To find the unit tangent vector to the curve $\mathbf{r}(t)$ at $t = 4$, follow these steps:

Step 1: Compute the Velocity Vector $\mathbf{v}(t)$

The tangent vector to the curve is given by the derivative of $\mathbf{r}(t)$:

$\mathbf{r}(t) = t^3 \mathbf{i} + 4t^2 \mathbf{j}$

Differentiate each component:

$\mathbf{v}(t) = \frac{d}{dt} (t^3) \mathbf{i} + \frac{d}{dt} (4t^2) \mathbf{j}$

$\mathbf{v}(t) = 3t^2 \mathbf{i} + 8t \mathbf{j}$

Step 2: Evaluate at $t = 4$

Substituting $t = 4$:

$\mathbf{v}(4) = 3(4)^2 \mathbf{i} + 8(4) \mathbf{j}$

$\mathbf{v}(4) = 3(16) \mathbf{i} + 32 \mathbf{j}$

$\mathbf{v}(4) = 48 \mathbf{i} + 32 \mathbf{j}$

Step 3: Compute the Magnitude of $\mathbf{v}(4)$

The unit tangent vector is obtained by dividing $\mathbf{v}(4)$ by its magnitude:

$\|\mathbf{v}(4)\| = \sqrt{(48)^2 + (32)^2}$

$= \sqrt{2304 + 1024}$

$= \sqrt{3328}$

$= 8\sqrt{52} = 8 \times 2\sqrt{13} = 16\sqrt{13}$

Step 4: Find the Unit Tangent Vector

$\mathbf{T}(4) = \frac{\mathbf{v}(4)}{\|\mathbf{v}(4)\|}$

$= \frac{48 \mathbf{i} + 32 \mathbf{j}}{16\sqrt{13}}$

$= \frac{48}{16\sqrt{13}} \mathbf{i} + \frac{32}{16\sqrt{13}} \mathbf{j}$

$= \frac{3}{\sqrt{13}} \mathbf{i} + \frac{2}{\sqrt{13}} \mathbf{j}$

Final Answer:

$\mathbf{T}(4) = \frac{3}{\sqrt{13}} \mathbf{i} + \frac{2}{\sqrt{13}} \mathbf{j}$

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Would you like a breakdown of any step? Here are some related questions:

1. How do you find the normal vector to a curve?
2. What is the geometric meaning of the unit tangent vector?
3. How do you compute the principal normal vector?
4. How do you find the curvature of a parametric curve?
5. What is the difference between the unit tangent vector and velocity?

Tip: The unit tangent vector always points in the direction of motion and has a magnitude of 1! 🚀