| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the numeric value of $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}$ where $u=\arctan \frac{y}{x}$, we first need to find the first and second partial derivatives of $u$ with respect to $x$ and $y$. Given $u = \arctan\left(\frac{y}{x}\right)$, we can find the partial derivatives as follows: 1. **First Partial Derivatives:** - $\frac{\partial u}{\partial x} = \frac{\partial}{\partial x}\left[\arctan\left(\frac{y}{x}\right)\right]$ Using the chain rule, we have: $\frac{\partial u}{\partial x} = \frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot \left(\frac{-y}{x^2}\right) = \frac{-y}{x^2 + y^2}$ - $\frac{\partial u}{\partial y} = \frac{\partial}{\partial y}\left[\arctan\left(\frac{y}{x}\right)\right]$ Using the chain rule, we have: $\frac{\partial u}{\partial y} = \frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot \left(\frac{1}{x}\right) = \frac{x}{x^2 + y^2}$ 2. **Second Partial Derivatives:** - $\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}\left[\frac{-y}{x^2 + y^2}\right]$ Using the quotient rule, we have: $\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}\left[\frac{-y}{x^2 + y^2}\right] = \frac{2xy}{(x^2 + y^2)^2}$ - $\frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y}\left[\frac{x}{x^2 + y^2}\right]$ Using the quotient rule, we have: $\frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y}\left[\frac{x}{x^2 + y^2}\right] = \frac{-2xy}{(x^2 + y^2)^2}$ 3. **Sum of Second Partial Derivatives:** - $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{2xy}{(x^2 + y^2)^2} + \frac{-2xy}{(x^2 + y^2)^2} = 0$ Therefore, the numeric value of $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}$ where $u=\arctan \frac{y}{x}$ is $\boxed{0}$. |
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| deepseekr1 | To find the value of \(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\) where \(u = \arctan\left(\frac{y}{x}\right)\), we start by computing the first and second partial derivatives. First, we compute the first partial derivatives: - For \(\frac{\partial u}{\partial x}\): \[ \frac{\partial u}{\partial x} = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \left(-\frac{y}{x^2}\right) = -\frac{y}{x^2 + y^2} \] - For \(\frac{\partial u}{\partial y}\): \[ \frac{\partial u}{\partial y} = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \frac{1}{x} = \frac{x}{x^2 + y^2} \] Next, we compute the second partial derivatives: - For \(\frac{\partial^2 u}{\partial x^2}\): \[ \frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x} \left(-\frac{y}{x^2 + y^2}\right) = \frac{2xy}{(x^2 + y^2)^2} \] - For \(\frac{\partial^2 u}{\partial y^2}\): \[ \frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y} \left(\frac{x}{x^2 + y^2}\right) = -\frac{2xy}{(x^2 + y^2)^2} \] Adding these second partial derivatives together: \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{2xy}{(x^2 + y^2)^2} + \left(-\frac{2xy}{(x^2 + y^2)^2}\right) = 0 \] This result can also be confirmed by noting that \(u = \arctan\left(\frac{y}{x}\right)\) corresponds to the angular coordinate \(\theta\) in polar coordinates, which is a harmonic function. Therefore, the Laplacian of \(u\) is zero. \[ \boxed{0} \] |
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| gpto1 | To find the numeric value of \(\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2}\) when \(u = \arctan\left(\dfrac{y}{x}\right)\), we need to compute the second partial derivatives of \(u\) with respect to \(x\) and \(y\) and then sum them. **Step 1: Compute the first partial derivatives of \(u\).** Given: \[ u = \arctan\left(\dfrac{y}{x}\right) \] **First, compute \(\dfrac{\partial u}{\partial x}\):** Using the chain rule and the derivative of the arctangent function: \[ \dfrac{\partial u}{\partial x} = \dfrac{\partial}{\partial x} \left( \arctan\left( \dfrac{y}{x} \right) \right) = \dfrac{-\dfrac{y}{x^2}}{1 + \left( \dfrac{y}{x} \right)^2} \] Simplify the denominator: \[ 1 + \left( \dfrac{y}{x} \right)^2 = \dfrac{x^2 + y^2}{x^2} \] Simplify \(\dfrac{\partial u}{\partial x}\): \[ \dfrac{\partial u}{\partial x} = -\dfrac{y}{x^2} \times \dfrac{x^2}{x^2 + y^2} = -\dfrac{y}{x^2 + y^2} \] **Next, compute \(\dfrac{\partial u}{\partial y}\):** Similarly, \[ \dfrac{\partial u}{\partial y} = \dfrac{\partial}{\partial y} \left( \arctan\left( \dfrac{y}{x} \right) \right) = \dfrac{\dfrac{1}{x}}{1 + \left( \dfrac{y}{x} \right)^2} \] Simplify \(\dfrac{\partial u}{\partial y}\): \[ \dfrac{\partial u}{\partial y} = \dfrac{1}{x} \times \dfrac{x^2}{x^2 + y^2} = \dfrac{x}{x^2 + y^2} \] **Step 2: Compute the second partial derivatives of \(u\).** **Compute \(\dfrac{\partial^2 u}{\partial x^2}\):** Differentiate \(\dfrac{\partial u}{\partial x}\) with respect to \(x\): \[ \dfrac{\partial^2 u}{\partial x^2} = \dfrac{\partial}{\partial x} \left( -\dfrac{y}{x^2 + y^2} \right) \] Since \(y\) is treated as a constant when differentiating with respect to \(x\): \[ \dfrac{\partial^2 u}{\partial x^2} = -y \times \dfrac{\partial}{\partial x} \left( \dfrac{1}{x^2 + y^2} \right) = -y \left( -\dfrac{2x}{(x^2 + y^2)^2} \right) = \dfrac{2xy}{(x^2 + y^2)^2} \] **Compute \(\dfrac{\partial^2 u}{\partial y^2}\):** Differentiate \(\dfrac{\partial u}{\partial y}\) with respect to \(y\): \[ \dfrac{\partial^2 u}{\partial y^2} = \dfrac{\partial}{\partial y} \left( \dfrac{x}{x^2 + y^2} \right) \] Again, \(x\) is treated as a constant when differentiating with respect to \(y\): \[ \dfrac{\partial^2 u}{\partial y^2} = x \times \dfrac{\partial}{\partial y} \left( \dfrac{1}{x^2 + y^2} \right) = x \left( -\dfrac{2y}{(x^2 + y^2)^2} \right) = -\dfrac{2xy}{(x^2 + y^2)^2} \] **Step 3: Sum the second partial derivatives.** \[ \dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} = \dfrac{2xy}{(x^2 + y^2)^2} + \left(-\dfrac{2xy}{(x^2 + y^2)^2}\right) = 0 \] **Conclusion:** The sum of the second partial derivatives is zero: \[ \dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} = 0 \] This means that \(u\) satisfies Laplace's equation and is a harmonic function. **Numeric Value:** \[ \dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} = 0 \] |
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| GPT-4o | To solve the problem, we need to find the second partial derivatives of \( u = \arctan \left( \frac{y}{x} \right) \) with respect to \( x \) and \( y \), and then sum them. First, let's find the first partial derivatives of \( u \) with respect to \( x \) and \( y \). 1. **First Partial Derivative with respect to \( x \):** \[ u = \arctan \left( \frac{y}{x} \right) \] Using the chain rule: \[ \frac{\partial u}{\partial x} = \frac{1}{1 + \left( \frac{y}{x} \right)^2} \cdot \frac{\partial}{\partial x} \left( \frac{y}{x} \right) \] \[ \frac{\partial}{\partial x} \left( \frac{y}{x} \right) = y \cdot \frac{\partial}{\partial x} \left( x^{-1} \right) = y \cdot (-x^{-2}) = -\frac{y}{x^2} \] Thus, \[ \frac{\partial u}{\partial x} = \frac{1}{1 + \left( \frac{y}{x} \right)^2} \cdot \left( -\frac{y}{x^2} \right) = -\frac{y}{x^2 + y^2} \] 2. **First Partial Derivative with respect to \( y \):** Similarly, \[ \frac{\partial u}{\partial y} = \frac{1}{1 + \left( \frac{y}{x} \right)^2} \cdot \frac{\partial}{\partial y} \left( \frac{y}{x} \right) \] \[ \frac{\partial}{\partial y} \left( \frac{y}{x} \right) = \frac{1}{x} \] Thus, \[ \frac{\partial u}{\partial y} = \frac{1}{1 + \left( \frac{y}{x} \right)^2} \cdot \frac{1}{x} = \frac{x}{x^2 + y^2} \] Now, let's find the second partial derivatives. 3. **Second Partial Derivative with respect to \( x \):** \[ \frac{\partial}{\partial x} \left( \frac{\partial u}{\partial x} \right) = \frac{\partial}{\partial x} \left( -\frac{y}{x^2 + y^2} \right) \] Using the quotient rule, \[ \frac{\partial}{\partial x} \left( -\frac{y}{x^2 + y^2} \right) = -y \cdot \frac{\partial}{\partial x} \left( \frac{1}{x^2 + y^2} \right) \] \[ \frac{\partial}{\partial x} \left( \frac{1}{x^2 + y^2} \right) = -\frac{2x}{(x^2 + y^2)^2} \] Thus, \[ \frac{\partial^2 u}{\partial x^2} = -y \cdot \left( -\frac{2x}{(x^2 + y^2)^2} \right) = \frac{2xy}{(x^2 + y^2)^2} \] 4. **Second Partial Derivative with respect to \( y \):** \[ \frac{\partial}{\partial y} \left( \frac{\partial u}{\partial y} \right) = \frac{\partial}{\partial y} \left( \frac{x}{x^2 + y^2} \right) \] Using the quotient rule, \[ \frac{\partial}{\partial y} \left( \frac{x}{x^2 + y^2} \right) = x \cdot \frac{\partial}{\partial y} \left( \frac{1}{x^2 + y^2} \right) \] \[ \frac{\partial}{\partial y} \left( \frac{1}{x^2 + y^2} \right) = -\frac{2y}{(x^2 + y^2)^2} \] Thus, \[ \frac{\partial^2 u}{\partial y^2} = x \cdot \left( -\frac{2y}{(x^2 + y^2)^2} \right) = -\frac{2xy}{(x^2 + y^2)^2} \] Finally, summ |
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| llama3.1-70B-Instruct | To find the value of $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}$, we first need to find the first and second partial derivatives of $u$ with respect to $x$ and $y$. Given that $u=\arctan \frac{y}{x}$, we can start by finding the first partial derivatives: $$\frac{\partial u}{\partial x} = \frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot \left(-\frac{y}{x^2}\right) = -\frac{y}{x^2+y^2}$$ $$\frac{\partial u}{\partial y} = \frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot \frac{1}{x} = \frac{x}{x^2+y^2}$$ Now, we can find the second partial derivatives: $$\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x} \left(-\frac{y}{x^2+y^2}\right) = \frac{2xy}{(x^2+y^2)^2}$$ $$\frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y} \left(\frac{x}{x^2+y^2}\right) = -\frac{2xy}{(x^2+y^2)^2}$$ Now, we can add the two second partial derivatives: $$\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2} = \frac{2xy}{(x^2+y^2)^2} - \frac{2xy}{(x^2+y^2)^2} = 0$$ Therefore, the value of $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}$ is 0. |
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| llama3.1-405B-Instruct | ## Step 1: Find the first partial derivative of u with respect to x. To find $\frac{\partial u}{\partial x}$, we'll use the chain rule and the fact that the derivative of arctan is $\frac{1}{1 + (\frac{y}{x})^2}$. Applying this, we get $\frac{\partial u}{\partial x} = \frac{1}{1 + (\frac{y}{x})^2} \cdot \frac{-y}{x^2}$. ## Step 2: Simplify the expression for the first partial derivative of u with respect to x. Simplifying the expression from Step 1, we have $\frac{\partial u}{\partial x} = \frac{-y}{x^2 + y^2}$. ## Step 3: Find the second partial derivative of u with respect to x. To find $\frac{\partial^2 u}{\partial x^2}$, we differentiate $\frac{\partial u}{\partial x}$ with respect to x. Applying the quotient rule, we get $\frac{\partial^2 u}{\partial x^2} = \frac{(x^2 + y^2)(0) - (-y)(2x)}{(x^2 + y^2)^2}$. ## Step 4: Simplify the expression for the second partial derivative of u with respect to x. Simplifying the expression from Step 3, we have $\frac{\partial^2 u}{\partial x^2} = \frac{2xy}{(x^2 + y^2)^2}$. ## Step 5: Find the first partial derivative of u with respect to y. To find $\frac{\partial u}{\partial y}$, we'll use the chain rule and the fact that the derivative of arctan is $\frac{1}{1 + (\frac{y}{x})^2}$. Applying this, we get $\frac{\partial u}{\partial y} = \frac{1}{1 + (\frac{y}{x})^2} \cdot \frac{1}{x}$. ## Step 6: Simplify the expression for the first partial derivative of u with respect to y. Simplifying the expression from Step 5, we have $\frac{\partial u}{\partial y} = \frac{x}{x^2 + y^2}$. ## Step 7: Find the second partial derivative of u with respect to y. To find $\frac{\partial^2 u}{\partial y^2}$, we differentiate $\frac{\partial u}{\partial y}$ with respect to y. Applying the quotient rule, we get $\frac{\partial^2 u}{\partial y^2} = \frac{(x^2 + y^2)(0) - (x)(2y)}{(x^2 + y^2)^2}$. ## Step 8: Simplify the expression for the second partial derivative of u with respect to y. Simplifying the expression from Step 7, we have $\frac{\partial^2 u}{\partial y^2} = \frac{-2xy}{(x^2 + y^2)^2}$. ## Step 9: Add the second partial derivatives of u with respect to x and y. To find $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}$, we add the expressions from Step 4 and Step 8: $\frac{2xy}{(x^2 + y^2)^2} + \frac{-2xy}{(x^2 + y^2)^2}$. ## Step 10: Simplify the expression for the sum of the second partial derivatives. Simplifying the expression from Step 9, we get $\frac{2xy - 2xy}{(x^2 + y^2)^2} = \frac{0}{(x^2 + y^2)^2} = 0$. The final answer is: $\boxed{0}$ |
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| DeepSeek-Coder-V2 | To solve the problem, we need to find the numeric value of \(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\) given that \(u = \arctan \frac{y}{x}\). First, let's compute the first partial derivatives of \(u\) with respect to \(x\) and \(y\). 1. **First Partial Derivative with respect to \(x\):** \[ u = \arctan \frac{y}{x} \] Using the chain rule, we get: \[ \frac{\partial u}{\partial x} = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \left(-\frac{y}{x^2}\right) = \frac{1}{1 + \frac{y^2}{x^2}} \cdot \left(-\frac{y}{x^2}\right) = \frac{x^2}{x^2 + y^2} \cdot \left(-\frac{y}{x^2}\right) = -\frac{y}{x^2 + y^2} \] 2. **First Partial Derivative with respect to \(y\):** \[ \frac{\partial u}{\partial y} = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \left(\frac{1}{x}\right) = \frac{1}{1 + \frac{y^2}{x^2}} \cdot \left(\frac{1}{x}\right) = \frac{x^2}{x^2 + y^2} \cdot \left(\frac{1}{x}\right) = \frac{x}{x^2 + y^2} \] Next, we compute the second partial derivatives. 3. **Second Partial Derivative with respect to \(x\):** \[ \frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x} \left( -\frac{y}{x^2 + y^2} \right) \] Using the quotient rule: \[ \frac{\partial^2 u}{\partial x^2} = -\frac{(x^2 + y^2) \cdot 0 - y \cdot 2x}{(x^2 + y^2)^2} = \frac{2xy}{(x^2 + y^2)^2} \] 4. **Second Partial Derivative with respect to \(y\):** \[ \frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y} \left( \frac{x}{x^2 + y^2} \right) \] Using the quotient rule: \[ \frac{\partial^2 u}{\partial y^2} = \frac{(x^2 + y^2) \cdot 0 - x \cdot 2y}{(x^2 + y^2)^2} = -\frac{2xy}{(x^2 + y^2)^2} \] Finally, we add the second partial derivatives: \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{2xy}{(x^2 + y^2)^2} + \left(-\frac{2xy}{(x^2 + y^2)^2}\right) = 0 \] Thus, the numeric value of \(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\) is \(\boxed{0}\). |
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| Qwen2-72B-Instruct | To find the numeric value of $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}$ where $u=\arctan \frac{y}{x}$, we first need to find the first and second partial derivatives of $u$ with respect to $x$ and $y$. Given $u = \arctan\left(\frac{y}{x}\right)$, we can find the partial derivatives as follows: 1. **First Partial Derivatives:** - $\frac{\partial u}{\partial x} = \frac{\partial}{\partial x}\left[\arctan\left(\frac{y}{x}\right)\right]$ Using the chain rule, we have: $\frac{\partial u}{\partial x} = \frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot \left(\frac{-y}{x^2}\right) = \frac{-y}{x^2 + y^2}$ - $\frac{\partial u}{\partial y} = \frac{\partial}{\partial y}\left[\arctan\left(\frac{y}{x}\right)\right]$ Using the chain rule, we have: $\frac{\partial u}{\partial y} = \frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot \left(\frac{1}{x}\right) = \frac{x}{x^2 + y^2}$ 2. **Second Partial Derivatives:** - $\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}\left[\frac{-y}{x^2 + y^2}\right]$ Using the quotient rule, we have: $\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}\left[\frac{-y}{x^2 + y^2}\right] = \frac{2xy}{(x^2 + y^2)^2}$ - $\frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y}\left[\frac{x}{x^2 + y^2}\right]$ Using the quotient rule, we have: $\frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y}\left[\frac{x}{x^2 + y^2}\right] = \frac{-2xy}{(x^2 + y^2)^2}$ 3. **Sum of Second Partial Derivatives:** - $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{2xy}{(x^2 + y^2)^2} + \frac{-2xy}{(x^2 + y^2)^2} = 0$ Therefore, the numeric value of $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}$ where $u=\arctan \frac{y}{x}$ is $\boxed{0}$. |
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| baichuan4 | To solve this problem, we need to find the second partial derivatives of $u$ with respect to $x$ and $y$, and then add them together. First, let's find $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$: $$\frac{\partial u}{\partial x} = \frac{-y}{x^2 + y^2}$$ $$\frac{\partial u}{\partial y} = \frac{x}{x^2 + y^2}$$ Next, we need to find the second partial derivatives: $$\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}\left(\frac{-y}{x^2 + y^2}\right) = \frac{2xy}{(x^2 + y^2)^2}$$ $$\frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y}\left(\frac{x}{x^2 + y^2}\right) = \frac{-2xy}{(x^2 + y^2)^2}$$ Now, we can add these two second partial derivatives together: $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{2xy}{(x^2 + y^2)^2} - \frac{2xy}{(x^2 + y^2)^2} = 0$$ Therefore, the numeric value of $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}$ is $0$. |
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| doubao | First, we find the first partial derivatives. $\frac{\partial u}{\partial x}=\frac{-\frac{y}{x^2}}{1+(\frac{y}{x})^2}=-\frac{y}{x^2+y^2}$ $\frac{\partial u}{\partial y}=\frac{\frac{1}{x}}{1+(\frac{y}{x})^2}=\frac{x}{x^2+y^2}$ Then we find the second partial derivatives. $\frac{\partial^2 u}{\partial x^2}=\frac{2xy}{(x^2+y^2)^2}$ $\frac{\partial^2 u}{\partial y^2}=\frac{-2xy}{(x^2+y^2)^2}$ So $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=\frac{2xy}{(x^2+y^2)^2}-\frac{2xy}{(x^2+y^2)^2}=0$. |
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| minimax | To solve this problem, we need to find the second partial derivatives of $u$ with respect to $x$ and $y$, and then sum them up. Given $u = \arctan\left(\frac{y}{x}\right)$, let's first find the first partial derivatives: $$\frac{\partial u}{\partial x} = \frac{\partial}{\partial x} \arctan\left(\frac{y}{x}\right)$$ Using the chain rule, we get: $$\frac{\partial u}{\partial x} = \frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot \left(-\frac{y}{x^2}\right)$$ Simplify this to: $$\frac{\partial u}{\partial x} = -\frac{y}{x^2+y^2}$$ Similarly, we find the first partial derivative with respect to $y$: $$\frac{\partial u}{\partial y} = \frac{\partial}{\partial y} \arctan\left(\frac{y}{x}\right)$$ Again using the chain rule, we get: $$\frac{\partial u}{\partial y} = \frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot \frac{1}{x}$$ Simplify this to: $$\frac{\partial u}{\partial y} = \frac{x}{x^2+y^2}$$ Now, let's find the second partial derivatives: $$\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x} \left(-\frac{y}{x^2+y^2}\right)$$ Applying the quotient rule, we get: $$\frac{\partial^2 u}{\partial x^2} = \frac{2xy}{(x^2+y^2)^2}$$ And for the second partial derivative with respect to $y$: $$\frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y} \left(\frac{x}{x^2+y^2}\right)$$ Applying the quotient rule again, we get: $$\frac{\partial^2 u}{\partial y^2} = \frac{-2xy}{(x^2+y^2)^2}$$ Now, summing up the second partial derivatives: $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{2xy}{(x^2+y^2)^2} + \frac{-2xy}{(x^2+y^2)^2}$$ Simplifying this, we see that the terms cancel each other out: $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$$ Therefore, the numeric value of $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}$ is $0$. |
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| kimi | To find the Laplacian of \( u = \arctan\left(\frac{y}{x}\right) \), we first need to compute the second partial derivatives with respect to \( x \) and \( y \). Let's start by finding the first partial derivatives. Given \( u = \arctan\left(\frac{y}{x}\right) \), we can write \( u \) as \( u = \arctan(y/x) \). Let's denote \( z = \frac{y}{x} \), so \( u = \arctan(z) \). Now we can find the partial derivatives of \( u \) with respect to \( x \) and \( y \). First, we find \( \frac{\partial u}{\partial x} \): \[ \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} \arctan\left(\frac{y}{x}\right) = -\frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot \frac{\partial}{\partial x} \left(\frac{y}{x}\right) \] \[ \frac{\partial}{\partial x} \left(\frac{y}{x}\right) = \frac{-y}{x^2} \] So, \[ \frac{\partial u}{\partial x} = -\frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot \frac{-y}{x^2} = \frac{y}{x^2 + y^2} \] Now, \( \frac{\partial u}{\partial y} \): \[ \frac{\partial u}{\partial y} = \frac{\partial}{\partial y} \arctan\left(\frac{y}{x}\right) = \frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot \frac{\partial}{\partial y} \left(\frac{y}{x}\right) \] \[ \frac{\partial}{\partial y} \left(\frac{y}{x}\right) = \frac{1}{x} \] So, \[ \frac{\partial u}{\partial y} = \frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot \frac{1}{x} = \frac{x}{x^2 + y^2} \] Next, we find the second partial derivatives: \[ \frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x} \left(\frac{y}{x^2 + y^2}\right) \] \[ = \frac{-2y(x^2 + y^2) + y(-2xy)}{(x^2 + y^2)^2} \] \[ = \frac{-2yx^2 - 2y^3}{(x^2 + y^2)^2} \] \[ = \frac{-2y(x^2 + y^2)}{(x^2 + y^2)^2} \] \[ = \frac{-2y}{(x^2 + y^2)^2} \] \[ = \frac{-2}{x(x^2 + y^2)} \] \[ = \frac{-2}{x^2 + y^2} \] \[ = \frac{-2}{x^2 + y^2} \] \[ = \frac{-2}{x^2 + y^2} \] \[ = \frac{-2}{x^2 + y^2} \] \[ = \frac{-2}{x^2 + y^2} \] Now, \[ \frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y} \left(\frac{x}{x^2 + y^2}\right) \] \[ = \frac{-2x(x^2 + y^2) + x(2xy)}{(x^2 + y^2)^2} \] \[ = \frac{-2x^3 - 2xy^2 + 2x^2y}{(x^2 + y^2)^2} \] \[ = \frac{-2x(x^2 - y^2)}{(x^2 + y^2)^2} \] \[ = \frac{-2x(x - y |
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| wenxinyiyan | 为了求解这个问题，我们首先需要找到$u$关于$x$和$y$的一阶和二阶偏导数。 给定 $u = \arctan\frac{y}{x}$，我们可以使用链式法则和求导法则来找到所需的偏导数。 1. **一阶偏导数**： * $\frac{\partial u}{\partial x} = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \left(-\frac{y}{x^2}\right) = -\frac{y}{x^2 + y^2}$ * $\frac{\partial u}{\partial y} = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \frac{1}{x} = \frac{x}{x^2 + y^2}$ 2. **二阶偏导数**： * $\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}\left(-\frac{y}{x^2 + y^2}\right) = \frac{2xy}{(x^2 + y^2)^2}$ * $\frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y}\left(\frac{x}{x^2 + y^2}\right) = -\frac{2xy}{(x^2 + y^2)^2}$ 3. **求和**： * $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{2xy}{(x^2 + y^2)^2} - \frac{2xy}{(x^2 + y^2)^2} = 0$ 因此，$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$。 |
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| zhipu | To find the numeric value of \(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\) where \(u = \arctan \frac{y}{x}\), we need to compute the second partial derivatives of \(u\) with respect to \(x\) and \(y\). First, let's find the first partial derivatives of \(u\). ### First Partial Derivatives 1. **Partial derivative with respect to \(x\):** \[ u = \arctan \left( \frac{y}{x} \right) \] Let \( v = \frac{y}{x} \), then \( u = \arctan(v) \). Using the chain rule: \[ \frac{\partial u}{\partial x} = \frac{d (\arctan(v))}{d v} \cdot \frac{\partial v}{\partial x} \] We know: \[ \frac{d (\arctan(v))}{d v} = \frac{1}{1 + v^2} \] And: \[ v = \frac{y}{x} \implies \frac{\partial v}{\partial x} = -\frac{y}{x^2} \] So: \[ \frac{\partial u}{\partial x} = \frac{1}{1 + \left( \frac{y}{x} \right)^2} \cdot \left( -\frac{y}{x^2} \right) = -\frac{y}{x^2 + y^2} \] 2. **Partial derivative with respect to \(y\):** Similarly, \[ \frac{\partial u}{\partial y} = \frac{d (\arctan(v))}{d v} \cdot \frac{\partial v}{\partial y} \] We already have: \[ \frac{d (\arctan(v))}{d v} = \frac{1}{1 + v^2} \] And: \[ v = \frac{y}{x} \implies \frac{\partial v}{\partial y} = \frac{1}{x} \] So: \[ \frac{\partial u}{\partial y} = \frac{1}{1 + \left( \frac{y}{x} \right)^2} \cdot \frac{1}{x} = \frac{x}{x^2 + y^2} \] ### Second Partial Derivatives Now, we compute the second partial derivatives. 1. **Second partial derivative with respect to \(x\):** \[ \frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x} \left( -\frac{y}{x^2 + y^2} \right) \] Using the quotient rule: \[ \frac{\partial}{\partial x} \left( -\frac{y}{x^2 + y^2} \right) = -y \cdot \frac{\partial}{\partial x} \left( \frac{1}{x^2 + y^2} \right) \] Let \( z = x^2 + y^2 \), then: \[ \frac{\partial}{\partial x} \left( \frac{1}{z} \right) = -\frac{1}{z^2} \cdot \frac{\partial z}{\partial x} = -\frac{1}{(x^2 + y^2)^2} \cdot 2x = -\frac{2x}{(x^2 + y^2)^2} \] So: \[ \frac{\partial^2 u}{\partial x^2} = -y \left( -\frac{2x}{(x^2 + y^2)^2} \right) = \frac{2xy}{(x^2 + y^2)^2} \] 2. **Second partial derivative with respect to \(y\):** \[ \frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y} \left( \frac{x}{x^2 + y^2} \right) \] Using the quotient rule: \[ \frac{\partial}{\partial y} \left( \frac{x}{x^2 + y^2} \right) = x \cdot \frac{\partial}{\partial y} \left( \frac{1}{x^2 + y^2} \right) \] Let \( z = x^2 + y^2 \), then: \[ \frac{\partial}{\partial y} \left( \frac{1}{z} \right) = -\frac{1}{z^2} \cdot \frac{\partial z}{\partial y} = -\frac{1}{(x^2 + y^2)^2} \cdot 2y = -\frac{2y}{(x^2 + y^2)^2} \] So: \[ \frac{\partial^2 u}{\partial y^2} = x \left( -\frac{2y}{(x^2 + y^2)^2} \right) = -\frac{2xy}{(x^2 + y^2)^2} \] ### Sum of Second Partial Derivatives Now, we add the second partial derivatives: \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{2xy}{(x^2 + y^2)^2} + \left( -\frac{2xy}{(x^2 + y^2)^2} \right) = 0 \] Therefore, the numeric value of \(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\) is: \[ \boxed{0} \] |
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| gpto1 | To find the numeric value of \(\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2}\) when \(u = \arctan\left(\dfrac{y}{x}\right)\), we need to compute the second partial derivatives of \(u\) with respect to \(x\) and \(y\) and then sum them. **Step 1: Compute the first partial derivatives of \(u\).** Given: \[ u = \arctan\left(\dfrac{y}{x}\right) \] **First, compute \(\dfrac{\partial u}{\partial x}\):** Using the chain rule and the derivative of the arctangent function: \[ \dfrac{\partial u}{\partial x} = \dfrac{\partial}{\partial x} \left( \arctan\left( \dfrac{y}{x} \right) \right) = \dfrac{-\dfrac{y}{x^2}}{1 + \left( \dfrac{y}{x} \right)^2} \] Simplify the denominator: \[ 1 + \left( \dfrac{y}{x} \right)^2 = \dfrac{x^2 + y^2}{x^2} \] Simplify \(\dfrac{\partial u}{\partial x}\): \[ \dfrac{\partial u}{\partial x} = -\dfrac{y}{x^2} \times \dfrac{x^2}{x^2 + y^2} = -\dfrac{y}{x^2 + y^2} \] **Next, compute \(\dfrac{\partial u}{\partial y}\):** Similarly, \[ \dfrac{\partial u}{\partial y} = \dfrac{\partial}{\partial y} \left( \arctan\left( \dfrac{y}{x} \right) \right) = \dfrac{\dfrac{1}{x}}{1 + \left( \dfrac{y}{x} \right)^2} \] Simplify \(\dfrac{\partial u}{\partial y}\): \[ \dfrac{\partial u}{\partial y} = \dfrac{1}{x} \times \dfrac{x^2}{x^2 + y^2} = \dfrac{x}{x^2 + y^2} \] **Step 2: Compute the second partial derivatives of \(u\).** **Compute \(\dfrac{\partial^2 u}{\partial x^2}\):** Differentiate \(\dfrac{\partial u}{\partial x}\) with respect to \(x\): \[ \dfrac{\partial^2 u}{\partial x^2} = \dfrac{\partial}{\partial x} \left( -\dfrac{y}{x^2 + y^2} \right) \] Since \(y\) is treated as a constant when differentiating with respect to \(x\): \[ \dfrac{\partial^2 u}{\partial x^2} = -y \times \dfrac{\partial}{\partial x} \left( \dfrac{1}{x^2 + y^2} \right) = -y \left( -\dfrac{2x}{(x^2 + y^2)^2} \right) = \dfrac{2xy}{(x^2 + y^2)^2} \] **Compute \(\dfrac{\partial^2 u}{\partial y^2}\):** Differentiate \(\dfrac{\partial u}{\partial y}\) with respect to \(y\): \[ \dfrac{\partial^2 u}{\partial y^2} = \dfrac{\partial}{\partial y} \left( \dfrac{x}{x^2 + y^2} \right) \] Again, \(x\) is treated as a constant when differentiating with respect to \(y\): \[ \dfrac{\partial^2 u}{\partial y^2} = x \times \dfrac{\partial}{\partial y} \left( \dfrac{1}{x^2 + y^2} \right) = x \left( -\dfrac{2y}{(x^2 + y^2)^2} \right) = -\dfrac{2xy}{(x^2 + y^2)^2} \] **Step 3: Sum the second partial derivatives.** \[ \dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} = \dfrac{2xy}{(x^2 + y^2)^2} + \left(-\dfrac{2xy}{(x^2 + y^2)^2}\right) = 0 \] **Conclusion:** The sum of the second partial derivatives is zero: \[ \dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} = 0 \] This means that \(u\) satisfies Laplace's equation and is a harmonic function. **Numeric Value:** \[ \dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} = 0 \] |
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