Let's break down the problem step by step.

The vector function is given as:

$\vec{r}(t) = \langle 3t^3 + 2, t^2 - 5t, 2 \ln{\left(\frac{t}{2}\right)} \rangle$

a) Coordinates of the point when $t = 2$

Substitute $t = 2$ into $\vec{r}(t)$:

1. $x(t) = 3t^3 + 2$: $x(2) = 3(2)^3 + 2 = 24 + 2 = 26$

2. $y(t) = t^2 - 5t$: $y(2) = (2)^2 - 5(2) = 4 - 10 = -6$

3. $z(t) = 2 \ln{\left(\frac{t}{2}\right)}$: $z(2) = 2 \ln{\left(\frac{2}{2}\right)} = 2 \ln{1} = 0$

Thus, the coordinates are: $(26, -6, 0)$

b) Direction vector for the tangent line at $t = 2$

The tangent vector is obtained by differentiating $\vec{r}(t)$ component-wise:

1. $\frac{dx}{dt} = \frac{d}{dt}(3t^3 + 2) = 9t^2$: $\frac{dx}{dt}(t = 2) = 9(2)^2 = 36$

2. $\frac{dy}{dt} = \frac{d}{dt}(t^2 - 5t) = 2t - 5$: $\frac{dy}{dt}(t = 2) = 2(2) - 5 = 4 - 5 = -1$

3. $\frac{dz}{dt} = \frac{d}{dt}\left(2 \ln{\left(\frac{t}{2}\right)}\right) = \frac{d}{dt}\left(2 \ln{t} - 2 \ln{2}\right) = \frac{2}{t}$: $\frac{dz}{dt}(t = 2) = \frac{2}{2} = 1$

The direction vector is: $\langle 36, -1, 1 \rangle$

c) Parametric equation for the tangent line at $t = 2$

The parametric equations are given by: $x(t) = x_0 + v_x t, \quad y(t) = y_0 + v_y t, \quad z(t) = z_0 + v_z t$

Using the point $(26, -6, 0)$ and direction vector $\langle 36, -1, 1 \rangle$:

1. $x(t) = 26 + 36t$
2. $y(t) = -6 - t$
3. $z(t) = 0 + t = t$

Thus: $x(t) = 26 + 36t, \quad y(t) = -6 - t, \quad z(t) = t$

Final Answers:

a) Point: $(26, -6, 0)$
b) Direction vector: $\langle 36, -1, 1 \rangle$
c) Parametric equations: $x(t) = 26 + 36t, \quad y(t) = -6 - t, \quad z(t) = t$

Do you need additional explanations or visualizations?

Follow-up Questions:

1. What is the significance of a direction vector in 3D space?
2. How do you compute derivatives of logarithmic functions in parametric form?
3. What is the geometric meaning of the tangent line in this context?
4. How would the solution change if a different point $t$ was used?
5. Can you extend this method to find the normal vector to the curve?

Tip:

Always check the differentiability of each component of the vector function before finding the tangent vector!