Find the curvature of: $\mathbf{r}(t) = \left\langle 3t, t^2, t^3 \right\rangle$ at the point $(3, 1, 1)$.

Neetesh Kumar | December 12, 2024

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This is the solution to Math 1C
Assignment: 13.3 Question Number 10
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Step-by-step solution:

To find the curvature $k$, we use the formula:

$k = \frac{|r'(t) \times r''(t)|}{|r'(t)|^3}$

Where:

• $r'(t)$ is the first derivative of $r(t)$
• $r''(t)$ is the second derivative of $r(t)$
• $\times$ denotes the cross product
• $| \cdot |$ denotes the magnitude of the vector

Step 1: Compute the first derivative $r'(t)$

The vector function is:

$r(t) = (3t, t^2, t^3)$

Differentiating each component with respect to $t$, we get:

$r'(t) = \left( \frac{d}{dt}(3t), \frac{d}{dt}(t^2), \frac{d}{dt}(t^3) \right) = (3, 2t, 3t^2)$

Step 2: Compute the second derivative $r''(t)$

Now, differentiating $r'(t)$:

$r''(t) = \left( \frac{d}{dt}(3), \frac{d}{dt}(2t), \frac{d}{dt}(3t^2) \right) = (0, 2, 6t)$

Step 3: Evaluate at the given point $(3, 1, 1)$

We are given that the point corresponds to $t = 1$, so we substitute $t = 1$ into $r'(t)$ and $r''(t)$:

$r'(1) = (3, 2 \times 1, 3 \times 1^2) = (3, 2, 3)$

$r''(1) = (0, 2, 6 \times 1) = (0, 2, 6)$

Step 4: Compute the cross product $r'(1) \times r''(1)$

Now, compute the cross product of $r'(1)$ and $r''(1)$:

$r'(1) \times r''(1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 3 \\ 0 & 2 & 6 \end{vmatrix}$

This gives:

$r'(1) \times r''(1) = \hat{i} \left( 2 \cdot 6 - 3 \cdot 2 \right) - \hat{j} \left( 3 \cdot 6 - 3 \cdot 0 \right) + \hat{k} \left( 3 \cdot 2 - 2 \cdot 0 \right)$

Simplifying:

$r'(1) \times r''(1) = (6, -18, 6)$

Step 5: Compute the magnitude of the cross product

Now, calculate the magnitude of the cross product:

$|r'(1) \times r''(1)| = \sqrt{6^2 + (-18)^2 + 6^2} = \sqrt{36 + 324 + 36} = \sqrt{396}$

$|r'(1) \times r''(1)| = 2\sqrt{99}$

Step 6: Compute the magnitude of $r'(1)$

Next, calculate the magnitude of $r'(1)$:

$|r'(1)| = \sqrt{3^2 + 2^2 + 3^2} = \sqrt{9 + 4 + 9} = \sqrt{22}$

Step 7: Compute the curvature $k$

Now, plug everything into the curvature formula:

$k = \frac{|r'(1) \times r''(1)|}{|r'(1)|^3}$

$k = \frac{2\sqrt{99}}{(\sqrt{22})^3}$

Simplify the denominator:

$|r'(1)|^3 = (\sqrt{22})^3 = 22\sqrt{22}$

Thus:

$k = \frac{2\sqrt{99}}{22\sqrt{22}}$

Simplifying further:

$k = \frac{\sqrt{99}}{11\sqrt{22}}$

Final Answer:

$k = \frac{\sqrt{99}}{11\sqrt{22}}$

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