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. **Find the position vector** \( \mathbf{r}(t) \). 2. **Calculate the derivative** \( \mathbf{r}'(t) \) to find the tangent vector. 3. **Normalize the tangent vector** to get the unit tangent vector \( \mathbf{T}(t) \). 4. **Find the derivative of the unit tangent vector** to get the normal vector. 5. **Normalize the normal vector** to get the unit normal vector \( \mathbf{N}(t) \). ### Step 1: Position Vector The position vector \( \mathbf{r}(t) \) is given by: \[ \mathbf{r}(t) = (t^2 - 1) \hat{i} + (4t - 3) \hat{j} + (2t^2 - 6t) \hat{k} \] ### Step 2: Derivative of the Position Vector Now, we calculate the derivative \( \mathbf{r}'(t) \): \[ \mathbf{r}'(t) = \frac{d}{dt}[(t^2 - 1) \hat{i} + (4t - 3) \hat{j} + (2t^2 - 6t) \hat{k}] \] Calculating each component: - \( \frac{d}{dt}(t^2 - 1) = 2t \) - \( \frac{d}{dt}(4t - 3) = 4 \) - \( \frac{d}{dt}(2t^2 - 6t) = 4t - 6 \) Thus, \[ \mathbf{r}'(t) = (2t) \hat{i} + (4) \hat{j} + (4t - 6) \hat{k} \] ### Step 3: Evaluate at \( t=2 \) Now, we evaluate \( \mathbf{r}'(2) \): \[ \mathbf{r}'(2) = (2 \cdot 2) \hat{i} + (4) \hat{j} + (4 \cdot 2 - 6) \hat{k} = 4 \hat{i} + 4 \hat{j} + 2 \hat{k} \] ### Step 4: Unit Tangent Vector Next, we find the magnitude of \( \mathbf{r}'(2) \): \[ \|\mathbf{r}'(2)\| = \sqrt{4^2 + 4^2 + 2^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6 \] Now, 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 5: Derivative of the Unit Tangent Vector Next, we need to find the derivative of the unit tangent vector \( \mathbf{T}(t) \). We can differentiate \( \mathbf{r}'(t) \) again to find the acceleration vector \( \mathbf{r}''(t) \): \[ \mathbf{r}''(t) = \frac{d}{dt}[(2t) \hat{i} + (4) \hat{j} + (4t - 6) \hat{k}] = (2) \hat{i} + (0) \hat{j} + (4) \hat{k} \] Evaluating at \( t=2 \): \[ \mathbf{r}''(2) = 2 \hat{i} + 0 \hat{j} + 4 \hat{k} \] ### Step 6: Unit Normal Vector To find the unit normal vector \( \mathbf{N}(t) \), we first need to find the magnitude of \( \mathbf{r}''(2) \): \[ \|\mathbf{r}''(2)\| = \sqrt{2^2 + 0^2 + 4^2} = \sqrt{4 + 0 + 16} = \sqrt{20} = 2\sqrt{5} \] Now, the unit normal vector \( \mathbf{N}(2) \) is: \[ \mathbf{N}(2) = \frac{\mathbf{r}''(2)}{\|\mathbf{r}''(2)\|} = \frac{2 \hat{i} + 0 \hat{j} + 4 \hat{k}}{2\sqrt{5}} = \frac{1}{\sqrt{5}} \hat{i} + 0 \hat{j} + \frac{2}{\sqrt{5}} \hat{k} \] ### Final Answer Thus, the unit tangent vector and unit normal vector at \( t=2 \) are: \[ \mathbf{T}(2) = \frac{1}{3}(2 \hat{i} + 2 \hat{j} + \hat{k}), \quad \mathbf{N}(2) = \frac{1}{3\sqrt{5}}(2 \hat{i} + 0 \hat{j} + 2 \hat{k}) \] This matches the provided answer.