - 2^2}} = \frac{-4}{4 \cdot 2} = \frac{-4}{8} = -\frac{1}{2} \] 4. **Calculate the curvature \( \kappa \)**: The curvature formula is: \[ \kappa = \frac{|f''(x)|}{\left( 1 + (f'(x))^2 \right)^{3/2}} \] At \( x = 2 \): \[ \kappa = \frac{\left| -\frac{1}{2} \right|}{\left( 1 + 0^2 \right)^{3/2}} = \frac{\frac{1}{2}}{1} = \frac{1}{2} \] Therefore, the curvature at \( x = 2 \) is \( \boxed{\frac{1}{2}} \). |
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| Qwen2-72B-Instruct | To find the curvature of the function \(f(x) = \sqrt{4x - x^2}\) at \(x = 2\), we need to use the formula for curvature, which is given by: \[ \kappa(x) = \frac{|f''(x)|}{(1 + [f'(x)]^2)^{3/2}} \] First, we need to find the first and second derivatives of \(f(x)\). Given: \[ f(x) = \sqrt{4x - x^2} = (4x - x^2)^{1/2} \] Find the first derivative, \(f'(x)\): \[ f'(x) = \frac{1}{2}(4x - x^2)^{-1/2}(4 - 2x) = \frac{2 - x}{\sqrt{4x - x^2}} \] Find the second derivative, \(f''(x)\): \[ f''(x) = \frac{d}{dx}\left( \frac{2 - x}{\sqrt{4x - x^2}} \right) = \frac{-\sqrt{4x - x^2} - (2 - x)\left( \frac{1}{2}(4x - x^2)^{-1/2}(-2x) \right)}{4x - x^2} \] Simplifying, \[ f''(x) = \frac{-\sqrt{4x - x^2} + (2 - x)\left( \frac{x}{\sqrt{4x - x^2}} \right)}{4x - x^2} = \frac{-\sqrt{4x - x^2} + \frac{x(2 - x)}{\sqrt{4x - x^2}}}{4x - x^2} \] \[ f''(x) = \frac{-\sqrt{4x - x^2} + \frac{2x - x^2}{\sqrt{4x - x^2}}}{4x - x^2} = \frac{-\sqrt{4x - x^2} + \frac{2x - x^2}{\sqrt{4x - x^2}}}{4x - x^2} \cdot \frac{\sqrt{4x - x^2}}{\sqrt{4x - x^2}} \] \[ f''(x) = \frac{-4x + x^2 + 2x - x^2}{(4x - x^2)^{3/2}} = \frac{-2x}{(4x - x^2)^{3/2}} \] Now, we evaluate \(f'(x)\) and \(f''(x)\) at \(x = 2\): \[ f'(2) = \frac{2 - 2}{\sqrt{4(2) - (2)^2}} = 0 \] \[ f''(2) = \frac{-2(2)}{(4(2) - (2)^2)^{3/2}} = \frac{-4}{(8 - 4)^{3/2}} = \frac{-4}{4^{3/2}} = \frac{-4}{8} = -\frac{1}{2} \] Then, substitute these values into the curvature formula: \[ \kappa(2) = \frac{|-\frac{1}{2}|}{(1 + 0^2)^{3/2}} = \frac{1/2}{1} = \frac{1}{2} \] Therefore, the curvature of \(f(x) = \sqrt{4x - x^2}\) at \(x = 2\) is \(\boxed{\frac{1}{2}}\). |
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| baichuan4 | To find the curvature of the function \(f(x) = \sqrt{4x - x^2}\) at \(x = 2\), we will follow these steps: 1. Find the first derivative \(f'(x)\). 2. Find the second derivative \(f''(x)\). 3. Calculate the curvature \(k(x)\) using the formula \(k(x) = \frac{|f''(x)|}{(1 + (f'(x))^2)^{3/2}}\). 4. Evaluate \(k(x)\) at \(x = 2\). Let's start by finding the derivatives: 1. First derivative \(f'(x)\): \[f'(x) = \frac{1}{2\sqrt{4x - x^2}} \cdot (4 - 2x)\] 2. Second derivative \(f''(x)\): \[f''(x) = \frac{-1}{4(4x - x^2)^{3/2}} \cdot (4 - 2x)^2 + \frac{1}{2\sqrt{4x - x^2}} \cdot (-2)\] Simplifying: \[f''(x) = \frac{-(4 - 2x)^2}{4(4x - x^2)^{3/2}} - \frac{1}{\sqrt{4x - x^2}}\] Now, let's calculate the curvature \(k(x)\): \[k(x) = \frac{|f''(x)|}{(1 + (f'(x))^2)^{3/2}}\] Substituting \(f'(x)\) and \(f''(x)\): \[k(x) = \frac{\left|\frac{-(4 - 2x)^2}{4(4x - x^2)^{3/2}} - \frac{1}{\sqrt{4x - x^2}}\right|}{\left(1 + \left(\frac{1}{2\sqrt{4x - x^2}} \cdot (4 - 2x)\right)^2\right)^{3/2}}\] Finally, we evaluate \(k(x)\) at \(x = 2\): \[k(2) = \frac{\left|\frac{-(4 - 2 \cdot 2)^2}{4(4 \cdot 2 - 2^2)^{3/2}} - \frac{1}{\sqrt{4 \cdot 2 - 2^2}}\right|}{\left(1 + \left(\frac{1}{2\sqrt{4 \cdot 2 - 2^2}} \cdot (4 - 2 \cdot 2)\right)^2\right)^{3/2}}\] Simplifying: \[k(2) = \frac{\left|0 - \frac{1}{2}\right|}{\left(1 + 0\right)^{3/2}}\] \[k(2) = \frac{1}{2}\] Therefore, the curvature for \(f(x) = \sqrt{4x - x^2}\) at \(x = 2\) is \(\frac{1}{2}\). |
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| doubao | First find the derivative of f(x): f(x) = √(4x - x^2) = (4x - x^2)^(1/2) f'(x) = (1/2)(4x - x^2)^(-1/2) * (4 - 2x) = (2 - x) / √(4x - x^2). Then find the second derivative: f''(x) = [√(4x - x^2) - (2 - x)*(2 - x) / √(4x - x^2)] / (4x - x^2) = [4x - x^2 - (4 - 4x + x^2)] / (4x - x^2)^(3/2) = (8x - 4 - 2x^2) / (4x - x^2)^(3/2). At x = 2, f'(2) = (2 - 2) / √(4*2 - 2^2) = 0. f''(2) = (8*2 - 4 - 2*2^2) / (4*2 - 2^2)^(3/2) = 0. The curvature κ is given by κ = |f''(x)| / (1 + f'(x)^2)^(3/2). Since f'(2) = 0 and f''(2) = 0, the curvature at x = 2 is 0. |
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| minimax | To find the curvature of the function \( f(x) = \sqrt{4x - x^2} \) at \( x = 2 \), we need to follow these steps: 1. Find the first derivative \( f'(x) \). 2. Find the second derivative \( f''(x) \). 3. Evaluate both derivatives at \( x = 2 \). 4. Use the formula for curvature \( \kappa \) at a point: \[ \kappa = \frac{|f'(x)f''(x) - [f'(x)]^3|}{[1 + (f'(x))^2]^{3/2}} \] Let's start with the first derivative: \[ f'(x) = \frac{d}{dx} \sqrt{4x - x^2} \] Using the chain rule, we get: \[ f'(x) = \frac{1}{2\sqrt{4x - x^2}} \cdot \frac{d}{dx}(4x - x^2) \] \[ f'(x) = \frac{1}{2\sqrt{4x - x^2}} \cdot (4 - 2x) \] \[ f'(x) = \frac{4 - 2x}{2\sqrt{4x - x^2}} \] Now, let's find the second derivative: \[ f''(x) = \frac{d}{dx} \left( \frac{4 - 2x}{2\sqrt{4x - x^2}} \right) \] This requires using the quotient rule and the chain rule again. Let's denote \( u = 4 - 2x \) and \( v = \sqrt{4x - x^2} \), so \( u' = -2 \) and \( v' = \frac{1}{2\sqrt{4x - x^2}} \cdot (4 - 2x) \). Applying the quotient rule: \[ f''(x) = \frac{u'v - uv'}{v^2} \] \[ f''(x) = \frac{-2 \cdot \sqrt{4x - x^2} - (4 - 2x) \cdot \frac{1}{2\sqrt{4x - x^2}} \cdot (4 - 2x)}{(4x - x^2)} \] Now we need to evaluate both \( f'(x) \) and \( f''(x) \) at \( x = 2 \): \[ f'(2) = \frac{4 - 2 \cdot 2}{2\sqrt{4 \cdot 2 - 2^2}} \] \[ f'(2) = \frac{0}{2\sqrt{8 - 4}} \] \[ f'(2) = 0 \] Since \( f'(2) = 0 \), the numerator of the curvature formula simplifies to: \[ |f'(2)f''(2) - [f'(2)]^3| = |0 \cdot f''(2) - 0^3| = 0 \] And the denominator simplifies to: \[ [1 + (f'(2))^2]^{3/2} = [1 + 0^2]^{3/2} = 1^{3/2} = 1 \] Therefore, the curvature \( \kappa \) at \( x = 2 \) is: \[ \kappa = \frac{0}{1} = 0 \] The curvature of the function \( f(x) = \sqrt{4x - x^2} \) at \( x = 2 \) is 0. |
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| kimi | To find the curvature of a function at a given point, we need to calculate the second derivative of the function and then apply the formula for curvature. The formula for curvature (k) of a function f(x) at a point x is given by: \[ k = \frac{|f''(x)|}{(1 + (f'(x))^2)^{3/2}} \] First, let's find the first and second derivatives of the function f(x) = \sqrt{4x - x^2}. 1. First derivative (f'(x)): \[ f(x) = \sqrt{4x - x^2} \] \[ f(x) = (4x - x^2)^{1/2} \] Using the chain rule, we get: \[ f'(x) = \frac{1}{2}(4x - x^2)^{-1/2} \cdot (4 - 2x) \] \[ f'(x) = \frac{4 - 2x}{2\sqrt{4x - x^2}} \] 2. Second derivative (f''(x)): \[ f''(x) = \frac{d}{dx} \left( \frac{4 - 2x}{2\sqrt{4x - x^2}} \right) \] Using the quotient rule, we get: \[ f''(x) = \frac{(2\sqrt{4x - x^2})(-2) - (4 - 2x)(\frac{1}{2}(4x - x^2)^{-1/2} \cdot (4 - 2x))}{(4x - x^2)} \] \[ f''(x) = \frac{-4\sqrt{4x - x^2} - (4 - 2x)^2}{2(4x - x^2)^{3/2}} \] Now, let's find the curvature at x = 2. 1. First derivative at x = 2: \[ f'(2) = \frac{4 - 2(2)}{2\sqrt{4(2) - (2)^2}} \] \[ f'(2) = \frac{0}{2\sqrt{8 - 4}} \] \[ f'(2) = \frac{0}{4} \] \[ f'(2) = 0 \] 2. Second derivative at x = 2: \[ f''(2) = \frac{-4\sqrt{4(2) - (2)^2} - (4 - 2(2))^2}{2(4(2) - (2)^2)^{3/2}} \] \[ f''(2) = \frac{-4\sqrt{8 - 4} - 0^2}{2(8 - 4)^{3/2}} \] \[ f''(2) = \frac{-4\sqrt{4}}{2(4)} \] \[ f''(2) = \frac{-4 \cdot 2}{8} \] \[ f''(2) = \frac{-8}{8} \] \[ f''(2) = -1 \] Now, we can find the curvature at x = 2: \[ k = \frac{|f''(2)|}{(1 + (f'(2))^2)^{3/2}} \] \[ k = \frac{|-1|}{(1 + 0^2)^{3/2}} \] \[ k = \frac{1}{1^{3/2}} \] \[ k = 1 \] So, the curvature of the function f(x) = \sqrt{4x - x^2} at x = 2 is 1. |
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| wenxinyiyan | 我们要找出函数$f(x) = \sqrt{4x - x^2}$在$x=2$处的曲率。 首先，我们需要知道曲率的计算公式是： $K = \frac{|f''(x)|}{(1 + [f'(x)]^2)^{\frac{3}{2}}}$ 现在，我们先来计算$f'(x)$和$f''(x)$。 计算$f'(x)$： $f'(x) = \frac{d}{dx}(\sqrt{4x - x^2})$ $= \frac{1}{2\sqrt{4x - x^2}} \times (4 - 2x)$ $= \frac{2 - x}{\sqrt{4x - x^2}}$ 接着，计算$f''(x)$： $f''(x) = \frac{d}{dx}(\frac{2 - x}{\sqrt{4x - x^2}})$ $= \frac{-1 \times \sqrt{4x - x^2} - (2 - x) \times \frac{1}{2\sqrt{4x - x^2}} \times (4 - 2x)}{(4x - x^2)}$ $= \frac{-1 \times \sqrt{4x - x^2} - (2 - x) \times \frac{2 - x}{\sqrt{4x - x^2}}}{(4x - x^2)}$ $= \frac{- (4x - x^2) - (2 - x)^2}{(4x - x^2)^{\frac{3}{2}}}$ $= \frac{- 4x + x^2 - 4 + 4x - x^2}{(4x - x^2)^{\frac{3}{2}}}$ $= \frac{- 4}{(4x - x^2)^{\frac{3}{2}}}$ 现在，我们将$x=2$代入$f'(x)$和$f''(x)$： $f'(2) = \frac{2 - 2}{\sqrt{4(2) - 2^2}} = 0$ $f''(2) = \frac{- 4}{(4(2) - 2^2)^{\frac{3}{2}}} = \frac{- 4}{8^{\frac{3}{2}}} = -\frac{1}{4}$ 最后，我们将$f'(2)$和$f''(2)$代入曲率的公式： $K = \frac{|- \frac{1}{4}|}{(1 + [0]^2)^{\frac{3}{2}}} = \frac{\frac{1}{4}}{1} = \frac{1}{4}$ 所以，函数$f(x) = \sqrt{4x - x^2}$在$x=2$处的曲率是$\frac{1}{4}$。 |
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| zhipu | To find the curvature \( \kappa \) of the function \( f(x) = \sqrt{4x - x^2} \) at \( x = 2 \), we will use the formula for the curvature of a function \( y = f(x) \) in the plane: \[ \kappa = \frac{|f''(x)|}{(1 + (f'(x))^2)^{3/2}} \] Here are the steps to solve the problem: ### Step 1: Find the first derivative \( f'(x) \) Given \( f(x) = \sqrt{4x - x^2} \), we apply the chain rule to find \( f'(x) \): \[ f'(x) = \frac{d}{dx} \left( \sqrt{4x - x^2} \right) = \frac{1}{2\sqrt{4x - x^2}} \cdot (4 - 2x) \] Simplifying this, we get: \[ f'(x) = \frac{4 - 2x}{2\sqrt{4x - x^2}} = \frac{2(2 - x)}{\sqrt{4x - x^2}} \] ### Step 2: Find the second derivative \( f''(x) \) Now, we differentiate \( f'(x) \) to find \( f''(x) \): \[ f''(x) = \frac{d}{dx} \left( \frac{2(2 - x)}{\sqrt{4x - x^2}} \right) \] We use the quotient rule \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \), where \( u = 2(2 - x) \) and \( v = \sqrt{4x - x^2} \). First, find \( u' \) and \( v' \): \[ u' = \frac{d}{dx} [2(2 - x)] = -2 \] \[ v' = \frac{d}{dx} [\sqrt{4x - x^2}] = \frac{1}{2\sqrt{4x - x^2}} \cdot (4 - 2x) = \frac{4 - 2x}{2\sqrt{4x - x^2}} \] Now, apply the quotient rule: \[ f''(x) = \frac{(-2)\sqrt{4x - x^2} - 2(2 - x)\left( \frac{4 - 2x}{2\sqrt{4x - x^2}} \right)}{(4x - x^2)} \] Simplify the numerator: \[ f''(x) = \frac{-2\sqrt{4x - x^2} - (2 - x)(4 - 2x)}{(4x - x^2)^{3/2}} \] \[ f''(x) = \frac{-2\sqrt{4x - x^2} - (8 - 4x - 4x + 2x^2)}{(4x - x^2)^{3/2}} \] \[ f''(x) = \frac{-2\sqrt{4x - x^2} - (8 - 8x + 2x^2)}{(4x - x^2)^{3/2}} \] ### Step 3: Evaluate \( f'(x) \) and \( f''(x) \) at \( x = 2 \) First, evaluate \( f'(2) \): \[ f'(2) = \frac{2(2 - 2)}{\sqrt{4(2) - 2^2}} = \frac{2(0)}{\sqrt{8 - 4}} = 0 \] Next, evaluate \( f''(2) \): \[ f''(2) = \frac{-2\sqrt{4(2) - 2^2} - (8 - 8(2) + 2(2^2))}{(4(2) - 2^2)^{3/2}} \] \[ f''(2) = \frac{-2\sqrt{8 - 4} - (8 - 16 + 8)}{(8 - 4)^{3/2}} \] \[ f''(2) = \frac{-2\sqrt{4} - (0)}{4^{3/2}} \] \[ f''(2) = \frac{-2 \cdot 2}{8} = -\frac{4}{8} = -\frac{1}{2} \] ### Step 4: Compute the curvature \( \kappa \) Now, substitute \( f'(2) \) and \( f''(2) \) into the curvature formula: \[ \kappa = \frac{|f''(2)|}{(1 + (f'(2))^2)^{3/2}} \] \[ \kappa = \frac{|-1/2|}{(1 + 0^2)^{3/2}} \] \[ \kappa = \frac{1/2}{1^{3/2}} \] \[ \kappa = \frac{1/2}{1} = \frac{1}{2} \] So, the curvature of \( f(x) = \sqrt{4x - x^2} \) at \( x = 2 \) is \( \frac{1}{2} \). |
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| gpto1 | To find the curvature \(\kappa\) of the function \(f(x) = \sqrt{4x - x^2}\) at \(x = 2\), we'll follow these steps: 1. **Compute the first derivative \(f'(x)\).** 2. **Compute the second derivative \(f''(x)\).** 3. **Use the curvature formula to find \(\kappa\) at \(x = 2\).** --- **Step 1: Compute the first derivative \(f'(x)\).** Start by expressing \(f(x)\) in exponential form: \[ f(x) = [4x - x^2]^{1/2} \] Using the chain rule: \[ f'(x) = \dfrac{1}{2} [4x - x^2]^{-1/2} \cdot (4 - 2x) \] Simplify: \[ f'(x) = \dfrac{4 - 2x}{2\sqrt{4x - x^2}} \] **At \(x = 2\):** \[ f'(2) = \dfrac{4 - 2 \cdot 2}{2\sqrt{4 \cdot 2 - 2^2}} = \dfrac{0}{2\sqrt{8 - 4}} = \dfrac{0}{2 \cdot 2} = 0 \] --- **Step 2: Compute the second derivative \(f''(x)\).** Using \(f'(x)\): \[ f'(x) = \dfrac{4 - 2x}{2[4x - x^2]^{1/2}} \] Compute \(f''(x)\) by differentiating \(f'(x)\): First, denote \(n(x) = 4 - 2x\) and \(d(x) = 2[4x - x^2]^{1/2}\). Then: \[ f''(x) = \dfrac{n'(x)d(x) - n(x)d'(x)}{[d(x)]^2} \] Compute derivatives: \[ n'(x) = -2 \] \[ d'(x) = 2 \cdot \dfrac{1}{2}[4x - x^2]^{-1/2}(4 - 2x) = [4x - x^2]^{-1/2}(4 - 2x) \] Simplify numerator: \[ n'(x)d(x) - n(x)d'(x) = -2 \cdot 2[4x - x^2]^{1/2} - (4 - 2x) \cdot [4 - 2x][4x - x^2]^{-1/2} \] Simplify further: \[ n'(x)d(x) - n(x)d'(x) = -4[4x - x^2]^{1/2} - (4 - 2x)^2[4x - x^2]^{-1/2} \] Factor out \([4x - x^2]^{-1/2}\): \[ = \left(-4[4x - x^2] - (4 - 2x)^2\right)[4x - x^2]^{-1/2} \] Simplify expressions: \[ -4(4x - x^2) - (4 - 2x)^2 = -16x + 4x^2 - (16 - 16x + 4x^2) = -16 \] Therefore: \[ f''(x) = \dfrac{-16[4x - x^2]^{-1/2}}{4[4x - x^2]} = \dfrac{-4}{[4x - x^2]^{3/2}} \] **At \(x = 2\):** \[ f''(2) = \dfrac{-4}{[4 \cdot 2 - 2^2]^{3/2}} = \dfrac{-4}{[8 - 4]^{3/2}} = \dfrac{-4}{[4]^{3/2}} = \dfrac{-4}{8} = -\dfrac{1}{2} \] --- **Step 3: Use the curvature formula to find \(\kappa\) at \(x = 2\).** The curvature \(\kappa\) of a function \(y = f(x)\) is given by: \[ \kappa(x) = \dfrac{|f''(x)|}{\left[1 + (f'(x))^2\right]^{3/2}} \] **At \(x = 2\):** Since \(f'(2) = 0\) and \(f''(2) = -\dfrac{1}{2}\): \[ \kappa(2) = \dfrac{\left|-\dfrac{1}{2}\right|}{\left[1 + (0)^2\right]^{3/2}} = \dfrac{\dfrac{1}{2}}{1^{3/2}} = \dfrac{1}{2} \] --- **Answer:** The curvature of \( f(x) = \sqrt{4x - x^2} \) at \( x = 2 \) is: \[ \kappa(2) = \dfrac{1}{2} \] **Therefore, the curvature at \( x = 2 \) is \( \dfrac{1}{2} \).** |
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| deepseekv3 | To find the curvature \( \kappa \) of the function \( f(x) = \sqrt{4x - x^2} \) at \( x = 2 \), we follow these steps: ### Step 1: Compute the first derivative \( f'(x) \) Given: \[ f(x) = \sqrt{4x - x^2} \] We can rewrite this as \( f(x) = (4x - x^2)^{1/2} \). Using the chain rule, the derivative is: \[ f'(x) = \frac{1}{2}(4x - x^2)^{-1/2} \cdot (4 - 2x) \] Simplify: \[ f'(x) = \frac{4 - 2x}{2\sqrt{4x - x^2}} = \frac{2 - x}{\sqrt{4x - x^2}} \] ### Step 2: Compute the second derivative \( f''(x) \) Using the quotient rule, the derivative of \( f'(x) \) is: \[ f''(x) = \frac{d}{dx}\left(\frac{2 - x}{\sqrt{4x - x^2}}\right) \] \[ f''(x) = \frac{(-1)\sqrt{4x - x^2} - (2 - x) \cdot \frac{4 - 2x}{2\sqrt{4x - x^2}}}{4x - x^2} \] Simplify the numerator: \[ (-1)\sqrt{4x - x^2} - \frac{(2 - x)(4 - 2x)}{2\sqrt{4x - x^2}} \] Combine the terms over a common denominator: \[ f''(x) = \frac{-2(4x - x^2) - (2 - x)(4 - 2x)}{2(4x - x^2)^{3/2}} \] Simplify further: \[ f''(x) = \frac{-8x + 2x^2 - (8 - 4x - 4x + 2x^2)}{2(4x - x^2)^{3/2}} \] \[ f''(x) = \frac{-8x + 2x^2 - (8 - 8x + 2x^2)}{2(4x - x^2)^{3/2}} \] \[ f''(x) = \frac{-8x + 2x^2 - 8 + 8x - 2x^2}{2(4x - x^2)^{3/2}} \] \[ f''(x) = \frac{-8}{2(4x - x^2)^{3/2}} = \frac{-4}{(4x - x^2)^{3/2}} \] ### Step 3: Evaluate \( f'(x) \) and \( f''(x) \) at \( x = 2 \) First, evaluate \( f'(2) \): \[ f'(2) = \frac{2 - 2}{\sqrt{4(2) - (2)^2}} = \frac{0}{\sqrt{8 - 4}} = 0 \] Next, evaluate \( f''(2) \): \[ f''(2) = \frac{-4}{(4(2) - (2)^2)^{3/2}} = \frac{-4}{(8 - 4)^{3/2}} = \frac{-4}{4^{3/2}} = \frac{-4}{8} = -\frac{1}{2} \] ### Step 4: Compute the curvature \( \kappa \) The formula for curvature is: \[ \kappa = \frac{|f''(x)|}{\left(1 + (f'(x))^2\right)^{3/2}} \] At \( x = 2 \): \[ \kappa = \frac{\left|-\frac{1}{2}\right|}{\left(1 + 0^2\right)^{3/2}} = \frac{\frac{1}{2}}{1} = \frac{1}{2} \] ### Final Answer: The curvature of \( f(x) = \sqrt{4x - x^2} \) at \( x = 2 \) is: \[ \boxed{\frac{1}{2}} \] |
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