Let's solve the problem step by step: Given curve: \[ \vec{r}(t) = \langle 4t, 4t^2, -\frac{8t^3}{3} \rangle \] We are to **find the curvature** at an arbitrary point \( t \). ## **Step 1: Find the first derivative \(\vec{r}'(t)\)** \[ \vec{r}'(t) = \left\langle \frac{d}{dt}(4t), \frac{d}{dt}(4t^2), \frac{d}{dt}\left(-\frac{8t^3}{3}\right) \right\rangle = \langle 4, 8t, -8t^2 \rangle \] ## **Step 2: Find the second derivative \(\vec{r}''(t)\)** \[ \vec{r}''(t) = \left\langle \frac{d}{dt}(4), \frac{d}{dt}(8t), \frac{d}{dt}(-8t^2) \right\rangle = \langle 0, 8, -16t \rangle \] ## **Step 3: Find the curvature formula** For a space curve: \[ \kappa = \frac{|\vec{r}'(t) \times \vec{r}''(t)|}{|\vec{r}'(t)|^3} \] --- ### **Step 4: Find the cross product \(\vec{r}'(t) \times \vec{r}''(t)\)** \[ \vec{r}'(t) = \langle 4, 8t, -8t^2 \rangle \\ \vec{r}''(t) = \langle 0, 8, -16t \rangle \] The cross product: \[ \vec{r}'(t) \times \vec{r}''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 8t & -8t^2 \\ 0 & 8 & -16t \end{vmatrix} \] Expanding the determinant: \[ = \mathbf{i} \left( (8t)(-16t) - (-8t^2)(8) \right) - \mathbf{j} \left( 4(-16t) - (-8t^2)(0) \right) + \mathbf{k} \left( 4 \cdot 8 - 8t \cdot 0 \right) \] Calculate: - \( (8t)(-16t) = -128t^2 \) - \( (-8t^2)(8) = -64t^2 \) - \( 4(-16t) = -64t \) - \( 4 \cdot 8 = 32 \) So, \[ = \mathbf{i} \left( -128t^2 - (-64t^2) \right) - \mathbf{j} (-64t) + \mathbf{k} (32) \] \[ = \mathbf{i}(-128t^2 + 64t^2) + \mathbf{j}(64t) + \mathbf{k}(32) \] \[ = \mathbf{i}(-64t^2) + \mathbf{j}(64t) + \mathbf{k}(32) \] \[ = \langle -64t^2, 64t, 32 \rangle \] --- ### **Step 5: Find the magnitude \(|\vec{r}'(t) \times \vec{r}''(t)|\)** \[ |\vec{r}'(t) \times \vec{r}''(t)| = \sqrt{(-64t^2)^2 + (64t)^2 + 32^2} = \sqrt{4096t^4 + 4096t^2 + 1024} \] \[ = \sqrt{4096(t^4 + t^2 + \frac{1}{4})} = 64\sqrt{t^4 + t^2 + \frac{1}{4}} \] --- ### **Step 6: Find \(|\vec{r}'(t)|\)** \[ |\vec{r}'(t)| = \sqrt{4^2 + (8t)^2 + (-8t^2)^2} = \sqrt{16 + 64t^2 + 64t^4} = \sqrt{16(1 + 4t^2 + 4t^4)} = 4\sqrt{1 + 4t^2 + 4t^4} \] --- ### **Step 7: Find curvature \(\kappa\) at arbitrary \(t\)** \[ \kappa = \frac{|\vec{r}'(t) \times \vec{r}''(t)|}{|\vec{r}'(t)|^3} = \frac{64\sqrt{t^4 + t^2 + \frac{1}{4}}}{\left(4\sqrt{1 + 4t^2 + 4t^4}\right)^3} \] \[ = \frac{64\sqrt{t^4 + t^2 + \frac{1}{4}}}{64 (1 + 4t^2 + 4t^4)^{3/2}} \] \[ = \frac{\sqrt{t^4 + t^2 + \frac{1}{4}}}{(1 + 4t^2 + 4t^4)^{3/2}} \] --- ## **Final Answer:** \[ \boxed{ \kappa(t) = \frac{\sqrt{t^4 + t^2 + \frac{1}{4}}}{(1 + 4t^2 + 4t^4)^{3/2}} } \] This is the **curvature** of the given curve at an arbitrary point \(t\).

Let's carefully go through the problem step-by-step, providing detailed calculations and explanations. --- ## **Given Curve:** \[ \vec{r}(t) = \langle 4t, 4t^2, -\frac{8t^3}{3} \rangle \] --- ## **Objective:** Find the **curvature \(\kappa(t)\)** at an arbitrary point \(t\). --- ## **Step 1: Compute the first derivative \(\vec{r}'(t)\)** Differentiate each component with respect to \(t\): - \(x(t) = 4t \implies x'(t) = 4\) - \(y(t) = 4t^2 \implies y'(t) = 8t\) - \(z(t) = -\frac{8t^3}{3} \implies z'(t) = -8t^2\) **Result:** \[ \boxed{ \vec{r}'(t) = \langle 4, \, 8t, \, -8t^2 \rangle } \] --- ## **Step 2: Compute the second derivative \(\vec{r}''(t)\)** Differentiate \(\vec{r}'(t)\): - \(4 \implies 0\) - \(8t \implies 8\) - \(-8t^2 \implies -16t\) **Result:** \[ \boxed{ \vec{r}''(t) = \langle 0, \, 8, \, -16t \rangle } \] --- ## **Step 3: Formula for curvature** \[ \boxed{ \kappa(t) = \frac{|\vec{r}'(t) \times \vec{r}''(t)|}{|\vec{r}'(t)|^3} } \] We need to compute the cross product numerator and the magnitude of \(\vec{r}'(t)\) in the denominator. --- ## **Step 4: Compute \(\vec{r}'(t) \times \vec{r}''(t)\)** Set up the determinant: \[ \vec{r}'(t) = \langle 4, \, 8t, \, -8t^2 \rangle \] \[ \vec{r}''(t) = \langle 0, \, 8, \, -16t \rangle \] Cross product: \[ \vec{r}'(t) \times \vec{r}''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 8t & -8t^2 \\ 0 & 8 & -16t \end{vmatrix} \] Expanding along the first row: \[ = \mathbf{i} \left( (8t)(-16t) - (-8t^2)(8) \right) - \mathbf{j} \left( 4(-16t) - (-8t^2)(0) \right) + \mathbf{k} \left( 4 \cdot 8 - 8t \cdot 0 \right) \] Calculate each part: - \( (8t)(-16t) = -128 t^2 \) - \( (-8t^2)(8) = -64 t^2 \) So, \[ \mathbf{i} \left( -128 t^2 - (-64 t^2) \right) = \mathbf{i}(-128 t^2 + 64 t^2) = \mathbf{i}(-64 t^2) \] Next, \[ - \mathbf{j} \left( 4(-16t) - 0 \right) = - \mathbf{j}(-64 t) = + \mathbf{j}(64 t) \] Finally, \[ \mathbf{k} (4 \times 8 - 0) = \mathbf{k}(32) \] **Result:** \[ \vec{r}'(t) \times \vec{r}''(t) = \langle -64 t^2, \, 64 t, \, 32 \rangle \] --- ## **Step 5: Magnitude of the cross product** \[ |\vec{r}'(t) \times \vec{r}''(t)| = \sqrt{(-64 t^2)^2 + (64 t)^2 + 32^2} \] Calculate each term: - \((-64 t^2)^2 = 4096 t^4\) - \((64 t)^2 = 4096 t^2\) - \(32^2 = 1024\) Sum: \[ = \sqrt{4096 t^4 + 4096 t^2 + 1024} \] Factor out 1024: \[ = \sqrt{1024 (4 t^4 + 4 t^2 + 1)} \] Since \(\sqrt{1024} = 32\): \[ |\vec{r}'(t) \times \vec{r}''(t)| = 32 \sqrt{4 t^4 + 4 t^2 + 1} \] --- ## **Step 6: Magnitude of \(\vec{r}'(t)\)** \[ |\vec{r}'(t)| = \sqrt{4^2 + (8t)^2 + (-8t^2)^2} \] \[ = \sqrt{16 + 64 t^2 + 64 t^4} \] Factor out 16: \[ = \sqrt{16 (1 + 4 t^2 + 4 t^4)} = 4 \sqrt{1 + 4 t^2 + 4 t^4} \] --- ## **Step 7: Compute curvature \(\kappa(t)\)** Recall: \[ \kappa(t) = \frac{|\vec{r}'(t) \times \vec{r}''(t)|}{|\vec{r}'(t)|^3} \] Substitute the magnitudes: \[ \kappa(t) = \frac{32 \sqrt{4 t^4 + 4 t^2 + 1}}{\left( 4 \sqrt{1 + 4 t^2 + 4 t^4} \right)^3} \] Simplify denominator: \[ (4)^3 \times (\sqrt{1 + 4 t^2 + 4 t^4})^3 = 64 \times (1 + 4 t^2 + 4 t^4)^{3/2} \] So, \[ \kappa(t) = \frac{32 \sqrt{4 t^4 + 4 t^2 + 1}}{64 (1 + 4 t^2 + 4 t^4)^{3/2}} \] Divide numerator and denominator by 32: \[ \kappa(t) = \frac{\sqrt{4 t^4 + 4 t^2 + 1}}{2 (1 + 4 t^2 + 4 t^4)^{3/2}} \] --- ## **Step 8: Final expression** Note that: \[ 4 t^4 + 4 t^2 + 1 = (2 t^2)^2 + 2 \times 2 t^2 \times 0 + 1 = \text{(no perfect square form), but we can leave as is.} \] Alternatively, observe the denominator: \[ 1 + 4 t^2 + 4 t^4 \] which is symmetric and can be written as: \[ (1 + 2 t^2)^2 \] since: \[ (1 + 2 t^2)^2 = 1 + 4 t^2 + 4 t^4 \] Similarly, numerator: \[ 4 t^4 + 4 t^2 + 1 = (2 t^2)^2 + 2 \times 2 t^2 \times 1 + 1^2 = (2 t^2 + 1)^2 \] Therefore, \[ \sqrt{4 t^4 + 4 t^2 + 1} = |2 t^2 + 1| \quad\text{(which is positive for all real \(t\))}, \] and \[ (1 + 4 t^2 + 4 t^4)^{3/2} = \left( (1 + 2 t^2)^2 \right)^{3/2} = (1 + 2 t^2)^3 \] **Final simplified form:** \[ \boxed{ \kappa(t) = \frac{2 t^2 + 1}{2 (1 + 2 t^2)^3} } \] --- ## **Final Answer:** \[ \boxed{ \boxed{ \textbf{Curvature}:\quad \kappa(t) = \frac{2 t^2 + 1}{2 (1 + 2 t^2)^3} } } \] This gives the curvature of the curve at any point \(t\). --- **Summary:** - Derived \(\vec{r}'(t)\) and \(\vec{r}''(t)\). - Computed their cross product and magnitudes. - Simplified the expression to a closed form. Let me know if you'd like further clarification!