3. Find the unit tangent and unit normal vector at \( t=2 \) on the curve \( x=t^{2}-1, y=4 t-3 \), \( z=2 t^{2}-6 t \) where \( t \) is any variable. Ans. \( \frac{1}{3}(2 \hat{i}+2 \hat{j}+\hat{k}), \frac{1}{3 \sqrt{5}}(2 \hat{i}+2 \hat{k}) \)

To find the unit tangent and unit normal vectors at \( t=2 \) on the given curve defined by the parametric equations: \[ x = t^2 - 1, \quad y = 4t - 3, \quad z = 2t^2 - 6t \] we will follow these steps: 1. **Calculate the derivative of the position vector** to find the tangent vector. 2. **Evaluate the tangent vector at \( t=2 \)**. 3. **Find the unit tangent vector** by normalizing the tangent vector. 4. **Calculate the derivative of the unit tangent vector** to find the normal vector. 5. **Evaluate the normal vector at \( t=2 \)** and normalize it to find the unit normal vector. ### Step 1: Calculate the derivative of the position vector The position vector \( \mathbf{r}(t) \) is given by: \[ \mathbf{r}(t) = (t^2 - 1) \hat{i} + (4t - 3) \hat{j} + (2t^2 - 6t) \hat{k} \] Now, we differentiate \( \mathbf{r}(t) \) with respect to \( t \): \[ \mathbf{r}'(t) = \frac{d}{dt}(t^2 - 1) \hat{i} + \frac{d}{dt}(4t - 3) \hat{j} + \frac{d}{dt}(2t^2 - 6t) \hat{k} \] Calculating the derivatives: \[ \mathbf{r}'(t) = (2t) \hat{i} + (4) \hat{j} + (4t - 6) \hat{k} \] ### Step 2: Evaluate the tangent vector at \( t=2 \) Now we evaluate \( \mathbf{r}'(2) \): \[ \mathbf{r}'(2) = (2 \cdot 2) \hat{i} + (4) \hat{j} + (4 \cdot 2 - 6) \hat{k} \] Calculating this gives: \[ \mathbf{r}'(2) = 4 \hat{i} + 4 \hat{j} + 2 \hat{k} \] ### Step 3: Find the unit tangent vector The magnitude of the tangent vector \( \mathbf{r}'(2) \) is: \[ \|\mathbf{r}'(2)\| = \sqrt{4^2 + 4^2 + 2^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6 \] Thus, the unit tangent vector \( \mathbf{T}(2) \) is: \[ \mathbf{T}(2) = \frac{\mathbf{r}'(2)}{\|\mathbf{r}'(2)\|} = \frac{4 \hat{i} + 4 \hat{j} + 2 \hat{k}}{6} = \frac{2}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{1}{3} \hat{k} \] ### Step 4: Calculate the derivative of the unit tangent vector Next, we need to find the derivative of the tangent vector \( \mathbf{T}(t) \): \[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} \] To find the unit normal vector, we first need to differentiate \( \mathbf{T}(t) \) and then evaluate it at \( t=2 \). ### Step 5: Evaluate the normal vector at \( t=2 \) We will calculate \( \mathbf{T}'(t) \) and then evaluate it at \( t=2 \). However, since we need the unit normal vector, we can directly find the normal vector from the derivative of the tangent vector. The normal vector \( \mathbf{N}(t) \) is given by: \[ \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|} \] To find \( \mathbf{T}'(t) \), we will differentiate \( \mathbf{T}(t) \) and evaluate it at \( t=2 \). Now, let's calculate the unit normal vector. We will calculate the derivative of \( \mathbf{r}'(t) \) and then find the unit normal vector. Calculating \( \mathbf{r}''(t) \): \[ \mathbf{r}''(t) = \frac{d}{dt}(2t) \hat{i} + \frac{d}{dt}(4) \hat{j} + \frac{d}{dt}(4t - 6) \hat{k} = 2 \hat{i} + 0 \hat{j} + 4 \hat{k} \] Now, we evaluate \( \mathbf{r}''(2) \): \[ \mathbf{r}''(2) = 2 \hat{i} + 0 \hat{j} + 4 \hat{k} \] ### Final Calculation Now we can find the unit normal vector \( \mathbf{N}(2) \): The magnitude of \( \mathbf{r}''(2) \) is: \[ \|\mathbf{r}''(2)\| = \sqrt{2^2 + 0^2 + 4^2} = \sqrt{4 + 0 + 16} = \sqrt{20} = 2\sqrt{5} \] Thus, the unit normal vector is: \[ \mathbf{N}(2) = \frac{\mathbf{r}''(2)}{\|\mathbf{r}''(2)\|} = \frac{2 \hat{i} + 0 \hat{j} + 4 \hat{k}}{2\sqrt{5}} = \frac{1}{\sqrt{5}}(2 \hat{i} + 0 \hat{j} + 4 \hat{k}) \] ### Summary of Results The unit tangent vector at \( t=2 \) is: \[ \mathbf{T}(2) = \frac{2}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{1}{3} \hat{k} \] The unit normal vector at \( t=2 \) is: \[ \mathbf{N}(2) = \frac{1}{\sqrt{5}}(2 \hat{i} + 0 \hat{j} + 4 \hat{k}) \] Thus, the final answers are: \[ \mathbf{T}(2) = \frac{1}{3}(2 \hat{i} + 2 \hat{j} +