A unit tangent vector at t=2 on the curve `x=t^(2)+2, y=4t-5` and `z=2t^(2)-6t` is

Options

A. `1/sqrt(3)(veci+vecj+veck)`
B. `1/3(2veci+2vecj+veck)`
C. `1/sqrt(6)(2veci+vecj+veck)`
D. `1/3(veci+vecj+veck)`

Explanation

To find the unit tangent vector at \( t = 2 \) on the given curve defined by the equations \( x = t^2 + 2 \), \( y = 4t - 5 \), and \( z = 2t^2 - 6t \), we will follow these steps:

### Step 1: Define the position vector
The position vector \( \mathbf{r}(t) \) can be expressed in terms of the components \( x, y, z \):
\[ \mathbf{r}(t) = (t^2 + 2) \mathbf{i} + (4t - 5) \mathbf{j} + (2t^2 - 6t) \mathbf{k} \]

### Step 2: Differentiate the position vector
To find the tangent vector, we need to differentiate \( \mathbf{r}(t) \) with respect to \( t \):
\[ \frac{d\mathbf{r}}{dt} = \frac{d}{dt}[(t^2 + 2) \mathbf{i} + (4t - 5) \mathbf{j} + (2t^2 - 6t) \mathbf{k}] \]

Calculating the derivatives:
- For \( x = t^2 + 2 \), \( \frac{dx}{dt} = 2t \)
- For \( y = 4t - 5 \), \( \frac{dy}{dt} = 4 \)
- For \( z = 2t^2 - 6t \), \( \frac{dz}{dt} = 4t - 6 \)

Thus, the derivative becomes:
\[ \frac{d\mathbf{r}}{dt} = (2t) \mathbf{i} + (4) \mathbf{j} + (4t - 6) \mathbf{k} \]

### Step 3: Evaluate the derivative at \( t = 2 \)
Now, we will substitute \( t = 2 \) into the derivative:
\[ \frac{d\mathbf{r}}{dt} \bigg|_{t=2} = (2 \cdot 2) \mathbf{i} + (4) \mathbf{j} + (4 \cdot 2 - 6) \mathbf{k} \]
Calculating this gives:
\[ = 4 \mathbf{i} + 4 \mathbf{j} + (8 - 6) \mathbf{k} = 4 \mathbf{i} + 4 \mathbf{j} + 2 \mathbf{k} \]

### Step 4: Find the magnitude of the tangent vector
Next, we need to find the magnitude of the tangent vector:
\[ \left\| \frac{d\mathbf{r}}{dt} \right\| = \sqrt{(4)^2 + (4)^2 + (2)^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6 \]

### Step 5: Calculate the unit tangent vector
The unit tangent vector \( \mathbf{T} \) is given by:
\[ \mathbf{T} = \frac{\frac{d\mathbf{r}}{dt}}{\left\| \frac{d\mathbf{r}}{dt} \right\|} = \frac{4 \mathbf{i} + 4 \mathbf{j} + 2 \mathbf{k}}{6} \]
This simplifies to:
\[ \mathbf{T} = \frac{2}{3} \mathbf{i} + \frac{2}{3} \mathbf{j} + \frac{1}{3} \mathbf{k} \]

### Final Answer
Thus, the unit tangent vector at \( t = 2 \) is:
\[ \mathbf{T} = \frac{2}{3} \mathbf{i} + \frac{2}{3} \mathbf{j} + \frac{1}{3} \mathbf{k} \]

---