| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the mean velocity of motion between times t=1 and t=10, we need to calculate the displacement vector and divide it by the time interval. At t=1, the position vector is: x(1) = 2(1) = 2 y(1) = ln(1) = 0 z(1) = (1)^2 = 1 So, the position vector at t=1 is (2, 0, 1). At t=10, the position vector is: x(10) = 2(10) = 20 y(10) = ln(10) ≈ 2.30259 z(10) = (10)^2 = 100 So, the position vector at t=10 is (20, 2.30259, 100). The displacement vector is the difference between the position vectors at t=10 and t=1: Δr = (20-2, 2.30259-0, 100-1) = (18, 2.30259, 99) The magnitude of the displacement vector is: |Δr| = √(18^2 + 2.30259^2 + 99^2) ≈ 100.09 The time interval between t=1 and t=10 is 9 seconds. Therefore, the mean velocity of motion is: v̄ = Δr / Δt = (18, 2.30259, 99) / 9 = (2, 0.25584, 11). So, the mean velocity of motion between times t=1 and t=10 is approximately (2, 0.25584, 11) m/s. |
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| deepseekr1 | To find the mean velocity of motion between times \( t = 1 \) and \( t = 10 \) for the point defined by the equations \( x = 2t \), \( y = \ln(t) \), and \( z = t^2 \), we need to calculate the displacement in each coordinate direction and divide by the time interval \( \Delta t = 10 - 1 = 9 \). 1. **Displacement in the x-direction:** - At \( t = 1 \): \( x = 2 \times 1 = 2 \) - At \( t = 10 \): \( x = 2 \times 10 = 20 \) - Displacement: \( 20 - 2 = 18 \) - Mean velocity in x-direction: \( \frac{18}{9} = 2 \) 2. **Displacement in the y-direction:** - At \( t = 1 \): \( y = \ln(1) = 0 \) - At \( t = 10 \): \( y = \ln(10) \) - Displacement: \( \ln(10) - 0 = \ln(10) \) - Mean velocity in y-direction: \( \frac{\ln(10)}{9} \) 3. **Displacement in the z-direction:** - At \( t = 1 \): \( z = 1^2 = 1 \) - At \( t = 10 \): \( z = 10^2 = 100 \) - Displacement: \( 100 - 1 = 99 \) - Mean velocity in z-direction: \( \frac{99}{9} = 11 \) Thus, the mean velocity vector is calculated by dividing each component of the displacement by the time interval \( \Delta t = 9 \). The result is: \[ \boxed{\left( 2, \dfrac{\ln(10)}{9}, 11 \right)} \] |
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| gpto1 | To find the mean velocity of the point between times \( t = 1 \) and \( t = 10 \), we'll first calculate the displacement vector over this time interval and then divide it by the total time elapsed. **Given:** - Position functions: \[ x(t) = 2t \] \[ y(t) = \ln(t) \] \[ z(t) = t^2 \] **Step 1: Calculate the initial and final positions.** At \( t = 1 \): \[ x(1) = 2 \times 1 = 2 \] \[ y(1) = \ln(1) = 0 \] \[ z(1) = (1)^2 = 1 \] So, the initial position vector is: \[ \mathbf{r}_1 = (2,\ 0,\ 1) \] At \( t = 10 \): \[ x(10) = 2 \times 10 = 20 \] \[ y(10) = \ln(10) \] \[ z(10) = (10)^2 = 100 \] So, the final position vector is: \[ \mathbf{r}_2 = (20,\ \ln(10),\ 100) \] **Step 2: Calculate the displacement vector.** \[ \Delta \mathbf{r} = \mathbf{r}_2 - \mathbf{r}_1 \] \[ \Delta \mathbf{r} = (20 - 2,\ \ln(10) - 0,\ 100 - 1) \] \[ \Delta \mathbf{r} = (18,\ \ln(10),\ 99) \] **Step 3: Calculate the total time elapsed.** \[ \Delta t = 10 - 1 = 9 \] **Step 4: Calculate the mean velocity vector.** \[ \text{Mean velocity} = \frac{\Delta \mathbf{r}}{\Delta t} \] \[ \text{Mean velocity} = \left( \frac{18}{9},\ \frac{\ln(10)}{9},\ \frac{99}{9} \right) \] \[ \text{Mean velocity} = (2,\ \frac{\ln(10)}{9},\ 11) \] So, the mean velocity of the point between \( t = 1 \) and \( t = 10 \) is: \[ \boxed{\begin{pmatrix} 2, & \dfrac{\ln(10)}{9}, & 11 \end{pmatrix}} \] **Answer:** An explicit vector: Mean velocity = (2 , [ln 10] ⁄ 9 , 11) |
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| GPT-4o | To find the mean velocity of the point between times \( t = 1 \) and \( t = 10 \), we need to determine the total displacement of the point over this time interval and then divide by the time interval. The position of the point at any time \( t \) is given by: \[ x = 2t \] \[ y = \ln(t) \] \[ z = t^2 \] First, we find the coordinates of the point at \( t = 1 \) and \( t = 10 \): At \( t = 1 \): \[ x(1) = 2(1) = 2 \] \[ y(1) = \ln(1) = 0 \] \[ z(1) = 1^2 = 1 \] So, the position at \( t = 1 \) is \( (2, 0, 1) \). At \( t = 10 \): \[ x(10) = 2(10) = 20 \] \[ y(10) = \ln(10) \] \[ z(10) = 10^2 = 100 \] So, the position at \( t = 10 \) is \( (20, \ln(10), 100) \). Next, we calculate the displacement vector from \( t = 1 \) to \( t = 10 \): \[ \Delta x = x(10) - x(1) = 20 - 2 = 18 \] \[ \Delta y = y(10) - y(1) = \ln(10) - 0 = \ln(10) \] \[ \Delta z = z(10) - z(1) = 100 - 1 = 99 \] The displacement vector is \( (18, \ln(10), 99) \). The magnitude of the displacement vector is given by: \[ \|\Delta \mathbf{r}\| = \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2} \] \[ \|\Delta \mathbf{r}\| = \sqrt{18^2 + (\ln(10))^2 + 99^2} \] \[ \|\Delta \mathbf{r}\| = \sqrt{324 + (\ln(10))^2 + 9801} \] \[ \|\Delta \mathbf{r}\| = \sqrt{10125 + (\ln(10))^2} \] The time interval is \( \Delta t = 10 - 1 = 9 \). The mean velocity is the displacement divided by the time interval: \[ \mathbf{v}_{\text{mean}} = \frac{\|\Delta \mathbf{r}\|}{\Delta t} \] \[ \mathbf{v}_{\text{mean}} = \frac{\sqrt{10125 + (\ln(10))^2}}{9} \] So, the mean velocity of the motion between times \( t = 1 \) and \( t = 10 \) is: \[ \mathbf{v}_{\text{mean}} = \frac{\sqrt{10125 + (\ln(10))^2}}{9} \] This completes the solution. |
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| llama3.1-70B-Instruct | To find the mean velocity of motion between times t=1 and t=10, we need to first find the displacement of the point during this time interval. The position of the point at time t=1 is: x(1) = 2(1) = 2 y(1) = ln(1) = 0 z(1) = (1)^2 = 1 The position of the point at time t=10 is: x(10) = 2(10) = 20 y(10) = ln(10) = 2.3026 z(10) = (10)^2 = 100 The displacement of the point during this time interval is: Δx = x(10) - x(1) = 20 - 2 = 18 Δy = y(10) - y(1) = 2.3026 - 0 = 2.3026 Δz = z(10) - z(1) = 100 - 1 = 99 The magnitude of the displacement is: |Δr| = √(Δx^2 + Δy^2 + Δz^2) = √(18^2 + 2.3026^2 + 99^2) = √(324 + 5.303 + 9801) = √10230.303 The time interval is: Δt = 10 - 1 = 9 The mean velocity is: v̄ = |Δr| / Δt = √10230.303 / 9 ≈ 33.33 Therefore, the mean velocity of motion between times t=1 and t=10 is approximately 33.33 units per unit time. |
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| llama3.1-405B-Instruct | To find the mean velocity of motion between times t=1 and t=10, we need to find the displacement of the point over this time interval and then divide it by the time interval. First, let's find the coordinates of the point at t=1 and t=10: At t=1: x(1) = 2(1) = 2 y(1) = ln(1) = 0 z(1) = (1)^2 = 1 At t=10: x(10) = 2(10) = 20 y(10) = ln(10) ≈ 2.3026 z(10) = (10)^2 = 100 Now, let's find the displacement of the point over the time interval: Δx = x(10) - x(1) = 20 - 2 = 18 Δy = y(10) - y(1) ≈ 2.3026 - 0 = 2.3026 Δz = z(10) - z(1) = 100 - 1 = 99 The displacement vector is: Δr = (18, 2.3026, 99) The magnitude of the displacement vector is: |Δr| = √(18^2 + 2.3026^2 + 99^2) ≈ √(324 + 5.3026 + 9801) ≈ √10130.3026 ≈ 100.65 The time interval is: Δt = 10 - 1 = 9 Now, we can find the mean velocity: v_mean = Δr / Δt ≈ 100.65 / 9 ≈ 11.18 So, the mean velocity of motion between times t=1 and t=10 is approximately 11.18 units per second. |
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| DeepSeek-Coder-V2 | To find the mean velocity of motion between times \( t = 1 \) and \( t = 10 \), we need to calculate the displacement vector and then determine the average velocity over the given time interval. 1. **Determine the position vectors at \( t = 1 \) and \( t = 10 \):** For \( t = 1 \): \[ \mathbf{r}(1) = \begin{pmatrix} x(1) \\ y(1) \\ z(1) \end{pmatrix} = \begin{pmatrix} 2 \cdot 1 \\ \ln(1) \\ 1^2 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} \] For \( t = 10 \): \[ \mathbf{r}(10) = \begin{pmatrix} x(10) \\ y(10) \\ z(10) \end{pmatrix} = \begin{pmatrix} 2 \cdot 10 \\ \ln(10) \\ 10^2 \end{pmatrix} = \begin{pmatrix} 20 \\ \ln(10) \\ 100 \end{pmatrix} \] 2. **Calculate the displacement vector:** The displacement vector \(\Delta \mathbf{r}\) is given by: \[ \Delta \mathbf{r} = \mathbf{r}(10) - \mathbf{r}(1) = \begin{pmatrix} 20 \\ \ln(10) \\ 100 \end{pmatrix} - \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 18 \\ \ln(10) \\ 99 \end{pmatrix} \] 3. **Determine the time interval:** The time interval \(\Delta t\) is: \[ \Delta t = 10 - 1 = 9 \] 4. **Calculate the mean velocity vector:** The mean velocity vector \(\mathbf{v}_{\text{mean}}\) is given by: \[ \mathbf{v}_{\text{mean}} = \frac{\Delta \mathbf{r}}{\Delta t} = \frac{1}{9} \begin{pmatrix} 18 \\ \ln(10) \\ 99 \end{pmatrix} = \begin{pmatrix} \frac{18}{9} \\ \frac{\ln(10)}{9} \\ \frac{99}{9} \end{pmatrix} = \begin{pmatrix} 2 \\ \frac{\ln(10)}{9} \\ 11 \end{pmatrix} \] Thus, the mean velocity of motion between times \( t = 1 \) and \( t = 10 \) is: \[ \boxed{\begin{pmatrix} 2 \\ \frac{\ln(10)}{9} \\ 11 \end{pmatrix}} \] |
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| Qwen2-72B-Instruct | To find the mean velocity of motion between times t=1 and t=10, we need to calculate the displacement vector and divide it by the time interval. At t=1, the position vector is: x(1) = 2(1) = 2 y(1) = ln(1) = 0 z(1) = (1)^2 = 1 So, the position vector at t=1 is (2, 0, 1). At t=10, the position vector is: x(10) = 2(10) = 20 y(10) = ln(10) ≈ 2.30259 z(10) = (10)^2 = 100 So, the position vector at t=10 is (20, 2.30259, 100). The displacement vector is the difference between the position vectors at t=10 and t=1: Δr = (20-2, 2.30259-0, 100-1) = (18, 2.30259, 99) The magnitude of the displacement vector is: |Δr| = √(18^2 + 2.30259^2 + 99^2) ≈ 100.09 The time interval between t=1 and t=10 is 9 seconds. Therefore, the mean velocity of motion is: v̄ = Δr / Δt = (18, 2.30259, 99) / 9 = (2, 0.25584, 11). So, the mean velocity of motion between times t=1 and t=10 is approximately (2, 0.25584, 11) m/s. |
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| baichuan4 | To find the mean velocity of motion between times t=1 and t=10, we first need to determine the position vector of the point at time t, which is given by: $$\mathbf{r}(t) = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} = 2t\mathbf{i} + \ln(t)\mathbf{j} + t^2\mathbf{k}$$ Next, we need to calculate the velocity vector by differentiating the position vector with respect to time: $$\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = 2\mathbf{i} + \frac{1}{t}\mathbf{j} + 2t\mathbf{k}$$ Now, we can find the mean velocity by taking the average of the velocity vectors at t=1 and t=10: $$\mathbf{v}_{\text{mean}} = \frac{\mathbf{v}(1) + \mathbf{v}(10)}{2}$$ Substituting the values of t=1 and t=10 into the velocity vector equation: $$\mathbf{v}(1) = 2\mathbf{i} + \mathbf{j} + 2\mathbf{k}$$ $$\mathbf{v}(10) = 2\mathbf{i} + \frac{1}{10}\mathbf{j} + 20\mathbf{k}$$ Taking the average: $$\mathbf{v}_{\text{mean}} = \frac{(2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) + (2\mathbf{i} + \frac{1}{10}\mathbf{j} + 20\mathbf{k})}{2}$$ Simplifying: $$\mathbf{v}_{\text{mean}} = 2\mathbf{i} + \left(\frac{1}{2} + \frac{1}{20}\right)\mathbf{j} + 11\mathbf{k}$$ $$\mathbf{v}_{\text{mean}} = 2\mathbf{i} + \frac{11}{20}\mathbf{j} + 11\mathbf{k}$$ Therefore, the mean velocity of motion between times t=1 and t=10 is: $$\mathbf{v}_{\text{mean}} = \left(2, \frac{11}{20}, 11\right)$$ |
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| doubao | First find the position at t=1: x1 = 2(1) = 2, y1 = ln(1) = 0, z1 = 1^2 = 1. So the position vector at t=1 is (2, 0, 1). Then find the position at t=10: x10 = 2(10) = 20, y10 = ln(10), z10 = 10^2 = 100. So the position vector at t=10 is (20, ln(10), 100). The displacement vector between t=1 and t=10 is (20-2, ln(10)-0, 100-1) = (18, ln(10), 99). The time interval is 10-1 = 9. The mean velocity is the displacement vector divided by the time interval: (18/9, ln(10)/9, 99/9) = (2, ln(10)/9, 11). |
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| minimax | To find the mean velocity of motion between times t=1 and t=10, we need to calculate the displacement of the point in each of the x, y, and z directions over this time interval and then divide by the time interval. The displacement in the x-direction is given by the change in x from t=1 to t=10: Δx = x(10) - x(1) = 2*10 - 2*1 = 20 - 2 = 18 The displacement in the y-direction is given by the change in y from t=1 to t=10: Δy = y(10) - y(1) = ln(10) - ln(1) = ln(10) - 0 = ln(10) The displacement in the z-direction is given by the change in z from t=1 to t=10: Δz = z(10) - z(1) = 10^2 - 1^2 = 100 - 1 = 99 The total time interval is Δt = 10 - 1 = 9. The mean velocity vector \(\vec{v}_{mean}\) is the displacement vector divided by the time interval: \[ \vec{v}_{mean} = \frac{\Delta \vec{r}}{\Delta t} = \left( \frac{\Delta x}{\Delta t}, \frac{\Delta y}{\Delta t}, \frac{\Delta z}{\Delta t} \right) \] So we calculate each component of the mean velocity vector: \[ v_{mean,x} = \frac{\Delta x}{\Delta t} = \frac{18}{9} = 2 \] \[ v_{mean,y} = \frac{\Delta y}{\Delta t} = \frac{\ln(10)}{9} \] \[ v_{mean,z} = \frac{\Delta z}{\Delta t} = \frac{99}{9} = 11 \] Therefore, the mean velocity vector is: \[ \vec{v}_{mean} = (2, \frac{\ln(10)}{9}, 11) \] This vector represents the mean velocity of the motion between times t=1 and t=10. |
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| kimi | To find the mean velocity of motion between times t=1 and t=10, we first need to calculate the displacement in each coordinate (x, y, z) over this time interval, and then find the overall displacement vector. Finally, we'll divide the total displacement by the time interval to get the mean velocity. Given the equations: \[ x = 2t \] \[ y = \ln(t) \] \[ z = t^2 \] Let's calculate the displacement in each coordinate from t=1 to t=10. For x: \[ \Delta x = x(10) - x(1) = 2(10) - 2(1) = 20 - 2 = 18 \] For y: \[ \Delta y = y(10) - y(1) = \ln(10) - \ln(1) = \ln(10) - 0 = \ln(10) \] Since \( \ln(10) \) is approximately 2.3026. For z: \[ \Delta z = z(10) - z(1) = 10^2 - 1^2 = 100 - 1 = 99 \] Now, we find the total displacement vector: \[ \Delta r = \Delta x \hat{i} + \Delta y \hat{j} + \Delta z \hat{k} \] \[ \Delta r = 18 \hat{i} + 2.3026 \hat{j} + 99 \hat{k} \] The magnitude of the total displacement vector is: \[ |\Delta r| = \sqrt{(18)^2 + (2.3026)^2 + (99)^2} \] \[ |\Delta r| = \sqrt{324 + 5.302 + 9801} \] \[ |\Delta r| = \sqrt{10130.302} \] The time interval is: \[ \Delta t = 10 - 1 = 9 \] The mean velocity \( \bar{v} \) is the total displacement divided by the time interval: \[ \bar{v} = \frac{|\Delta r|}{\Delta t} \] \[ \bar{v} = \frac{\sqrt{10130.302}}{9} \] Now, let's calculate the numerical value: \[ \bar{v} = \frac{\sqrt{10130.302}}{9} \] \[ \bar{v} \approx \frac{100.65}{9} \] \[ \bar{v} \approx 11.183 \] So, the mean velocity of motion between times t=1 and t=10 is approximately 11.183 units per time unit. |
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| wenxinyiyan | To find the mean velocity of motion between times t=1 and t=10, we need to calculate the change in position of the point and divide by the change in time. Given: x = 2t y = ln(t) z = t^2 First, let's find the position of the point at t=1 and t=10. At t=1: x1 = 2*1 = 2 y1 = ln(1) = 0 z1 = 1^2 = 1 So, the position vector at t=1 is r1 = <2, 0, 1> At t=10: x2 = 2*10 = 20 y2 = ln(10) z2 = 10^2 = 100 So, the position vector at t=10 is r2 = <20, ln(10), 100> Now, let's find the change in position vector (Δr) and the change in time (Δt). Δr = r2 - r1 = <20-2, ln(10)-0, 100-1> = <18, ln(10), 99> Δt = 10 - 1 = 9 The mean velocity (V_mean) is given by: V_mean = Δr / Δt V_mean = <18/9, ln(10)/9, 99/9> V_mean = <2, ln(10)/9, 11> So, the mean velocity of motion between times t=1 and t=10 is given by the vector <2, ln(10)/9, 11>. |
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| zhipu | To find the mean velocity of motion between times \( t = 1 \) and \( t = 10 \), we need to calculate the displacement vector and then divide it by the time interval. The position of the point at any time \( t \) is given by: \[ x(t) = 2t \] \[ y(t) = \ln(t) \] \[ z(t) = t^2 \] First, we find the positions at \( t = 1 \) and \( t = 10 \): At \( t = 1 \): \[ x(1) = 2 \cdot 1 = 2 \] \[ y(1) = \ln(1) = 0 \] \[ z(1) = 1^2 = 1 \] At \( t = 10 \): \[ x(10) = 2 \cdot 10 = 20 \] \[ y(10) = \ln(10) \] \[ z(10) = 10^2 = 100 \] Next, we calculate the displacement vector \(\Delta \mathbf{r}\) between these two points: \[ \Delta \mathbf{r} = \langle x(10) - x(1), y(10) - y(1), z(10) - z(1) \rangle \] \[ \Delta \mathbf{r} = \langle 20 - 2, \ln(10) - 0, 100 - 1 \rangle \] \[ \Delta \mathbf{r} = \langle 18, \ln(10), 99 \rangle \] The time interval \(\Delta t\) is: \[ \Delta t = 10 - 1 = 9 \] The mean velocity \(\mathbf{v}_{\text{mean}}\) is the displacement vector divided by the time interval: \[ \mathbf{v}_{\text{mean}} = \frac{\Delta \mathbf{r}}{\Delta t} \] \[ \mathbf{v}_{\text{mean}} = \left\langle \frac{18}{9}, \frac{\ln(10)}{9}, \frac{99}{9} \right\rangle \] \[ \mathbf{v}_{\text{mean}} = \left\langle 2, \frac{\ln(10)}{9}, 11 \right\rangle \] So, the mean velocity of motion between times \( t = 1 \) and \( t = 10 \) is: \[ \mathbf{v}_{\text{mean}} = \left\langle 2, \frac{\ln(10)}{9}, 11 \right\rangle \] This vector gives the components of the mean velocity in the x, y, and z directions, respectively. |
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| gpto1 | To find the mean velocity of the point between times \( t = 1 \) and \( t = 10 \), we'll first calculate the displacement vector over this time interval and then divide it by the total time elapsed. **Given:** - Position functions: \[ x(t) = 2t \] \[ y(t) = \ln(t) \] \[ z(t) = t^2 \] **Step 1: Calculate the initial and final positions.** At \( t = 1 \): \[ x(1) = 2 \times 1 = 2 \] \[ y(1) = \ln(1) = 0 \] \[ z(1) = (1)^2 = 1 \] So, the initial position vector is: \[ \mathbf{r}_1 = (2,\ 0,\ 1) \] At \( t = 10 \): \[ x(10) = 2 \times 10 = 20 \] \[ y(10) = \ln(10) \] \[ z(10) = (10)^2 = 100 \] So, the final position vector is: \[ \mathbf{r}_2 = (20,\ \ln(10),\ 100) \] **Step 2: Calculate the displacement vector.** \[ \Delta \mathbf{r} = \mathbf{r}_2 - \mathbf{r}_1 \] \[ \Delta \mathbf{r} = (20 - 2,\ \ln(10) - 0,\ 100 - 1) \] \[ \Delta \mathbf{r} = (18,\ \ln(10),\ 99) \] **Step 3: Calculate the total time elapsed.** \[ \Delta t = 10 - 1 = 9 \] **Step 4: Calculate the mean velocity vector.** \[ \text{Mean velocity} = \frac{\Delta \mathbf{r}}{\Delta t} \] \[ \text{Mean velocity} = \left( \frac{18}{9},\ \frac{\ln(10)}{9},\ \frac{99}{9} \right) \] \[ \text{Mean velocity} = (2,\ \frac{\ln(10)}{9},\ 11) \] So, the mean velocity of the point between \( t = 1 \) and \( t = 10 \) is: \[ \boxed{\begin{pmatrix} 2, & \dfrac{\ln(10)}{9}, & 11 \end{pmatrix}} \] **Answer:** An explicit vector: Mean velocity = (2 , [ln 10] ⁄ 9 , 11) |
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| deepseekv3 | To find the **mean velocity** of the motion between times \( t=1 \) and \( t=10 \), we need to calculate the **displacement vector** and divide it by the **time interval**. ### Step 1: Find the position vectors at \( t=1 \) and \( t=10 \) The position of the point is given by: \[ \vec{r}(t) = (x(t), y(t), z(t)) = (2t, \ln(t), t^2) \] At \( t=1 \): \[ \vec{r}(1) = (2 \cdot 1, \ln(1), 1^2) = (2, 0, 1) \] At \( t=10 \): \[ \vec{r}(10) = (2 \cdot 10, \ln(10), 10^2) = (20, \ln(10), 100) \] ### Step 2: Calculate the displacement vector The displacement vector is the difference between the position vectors at \( t=10 \) and \( t=1 \): \[ \Delta \vec{r} = \vec{r}(10) - \vec{r}(1) = (20 - 2, \ln(10) - 0, 100 - 1) = (18, \ln(10), 99) \] ### Step 3: Calculate the time interval The time interval is: \[ \Delta t = 10 - 1 = 9 \] ### Step 4: Compute the mean velocity The mean velocity is the displacement vector divided by the time interval: \[ \vec{v}_{\text{mean}} = \frac{\Delta \vec{r}}{\Delta t} = \left( \frac{18}{9}, \frac{\ln(10)}{9}, \frac{99}{9} \right) = (2, \frac{\ln(10)}{9}, 11) \] ### Final Answer: The mean velocity of the motion between \( t=1 \) and \( t=10 \) is: \[ \boxed{\left(2, \frac{\ln(10)}{9}, 11\right)} \] |
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