We are given the vector-valued function:

$\mathbf{f}(t) = \langle 3t^2, \frac{2}{t}, t - 4 \rangle$

The goal is to find the vector form of the tangent line to the image of $f(t)$ at $t = 3$.

Step 1: Find the point of tangency

To get the point of tangency, we evaluate $\mathbf{f}(t)$ at $t = 3$:

$\mathbf{f}(3) = \langle 3(3)^2, \frac{2}{3}, 3 - 4 \rangle = \langle 27, \frac{2}{3}, -1 \rangle$

Thus, the point of tangency is $\mathbf{P} = (27, \frac{2}{3}, -1)$.

Step 2: Find the derivative of $\mathbf{f}(t)$

The derivative of $\mathbf{f}(t)$, denoted as $\mathbf{f'}(t)$, gives us the direction of the tangent line. We compute the derivative component-wise:

• $\frac{d}{dt} \left( 3t^2 \right) = 6t$
• $\frac{d}{dt} \left( \frac{2}{t} \right) = -\frac{2}{t^2}$
• $\frac{d}{dt} \left( t - 4 \right) = 1$

Thus:

$\mathbf{f'}(t) = \langle 6t, -\frac{2}{t^2}, 1 \rangle$

Step 3: Evaluate the derivative at $t = 3$

Now, we evaluate $\mathbf{f'}(t)$ at $t = 3$:

$\mathbf{f'}(3) = \langle 6(3), -\frac{2}{(3)^2}, 1 \rangle = \langle 18, -\frac{2}{9}, 1 \rangle$

Thus, the tangent direction vector is $\mathbf{v} = \langle 18, -\frac{2}{9}, 1 \rangle$.

Step 4: Write the equation of the tangent line

The vector form of the tangent line at $t = 3$ is given by:

$\mathbf{l}(t) = \mathbf{P} + \mathbf{v}(t - 3)$

Substituting $\mathbf{P} = \langle 27, \frac{2}{3}, -1 \rangle$ and $\mathbf{v} = \langle 18, -\frac{2}{9}, 1 \rangle$, we get:

$\mathbf{l}(t) = \langle 27, \frac{2}{3}, -1 \rangle + \langle 18, -\frac{2}{9}, 1 \rangle (t - 3)$

Or more explicitly:

$\mathbf{l}(t) = \langle 27 + 18(t - 3), \frac{2}{3} - \frac{2}{9}(t - 3), -1 + (t - 3) \rangle$

This is the vector form of the tangent line to the image of $f(t)$ at $t = 3$.

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Would you like further clarification or more details on this topic?

Here are some related questions to expand your understanding:

1. How do you calculate the tangent vector for parametric curves in higher dimensions?
2. What is the geometric interpretation of the derivative of a vector-valued function?
3. How can you find the equation of a normal line to a parametric curve?
4. What are the conditions for a vector field to be parallel to a tangent line?
5. How does the concept of curvature relate to the tangent line in vector calculus?

Tip: Remember that the tangent line to a curve at a given point can be found using the derivative of the curve at that point, and that the direction vector of the tangent line is essentially the derivative of the vector-valued function.