| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the curvature of a curve given by a vector function \( \mathbf{r}(t) \), we use the formula for curvature, which is given by: \[ \kappa(t) = \frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3} \] where \( \mathbf{r}'(t) \) is the first derivative of \( \mathbf{r}(t) \) with respect to \( t \), \( \mathbf{r}''(t) \) is the second derivative of \( \mathbf{r}(t) \) with respect to \( t \), and \( \times \) is the cross product. First, we find the first and second derivatives of \( \mathbf{r}(t) \): \[ \mathbf{r}'(t) = -5\sin(t)\mathbf{i} + 4\cos(t)\mathbf{j} + 3\mathbf{k} \] \[ \mathbf{r}''(t) = -5\cos(t)\mathbf{i} - 4\sin(t)\mathbf{j} \] Next, we find the cross product of \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \): \[ \mathbf{r}'(t) \times \mathbf{r}''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -5\sin(t) & 4\cos(t) & 3 \\ -5\cos(t) & -4\sin(t) & 0 \end{vmatrix} \] \[ = \mathbf{i} \left( 4\cos(t) \cdot 0 - 3 \cdot (-4\sin(t)) \right) - \mathbf{j} \left( -5\sin(t) \cdot 0 - 3 \cdot (-5\cos(t)) \right) + \mathbf{k} \left( -5\sin(t) \cdot (-4\sin(t)) - 4\cos(t) \cdot (-5\cos(t)) \right) \] \[ = \mathbf{i} (12\sin(t)) - \mathbf{j} (15\cos(t)) + \mathbf{k} (20\sin^2(t) + 20\cos^2(t)) \] \[ = 12\sin(t)\mathbf{i} - 15\cos(t)\mathbf{j} + 20\mathbf{k} \] Now, we evaluate the cross product at \( t = \frac{4\pi}{3} \): \[ \mathbf{r}' \left( \frac{4\pi}{3} \right) \times \mathbf{r}'' \left( \frac{4\pi}{3} \right) = 12\sin \left( \frac{4\pi}{3} \right) \mathbf{i} - 15\cos \left( \frac{4\pi}{3} \right) \mathbf{j} + 20\mathbf{k} \] \[ = 12 \left( -\frac{\sqrt{3}}{2} \right) \mathbf{i} - 15 \left( -\frac{1}{2} \right) \mathbf{j} + 20\mathbf{k} \] \[ = -6\sqrt{3}\mathbf{i} + \frac{15}{2}\mathbf{j} + 20\mathbf{k} \] Next, we find the magnitude of \( \mathbf{r}' \left( \frac{4\pi}{3} \right) \times \mathbf{r}'' \left( \frac{4\pi}{3} \right) \): \[ \|\mathbf{r}' \left( \frac{4\pi}{3} \right) \times \mathbf{r}'' \left( \frac{4\pi}{3} \right)\| = \sqrt{ \left( -6\sqrt{3} \right)^2 + \left( \frac{15}{2} \right)^2 + 20^2 } \] \[ = \sqrt{ 108 + \frac{225}{4} + 400 } \] \[ = \sqrt{ 108 + 56.25 + 400 } \] \[ = \sqrt{ 564.25 } \] Now, we find the magnitude of \( \mathbf{r}'(t) \): \[ \|\mathbf{r}'(t)\| = \sqrt{ \left( -5\sin(t) \right)^2 + \left( 4\cos(t) \right)^2 + 3^2 } \] \[ = \sqrt{ 25\sin^2(t) + 16\cos^2(t) + 9 } \] \[ = \sqrt{ 25 \left( \frac{3}{4} \right) + 16 \left( \frac{1}{4} \right) + 9 } \] \[ = \sqrt{ \frac{75}{4} + 4 + 9 } \] \[ = \sqrt{ \frac{127}{4} } \] Finally, we find the curvature at \( t = \frac{4\pi}{3} \): \[ \kappa \left( \frac{4\pi}{3} \right) = \frac{\|\mathbf{r}' \left( \frac{4\pi}{3} \right) \times \mathbf{r}'' \left( \frac{4\pi}{3} \right)\|}{\|\mathbf{r}' \left( \frac{4\pi}{3} \right)\|^3} \] \[ = \frac{\sqrt{564.25}}{\left( \sqrt{\frac{127}{4}} \right)^3} \] \[ = \frac{\sqrt{564.25}}{\left( \frac{127}{4} \right)^{3/2}} \] \[ = \frac{\sqrt{564.25}}{\frac{127^{3/2}}{8}} \] \[ = \frac{8\sqrt{564.25}}{127^{3/2}} \] So, the curvature at \( t = \frac{4\pi}{3} \) is: \[ \kappa \left( \frac{4\pi}{3} \right) = \frac{8\sqrt{564.25}}{127^{3/2}} \] |
0.16 |
| deepseekr1 | To find the curvature of the vector function \( \mathbf{r}(t) = 5\cos(t)\mathbf{i} + 4\sin(t)\mathbf{j} + 3t\mathbf{k} \) at \( t = \frac{4\pi}{3} \): 1. **Compute the first derivative \(\mathbf{r}'(t)\):** \[ \mathbf{r}'(t) = -5\sin(t)\mathbf{i} + 4\cos(t)\mathbf{j} + 3\mathbf{k} \] 2. **Compute the second derivative \(\mathbf{r}''(t)\):** \[ \mathbf{r}''(t) = -5\cos(t)\mathbf{i} - 4\sin(t)\mathbf{j} \] 3. **Compute the cross product \(\mathbf{r}'(t) \times \mathbf{r}''(t)\):** \[ \mathbf{r}'(t) \times \mathbf{r}''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -5\sin(t) & 4\cos(t) & 3 \\ -5\cos(t) & -4\sin(t) & 0 \end{vmatrix} \] Expanding the determinant: \[ \mathbf{r}'(t) \times \mathbf{r}''(t) = 12\sin(t)\mathbf{i} - 15\cos(t)\mathbf{j} + 20\mathbf{k} \] 4. **Evaluate the cross product at \( t = \frac{4\pi}{3} \):** - \( \sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2} \) - \( \cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2} \) \[ \mathbf{r}'\left(\frac{4\pi}{3}\right) \times \mathbf{r}''\left(\frac{4\pi}{3}\right) = -6\sqrt{3}\mathbf{i} + 7.5\mathbf{j} + 20\mathbf{k} \] 5. **Compute the magnitude of the cross product:** \[ \left| \mathbf{r}'\left(\frac{4\pi}{3}\right) \times \mathbf{r}''\left(\frac{4\pi}{3}\right) \right| = \sqrt{(-6\sqrt{3})^2 + (7.5)^2 + 20^2} = \sqrt{108 + 56.25 + 400} = \sqrt{564.25} = \frac{\sqrt{2257}}{2} \] 6. **Compute the magnitude of \(\mathbf{r}'(t)\) at \( t = \frac{4\pi}{3} \):** \[ \left| \mathbf{r}'\left(\frac{4\pi}{3}\right) \right| = \sqrt{25\sin^2\left(\frac{4\pi}{3}\right) + 16\cos^2\left(\frac{4\pi}{3}\right) + 9} = \sqrt{\frac{75}{4} + 4 + 9} = \sqrt{\frac{127}{4}} = \frac{\sqrt{127}}{2} \] 7. **Compute the curvature \(\kappa\):** \[ \kappa = \frac{\left| \mathbf{r}'(t) \times \mathbf{r}''(t) \right|}{\left| \mathbf{r}'(t) \right|^3} = \frac{\frac{\sqrt{2257}}{2}}{\left( \frac{\sqrt{127}}{2} \right)^3} = \frac{\frac{\sqrt{2257}}{2}}{\frac{127\sqrt{127}}{8}} = \frac{4\sqrt{2257}}{127\sqrt{127}} \] The curvature at \( t = \frac{4\pi}{3} \) is: \[ \boxed{\dfrac{4\sqrt{2257}}{127\sqrt{127}}} \] |
0.16 |
| gpto1 | To find the curvature \(\kappa(t)\) of the space curve defined by \[ \mathbf{r}(t) = 5\cos(t)\, \mathbf{i} + 4\sin(t)\, \mathbf{j} + 3t\, \mathbf{k} \] at \(t = \dfrac{4\pi}{3}\), we'll follow these steps: **1. Find the first derivative \(\mathbf{r}'(t)\):** Differentiate each component of \(\mathbf{r}(t)\) with respect to \(t\): \[ \begin{aligned} \mathbf{r}'(t) &= \dfrac{d}{dt}[5\cos(t)]\, \mathbf{i} + \dfrac{d}{dt}[4\sin(t)]\, \mathbf{j} + \dfrac{d}{dt}[3t]\, \mathbf{k} \\ &= -5\sin(t)\, \mathbf{i} + 4\cos(t)\, \mathbf{j} + 3\, \mathbf{k} \end{aligned} \] **2. Find the second derivative \(\mathbf{r}''(t)\):** Differentiate \(\mathbf{r}'(t)\) with respect to \(t\): \[ \begin{aligned} \mathbf{r}''(t) &= \dfrac{d}{dt}[-5\sin(t)]\, \mathbf{i} + \dfrac{d}{dt}[4\cos(t)]\, \mathbf{j} + \dfrac{d}{dt}[3]\, \mathbf{k} \\ &= -5\cos(t)\, \mathbf{i} - 4\sin(t)\, \mathbf{j} + 0\, \mathbf{k} \end{aligned} \] **3. Compute the cross product \(\mathbf{r}'(t) \times \mathbf{r}''(t)\):** Let \(\mathbf{r}'(t) = (a_1, a_2, a_3)\) and \(\mathbf{r}''(t) = (b_1, b_2, b_3)\), where: \[ \begin{aligned} a_1 &= -5\sin(t), \quad a_2 = 4\cos(t), \quad a_3 = 3 \\ b_1 &= -5\cos(t), \quad b_2 = -4\sin(t), \quad b_3 = 0 \end{aligned} \] Compute the cross product components: \[ \begin{aligned} c_1 &= a_2 b_3 - a_3 b_2 = (4\cos(t))(0) - (3)(-4\sin(t)) = 12\sin(t) \\ c_2 &= a_3 b_1 - a_1 b_3 = (3)(-5\cos(t)) - (-5\sin(t))(0) = -15\cos(t) \\ c_3 &= a_1 b_2 - a_2 b_1 = (-5\sin(t))(-4\sin(t)) - (4\cos(t))(-5\cos(t)) = 20\sin^2(t) + 20\cos^2(t) = 20(\sin^2 t + \cos^2 t) = 20 \end{aligned} \] So, \[ \mathbf{r}'(t) \times \mathbf{r}''(t) = \left(12\sin(t),\ -15\cos(t),\ 20\right) \] **4. Compute the magnitude \(|\mathbf{r}'(t) \times \mathbf{r}''(t)|\):** \[ \begin{aligned} |\mathbf{r}'(t) \times \mathbf{r}''(t)| &= \sqrt{[12\sin(t)]^2 + [-15\cos(t)]^2 + 20^2} \\ &= \sqrt{144\sin^2 t + 225\cos^2 t + 400} \end{aligned} \] **5. Compute the magnitude \(|\mathbf{r}'(t)|\):** \[ \begin{aligned} |\mathbf{r}'(t)| &= \sqrt{[-5\sin(t)]^2 + [4\cos(t)]^2 + 3^2} \\ &= \sqrt{25\sin^2 t + 16\cos^2 t + 9} \end{aligned} \] **6. Compute the curvature \(\kappa(t)\):** The curvature of a space curve is given by: \[ \kappa(t) = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3} \] **7. Evaluate at \(t = \dfrac{4\pi}{3}\):** First, compute \(\sin\left(\dfrac{4\pi}{3}\right)\) and \(\cos\left(\dfrac{4\pi}{3}\right)\): \[ \sin\left(\dfrac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}, \quad \cos\left(\dfrac{4\pi}{3}\right) = -\frac{1}{2} \] **Compute \(|\mathbf{r}'(t) \times \mathbf{r}''(t)|\) at \(t = \dfrac{4\pi}{3}\):** \[ \begin{aligned} |\mathbf{r}'\times \mathbf{r}''| &= \sqrt{144\left(\frac{3}{4}\right) + 225\left(\frac{1}{4}\right) + 400} \\ &= \sqrt{108 + 56.25 + 400} \\ &= \sqrt{564.25} \end{aligned} \] **Compute \(|\mathbf{r}'(t)|\) at \(t = \dfrac{4\pi}{3}\):** \[ \begin{aligned} |\mathbf{r}'(t)| &= \sqrt{25\left(\frac{3}{4}\right) + 16\left(\frac{1}{4}\right) + 9} \\ &= \sqrt{18.75 + 4 + 9} \\ &= \sqrt{31.75} \end{aligned} \] **Compute \(|\mathbf{r}'(t)|^3\):** \[ |\mathbf{r}'(t)|^3 = \left(\sqrt{31.75}\,\right)^3 = (31.75) \sqrt{31.75} \] **Compute the curvature \(\kappa\):** \[ \kappa = \frac{|\mathbf{r}'\times \mathbf{r}''|}{|\mathbf{r}'(t)|^3} = \frac{\sqrt{564.25}}{(31.75)\sqrt{31.75}} = \frac{\sqrt{564.25}}{(31.75)^{3/2}} \] **Simplify the expression:** Note that \(564.25 = \frac{2257}{4}\) and \(31.75 = \frac{127}{4}\). Therefore, \[ \sqrt{564.25} = \sqrt{\frac{2257}{4}} = \frac{\sqrt{2257}}{2}, \quad |\mathbf{r}'(t)|^3 = \left(\frac{127}{4}\right)^{3/2} = \left(\frac{127}{4}\right)^{1} \cdot \sqrt{\frac{127}{4}} = \left(\frac{127}{4}\right) \cdot \frac{\sqrt{127}}{2} = \frac{127\sqrt{127}}{8} \] **Compute the curvature:** \[ \begin{aligned} \kappa &= \frac{\frac{\sqrt{2257}}{2}}{\frac{127\sqrt{127}}{8}} = \frac{\sqrt{2257}}{2} \times \frac{8}{127\sqrt{127}} \\ &= \frac{4\sqrt{2257}}{127\sqrt{127}} \end{aligned} \] **Simplify further:** Since \(\sqrt{127^3} = \sqrt{127 \cdot 127 \cdot 127} = 127 \sqrt{127}\), the denominator simplifies to \(127 \sqrt{127}\). Thus, the curvature at \(t = \dfrac{4\pi}{3}\) is: \[ \kappa = \frac{4\sqrt{2257}}{127\sqrt{127}} \] This is the exact value of the curvature at the given point on the curve. **Answer:** \(\displaystyle \kappa = \dfrac{4\,\sqrt{2257}}{127\,\sqrt{127}}\) |
0.16 |
| GPT-4o | To find the curvature \(\kappa\) of a vector function \(\mathbf{r}(t)\), we follow these steps: 1. **Compute the first derivative \(\mathbf{r}'(t)\):** \[ \mathbf{r}(t) = 5\cos(t)\mathbf{i} + 4\sin(t)\mathbf{j} + 3t\mathbf{k} \] \[ \mathbf{r}'(t) = \frac{d}{dt}[5\cos(t)\mathbf{i} + 4\sin(t)\mathbf{j} + 3t\mathbf{k}] = -5\sin(t)\mathbf{i} + 4\cos(t)\mathbf{j} + 3\mathbf{k} \] 2. **Compute the second derivative \(\mathbf{r}''(t)\):** \[ \mathbf{r}''(t) = \frac{d}{dt}[-5\sin(t)\mathbf{i} + 4\cos(t)\mathbf{j} + 3\mathbf{k}] = -5\cos(t)\mathbf{i} - 4\sin(t)\mathbf{j} \] 3. **Evaluate \(\mathbf{r}'(t)\) and \(\mathbf{r}''(t)\) at \(t = \frac{4\pi}{3}\):** \[ \mathbf{r}'\left(\frac{4\pi}{3}\right) = -5\sin\left(\frac{4\pi}{3}\right)\mathbf{i} + 4\cos\left(\frac{4\pi}{3}\right)\mathbf{j} + 3\mathbf{k} \] \[ \sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}, \quad \cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2} \] \[ \mathbf{r}'\left(\frac{4\pi}{3}\right) = -5\left(-\frac{\sqrt{3}}{2}\right)\mathbf{i} + 4\left(-\frac{1}{2}\right)\mathbf{j} + 3\mathbf{k} = \frac{5\sqrt{3}}{2}\mathbf{i} - 2\mathbf{j} + 3\mathbf{k} \] \[ \mathbf{r}''\left(\frac{4\pi}{3}\right) = -5\cos\left(\frac{4\pi}{3}\right)\mathbf{i} - 4\sin\left(\frac{4\pi}{3}\right)\mathbf{j} \] \[ \mathbf{r}''\left(\frac{4\pi}{3}\right) = -5\left(-\frac{1}{2}\right)\mathbf{i} - 4\left(-\frac{\sqrt{3}}{2}\right)\mathbf{j} = \frac{5}{2}\mathbf{i} + 2\sqrt{3}\mathbf{j} \] 4. **Compute the cross product \(\mathbf{r}'(t) \times \mathbf{r}''(t)\):** \[ \mathbf{r}'\left(\frac{4\pi}{3}\right) \times \mathbf{r}''\left(\frac{4\pi}{3}\right) = \left(\frac{5\sqrt{3}}{2}\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}\right) \times \left(\frac{5}{2}\mathbf{i} + 2\sqrt{3}\mathbf{j}\right) \] Using the determinant method for the cross product: \[ \mathbf{r}'(t) \times \mathbf{r}''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{5\sqrt{3}}{2} & -2 & 3 \\ \frac{5}{2} & 2\sqrt{3} & 0 \end{vmatrix} \] \[ = \mathbf{i}\left(-2 \cdot 0 - 3 \cdot 2\sqrt{3}\right) - \mathbf{j}\left(\frac{5\sqrt{3}}{2} \cdot 0 - 3 \cdot \frac{5}{2}\right) + \mathbf{k}\left(\frac{5\ |
0.16 |
| llama3.1-70B-Instruct | To find the curvature of the curve given by r(t) = 5cos(t)i + 4sin(t)j + 3tk at t = 4π/3, we'll follow these steps: 1. **Find the first derivative of r(t)**: r'(t) = -5sin(t)i + 4cos(t)j + 3k 2. **Find the second derivative of r(t)**: r''(t) = -5cos(t)i - 4sin(t)j 3. **Compute the cross product of r'(t) and r''(t)** to find the normal vector n(t): \[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -5\sin(t) & 4\cos(t) & 3 \\ -5\cos(t) & -4\sin(t) & 0 \\ \end{vmatrix} \] n(t) = (12sin(t))i - (15cos(t))j + (20sin^2(t) + 20cos^2(t))k Simplifying, n(t) = (12sin(t))i - (15cos(t))j + 20k 4. **Find the magnitude of r'(t)**: |r'(t)| = sqrt((-5sin(t))^2 + (4cos(t))^2 + 3^2) = sqrt(25sin^2(t) + 16cos^2(t) + 9) 5. **Find the magnitude of the cross product n(t)**: |n(t)| = sqrt((12sin(t))^2 + (-15cos(t))^2 + 20^2) = sqrt(144sin^2(t) + 225cos^2(t) + 400) 6. **Compute the curvature κ(t)** using the formula κ(t) = |n(t)| / |r'(t)|^3: κ(t) = sqrt(144sin^2(t) + 225cos^2(t) + 400) / (sqrt(25sin^2(t) + 16cos^2(t) + 9))^3 7. **Evaluate the curvature at t = 4π/3**: At t = 4π/3, sin(t) = -sqrt(3)/2 and cos(t) = -1/2. Plug these values into the expression for κ(t) and simplify. κ(4π/3) = sqrt(144(-sqrt(3)/2)^2 + 225(-1/2)^2 + 400) / (sqrt(25(-sqrt(3)/2)^2 + 16(-1/2)^2 + 9))^3 κ(4π/3) = sqrt(207 + 400) / (sqrt(75/4 + 4 + 9))^3 κ(4π/3) = sqrt(607) / (sqrt(121/4))^3 κ(4π/3) = sqrt(607) / (11/2)^3 κ(4π/3) = (2^3 * sqrt(607)) / 11^3 κ(4π/3) = 8 * sqrt(607) / 1331 Therefore, the curvature of the curve at t = 4π/3 is 8 * sqrt(607) / 1331. |
0.16 |
| llama3.1-405B-Instruct | To find the curvature of the curve r(t) = 5cos(t)i + 4sin(t)j + 3tk at t = 4π/3, we need to follow these steps: 1. Find the derivative of r(t) with respect to t: r'(t) = -5sin(t)i + 4cos(t)j + 3k 2. Find the magnitude of r'(t): |r'(t)| = √((-5sin(t))^2 + (4cos(t))^2 + 3^2) = √(25sin^2(t) + 16cos^2(t) + 9) 3. Find the unit tangent vector T(t): T(t) = r'(t) / |r'(t)| 4. Find the derivative of T(t) with respect to t: T'(t) = (r''(t) / |r'(t)|) - (r'(t) \* (r'(t) \* r''(t)) / |r'(t)|^3) 5. Find the curvature κ(t): κ(t) = |T'(t)| Now, let's evaluate these expressions at t = 4π/3: 1. r'(4π/3) = -5sin(4π/3)i + 4cos(4π/3)j + 3k = -5(-√3/2)i + 4(-1/2)j + 3k = (5√3/2)i - 2j + 3k 2. |r'(4π/3)| = √((5√3/2)^2 + (-2)^2 + 3^2) = √(75/4 + 4 + 9) = √(75/4 + 13) = √(75 + 52)/4 = √127/4 3. T(4π/3) = r'(4π/3) / |r'(4π/3)| = ((5√3/2)i - 2j + 3k) / (√127/4) 4. T'(4π/3) = (r''(4π/3) / |r'(4π/3)|) - (r'(4π/3) \* (r'(4π/3) \* r''(4π/3)) / |r'(4π/3)|^3) To find r''(4π/3), we need to find the second derivative of r(t): r''(t) = -5cos(t)i - 4sin(t)j r''(4π/3) = -5cos(4π/3)i - 4sin(4π/3)j = -5(-1/2)i - 4(-√3/2)j = (5/2)i + 2√3j Now, we can plug in the values: T'(4π/3) = ((5/2)i + 2√3j) / (√127/4) - (((5√3/2)i - 2j + 3k) \* ((5√3/2)i - 2j + 3k) \* ((5/2)i + 2√3j)) / (√127/4)^3 5. κ(4π/3) = |T'(4π/3)| After simplifying the expression, we get: κ(4π/3) = √(25 + 48 + 36) / (√127/4) = √109 / (√127/4) = 4√109 / √127 Therefore, the curvature of the curve r(t) = 5cos(t)i + 4sin(t)j + 3tk at t = 4π/3 is 4√109 / √127. |
0.16 |
| DeepSeek-Coder-V2 | To find the curvature of the vector-valued function \( \mathbf{r}(t) = 5\cos(t)\mathbf{i} + 4\sin(t)\mathbf{j} + 3t\mathbf{k} \) at \( t = \frac{4\pi}{3} \), we will follow these steps: 1. **Compute the first derivative \( \mathbf{r}'(t) \)**: \[ \mathbf{r}'(t) = \frac{d}{dt} \left( 5\cos(t)\mathbf{i} + 4\sin(t)\mathbf{j} + 3t\mathbf{k} \right) \] \[ \mathbf{r}'(t) = -5\sin(t)\mathbf{i} + 4\cos(t)\mathbf{j} + 3\mathbf{k} \] 2. **Compute the second derivative \( \mathbf{r}''(t) \)**: \[ \mathbf{r}''(t) = \frac{d}{dt} \left( -5\sin(t)\mathbf{i} + 4\cos(t)\mathbf{j} + 3\mathbf{k} \right) \] \[ \mathbf{r}''(t) = -5\cos(t)\mathbf{i} - 4\sin(t)\mathbf{j} \] 3. **Evaluate \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \) at \( t = \frac{4\pi}{3} \)**: \[ \mathbf{r}' \left( \frac{4\pi}{3} \right) = -5\sin \left( \frac{4\pi}{3} \right) \mathbf{i} + 4\cos \left( \frac{4\pi}{3} \right) \mathbf{j} + 3\mathbf{k} \] \[ \sin \left( \frac{4\pi}{3} \right) = -\frac{\sqrt{3}}{2}, \quad \cos \left( \frac{4\pi}{3} \right) = -\frac{1}{2} \] \[ \mathbf{r}' \left( \frac{4\pi}{3} \right) = -5 \left( -\frac{\sqrt{3}}{2} \right) \mathbf{i} + 4 \left( -\frac{1}{2} \right) \mathbf{j} + 3\mathbf{k} \] \[ \mathbf{r}' \left( \frac{4\pi}{3} \right) = \frac{5\sqrt{3}}{2} \mathbf{i} - 2 \mathbf{j} + 3 \mathbf{k} \] \[ \mathbf{r}'' \left( \frac{4\pi}{3} \right) = -5\cos \left( \frac{4\pi}{3} \right) \mathbf{i} - 4\sin \left( \frac{4\pi}{3} \right) \mathbf{j} \] \[ \mathbf{r}'' \left( \frac{4\pi}{3} \right) = -5 \left( -\frac{1}{2} \right) \mathbf{i} - 4 \left( -\frac{\sqrt{3}}{2} \right) \mathbf{j} \] \[ \mathbf{r}'' \left( \frac{4\pi}{3} \right) = \frac{5}{2} \mathbf{i} + 2\sqrt{3} \mathbf{j} \] 4. **Compute the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \)**: \[ \mathbf{r}' \left( \frac{4\pi}{3} \right) \times \mathbf{r}'' \left( \frac{4\pi}{3} \right) = \left( \frac{5\sqrt{3}}{2} \mathbf{i} - 2 \mathbf{j} + 3 \mathbf{k} \right) \times \left( \frac{5}{2} \mathbf{i} + 2\sqrt{3} \mathbf{j} \right) \] Using the determinant method for the cross product: \[ \mathbf{r}' \left( \frac{4\pi}{3} \right) \times \mathbf{r}'' \left( \frac{4\pi}{3} \right) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{5\sqrt{3}}{2} & -2 & 3 \\ \frac{5}{2} & 2\sqrt{3} & 0 \end{vmatrix} \] \[ = \mathbf{i} \left( -2 \cdot 0 - 3 \cdot 2\sqrt{3} \right) - \mathbf{j} \left( \frac{5\sqrt{3}}{2} \cdot 0 - 3 \cdot \frac{5}{2} \right) + \mathbf{k} \left( \frac{5\sqrt{3}}{2} \cdot 2\sqrt{3} - (-2) \cdot \frac{5}{2} \right) \] \[ = \mathbf{i} \left( -6\sqrt{3} \right) - \mathbf{j} \left( -\frac{15}{2} \right) + \mathbf{k} \left( 15 + 5 \right) \] \[ = -6\sqrt{3} \mathbf{i} + \frac{15}{2} \mathbf{j} + 20 \mathbf{k} \] 5. **Compute the magnitude of the cross product**: \[ \| \mathbf{r}' \left( \frac{4\pi}{3} \right) \times \mathbf{r}'' \left( \frac{4\pi}{3} \right) \| = \sqrt{ \left( -6\sqrt{3} \right)^2 + \left( \frac{15}{2} \right)^2 + 20^2 } \] \[ = \sqrt{ 108 + \frac{225}{4} + 400 } \] \[ = \sqrt{ 108 + 56.25 + 400 } \] \[ = \sqrt{ 564.25 } \] \[ = \sqrt{ 2257/4 } \] \[ = \frac{\sqrt{2257}}{2} \] 6. **Compute the magnitude of \( \mathbf{r}'(t) \)**: \[ \| \mathbf{r}' \left( \frac{4\pi}{3} \right) \| = \sqrt{ \left( \frac{5\sqrt{3}}{2} \right)^2 + (-2)^2 + 3^2 } \] \[ = \sqrt{ \frac{75}{4} + 4 + 9 } \] \[ = \sqrt{ \frac{75}{4} + 13 } \] \[ = \sqrt{ \frac{75}{4} + \frac{52}{4} } \] \[ = \sqrt{ \frac{127}{4} } \] \[ = \frac{\sqrt{127}}{2} \] 7. **Compute the curvature \( \kappa \)**: \[ \kappa = \frac{ \| \mathbf{r}' \left( \frac{4\pi}{3} \right) \times \mathbf{r}'' \left( \frac{4\pi}{3} \right) \| }{ \| \mathbf{r}' \left( \frac{4\pi}{3} \right) \|^3 } \] \[ = \frac{ \frac{\sqrt{2257}}{2} }{ \left( \frac{\sqrt{127}}{2} \right)^3 } \] \[ = \frac{ \frac{\sqrt{2257}}{2} }{ \frac{127\sqrt{127}}{8} } \] \[ = \frac{ 4\sqrt{2257} }{ 127\sqrt{127} } \] \[ = \frac{ 4\sqrt{2257} }{ 127\sqrt{127} } \] Thus, the curvature at \( t = \frac{4\pi}{3} \) is: \[ \kappa = \frac{ 4\sqrt{2257} }{ 127\sqrt{127} } \] |
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| Qwen2-72B-Instruct | To find the curvature of a curve given by a vector function r(t), we use the formula for curvature, which is given by: κ(t) = ||r'(t) × r''(t)|| / ||r'(t)||³ where r'(t) is the first derivative of r(t) with respect to t, r''(t) is the second derivative of r(t) with respect to t, and × is the cross product. First, we find the first and second derivatives of r(t): r'(t) = -5sin(t)i + 4cos(t)j + 3k r''(t) = -5cos(t)i - 4sin(t)j Next, we find the cross product of r'(t) and r''(t): r'(t) × r''(t) = (-5sin(t) * (-4sin(t)) - 4cos(t) * (-5cos(t)))i - ((-5sin(t) * 3) - 3 * (-5cos(t)))j + ((-5sin(t) * (-4sin(t))) - 4cos(t) * 4cos(t))k = (20sin²(t) + 20cos²(t))i + (15sin(t) - 15cos(t))j + (20sin²(t) - 16cos²(t))k = 20i + 15(sin(t) - cos(t))j + (20sin²(t) - 16cos²(t))k Now, we evaluate the cross product at t = 4π/3: r'(t) × r''(t) = 20i + 15(-√3/2 - (-1/2))j + (20(-√3/2)² - 16(-1/2)²)k = 20i + 15(-√3/2 + 1/2)j + (20(3/4) - 16(1/4))k = 20i + 15(1 - √3/2)j + 11k Next, we find the magnitude of r'(t) × r''(t): ||r'(t) × r''(t)|| = √(20² + 15²(1 - √3/2)² + 11²) = √(400 + 225(1 - 3 + 2√3/2) + 121) = √(400 + 225(-2 + √3) + 121) = √(521 - 450 + 225√3) = √(71 + 225√3) Now, we find the magnitude of r'(t): ||r'(t)|| = √((-5sin(t))² + (4cos(t))² + 3²) = √(25sin²(t) + 16cos²(t) + 9) = √(25(3/4) + 16(1/4) + 9) = √(75/4 + 16/4 + 36/4) = √(127/4) Finally, we find the curvature at t = 4π/3: κ(4π/3) = ||r'(t) × r''(t)|| / ||r'(t)||³ = (√(71 + 225√3)) / (√(127/4))³ = (√(71 + 225√3)) / (127/4)^(3/2) = (√(71 + 225√3)) / (127^(3/2) / 8) = 8√(71 + 225√3) / 127^(3/2) So, the curvature at t = 4π/3 is: κ(4π/3) = 8√(71 + 225√3) / 127^(3/2) |
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| baichuan4 | To find the curvature of the given vector-valued function \( r(t) = 5\cos(t)\mathbf{i} + 4\sin(t)\mathbf{j} + 3t\mathbf{k} \) at the specific value \( t = \frac{4\pi}{3} \), we need to follow these steps: 1. Calculate the derivative of \( r(t) \) with respect to \( t \), which gives us the unit tangent vector \( T(t) \). 2. Find the second derivative of \( r(t) \) with respect to \( t \). 3. Calculate the cross product of \( T(t) \) and the second derivative to get the curvature vector. 4. Divide the magnitude of the curvature vector by the square of the magnitude of \( T(t) \) to obtain the curvature \( \kappa \). Let's go through these steps: ### Step 1: First Derivative (Tangent Vector) \[ r'(t) = -5\sin(t)\mathbf{i} + 4\cos(t)\mathbf{j} + 3\mathbf{k} \] ### Step 2: Second Derivative \[ r''(t) = -5\cos(t)\mathbf{i} - 4\sin(t)\mathbf{j} \] ### Step 3: Cross Product (Curvature Vector) The cross product \( r'(t) \times r''(t) \) gives us the curvature vector: \[ r'(t) \times r''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -5\sin(t) & 4\cos(t) & 3 \\ -5\cos(t) & -4\sin(t) & 0 \end{vmatrix} \] Expanding the determinant, we get: \[ = (12\sin(t)\mathbf{i} - (15\cos(t)\mathbf{j}) + (20\cos(t)\mathbf{k}) \] ### Step 4: Curvature Calculation The curvature \( \kappa \) is: \[ \kappa(t) = \frac{|r'(t) \times r''(t)|}{|r'(t)|^3} \] We need to find the magnitudes of both the cross product and the first derivative: - Magnitude of the cross product: \[ |r'(t) \times r''(t)| = \sqrt{(12\sin(t))^2 + (-15\cos(t))^2 + (20\cos(t))^2} \] - Magnitude of the first derivative: \[ |r'(t)| = \sqrt{(-5\sin(t))^2 + (4\cos(t))^2 + 3^2} \] Now, plug in \( t = \frac{4\pi}{3} \): - Magnitude of the cross product at \( t = \frac{4\pi}{3} \): \[ |r'(\frac{4\pi}{3}) \times r''(\frac{4\pi}{3})| = \sqrt{(12\sin(\frac{4\pi}{3}))^2 + (-15\cos(\frac{4\pi}{3}))^2 + (20\cos(\frac{4\pi}{3}))^2} \] - Magnitude of the first derivative at \( t = \frac{4\pi}{3} \): \[ |r'(\frac{4\pi}{3})| = \sqrt{(-5\sin(\frac{4\pi}{3}))^2 + (4\cos(\frac{4\pi}{3}))^2 + 3^2} \] Calculating these values: - \( \sin(\frac{4\pi}{3}) = -\frac{\sqrt{3}}{2} \), \( \cos(\frac{4\pi}{3}) = -\frac{1}{2} \) - \( |r'(\frac{4\pi}{3}) \times r''(\frac{4\pi}{3})| = \sqrt{12^2(\frac{-\sqrt{3}}{2})^2 + (-15)(-\frac{1}{2})^2 + 20^2(-\frac{1}{2})^2} = \sqrt{36 + \frac{225}{4} + 200} \) - \( |r'(\frac{4\pi}{3})| = \sqrt{25(\frac{-\sqrt{3}}{2})^2 + 16(-\frac{1}{2})^2 + 9} = \sqrt{\frac{75}{4} + 4 + 9} \) Finally, we can calculate the curvature: \[ \kappa(\frac{4\pi}{3}) = \frac{\sqrt{36 + \frac{225}{4} + 200}}{(\sqrt{\frac{75}{4} + 4 + 9})^3} \] Simplifying: \[ \kappa(\frac{4\pi}{3}) = \frac{\sqrt{236 + \frac{225}{4}}}{(\sqrt{\frac{87}{4}})^3} = \frac{\sqrt{236 + 56.25}}{(\frac{\sqrt{87}}{2})^3} = \frac{\sqrt{292.25}}{(\frac{9.325}{2})^3} = \frac{17.07}{(\frac{9.325}{2})^3} \] \[ \kappa(\frac{4\pi}{3}) \approx \frac{17.07}{11.44} \approx 1.492 \] So the curvature at \( t = \frac{4\pi}{3} \) is approximately \( 1.492 \). |
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| doubao | The curvature κ of a curve r(t) is given by: κ = |r'(t) × r''(t)| / |r'(t)|³ First find r'(t) = -5sin(t)i + 4cos(t)j + 3k and r''(t) = -5cos(t)i - 4sin(t)j. At t = 4π/3, r'(4π/3) = -5sin(4π/3)i + 4cos(4π/3)j + 3k = 5√3/2 i - 2j + 3k and r''(4π/3) = -5cos(4π/3)i - 4sin(4π/3)j = 5/2 i + 2√3 j. The cross product r'(4π/3) × r''(4π/3) = (-6 - 5√3)i - 15/2 j + 5√3/2 k. The magnitude of r'(4π/3) × r''(4π/3) is √((-6 - 5√3)² + (-15/2)² + (5√3/2)²) = √(126 + 60√3). The magnitude of r'(4π/3) is √((5√3/2)² + (-2)² + 3²) = √(75/4 + 4 + 9) = √(121/4) = 11/2. So the curvature at t = 4π/3 is κ = (√(126 + 60√3)) / (11/2)³ = 8(√(126 + 60√3)) / 1331. |
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| minimax | To find the curvature of a vector function \( \mathbf{r}(t) \), we can use the formula for curvature \( \kappa \) given by: \[ \kappa = \frac{||\mathbf{r}'(t) \times \mathbf{r}''(t)||}{||\mathbf{r}'(t)||^3} \] where \( \mathbf{r}'(t) \) is the first derivative of \( \mathbf{r}(t) \) with respect to \( t \), and \( \mathbf{r}''(t) \) is the second derivative. The cross product \( \times \) is between \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \), and \( ||\cdot|| \) denotes the magnitude of a vector. Given the vector function \( \mathbf{r}(t) = 5\cos(t)\mathbf{i} + 4\sin(t)\mathbf{j} + 3t\mathbf{k} \), let's find the first and second derivatives: First derivative \( \mathbf{r}'(t) \): \[ \mathbf{r}'(t) = -5\sin(t)\mathbf{i} + 4\cos(t)\mathbf{j} + 3\mathbf{k} \] Second derivative \( \mathbf{r}''(t) \): \[ \mathbf{r}''(t) = -5\cos(t)\mathbf{i} - 4\sin(t)\mathbf{j} \] Now we need to evaluate these derivatives at \( t = \frac{4\pi}{3} \): \[ \mathbf{r}'\left(\frac{4\pi}{3}\right) = -5\sin\left(\frac{4\pi}{3}\right)\mathbf{i} + 4\cos\left(\frac{4\pi}{3}\right)\mathbf{j} + 3\mathbf{k} \] \[ \mathbf{r}''\left(\frac{4\pi}{3}\right) = -5\cos\left(\frac{4\pi}{3}\right)\mathbf{i} - 4\sin\left(\frac{4\pi}{3}\right)\mathbf{j} \] Using the unit circle, we know that \( \sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2} \) and \( \cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2} \). Substituting these values in: \[ \mathbf{r}'\left(\frac{4\pi}{3}\right) = -5\left(-\frac{\sqrt{3}}{2}\right)\mathbf{i} + 4\left(-\frac{1}{2}\right)\mathbf{j} + 3\mathbf{k} \] \[ \mathbf{r}'\left(\frac{4\pi}{3}\right) = \frac{5\sqrt{3}}{2}\mathbf{i} - 2\mathbf{j} + 3\mathbf{k} \] \[ \mathbf{r}''\left(\frac{4\pi}{3}\right) = -5\left(-\frac{1}{2}\right)\mathbf{i} - 4\left(-\frac{\sqrt{3}}{2}\right)\mathbf{j} \] \[ \mathbf{r}''\left(\frac{4\pi}{3}\right) = \frac{5}{2}\mathbf{i} + 2\sqrt{3}\mathbf{j} \] Next, we calculate the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \) at \( t = \frac{4\pi}{3} \): \[ \mathbf{r}'\left(\frac{4\pi}{3}\right) \times \mathbf{r}''\left(\frac{4\pi}{3}\right) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{5\sqrt{3}}{2} & -2 & 3 \\ \frac{5}{2} & 2\sqrt{3} & 0 \\ \end{vmatrix} \] \[ = \left( -2 \cdot 0 - 3 \cdot 2\sqrt{3} \right)\mathbf{i} - \left( \frac{5\sqrt{3}}{2} \cdot 0 - 3 \cdot \frac{5}{2} \right)\mathbf{j} + \left( \frac{5\sqrt{3}}{2} \cdot 2\sqrt{3} - (-2) \cdot \frac{5}{2} \right)\mathbf{k} \] \[ = \left( -6\sqrt{3} \right)\mathbf{i} + \left( \frac{15}{2} \right)\mathbf{j} + \left( 15 - (-5) \right)\mathbf{k} \] \[ = -6\sqrt{3}\mathbf{i} + \frac{15}{2}\mathbf{j} + 20\mathbf{k} \] Now we find the magnitudes of \( \mathbf{r}'\left(\frac{4\pi}{3}\right) \) and \( \mathbf{r}'\left(\frac{4\pi}{3}\right) \times \mathbf{r}''\left(\frac{4\pi}{3}\right) \): \[ ||\mathbf{r}'\left(\frac{4\pi}{3}\right)|| = \sqrt{\left(\frac{5\sqrt{3}}{2}\right)^2 + (-2)^2 + 3^2} \] \[ ||\mathbf{r}'\left(\frac{4\pi}{3}\right)|| = \sqrt{\frac{75}{4} + 4 + 9} \] \[ ||\mathbf{r}'\left(\frac{4\pi}{3}\right)|| = \sqrt{\frac{75}{4} + \frac{16}{4} + \frac{36}{4}} \] \[ ||\mathbf{r}'\left(\frac{4\pi}{3}\right)|| = \sqrt{\frac{127}{4}} \] \[ ||\mathbf{r}'\left(\frac{4\pi}{3}\right)|| = \frac{\sqrt{127}}{2} \] \[ ||\mathbf{r}'\left(\frac{4\pi}{3}\right) \times \mathbf{r}''\left(\frac{4\pi}{3}\right)|| = \sqrt{\left(-6\sqrt{3}\right)^2 + \left(\frac{15}{2}\right)^2 + 20^2} \] \[ ||\mathbf{r}'\left(\frac{4\pi}{3}\right) \times \mathbf{r}''\left(\frac{4\pi}{3}\right)|| = \sqrt{108 + \frac{225}{4} + 400} \] \[ ||\mathbf{r}'\left(\frac{4\pi}{3}\right) \times \mathbf{r}''\left(\frac{4\pi}{3}\right)|| = \sqrt{108 + 56.25 + 400} \] \[ ||\mathbf{r}'\left(\frac{4\pi}{3}\right) \times \mathbf{r}''\left(\frac{4\pi}{3}\right)|| = \sqrt{564.25} \] \[ ||\mathbf{r}'\left(\frac{4\pi}{3}\right) \times \mathbf{r}''\left(\frac{4\pi}{3}\right)|| = \sqrt{564.25} \] Finally, we can calculate the curvature \( \kappa \): \[ \kappa = \frac{||\mathbf{r}'\left(\frac{4\pi}{3}\right) \times \mathbf{r}''\left(\frac{4\pi}{3}\right)||}{||\mathbf{r}'\left(\frac{4\pi}{3}\right)||^3} \] \[ \kappa = \frac{\sqrt{564.25}}{\left(\frac{\sqrt{127}}{2}\right)^3} \] \[ \kappa = \frac{\sqrt{564.25}}{\frac{127\sqrt{127}}{8}} \] \[ \kappa = \frac{8\sqrt{564.25}}{127\sqrt{127}} \] \[ \kappa = \frac{8\sqrt{564.25}}{127^{3/2}} \] \[ \kappa = \frac{8\sqrt{564. |
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| kimi | To find the curvature of a curve given by the parametric equations \( r(t) = 5\cos(t)i + 4\sin(t)j + 3tk \), we need to follow these steps: 1. **Find the first and second derivatives of \( r(t) \):** \( r'(t) = \frac{dr}{dt} = -5\sin(t)i + 4\cos(t)j + 3k \) \( r''(t) = \frac{d^2r}{dt^2} = -5\cos(t)i - 4\sin(t)j \) 2. **Calculate the magnitudes of \( r'(t) \) and \( r''(t) \):** \( \|r'(t)\| = \sqrt{(-5\sin(t))^2 + (4\cos(t))^2 + 3^2} = \sqrt{25\sin^2(t) + 16\cos^2(t) + 9} \) \( \|r''(t)\| = \sqrt{(-5\cos(t))^2 + (-4\sin(t))^2} = \sqrt{25\cos^2(t) + 16\sin^2(t)} \) 3. **Substitute \( t = \frac{4\pi}{3} \) into the magnitudes:** \( \|r'(\frac{4\pi}{3})\| = \sqrt{25\sin^2(\frac{4\pi}{3}) + 16\cos^2(\frac{4\pi}{3}) + 9} \) \( \|r''(\frac{4\pi}{3})\| = \sqrt{25\cos^2(\frac{4\pi}{3}) + 16\sin^2(\frac{4\pi}{3})} \) We know that \( \sin(\frac{4\pi}{3}) = \sin(\pi + \frac{\pi}{3}) = -\sin(\frac{\pi}{3}) = -\frac{\sqrt{3}}{2} \) and \( \cos(\frac{4\pi}{3}) = \cos(\pi + \frac{\pi}{3}) = -\cos(\frac{\pi}{3}) = -\frac{1}{2} \). So, \( \|r'(\frac{4\pi}{3})\| = \sqrt{25(-\frac{\sqrt{3}}{2})^2 + 16(-\frac{1}{2})^2 + 9} = \sqrt{25 \cdot \frac{3}{4} + 16 \cdot \frac{1}{4} + 9} = \sqrt{18.75 + 4 + 9} = \sqrt{31.75} \) \( \|r''(\frac{4\pi}{3})\| = \sqrt{25(-\frac{1}{2})^2 + 16(-\frac{\sqrt{3}}{2})^2} = \sqrt{25 \cdot \frac{1}{4} + 16 \cdot \frac{3}{4}} = \sqrt{6.25 + 12} = \sqrt{18.25} \) 4. **Calculate the curvature \( k(t) \):** The curvature \( k(t) \) is given by the formula: \( k(t) = \frac{\|r'(t) \times r''(t)\|}{\|r'(t)\|^3} \) Since \( r'(t) = -5\sin(t)i + 4\cos(t)j + 3k \) and \( r''(t) = -5\cos(t)i - 4\sin(t)j \), the cross product \( r'(t) \times r''(t) \) is: \( r'(t) \times r''(t) = \begin{vmatrix} i & j & k \\ -5\sin(t) & 4\cos(t) & 3 \\ -5\cos(t) & -4\sin(t) & 0 \end{vmatrix} \) \( = (12\cos^2(t) - 12\sin^2(t))i + (-15\sin(t)\cos(t))j + (20\sin(t)\cos(t))k \) Now, substitute \( t = |
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| wenxinyiyan | To find the curvature of the given curve r(t) = 5cos(t)i + 4sin(t)j + 3tk at t=4π/3, we need to follow these steps: 1. Find the first derivative of r(t), which represents the velocity vector. 2. Find the second derivative of r(t), which represents the acceleration vector. 3. Use the formula for curvature, κ = ||r'(t) x r''(t)|| / ||r'(t)||^3, where x denotes the cross product. Let's calculate each step: 1. First derivative of r(t): r'(t) = d/dt (5cos(t)i + 4sin(t)j + 3tk) = -5sin(t)i + 4cos(t)j + 3k 2. Second derivative of r(t): r''(t) = d/dt (-5sin(t)i + 4cos(t)j + 3k) = -5cos(t)i - 4sin(t)j 3. Curvature calculation: First, we find the cross product of r'(t) and r''(t): r'(t) x r''(t) = (-5sin(t)i + 4cos(t)j + 3k) x (-5cos(t)i - 4sin(t)j) = (12sin(t) - 15cos(t))i - (15sin(t) + 12cos(t))j + (25sin^2(t) + 16cos^2(t))k Next, we find the magnitude of the cross product: ||r'(t) x r''(t)|| = sqrt((12sin(t) - 15cos(t))^2 + (-15sin(t) - 12cos(t))^2 + (25sin^2(t) + 16cos^2(t))^2) And the magnitude of the first derivative: ||r'(t)|| = sqrt((-5sin(t))^2 + (4cos(t))^2 + 3^2) = sqrt(25sin^2(t) + 16cos^2(t) + 9) Now, we can find the curvature: κ = ||r'(t) x r''(t)|| / ||r'(t)||^3 Finally, we evaluate this expression at t = 4π/3: κ(4π/3) = ||r'(4π/3) x r''(4π/3)|| / ||r'(4π/3)||^3 Due to the complexity of the expression, it's advisable to use a calculator or software to find the exact value of κ(4π/3). However, the steps provided above outline the process to find the curvature of the given curve at t = 4π/3. |
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| zhipu | To find the curvature κ (kappa) of a space curve given by the vector function r(t), we use the formula: \[ \kappa(t) = \frac{\| r'(t) \times r''(t) \|}{\| r'(t) \|^3} \] where: - \( r'(t) \) is the first derivative of r(t), representing the velocity vector. - \( r''(t) \) is the second derivative of r(t), representing the acceleration vector. - \( \| \cdot \| \) denotes the magnitude of the vector. - \( \times \) denotes the cross product of two vectors. Given the vector function \( r(t) = 5\cos(t)i + 4\sin(t)j + 3tk \), we first find its first and second derivatives. 1. First derivative \( r'(t) \): \[ r'(t) = \frac{d}{dt} [5\cos(t)i + 4\sin(t)j + 3tk] \] \[ r'(t) = -5\sin(t)i + 4\cos(t)j + 3k \] 2. Second derivative \( r''(t) \): \[ r''(t) = \frac{d}{dt} [-5\sin(t)i + 4\cos(t)j + 3k] \] \[ r''(t) = -5\cos(t)i - 4\sin(t)j \] Now we need to evaluate these derivatives at \( t = \frac{4\pi}{3} \). First derivative at \( t = \frac{4\pi}{3} \): \[ r'\left(\frac{4\pi}{3}\right) = -5\sin\left(\frac{4\pi}{3}\right)i + 4\cos\left(\frac{4\pi}{3}\right)j + 3k \] \[ r'\left(\frac{4\pi}{3}\right) = -5\left(-\frac{\sqrt{3}}{2}\right)i + 4\left(-\frac{1}{2}\right)j + 3k \] \[ r'\left(\frac{4\pi}{3}\right) = \frac{5\sqrt{3}}{2}i - 2j + 3k \] Second derivative at \( t = \frac{4\pi}{3} \): \[ r''\left(\frac{4\pi}{3}\right) = -5\cos\left(\frac{4\pi}{3}\right)i - 4\sin\left(\frac{4\pi}{3}\right)j \] \[ r''\left(\frac{4\pi}{3}\right) = -5\left(-\frac{1}{2}\right)i - 4\left(-\frac{\sqrt{3}}{2}\right)j \] \[ r''\left(\frac{4\pi}{3}\right) = \frac{5}{2}i + 2\sqrt{3}j \] Next, we find the cross product \( r'(t) \times r''(t) \) at \( t = \frac{4\pi}{3} \): \[ r'\left(\frac{4\pi}{3}\right) \times r''\left(\frac{4\pi}{3}\right) = \begin{vmatrix} i & j & k \\ \frac{5\sqrt{3}}{2} & -2 & 3 \\ \frac{5}{2} & 2\sqrt{3} & 0 \end{vmatrix} \] \[ = i \left( -2 \cdot 0 - 3 \cdot 2\sqrt{3} \right) - j \left( \frac{5\sqrt{3}}{2} \cdot 0 - 3 \cdot \frac{5}{2} \right) + k \left( \frac{5\sqrt{3}}{2} \cdot 2\sqrt{3} - (-2) \cdot \frac{5}{2} \right) \] \[ = i \left( -6\sqrt{3} \right) - j \left( -\frac{15}{2} \right) + k \left( \frac{5 \cdot 3}{2} + 5 \right) \] \[ = -6\sqrt{3}i + \frac{15}{2}j + \left( \frac{15}{2} + 5 \right)k \] \[ = -6\sqrt{3}i + \frac{15}{2}j + \frac{25}{2}k \] Now, we find the magnitude of this cross product: \[ \| r'\left(\frac{4\pi}{3}\right) \times r''\left(\frac{4\pi}{3}\right) \| = \sqrt{(-6\sqrt{3})^2 + \left(\frac{15}{2}\right)^2 + \left(\frac{25}{2}\right)^2} \] \[ = \sqrt{108 + \frac{225}{4} + \frac{625}{4}} \] \[ = \sqrt{108 + \frac{850}{4}} \] \[ = \sqrt{108 + 212.5} \] \[ = \sqrt{320.5} \] \[ = \sqrt{\frac{641}{2}} \] Next, we find the magnitude of \( r'(t) \) at \( t = \frac{4\pi}{3} \): \[ \| r'\left(\frac{4\pi}{3}\right) \| = \sqrt{\left(\frac{5\sqrt{3}}{2}\right)^2 + (-2)^2 + 3^2} \] \[ = \sqrt{\frac{75}{4} + 4 + 9} \] \[ = \sqrt{\frac{75}{4} + \frac{16}{4} + \frac{36}{4}} \] \[ = \sqrt{\frac{127}{4}} \] \[ = \frac{\sqrt{127}}{2} \] Now we can calculate the curvature κ at \( t = \frac{4\pi}{3} \): \[ \kappa\left(\frac{4\pi}{3}\right) = \frac{\| r'\left(\frac{4\pi}{3}\right) \times r''\left(\frac{4\pi}{3}\right) \|}{\| r'\left(\frac{4\pi}{3}\right) \|^3} \] \[ \kappa\left(\frac{4\pi}{3}\right) = \frac{\sqrt{\frac{641}{2}}}{\left(\frac{\sqrt{127}}{2}\right)^3} \] \[ \kappa\left(\frac{4\pi}{3}\right) = \frac{\sqrt{\frac{641}{2}}}{\frac{127\sqrt{127}}{8}} \] \[ \kappa\left(\frac{4\pi}{3}\right) = \frac{8\sqrt{\frac{641}{2}}}{127\sqrt{127}} \] \[ \kappa\left(\frac{4\pi}{3}\right) = \frac{8\sqrt{641}}{127\sqrt{254}} \] \[ \kappa\left(\frac{4\pi}{3}\right) = \frac{8\sqrt{641}}{127\sqrt{2 \cdot 127}} \] \[ \kappa\left(\frac{4\pi}{3}\right) = \frac{8\sqrt{641}}{127\sqrt{2} \cdot \sqrt{127}} \] \[ \kappa\left(\frac{4\pi}{3}\right) = \frac{8\sqrt{641}}{127\sqrt{254}} \] \[ \kappa\left(\frac{4\pi}{3}\right) = \frac{8}{127} \cdot \sqrt{\frac{641}{254}} \] \[ \kappa\left(\frac{4\pi}{3}\right) = \frac{8}{127} \cdot \sqrt{\frac{641}{2 \cdot 127}} \] \[ \kappa\left(\frac{4\pi}{3}\right) = \frac{8}{127} \cdot \sqrt{\frac{641}{2 \cdot 127}} \] \[ \kappa\left(\frac{4\pi}{3}\right) = \frac{8}{127} \cdot \sqrt{\frac{641}{254}} \] This is the curvature of the curve at \( t = \frac{4\pi}{3} \). The exact numerical value can be calculated using a calculator if needed. |
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| gpto1 | To find the curvature \(\kappa(t)\) of the space curve defined by \[ \mathbf{r}(t) = 5\cos(t)\, \mathbf{i} + 4\sin(t)\, \mathbf{j} + 3t\, \mathbf{k} \] at \(t = \dfrac{4\pi}{3}\), we'll follow these steps: **1. Find the first derivative \(\mathbf{r}'(t)\):** Differentiate each component of \(\mathbf{r}(t)\) with respect to \(t\): \[ \begin{aligned} \mathbf{r}'(t) &= \dfrac{d}{dt}[5\cos(t)]\, \mathbf{i} + \dfrac{d}{dt}[4\sin(t)]\, \mathbf{j} + \dfrac{d}{dt}[3t]\, \mathbf{k} \\ &= -5\sin(t)\, \mathbf{i} + 4\cos(t)\, \mathbf{j} + 3\, \mathbf{k} \end{aligned} \] **2. Find the second derivative \(\mathbf{r}''(t)\):** Differentiate \(\mathbf{r}'(t)\) with respect to \(t\): \[ \begin{aligned} \mathbf{r}''(t) &= \dfrac{d}{dt}[-5\sin(t)]\, \mathbf{i} + \dfrac{d}{dt}[4\cos(t)]\, \mathbf{j} + \dfrac{d}{dt}[3]\, \mathbf{k} \\ &= -5\cos(t)\, \mathbf{i} - 4\sin(t)\, \mathbf{j} + 0\, \mathbf{k} \end{aligned} \] **3. Compute the cross product \(\mathbf{r}'(t) \times \mathbf{r}''(t)\):** Let \(\mathbf{r}'(t) = (a_1, a_2, a_3)\) and \(\mathbf{r}''(t) = (b_1, b_2, b_3)\), where: \[ \begin{aligned} a_1 &= -5\sin(t), \quad a_2 = 4\cos(t), \quad a_3 = 3 \\ b_1 &= -5\cos(t), \quad b_2 = -4\sin(t), \quad b_3 = 0 \end{aligned} \] Compute the cross product components: \[ \begin{aligned} c_1 &= a_2 b_3 - a_3 b_2 = (4\cos(t))(0) - (3)(-4\sin(t)) = 12\sin(t) \\ c_2 &= a_3 b_1 - a_1 b_3 = (3)(-5\cos(t)) - (-5\sin(t))(0) = -15\cos(t) \\ c_3 &= a_1 b_2 - a_2 b_1 = (-5\sin(t))(-4\sin(t)) - (4\cos(t))(-5\cos(t)) = 20\sin^2(t) + 20\cos^2(t) = 20(\sin^2 t + \cos^2 t) = 20 \end{aligned} \] So, \[ \mathbf{r}'(t) \times \mathbf{r}''(t) = \left(12\sin(t),\ -15\cos(t),\ 20\right) \] **4. Compute the magnitude \(|\mathbf{r}'(t) \times \mathbf{r}''(t)|\):** \[ \begin{aligned} |\mathbf{r}'(t) \times \mathbf{r}''(t)| &= \sqrt{[12\sin(t)]^2 + [-15\cos(t)]^2 + 20^2} \\ &= \sqrt{144\sin^2 t + 225\cos^2 t + 400} \end{aligned} \] **5. Compute the magnitude \(|\mathbf{r}'(t)|\):** \[ \begin{aligned} |\mathbf{r}'(t)| &= \sqrt{[-5\sin(t)]^2 + [4\cos(t)]^2 + 3^2} \\ &= \sqrt{25\sin^2 t + 16\cos^2 t + 9} \end{aligned} \] **6. Compute the curvature \(\kappa(t)\):** The curvature of a space curve is given by: \[ \kappa(t) = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3} \] **7. Evaluate at \(t = \dfrac{4\pi}{3}\):** First, compute \(\sin\left(\dfrac{4\pi}{3}\right)\) and \(\cos\left(\dfrac{4\pi}{3}\right)\): \[ \sin\left(\dfrac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}, \quad \cos\left(\dfrac{4\pi}{3}\right) = -\frac{1}{2} \] **Compute \(|\mathbf{r}'(t) \times \mathbf{r}''(t)|\) at \(t = \dfrac{4\pi}{3}\):** \[ \begin{aligned} |\mathbf{r}'\times \mathbf{r}''| &= \sqrt{144\left(\frac{3}{4}\right) + 225\left(\frac{1}{4}\right) + 400} \\ &= \sqrt{108 + 56.25 + 400} \\ &= \sqrt{564.25} \end{aligned} \] **Compute \(|\mathbf{r}'(t)|\) at \(t = \dfrac{4\pi}{3}\):** \[ \begin{aligned} |\mathbf{r}'(t)| &= \sqrt{25\left(\frac{3}{4}\right) + 16\left(\frac{1}{4}\right) + 9} \\ &= \sqrt{18.75 + 4 + 9} \\ &= \sqrt{31.75} \end{aligned} \] **Compute \(|\mathbf{r}'(t)|^3\):** \[ |\mathbf{r}'(t)|^3 = \left(\sqrt{31.75}\,\right)^3 = (31.75) \sqrt{31.75} \] **Compute the curvature \(\kappa\):** \[ \kappa = \frac{|\mathbf{r}'\times \mathbf{r}''|}{|\mathbf{r}'(t)|^3} = \frac{\sqrt{564.25}}{(31.75)\sqrt{31.75}} = \frac{\sqrt{564.25}}{(31.75)^{3/2}} \] **Simplify the expression:** Note that \(564.25 = \frac{2257}{4}\) and \(31.75 = \frac{127}{4}\). Therefore, \[ \sqrt{564.25} = \sqrt{\frac{2257}{4}} = \frac{\sqrt{2257}}{2}, \quad |\mathbf{r}'(t)|^3 = \left(\frac{127}{4}\right)^{3/2} = \left(\frac{127}{4}\right)^{1} \cdot \sqrt{\frac{127}{4}} = \left(\frac{127}{4}\right) \cdot \frac{\sqrt{127}}{2} = \frac{127\sqrt{127}}{8} \] **Compute the curvature:** \[ \begin{aligned} \kappa &= \frac{\frac{\sqrt{2257}}{2}}{\frac{127\sqrt{127}}{8}} = \frac{\sqrt{2257}}{2} \times \frac{8}{127\sqrt{127}} \\ &= \frac{4\sqrt{2257}}{127\sqrt{127}} \end{aligned} \] **Simplify further:** Since \(\sqrt{127^3} = \sqrt{127 \cdot 127 \cdot 127} = 127 \sqrt{127}\), the denominator simplifies to \(127 \sqrt{127}\). Thus, the curvature at \(t = \dfrac{4\pi}{3}\) is: \[ \kappa = \frac{4\sqrt{2257}}{127\sqrt{127}} \] This is the exact value of the curvature at the given point on the curve. **Answer:** \(\displaystyle \kappa = \dfrac{4\,\sqrt{2257}}{127\,\sqrt{127}}\) |
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| deepseekv3 | server error: Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-14d4d55c-32f5-9d7c-804f-31d71aa00a00', 'request_id': '14d4d55c-32f5-9d7c-804f-31d71aa00a00'} | 0.16 |