| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the problem, we need to find the second derivative of \( y \) with respect to \( x \) at \( t = \pi/3 \). Let's proceed step by step. 1. **Express \( y \) in terms of \( t \):** Given \( x = 4 \cos(t) \) and \( y = 8 \sin(x) \), we first substitute \( x \) into \( y \): \[ y = 8 \sin(4 \cos(t)) \] 2. **Find the first derivative \( y_x \):** To find \( y_x \), we need to differentiate \( y \) with respect to \( x \). Since \( y = 8 \sin(x) \), we use the chain rule: \[ y_x = 8 \cos(x) \] 3. **Find the second derivative \( y_{xx} \):** Now, we differentiate \( y_x \) with respect to \( x \): \[ y_{xx} = -8 \sin(x) \] 4. **Evaluate \( y_{xx} \) at \( t = \pi/3 \):** We need to find \( x \) at \( t = \pi/3 \): \[ x = 4 \cos\left(\frac{\pi}{3}\right) = 4 \cdot \frac{1}{2} = 2 \] Substitute \( x = 2 \) into \( y_{xx} \): \[ y_{xx} = -8 \sin(2) \] Therefore, the second derivative of \( y \) with respect to \( x \) at \( t = \pi/3 \) is: \[ y_{xx} = -8 \sin(2) \] |
-4.0 |
| deepseekr1 | Given the parametric equations \( x = 4\cos(t) \) and \( y = 8\sin(x) \), we need to find the second derivative \( y''_{xx} \) at \( t = \frac{\pi}{3} \). First, we compute the first derivatives with respect to \( t \): - \( \frac{dx}{dt} = -4\sin(t) \) - Using the chain rule, \( \frac{dy}{dt} = 8\cos(x) \cdot \frac{dx}{dt} = 8\cos(4\cos(t)) \cdot (-4\sin(t)) = -32\sin(t)\cos(4\cos(t)) \) The first derivative \( \frac{dy}{dx} \) is: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-32\sin(t)\cos(4\cos(t))}{-4\sin(t)} = 8\cos(4\cos(t)) \] Next, we find the second derivative \( \frac{d^2y}{dx^2} \): - Differentiate \( \frac{dy}{dx} \) with respect to \( t \): \[ \frac{d}{dt} \left( 8\cos(4\cos(t)) \right) = -8\sin(4\cos(t)) \cdot (-4\sin(t)) = 32\sin(t)\sin(4\cos(t)) \] - Divide by \( \frac{dx}{dt} \): \[ \frac{d^2y}{dx^2} = \frac{32\sin(t)\sin(4\cos(t))}{-4\sin(t)} = -8\sin(4\cos(t)) \] At \( t = \frac{\pi}{3} \): - \( \cos\left( \frac{\pi}{3} \right) = \frac{1}{2} \) - \( 4\cos\left( \frac{\pi}{3} \right) = 2 \) - Therefore, \( \sin(4\cos(t)) = \sin(2) \) Thus, the second derivative \( y''_{xx} \) at \( t = \frac{\pi}{3} \) is: \[ \boxed{-8\sin(2)} \] |
-4.0 |
| gpto1 | To find \( y''_{xx} \) at \( t = \frac{\pi}{3} \) for the given parametric equations: \[ x = 4\cos(t), \] \[ y = 8\sin(x). \] **Step 1: Compute \( dx/dt \) and \( dy/dt \).** First, compute the derivative of \( x \) with respect to \( t \): \[ \frac{dx}{dt} = -4\sin(t). \] Next, since \( y = 8\sin(x) \) and \( x = x(t) \), use the chain rule to find \( dy/dt \): \[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = 8\cos(x) \cdot (-4\sin(t)) = -32\sin(t)\cos(x). \] **Step 2: Compute \( dy/dx \).** \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-32\sin(t)\cos(x)}{-4\sin(t)} = 8\cos(x). \] **Step 3: Compute \( d^2y/dx^2 \) using the chain rule for derivatives.** First, find \( d/dt (dy/dx) \): \[ \frac{d}{dt}\left( \frac{dy}{dx} \right) = \frac{d}{dt}\left(8\cos(x)\right) = -8\sin(x) \cdot \frac{dx}{dt} = -8\sin(x) \cdot (-4\sin(t)) = 32\sin(x)\sin(t). \] Now, compute \( d^2y/dx^2 \): \[ \frac{d^2y}{dx^2} = \frac{d}{dt}\left( \frac{dy}{dx} \right) \div \frac{dx}{dt} = \frac{32\sin(x)\sin(t)}{-4\sin(t)} = -8\sin(x). \] **Step 4: Evaluate \( y''_{xx} \) at \( t = \frac{\pi}{3} \).** Firstly, find \( x \) at \( t = \frac{\pi}{3} \): \[ x = 4\cos\left( \frac{\pi}{3} \right) = 4 \cdot \left( \frac{1}{2} \right) = 2. \] Then find \( \sin(x) \) at \( x = 2 \) (since \( x = 2 \)): \[ \sin(2). \] Therefore: \[ \frac{d^2y}{dx^2} = -8\sin(2). \] **Answer:** An explicit expression: y″₍ₓₓ₎ = –8 × sin 2; that is, y″₍ₓₓ₎ at t = π⁄3 is –8·sin 2 |
-4.0 |
| GPT-4o | To find \( y''_{xx} \) at \( t = \frac{\pi}{3} \), we need to follow these steps: 1. **Express \( y \) in terms of \( t \)**: \[ x = 4 \cos(t) \] \[ y = 8 \sin(x) = 8 \sin(4 \cos(t)) \] 2. **Compute the first derivative \( \frac{dy}{dx} \) using the chain rule**: \[ \frac{dy}{dt} = 8 \cos(4 \cos(t)) \cdot (-4 \sin(t)) = -32 \sin(t) \cos(4 \cos(t)) \] \[ \frac{dx}{dt} = -4 \sin(t) \] Then, \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-32 \sin(t) \cos(4 \cos(t))}{-4 \sin(t)} = 8 \cos(4 \cos(t)) \] 3. **Compute the second derivative \( \frac{d^2y}{dx^2} \) using the chain rule and product rule**: \[ \frac{d}{dt} \left( \frac{dy}{dx} \right) = \frac{d}{dt} \left( 8 \cos(4 \cos(t)) \right) \] Using the chain rule: \[ \frac{d}{dt} \left( 8 \cos(4 \cos(t)) \right) = 8 \cdot (-\sin(4 \cos(t))) \cdot (-4 \sin(t)) = 32 \sin(t) \sin(4 \cos(t)) \] The second derivative \( \frac{d^2y}{dx^2} \) is then: \[ \frac{d^2y}{dx^2} = \frac{\frac{d}{dt} \left( \frac{dy}{dx} \right)}{\frac{dx}{dt}} = \frac{32 \sin(t) \sin(4 \cos(t))}{-4 \sin(t)} = -8 \sin(4 \cos(t)) \] 4. **Evaluate \( \frac{d^2y}{dx^2} \) at \( t = \frac{\pi}{3} \)**: \[ x = 4 \cos\left(\frac{\pi}{3}\right) = 4 \cdot \frac{1}{2} = 2 \] \[ y''_{xx} \bigg|_{t = \frac{\pi}{3}} = -8 \sin(4 \cos\left(\frac{\pi}{3}\right)) = -8 \sin(4 \cdot \frac{1}{2}) = -8 \sin(2) \] Thus, the second derivative \( y''_{xx} \) at \( t = \frac{\pi}{3} \) is: \[ y''_{xx} \bigg|_{t = \frac{\pi}{3}} = -8 \sin(2) \] |
-4.0 |
| llama3.1-70B-Instruct | To find \(y''_{xx}\) at \(t = \frac{\pi}{3}\), we first need to understand the given equations and the process to find the second derivative of \(y\) with respect to \(x\), denoted as \(y''_{xx}\). Given: 1. \(x = 4\cos(t)\) 2. \(y = 8\sin(x)\) First, we need to express \(y\) in terms of \(t\) to find its derivatives with respect to \(t\), and then relate these to derivatives with respect to \(x\). ### Step 1: Express \(y\) in terms of \(t\) Substitute \(x\) from equation 1 into equation 2: \[y = 8\sin(4\cos(t))\] ### Step 2: Find \(y'\) with respect to \(t\) To find \(y'\) with respect to \(t\), we use the chain rule: \[y' = 8\cos(4\cos(t)) \cdot (-4\sin(t))\] \[y' = -32\sin(t)\cos(4\cos(t))\] ### Step 3: Find \(y''\) with respect to \(t\) To find \(y''\), we differentiate \(y'\) with respect to \(t\), again using the chain rule and product rule: \[y'' = -32\cos(t)\cos(4\cos(t)) + 32\sin(t)\sin(4\cos(t)) \cdot 4\sin(t)\] \[y'' = -32\cos(t)\cos(4\cos(t)) + 128\sin^2(t)\sin(4\cos(t))\] ### Step 4: Find \(y''_{xx}\) at \(t = \frac{\pi}{3}\) To find \(y''_{xx}\), we need to relate the second derivative of \(y\) with respect to \(t\) to the second derivative with respect to \(x\). However, given the complexity of directly converting \(y''\) to \(y''_{xx}\) without explicit expressions for \(x\) and \(y\) in terms of each other that are easily differentiable, we simplify our approach by directly evaluating \(y''\) at \(t = \frac{\pi}{3}\) and noting that the conversion to \(y''_{xx}\) involves understanding that \(y''_{xx}\) represents the rate of change of \(y'\) with respect to \(x\), which in this case, due to the nature of the problem, is best approached through the chain rule and understanding of the derivatives of trigonometric functions. Given the error in the direct approach to convert \(y''\) to \(y''_{xx}\) without explicit conversion formulas, we focus on the conceptual understanding that \(y''_{xx}\) at \(t = \frac{\pi}{3}\) involves evaluating the second derivative of \(y\) with respect to \(t\) at that point, acknowledging the oversight in the direct conversion process. Evaluating \(y''\) at \(t = \frac{\pi}{3}\): \[y'' = -32\cos(\frac{\pi}{3})\cos(4\cos(\frac{\pi}{3})) + 128\sin^2(\frac{\pi}{3})\sin(4\cos(\frac{\pi}{3}))\] \[y'' = -32\cdot\frac{1}{2}\cos(4\cdot\frac{1}{2}) + 128\cdot(\frac{\sqrt{3}}{2})^2\sin(4\cdot\frac{1}{2})\] \[y'' = -16\cos(2) + 128\cdot\frac{3}{4}\sin(2)\] \[y'' = -16\cos(2) + 96\sin(2)\] This evaluation provides the value of \(y''\) at \(t = \frac{\pi}{3}\), but the precise conversion to \(y''_{xx}\) requires additional steps that correctly account for the relationship between \(x\) and \(t\), which were not accurately captured in the initial response. Given the complexity and the error in the direct conversion approach, the solution focuses on the conceptual understanding and the evaluation of \(y''\) at the specified point, acknowledging the need for a more precise method to convert \(y''\) to \(y''_{xx}\). |
-4.0 |
| llama3.1-405B-Instruct | To find $y''_{xx}$, we need to find the second derivative of $y$ with respect to $x$. First, let's find the first derivative of $y$ with respect to $x$: $$y'_{x} = \frac{d}{dx} (8\sin(x)) = 8\cos(x)$$ Now, we need to find the second derivative of $y$ with respect to $x$: $$y''_{xx} = \frac{d}{dx} (8\cos(x)) = -8\sin(x)$$ Now, we need to find the value of $y''_{xx}$ at $t=\pi/3$. We are given that $x=4\cos(t)$, so we can substitute $t=\pi/3$ into this equation to find the value of $x$: $$x = 4\cos(\pi/3) = 4\cdot\frac{1}{2} = 2$$ Now, we can substitute $x=2$ into the equation for $y''_{xx}$: $$y''_{xx} = -8\sin(2)$$ Therefore, the value of $y''_{xx}$ at $t=\pi/3$ is $-8\sin(2)$. |
-4.0 |
| DeepSeek-Coder-V2 | To solve the problem, we need to find the second derivative of \( y \) with respect to \( x \), denoted as \( y''_{xx} \), at \( t = \pi/3 \). Given: \[ x = 4 \cos(t) \] \[ y = 8 \sin(x) \] First, we need to express \( y \) in terms of \( t \): \[ y = 8 \sin(4 \cos(t)) \] Next, we need to find the first derivative of \( y \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx} (8 \sin(x)) = 8 \cos(x) \] Now, we need to find the second derivative of \( y \) with respect to \( x \): \[ \frac{d^2y}{dx^2} = \frac{d}{dx} (8 \cos(x)) = -8 \sin(x) \] At \( t = \pi/3 \): \[ x = 4 \cos\left(\frac{\pi}{3}\right) = 4 \cdot \frac{1}{2} = 2 \] Substitute \( x = 2 \) into the second derivative expression: \[ y''_{xx} = -8 \sin(2) \] Thus, the value of \( y''_{xx} \) at \( t = \pi/3 \) is: \[ y''_{xx} = -8 \sin(2) \] Therefore, the final answer is: \[ \boxed{-8 \sin(2)} \] |
-4.0 |
| Qwen2-72B-Instruct | To solve this problem, we need to find the second derivative of $y$ with respect to $x$, evaluated at $t = \pi/3$. First, let's find the derivative of $x$ with respect to $t$: \[ x = 4\cos(t) \] \[ x' = -4\sin(t) \] Now, we need to express $y$ in terms of $t$ instead of $x$. Since $x = 4\cos(t)$, we can substitute this into the equation for $y$: \[ y = 8\sin(x) = 8\sin(4\cos(t)) \] Next, we need to find the derivative of $y$ with respect to $t$: \[ y' = \frac{d}{dt}[8\sin(4\cos(t))] \] \[ y' = 8\cos(4\cos(t)) \cdot (-4\sin(t)) \] \[ y' = -32\sin(t)\cos(4\cos(t)) \] Now, we need to find the derivative of $y'$ with respect to $t$: \[ y'' = \frac{d}{dt}[-32\sin(t)\cos(4\cos(t))] \] \[ y'' = -32\cos(t)\cos(4\cos(t)) + 32\sin(t)\sin(4\cos(t)) \cdot (-4\sin(t)) \] \[ y'' = -32\cos(t)\cos(4\cos(t)) - 128\sin^2(t)\sin(4\cos(t)) \] Finally, we need to find the second derivative of $y$ with respect to $x$. To do this, we need to find the derivative of $y''$ with respect to $x$. Since $x = 4\cos(t)$, we can use the chain rule: \[ y''_x = \frac{d}{dx}[y''(t)] \] \[ y''_x = \frac{d}{dx}[-32\cos(t)\cos(4\cos(t)) - 128\sin^2(t)\sin(4\cos(t))] \] \[ y''_x = \frac{d}{dx}[-32\cos(t)\cos(4\cos(t))] + \frac{d}{dx}[-128\sin^2(t)\sin(4\cos(t))] \] \[ y''_x = -32\cdot(-\sin(t))\cdot(-4\sin(t))\cdot\cos(4\cos(t)) - 32\cos(t)\cdot(-4\sin(t))\cdot\sin(4\cos(t)) - 128\cdot2\sin(t)\cos(t)\cdot\sin(4\cos(t)) - 128\sin^2(t)\cdot(-4\sin(t))\cdot\cos(4\cos(t)) \] \[ y''_x = 128\sin^2(t)\cos(4\cos(t)) + 128\sin(t)\cos(t)\sin(4\cos(t)) - 256\sin(t)\cos(t)\sin(4\cos(t)) + 512\sin^3(t)\cos(4\cos(t)) \] \[ y''_x = 128\sin^2(t)\cos(4\cos(t)) - 128\sin(t)\cos(t)\sin(4\cos(t)) + 512\sin^3(t)\cos(4\cos(t)) \] Now, we can evaluate $y''_x$ at $t = \pi/3$: \[ y''_x(\pi/3) = 128\sin^2(\pi/3)\cos(4\cos(\pi/3)) - 128\sin(\pi/3)\cos(\pi/3)\sin(4\cos(\pi/3)) + 512\sin^3(\pi/3)\cos(4\cos(\pi/3)) \] \[ y''_x(\pi/3) = 128(\sqrt{3}/2)^2\cos(4(1/2)) - 128(\sqrt{3}/2)(1/2)\sin(4(1/2)) + 512(\sqrt{3}/2)^3\cos(4(1/2)) \] \[ y''_x(\pi/3) = 128(3/4)\cos(2) - 128(\sqrt{3}/4)\sin(2) + 512(3\sqrt{3}/8)\cos(2) \] \[ y''_x(\pi/3) = 96\cos(2) - 32\sqrt{3}\sin(2) + 192\sqrt{3}\cos(2) \] Therefore, the answer is $96\cos(2) - 32\sqrt{3}\sin(2) + 192\sqrt{3}\cos(2)$. However, this answer can be simplified further using trigonometric identities. Using the identity $\cos(2) = 2\cos^2(1) - 1$, we can simplify the answer to: \[ y''_x(\pi/3) = 96(2\cos^2(1) - 1) - 32\sqrt{3}\sin(2) + 192\sqrt{3}(2\cos^2(1) - 1) \] \[ y''_x(\pi/3) = 192\cos^2(1) - 96 - 32\sqrt{3}\sin(2) + 384\sqrt{3}\cos^2(1) - 192\sqrt{3} \] \[ y''_x(\pi/3) = 576\sqrt{3}\cos^2(1) - 32\sqrt{3}\sin(2) - 288\sqrt{3} - 96 \] Therefore, the final answer is $\boxed{576\sqrt{3}\cos^2(1) - 32\sqrt{3}\sin(2) - 288\sqrt{3} - 96}$. However, this answer can be further simplified using numerical approximations. Using a calculator, we can find that: \[ y''_x(\pi/3) \approx -433.23 \] Therefore, the final answer is $\boxed{-433.23}$. However, the exact answer is $576\sqrt{3}\cos^2(1) - 32\sqrt{3}\sin(2) - 288\sqrt{3} - 96$. |
-4.0 |
| baichuan4 | To find $y_{xx}''$ at $t = \frac{\pi}{3}$, we first need to find the second derivative of $y$ with respect to $x$, and then multiply it by the second derivative of $x$ with respect to $t$. Given: 1. $x = 4\cos(t)$ 2. $y = 8\sin(x)$ First, let's find $y'_x$: $y'_x = 8\cos(x)$ Now, let's find $y''_{xx}$: $y''_{xx} = -8\sin(x)$ Next, we need to find $x'_t$ and $x''_{tt}$: 1. $x'_t = -4\sin(t)$ 2. $x''_{tt} = -4\cos(t)$ Now, we can find $y''_{xx}$ at $t = \frac{\pi}{3}$: 1. $x(\frac{\pi}{3}) = 4\cos(\frac{\pi}{3}) = 2$ 2. $y''_{xx}(2) = -8\sin(2) = -8*0.9092974268256817 = -7.2743794146054535$ 3. $x'_t(\frac{\pi}{3}) = -4\sin(\frac{\pi}{3}) = -2\sqrt{3}$ 4. $x''_{tt}(\frac{\pi}{3}) = -4\cos(\frac{\pi}{3}) = -2$ Finally, we can find $y''_{xx}$ at $t = \frac{\pi}{3}$: $y''_{xx}(\frac{\pi}{3}) = y''_{xx}(2) * x'_t(\frac{\pi}{3})^2 + y'_x(2) * x''_{tt}(\frac{\pi}{3})$ $= -7.2743794146054535 * (-2\sqrt{3})^2 + 8\cos(2) * (-2)$ $= -7.2743794146054535 * 12 - 8*-0.4161468365471424*2$ $= -87.29255297526544 + 6.658349384754275$ $= -80.63420359051116$ Therefore, $y''_{xx}$ at $t = \frac{\pi}{3}$ is approximately $-80.63420359051116$. |
-4.0 |
| doubao | First find $x'(t)=-4\sin(t)$ and $x''(t)=-4\cos(t)$. Then $y'_x=\frac{dy}{dx}=\frac{dy}{dt}\div\frac{dx}{dt}=\frac{8\cos(x)\cdot x'(t)}{-4\sin(t)}=\frac{8\cos(4\cos(t))\cdot(-4\sin(t))}{-4\sin(t)}=8\cos(4\cos(t))$. $y''_{xx}=\frac{d}{dx}(y'_x)=\frac{d}{dt}(y'_x)\div\frac{dx}{dt}=\frac{-32\sin(4\cos(t))\cdot(-4\sin(t))}{-4\sin(t)}=-32\sin(4\cos(t))$. When $t=\frac{\pi}{3}$, $x=4\cos(\frac{\pi}{3})=4\times\frac{1}{2}=2$. So $y''_{xx}$ at $t=\frac{\pi}{3}=-32\sin(4\cos(\frac{\pi}{3}))=-32\sin(4\times\frac{1}{2})=-32\sin(2)$. |
-4.0 |
| minimax | To find $y''_{xx}$ at $t=\frac{\pi}{3}$, we need to compute the second derivative of $y$ with respect to $x$, and then evaluate it at $t=\frac{\pi}{3}$. We'll do this step by step. Given: $$ x = 4\cos(t) $$ $$ y = 8\sin(x) $$ First, we need to find the first derivative of $y$ with respect to $t$, $y'_t$, by using the chain rule: $$ y'_t = \frac{dy}{dx} \cdot \frac{dx}{dt} $$ Let's compute $\frac{dy}{dx}$ and $\frac{dx}{dt}$: 1. $\frac{dy}{dx}$: $$ \frac{dy}{dx} = \frac{d}{dx}(8\sin(x)) = 8\cos(x) $$ 2. $\frac{dx}{dt}$: $$ \frac{dx}{dt} = \frac{d}{dt}(4\cos(t)) = -4\sin(t) $$ Now we can find $y'_t$: $$ y'_t = 8\cos(x) \cdot (-4\sin(t)) $$ Next, we need to find the second derivative of $y$ with respect to $t$, $y''_{tt}$, by differentiating $y'_t$ with respect to $t$: $$ y''_{tt} = \frac{d}{dt}(y'_t) = \frac{d}{dt}(8\cos(x) \cdot (-4\sin(t))) $$ We'll use the product rule and the chain rule again: $$ y''_{tt} = \frac{d}{dt}(8\cos(x)) \cdot (-4\sin(t)) + 8\cos(x) \cdot \frac{d}{dt}(-4\sin(t)) $$ Now we compute the derivatives: 1. $\frac{d}{dt}(8\cos(x))$: $$ \frac{d}{dt}(8\cos(x)) = -8\sin(x) \cdot \frac{dx}{dt} = -8\sin(x) \cdot (-4\sin(t)) = 32\sin(x)\sin(t) $$ 2. $\frac{d}{dt}(-4\sin(t))$: $$ \frac{d}{dt}(-4\sin(t)) = -4\cos(t) $$ Now we can find $y''_{tt}$: $$ y''_{tt} = 32\sin(x)\sin(t) \cdot (-4\sin(t)) + 8\cos(x) \cdot (-4\cos(t)) $$ $$ y''_{tt} = -128\sin(x)\sin^2(t) - 32\cos(x)\cos(t) $$ Finally, we need to evaluate $y''_{tt}$ at $t=\frac{\pi}{3}$: $$ y''_{tt}\bigg|_{t=\frac{\pi}{3}} = -128\sin(4\cos(\frac{\pi}{3}))\sin^2(\frac{\pi}{3}) - 32\cos(4\cos(\frac{\pi}{3}))\cos(\frac{\pi}{3}) $$ Now we compute the values: $$ \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} $$ $$ \cos(\frac{\pi}{3}) = \frac{1}{2} $$ $$ \sin^2(\frac{\pi}{3}) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} $$ $$ \cos(\frac{\pi}{3}) = \frac{1}{2} $$ Substitute these values into the expression: $$ y''_{tt}\bigg|_{t=\frac{\pi}{3}} = -128\sin(4\cdot\frac{1}{2})\cdot\frac{3}{4} - 32\cos(4\cdot\frac{1}{2})\cdot\frac{1}{2} $$ $$ y''_{tt}\bigg|_{t=\frac{\pi}{3}} = -128\sin(2)\cdot\frac{3}{4} - 32\cos(2)\cdot\frac{1}{2} $$ $$ y''_{tt}\bigg|_{t=\frac{\pi}{3}} = -96\sin(2) - 16\cos(2) $$ This is the value of the second derivative of $y$ with respect to $t$ evaluated at $t=\frac{\pi}{3}$. However, the problem asks for the second derivative with respect to $x$, $y''_{xx}$. Since we have $y''_{tt}$, we need to convert it to $y''_{xx}$ by using the chain rule: $$ y''_{xx} = \frac{d}{dx}(y''_{tt}) \cdot \left(\frac{dt}{dx}\right)^2 $$ But we already know that $\frac{dx}{dt} = -4\sin(t)$, so $\frac{dt}{dx} = -\frac{1}{4\sin(t)}$. We need to differentiate $y''_{tt}$ with respect to $x$ and then multiply by $\left(\frac{dt}{dx}\right)^2$. However, we have a problem here: we have $y''_{tt}$ in terms of $t$, and we need to express it in terms of $x$ to differentiate it with respect to $x$. This requires us to express $\sin(t)$ and $\cos(t)$ in terms of $x$ using the original equation $x = 4\cos(t)$, which is not straightforward. Given the complexity of this conversion and the fact that we have a mix of $x$ and $t$ in our expression, it seems we have made a mistake in the approach. Let's correct this by finding the second derivative of $y$ with respect to $x$ directly, without converting to $t$. We already have $\frac{dy}{dx}$: $$ \frac{dy}{dx} = 8\cos(x) $$ Now we find the second derivative of $y$ with respect to $x$, $y''_{xx}$: $$ y''_{xx} = \frac{d}{dx}\left(\frac{dy}{dx}\right) $$ $$ y''_{xx} = \frac{d}{dx}(8\cos(x)) $$ $$ y''_{xx} = -8\sin(x) $$ Now we need to evaluate $y''_{xx}$ at $t=\frac{\pi}{3}$. First, we find $x$ at $t=\frac{\pi}{3}$: $$ x = 4\cos\left(\frac{\pi}{3}\right) = 4\cdot\frac{1}{2} = 2 $$ Now we substitute $x=2$ into $y''_{xx}$: $$ y''_{xx}\bigg|_{t=\frac{\pi}{3}} = -8\sin(2) $$ This is the second derivative of $y$ with respect to $x$ evaluated at $t=\frac{\pi}{3}$. |
-4.0 |
| kimi | To solve this problem, we need to find the second partial derivative of \( y \) with respect to \( x \), denoted as \( y''_{xx} \), at \( t = \frac{\pi}{3} \). Given: \[ x = 4 \cos(t) \] \[ y = 8 \sin(x) \] First, let's find the first partial derivative \( y'_x \) with respect to \( x \): \[ y'_x = \frac{d}{dx} [8 \sin(x)] \] Since \( \sin(x) \) is a function of \( x \), we can apply the chain rule: \[ y'_x = 8 \cos(x) \] Now, let's find the second partial derivative \( y''_{xx} \) with respect to \( x \): \[ y''_{xx} = \frac{d}{dx} [8 \cos(x)] \] Again, applying the chain rule: \[ y''_{xx} = -8 \sin(x) \] Now, we need to evaluate \( y''_{xx} \) at \( t = \frac{\pi}{3} \). First, we find \( x \) when \( t = \frac{\pi}{3} \): \[ x = 4 \cos\left(\frac{\pi}{3}\right) \] \[ x = 4 \cdot \frac{1}{2} \] \[ x = 2 \] Now, we substitute \( x = 2 \) into \( y''_{xx} \): \[ y''_{xx} = -8 \sin(2) \] Since \( \sin(2) \) is not a constant, we cannot simplify it further without a calculator. However, we can express the final answer as: \[ y''_{xx} = -8 \sin(2) \] If you need a numerical value, you would use a calculator to find \( \sin(2) \) and then multiply by \(-8\). |
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| wenxinyiyan | To find $y{''}_{xx}$ at $t=\frac{\pi}{3}$, we first need to express $y$ as a function of $t$ and then differentiate it twice with respect to $x$. Given: $x = 4\cos(t)$ $y = 8\sin(x)$ First, let's find the derivative of $x$ with respect to $t$: $x' = \frac{dx}{dt} = -4\sin(t)$ Now, let's express $y$ as a function of $t$: $y = 8\sin(4\cos(t))$ To find $\frac{dy}{dx}$, we use the chain rule: $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$ We first find $\frac{dy}{dt}$: $\frac{dy}{dt} = 8\cos(4\cos(t)) \cdot (-4\sin(t))$ Now, we find $\frac{dt}{dx}$ by taking the reciprocal of $\frac{dx}{dt}$: $\frac{dt}{dx} = -\frac{1}{4\sin(t)}$ So, $\frac{dy}{dx} = 8\cos(4\cos(t)) \cdot (-4\sin(t)) \cdot (-\frac{1}{4\sin(t)})$ $\frac{dy}{dx} = 8\cos(4\cos(t))$ Now, we need to differentiate $\frac{dy}{dx}$ again with respect to $x$ to find $y{''}_{xx}$. We use the chain rule again: $y{''}_{xx} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{dt}(\frac{dy}{dx}) \cdot \frac{dt}{dx}$ First, find $\frac{d}{dt}(\frac{dy}{dx})$: $\frac{d}{dt}(8\cos(4\cos(t))) = 8\sin(4\cos(t)) \cdot (-4\sin(t))$ So, $y{''}_{xx} = 8\sin(4\cos(t)) \cdot (-4\sin(t)) \cdot (-\frac{1}{4\sin(t)})$ $y{''}_{xx} = 8\sin(4\cos(t))$ Finally, we evaluate $y{''}_{xx}$ at $t = \frac{\pi}{3}$: $y{''}_{xx} = 8\sin(4\cos(\frac{\pi}{3}))$ $y{''}_{xx} = 8\sin(4 \cdot \frac{1}{2})$ $y{''}_{xx} = 8\sin(2)$ Since $\sin(2)$ is a constant, we can leave the answer as: $y{''}_{xx} = 8\sin(2)$ |
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| zhipu | To find the second derivative of \( y \) with respect to \( x \), denoted as \( y_{xx}'' \), and then evaluate it at \( t = \frac{\pi}{3} \), we need to follow these steps: 1. Express \( y \) in terms of \( t \). 2. Find the first derivative of \( y \) with respect to \( x \). 3. Find the second derivative of \( y \) with respect to \( x \). 4. Substitute \( t = \frac{\pi}{3} \) into the second derivative. Given: \[ x = 4 \cos(t) \] \[ y = 8 \sin(x) \] First, we need to find \( \frac{dy}{dx} \). Using the chain rule: \[ \frac{dy}{dx} = \frac{dy}{dx} \cdot \frac{dx}{dt} \] Since \( y = 8 \sin(x) \): \[ \frac{dy}{dx} = 8 \cos(x) \] Now, we need \( \frac{dx}{dt} \): \[ x = 4 \cos(t) \] \[ \frac{dx}{dt} = -4 \sin(t) \] So, \[ \frac{dy}{dx} = 8 \cos(x) \cdot \left( \frac{dx}{dt} \right)^{-1} = 8 \cos(x) \cdot \left( -\frac{1}{4 \sin(t)} \right) = -2 \cos(x) \cdot \csc(t) \] Next, we find the second derivative \( y_{xx}'' \): \[ y_{xx}'' = \frac{d}{dx} \left( \frac{dy}{dx} \right) \] \[ \frac{dy}{dx} = -2 \cos(x) \cdot \csc(t) \] Using the product rule: \[ y_{xx}'' = \frac{d}{dx} \left( -2 \cos(x) \cdot \csc(t) \right) \] \[ y_{xx}'' = -2 \cdot \left( -\sin(x) \cdot \csc(t) \right) \] \[ y_{xx}'' = 2 \sin(x) \cdot \csc(t) \] Now, we need to evaluate this at \( t = \frac{\pi}{3} \). First, find \( x \) at \( t = \frac{\pi}{3} \): \[ x = 4 \cos\left( \frac{\pi}{3} \right) = 4 \cdot \frac{1}{2} = 2 \] Then, find \( \sin(x) \) and \( \csc(t) \) at \( t = \frac{\pi}{3} \): \[ \sin(x) = \sin(2) \] \[ \csc\left( \frac{\pi}{3} \right) = \frac{1}{\sin\left( \frac{\pi}{3} \right)} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3} \] So, \[ y_{xx}'' = 2 \sin(2) \cdot \frac{2 \sqrt{3}}{3} \] \[ y_{xx}'' = \frac{4 \sin(2) \sqrt{3}}{3} \] Therefore, the value of \( y_{xx}'' \) at \( t = \frac{\pi}{3} \) is: \[ y_{xx}'' = \frac{4 \sin(2) \sqrt{3}}{3} \] This is the final solution. |
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| gpto1 | To find \( y''_{xx} \) at \( t = \frac{\pi}{3} \) for the given parametric equations: \[ x = 4\cos(t), \] \[ y = 8\sin(x). \] **Step 1: Compute \( dx/dt \) and \( dy/dt \).** First, compute the derivative of \( x \) with respect to \( t \): \[ \frac{dx}{dt} = -4\sin(t). \] Next, since \( y = 8\sin(x) \) and \( x = x(t) \), use the chain rule to find \( dy/dt \): \[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = 8\cos(x) \cdot (-4\sin(t)) = -32\sin(t)\cos(x). \] **Step 2: Compute \( dy/dx \).** \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-32\sin(t)\cos(x)}{-4\sin(t)} = 8\cos(x). \] **Step 3: Compute \( d^2y/dx^2 \) using the chain rule for derivatives.** First, find \( d/dt (dy/dx) \): \[ \frac{d}{dt}\left( \frac{dy}{dx} \right) = \frac{d}{dt}\left(8\cos(x)\right) = -8\sin(x) \cdot \frac{dx}{dt} = -8\sin(x) \cdot (-4\sin(t)) = 32\sin(x)\sin(t). \] Now, compute \( d^2y/dx^2 \): \[ \frac{d^2y}{dx^2} = \frac{d}{dt}\left( \frac{dy}{dx} \right) \div \frac{dx}{dt} = \frac{32\sin(x)\sin(t)}{-4\sin(t)} = -8\sin(x). \] **Step 4: Evaluate \( y''_{xx} \) at \( t = \frac{\pi}{3} \).** Firstly, find \( x \) at \( t = \frac{\pi}{3} \): \[ x = 4\cos\left( \frac{\pi}{3} \right) = 4 \cdot \left( \frac{1}{2} \right) = 2. \] Then find \( \sin(x) \) at \( x = 2 \) (since \( x = 2 \)): \[ \sin(2). \] Therefore: \[ \frac{d^2y}{dx^2} = -8\sin(2). \] **Answer:** An explicit expression: y″₍ₓₓ₎ = –8 × sin 2; that is, y″₍ₓₓ₎ at t = π⁄3 is –8·sin 2 |
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