| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the equation of the line tangent to the curve defined by F(t) = (t^2 + 1, 2^t) at the point where y = 4, we need to follow these steps: 1. Find the value of t that corresponds to y = 4. 2. Find the coordinates of the point on the curve where y = 4. 3. Find the derivative of the function F(t) with respect to t. 4. Evaluate the derivative at the value of t found in step 1 to get the slope of the tangent line. 5. Use the point-slope form of a line to write the equation of the tangent line. Step 1: Find the value of t that corresponds to y = 4. We are given that y = 4, so we need to find the value of t that satisfies the equation 2^t = 4. 2^t = 4 t = log2(4) t = 2 Step 2: Find the coordinates of the point on the curve where y = 4. Now that we have the value of t, we can find the coordinates of the point on the curve where y = 4. F(t) = (t^2 + 1, 2^t) F(2) = (2^2 + 1, 2^2) F(2) = (5, 4) So, the point on the curve where y = 4 is (5, 4). Step 3: Find the derivative of the function F(t) with respect to t. To find the derivative of F(t), we need to find the derivatives of the x and y components separately. dx/dt = d(t^2 + 1)/dt = 2t dy/dt = d(2^t)/dt = 2^t \* ln(2) Step 4: Evaluate the derivative at the value of t found in step 1 to get the slope of the tangent line. Now that we have the derivatives, we can evaluate them at t = 2 to get the slope of the tangent line. dx/dt(2) = 2(2) = 4 dy/dt(2) = 2^2 \* ln(2) = 4 \* ln(2) The slope of the tangent line is dy/dx = (dy/dt) / (dx/dt) = (4 \* ln(2)) / 4 = ln(2). Step 5: Use the point-slope form of a line to write the equation of the tangent line. Now that we have the slope and the point on the curve, we can use the point-slope form of a line to write the equation of the tangent line. y - 4 = ln(2)(x - 5) So, the equation of the tangent line is y - 4 = ln(2)(x - 5). The correct answer is D. y - 4 = ln 2(x - 5). |
D |
| deepseekr1 | To find the tangent line to the curve \( F(t) = (t^2 + 1, 2^t) \) at \( y = 4 \): 1. **Determine \( t \) when \( y = 4 \):** \( 2^t = 4 \Rightarrow t = 2 \). 2. **Find the point on the curve at \( t = 2 \):** \( x = t^2 + 1 = 5 \), so the point is \( (5, 4) \). 3. **Compute derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):** \( \frac{dx}{dt} = 2t \) and \( \frac{dy}{dt} = \ln 2 \cdot 2^t \). 4. **Calculate \( \frac{dy}{dx} \) at \( t = 2 \):** \( \frac{dy}{dx} = \frac{\ln 2 \cdot 4}{4} = \ln 2 \). **Slope** = \( \ln 2 \). 5. **Use point-slope form with \( (5, 4) \):** \( y - 4 = \ln 2 \cdot (x - 5) \). **Answer:** D. \( y – 4 = \ln 2(x – 5) \). |
D |
| gpto1 | D | D |
| GPT-4o | To determine the equation of the tangent line, we need to find the point of tangency and the slope of the tangent line at that point. Given the function \( F(t) = (t^2 + 1, 2^t) \), we need to find the value of \( t \) such that \( y = 4 \). From \( 2^t = 4 \), we can solve for \( t \): \[ 2^t = 4 \] \[ 2^t = 2^2 \] \[ t = 2 \] So, when \( t = 2 \), \( y = 4 \). Now, let's find the corresponding \( x \) value: \[ x = t^2 + 1 \] \[ x = 2^2 + 1 \] \[ x = 4 + 1 \] \[ x = 5 \] Thus, the point of tangency is \( (5, 4) \). Next, we need to find the slope of the tangent line at this point. The slope of the tangent line is given by the derivative of \( y \) with respect to \( x \). We can find this by using the derivatives of the parametric functions: \[ \frac{dx}{dt} = \frac{d}{dt}(t^2 + 1) = 2t \] \[ \frac{dy}{dt} = \frac{d}{dt}(2^t) = 2^t \ln 2 \] At \( t = 2 \): \[ \frac{dx}{dt} = 2(2) = 4 \] \[ \frac{dy}{dt} = 2^2 \ln 2 = 4 \ln 2 \] The slope \( m \) of the tangent line is then: \[ m = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4 \ln 2}{4} = \ln 2 \] Now, we can write the equation of the tangent line in point-slope form, \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) = (5, 4) \) and \( m = \ln 2 \): \[ y - 4 = \ln 2(x - 5) \] Thus, the correct answer is: D. y – 4 = ln 2(x – 5) |
D |
| llama3.1-70B-Instruct | The correct answer is A. |
D |
| llama3.1-405B-Instruct | A The best answer is A. |
D |
| DeepSeek-Coder-V2 | B | D |
| Qwen2-72B-Instruct | B. y – 4 = 4 ln 2(x – 2) | D |
| baichuan4 | To find the equation of the tangent line to the curve \(F(t) = (t^2 + 1, 2^t)\) at the point where \(y = 4\), we first need to find the slope of the tangent line by taking the derivative of the function with respect to \(t\). The derivative of \(F(t)\) with respect to \(t\) is: \[F'(t) = \left(\frac{d}{dt}(t^2 + 1), \frac{d}{dt}(2^t)\right) = (2t, 2^t \ln(2))\] Now, we need to find the value of \(t\) where \(y = 4\), which means solving the equation \(2^t = 4\). Since \(4 = 2^2\), we have \(t = 2\). Next, we evaluate \(F'(t)\) at \(t = 2\) to find the slope of the tangent line: \[F'(2) = (2*2, 2^2 \ln(2)) = (4, 4\ln(2))\] The slope of the tangent line is \(4\ln(2)\). Now, we need to find a point on the curve where \(y = 4\). We already know that this happens at \(t = 2\), so we have the point \((2^2 + 1, 2^2) = (5, 4)\) on the curve. Finally, we use the point-slope formula for a line to write the equation of the tangent line: \[y - y_1 = m(x - x_1)\] \[y - 4 = 4\ln(2)(x - 5)\] Therefore, the correct answer is: B. \(y - 4 = 4\ln(2)(x - 5)\) |
D |
| doubao | First find the value of t when y = 2^t = 4, so 2^t = 2^2, t = 2. Then F(2) = (5, 4). The derivative of the x-component is 2t, so the slope at t = 2 is 4. The equation of the tangent line is y - 4 = 4(x - 5). So the answer is C. | D |
| minimax | To find the equation of the tangent line to the curve at the point where \( y = 4 \), we need to find the derivative of \( F(t) \) with respect to \( t \), evaluate it at the point where \( y = 4 \), and then use the point-slope form of the line equation. Given \( F(t) = (t^2 + 1, 2^t) \), we have two components: \( x(t) = t^2 + 1 \) and \( y(t) = 2^t \). First, we find the derivatives of \( x(t) \) and \( y(t) \) with respect to \( t \): \( x'(t) = \frac{d}{dt}(t^2 + 1) = 2t \) \( y'(t) = \frac{d}{dt}(2^t) = 2^t \ln(2) \) The slope of the tangent line at any point \( t \) is given by the derivative of \( y \) with respect to \( x \), which is \( \frac{dy}{dx} = \frac{y'(t)}{x'(t)} \). Now, we need to find the value of \( t \) where \( y = 4 \): \( 2^t = 4 \) \( 2^t = 2^2 \) \( t = 2 \) Now we can find the slope of the tangent line at \( t = 2 \): \( \frac{dy}{dx} \bigg|_{t=2} = \frac{y'(2)}{x'(2)} = \frac{2^2 \ln(2)}{2 \cdot 2} = \frac{4 \ln(2)}{4} = \ln(2) \) The point on the curve where \( y = 4 \) is \( (t^2 + 1, 2^t) \) when \( t = 2 \), which is \( (2^2 + 1, 2^2) = (5, 4) \). Using the point-slope form of the line equation \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point on the line, we get: \( y - 4 = \ln(2)(x - 5) \) Therefore, the correct answer is: D. \( y – 4 = \ln 2(x – 5) \) |
D |
| kimi | To find the equation of the tangent line to the curve at the point where \( y = 4 \), we first need to find the point on the curve where \( y = 4 \). The curve is defined by \( F(t) = (t^2 + 1, 2^t) \), so we set \( 2^t = 4 \) to find the corresponding \( t \) value. \[ 2^t = 4 \] \[ 2^t = 2^2 \] \[ t = 2 \] Now we have the point \( (t^2 + 1, 2^t) = (2^2 + 1, 2^2) = (5, 4) \) on the curve. Next, we need to find the derivative of \( F(t) \) to determine the slope of the tangent line at this point. The derivative \( F'(t) \) is the rate of change of \( F(t) \) with respect to \( t \), and it will give us the slope of the tangent line at any point on the curve. \[ F(t) = (t^2 + 1, 2^t) \] \[ F'(t) = (2t, 2^t \ln 2) \] At \( t = 2 \), the slope \( m \) of the tangent line is: \[ m = F'(2) = (2 \cdot 2, 2^2 \ln 2) = (4, 4 \ln 2) \] The slope of the tangent line is the \( y \)-component of \( F'(2) \), which is \( 4 \ln 2 \). Now, we can use the point-slope form of the equation of a line to find the equation of the tangent line: \[ y - y_1 = m(x - x_1) \] \[ y - 4 = 4 \ln 2 (x - 5) \] So the correct answer is: D. \( y - 4 = 4 \ln 2 (x - 5) \) |
D |
| wenxinyiyan | B | D |
| zhipu | To find the equation of the line tangent to the curve \( F(t) = (t^2 + 1, 2^t) \) at the point where \( y = 4 \), we need to follow these steps: 1. **Determine the point of tangency:** - We know \( y = 4 \), so we need to find the corresponding \( t \) such that \( 2^t = 4 \). - Solving \( 2^t = 4 \), we get \( t = 2 \) because \( 2^2 = 4 \). 2. **Find the coordinates of the point of tangency:** - Substitute \( t = 2 \) into \( F(t) \): \[ F(2) = (2^2 + 1, 2^2) = (4 + 1, 4) = (5, 4) \] - So, the point of tangency is \( (5, 4) \). 3. **Find the slope of the tangent line:** - The slope of the tangent line to the curve at a point is given by the derivative \( \frac{dy}{dx} \). - We have \( x = t^2 + 1 \) and \( y = 2^t \). - Differentiate both with respect to \( t \): \[ \frac{dx}{dt} = 2t \quad \text{and} \quad \frac{dy}{dt} = 2^t \ln 2 \] - The slope \( \frac{dy}{dx} \) is then: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2^t \ln 2}{2t} \] - At \( t = 2 \): \[ \frac{dy}{dx} \bigg|_{t=2} = \frac{2^2 \ln 2}{2 \cdot 2} = \frac{4 \ln 2}{4} = \ln 2 \] 4. **Write the equation of the tangent line:** - The point-slope form of the equation of a line is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency. - Here, \( m = \ln 2 \) and the point is \( (5, 4) \). - Substituting these values, we get: \[ y - 4 = \ln 2 (x - 5) \] So, the correct equation of the tangent line is \( y - 4 = \ln 2 (x - 5) \). Therefore, the correct answer is: **D. y – 4 = ln 2(x – 5)** |
D |
| gpto1 | D | D |
| deepseekv3 | To solve this problem, we need to find the equation of the tangent line to the curve defined by \( F(t) = (t^2 + 1, 2^t) \) at the point where \( y = 4 \). ### Step 1: Find the value of \( t \) where \( y = 4 \) Given \( y = 2^t \), set \( y = 4 \): \[ 2^t = 4 \implies t = 2 \] ### Step 2: Find the corresponding \( x \)-coordinate Using \( t = 2 \) in \( x = t^2 + 1 \): \[ x = 2^2 + 1 = 5 \] So, the point is \( (5, 4) \). ### Step 3: Compute the derivative \( F'(t) \) The derivative of \( F(t) = (t^2 + 1, 2^t) \) is: \[ F'(t) = \left( 2t, 2^t \ln 2 \right) \] At \( t = 2 \): \[ F'(2) = \left( 4, 4 \ln 2 \right) \] The slope of the tangent line is the ratio of the \( y \)-component to the \( x \)-component of \( F'(2) \): \[ \text{Slope} = \frac{4 \ln 2}{4} = \ln 2 \] ### Step 4: Write the equation of the tangent line Using the point-slope form \( y - y_1 = m(x - x_1) \) with \( (x_1, y_1) = (5, 4) \) and \( m = \ln 2 \): \[ y - 4 = \ln 2 (x - 5) \] Thus, the correct answer is: **D. \( y - 4 = \ln 2(x - 5) \)** |
D |