Answer :
To find the curvature [tex]\(\kappa(t)\)[/tex] of the curve given by the vector function [tex]\(r(t) = \sqrt{6} t^2 \mathbf{i} + 2t \mathbf{j} + 2t^3 \mathbf{k}\)[/tex], we will follow these steps:
1. Calculate the first derivative [tex]\(r'(t)\)[/tex].
2. Calculate the second derivative [tex]\(r''(t)\)[/tex].
3. Compute the cross product [tex]\(r'(t) \times r''(t)\)[/tex].
4. Determine the magnitudes of [tex]\(r'(t) \times r''(t)\)[/tex] and [tex]\(r'(t)\)[/tex].
5. Apply the curvature formula [tex]\(\kappa(t) = \frac{|r'(t) \times r''(t)|}{|r'(t)|^3}\)[/tex].
### Step 1: Calculate the first derivative [tex]\(r'(t)\)[/tex]
Given:
[tex]\[ r(t) = \sqrt{6} t^2 \mathbf{i} + 2t \mathbf{j} + 2t^3 \mathbf{k} \][/tex]
The first derivative with respect to [tex]\(t\)[/tex] is:
[tex]\[ r'(t) = \frac{d}{dt}(\sqrt{6} t^2 \mathbf{i} + 2t \mathbf{j} + 2t^3 \mathbf{k}) \][/tex]
[tex]\[ r'(t) = 2 \sqrt{6} t \mathbf{i} + 2 \mathbf{j} + 6t^2 \mathbf{k} \][/tex]
### Step 2: Calculate the second derivative [tex]\(r''(t)\)[/tex]
The second derivative with respect to [tex]\(t\)[/tex] is:
[tex]\[ r''(t) = \frac{d}{dt}(2 \sqrt{6} t \mathbf{i} + 2 \mathbf{j} + 6t^2 \mathbf{k}) \][/tex]
[tex]\[ r''(t) = 2 \sqrt{6} \mathbf{i} + 0 \mathbf{j} + 12t \mathbf{k} \][/tex]
### Step 3: Compute the cross product [tex]\(r'(t) \times r''(t)\)[/tex]
Using the components of [tex]\(r'(t)\)[/tex] and [tex]\(r''(t)\)[/tex]:
[tex]\[ r'(t) = (2 \sqrt{6} t, 2, 6t^2) \][/tex]
[tex]\[ r''(t) = (2 \sqrt{6}, 0, 12t) \][/tex]
The cross product is given by the determinant of the following matrix:
[tex]\[ r'(t) \times r''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2\sqrt{6}t & 2 & 6t^2 \\ 2\sqrt{6} & 0 & 12t \\ \end{vmatrix} \][/tex]
Calculating the determinant:
[tex]\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2\sqrt{6}t & 2 & 6t^2 \\ 2\sqrt{6} & 0 & 12t \\ \end{vmatrix} = \mathbf{i}(2 \cdot 12t - 6t^2 \cdot 0) - \mathbf{j}(2\sqrt{6}t \cdot 12t - 6t^2 \cdot 2\sqrt{6}) + \mathbf{k}(2\sqrt{6}t \cdot 0 - 2\sqrt{6} \cdot 2) \][/tex]
[tex]\[ = \mathbf{i}(24t) - \mathbf{j}(24\sqrt{6}t^2 - 12\sqrt{6}t^2) + \mathbf{k}(-4\sqrt{6}) \][/tex]
[tex]\[ = 24t \mathbf{i} - 12\sqrt{6}t^2 \mathbf{j} - 4\sqrt{6} \mathbf{k} \][/tex]
### Step 4: Determine the magnitudes
The magnitude of the cross product [tex]\(r'(t) \times r''(t)\)[/tex] is:
[tex]\[ |r'(t) \times r''(t)| = \sqrt{(24t)^2 + (-12\sqrt{6}t^2)^2 + (-4\sqrt{6})^2} \][/tex]
[tex]\[ = \sqrt{576t^2 + 144t^4 \cdot 6 + 96} \][/tex]
[tex]\[ = \sqrt{576t^2 + 864t^4 + 96} \][/tex]
[tex]\[ = \sqrt{864t^4 + 576t^2 + 96} \][/tex]
The magnitude of [tex]\(r'(t)\)[/tex] is:
[tex]\[ |r'(t)| = \sqrt{(2\sqrt{6}t)^2 + (2)^2 + (6t^2)^2} \][/tex]
[tex]\[ = \sqrt{24t^2 + 4 + 36t^4} \][/tex]
[tex]\[ = \sqrt{36t^4 + 24t^2 + 4} \][/tex]
### Step 5: Apply the curvature formula
Finally, we compute the curvature [tex]\(\kappa(t)\)[/tex]:
[tex]\[ \kappa(t) = \frac{|r'(t) \times r''(t)|}{|r'(t)|^3} \][/tex]
Substituting the magnitudes:
[tex]\[ \kappa(t) = \frac{\sqrt{864t^4 + 576t^2 + 96}}{(\sqrt{36t^4 + 24t^2 + 4})^3} \][/tex]
Simplify the denominator:
[tex]\[ (\sqrt{36t^4 + 24t^2 + 4})^3 = (36t^4 + 24t^2 + 4)^{3/2} \][/tex]
Thus the curvature is:
[tex]\[ \kappa(t) = \frac{\sqrt{864t^4 + 576t^2 + 96}}{(36t^4 + 24t^2 + 4)^{3/2}} \][/tex]
1. Calculate the first derivative [tex]\(r'(t)\)[/tex].
2. Calculate the second derivative [tex]\(r''(t)\)[/tex].
3. Compute the cross product [tex]\(r'(t) \times r''(t)\)[/tex].
4. Determine the magnitudes of [tex]\(r'(t) \times r''(t)\)[/tex] and [tex]\(r'(t)\)[/tex].
5. Apply the curvature formula [tex]\(\kappa(t) = \frac{|r'(t) \times r''(t)|}{|r'(t)|^3}\)[/tex].
### Step 1: Calculate the first derivative [tex]\(r'(t)\)[/tex]
Given:
[tex]\[ r(t) = \sqrt{6} t^2 \mathbf{i} + 2t \mathbf{j} + 2t^3 \mathbf{k} \][/tex]
The first derivative with respect to [tex]\(t\)[/tex] is:
[tex]\[ r'(t) = \frac{d}{dt}(\sqrt{6} t^2 \mathbf{i} + 2t \mathbf{j} + 2t^3 \mathbf{k}) \][/tex]
[tex]\[ r'(t) = 2 \sqrt{6} t \mathbf{i} + 2 \mathbf{j} + 6t^2 \mathbf{k} \][/tex]
### Step 2: Calculate the second derivative [tex]\(r''(t)\)[/tex]
The second derivative with respect to [tex]\(t\)[/tex] is:
[tex]\[ r''(t) = \frac{d}{dt}(2 \sqrt{6} t \mathbf{i} + 2 \mathbf{j} + 6t^2 \mathbf{k}) \][/tex]
[tex]\[ r''(t) = 2 \sqrt{6} \mathbf{i} + 0 \mathbf{j} + 12t \mathbf{k} \][/tex]
### Step 3: Compute the cross product [tex]\(r'(t) \times r''(t)\)[/tex]
Using the components of [tex]\(r'(t)\)[/tex] and [tex]\(r''(t)\)[/tex]:
[tex]\[ r'(t) = (2 \sqrt{6} t, 2, 6t^2) \][/tex]
[tex]\[ r''(t) = (2 \sqrt{6}, 0, 12t) \][/tex]
The cross product is given by the determinant of the following matrix:
[tex]\[ r'(t) \times r''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2\sqrt{6}t & 2 & 6t^2 \\ 2\sqrt{6} & 0 & 12t \\ \end{vmatrix} \][/tex]
Calculating the determinant:
[tex]\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2\sqrt{6}t & 2 & 6t^2 \\ 2\sqrt{6} & 0 & 12t \\ \end{vmatrix} = \mathbf{i}(2 \cdot 12t - 6t^2 \cdot 0) - \mathbf{j}(2\sqrt{6}t \cdot 12t - 6t^2 \cdot 2\sqrt{6}) + \mathbf{k}(2\sqrt{6}t \cdot 0 - 2\sqrt{6} \cdot 2) \][/tex]
[tex]\[ = \mathbf{i}(24t) - \mathbf{j}(24\sqrt{6}t^2 - 12\sqrt{6}t^2) + \mathbf{k}(-4\sqrt{6}) \][/tex]
[tex]\[ = 24t \mathbf{i} - 12\sqrt{6}t^2 \mathbf{j} - 4\sqrt{6} \mathbf{k} \][/tex]
### Step 4: Determine the magnitudes
The magnitude of the cross product [tex]\(r'(t) \times r''(t)\)[/tex] is:
[tex]\[ |r'(t) \times r''(t)| = \sqrt{(24t)^2 + (-12\sqrt{6}t^2)^2 + (-4\sqrt{6})^2} \][/tex]
[tex]\[ = \sqrt{576t^2 + 144t^4 \cdot 6 + 96} \][/tex]
[tex]\[ = \sqrt{576t^2 + 864t^4 + 96} \][/tex]
[tex]\[ = \sqrt{864t^4 + 576t^2 + 96} \][/tex]
The magnitude of [tex]\(r'(t)\)[/tex] is:
[tex]\[ |r'(t)| = \sqrt{(2\sqrt{6}t)^2 + (2)^2 + (6t^2)^2} \][/tex]
[tex]\[ = \sqrt{24t^2 + 4 + 36t^4} \][/tex]
[tex]\[ = \sqrt{36t^4 + 24t^2 + 4} \][/tex]
### Step 5: Apply the curvature formula
Finally, we compute the curvature [tex]\(\kappa(t)\)[/tex]:
[tex]\[ \kappa(t) = \frac{|r'(t) \times r''(t)|}{|r'(t)|^3} \][/tex]
Substituting the magnitudes:
[tex]\[ \kappa(t) = \frac{\sqrt{864t^4 + 576t^2 + 96}}{(\sqrt{36t^4 + 24t^2 + 4})^3} \][/tex]
Simplify the denominator:
[tex]\[ (\sqrt{36t^4 + 24t^2 + 4})^3 = (36t^4 + 24t^2 + 4)^{3/2} \][/tex]
Thus the curvature is:
[tex]\[ \kappa(t) = \frac{\sqrt{864t^4 + 576t^2 + 96}}{(36t^4 + 24t^2 + 4)^{3/2}} \][/tex]
