The question asks for the angle between the tangents to the vector curve $\vec{R} = t^2\hat{i} + 2t\hat{j}$ at the points where $t = 1$ and $t = -1$. We can find this angle by calculating the tangent vectors at these points and then using the dot product formula.

Tangent Vector Calculation

The tangent vector $\vec{T}(t)$ to the curve $\vec{R}(t)$ is its derivative with respect to the parameter $t$.
Given curve:
$\vec{R}(t) = t^2\hat{i} + 2t\hat{j}$
Differentiating with respect to $t$:
$\vec{T}(t) = \frac{d\vec{R}}{dt} = \frac{d}{dt}(t^2)\hat{i} + \frac{d}{dt}(2t)\hat{j}$
$\vec{T}(t) = 2t\hat{i} + 2\hat{j}$

Tangent Vectors at $t = \pm 1$

• Tangent vector at $t = 1$:
  $\vec{T}(1) = 2(1)\hat{i} + 2\hat{j} = 2\hat{i} + 2\hat{j}$
• Tangent vector at $t = -1$:
  $\vec{T}(-1) = 2(-1)\hat{i} + 2\hat{j} = -2\hat{i} + 2\hat{j}$

Angle Between Tangent Vectors

Let $\vec{a} = \vec{T}(1) = 2\hat{i} + 2\hat{j}$ and $\vec{b} = \vec{T}(-1) = -2\hat{i} + 2\hat{j}$.
The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by the formula:
$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$

• Calculate the dot product $\vec{a} \cdot \vec{b}$:
  $\vec{a} \cdot \vec{b} = (2)(-2) + (2)(2) = -4 + 4 = 0$
• Calculate the magnitudes $|\vec{a}|$ and $|\vec{b}|$:
  $|\vec{a}| = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8}$
  $|\vec{b}| = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8}$
• Calculate $\cos \theta$:
  $\cos \theta = \frac{0}{\sqrt{8} \times \sqrt{8}} = \frac{0}{8} = 0$
• Find the angle $\theta$:
  Since $\cos \theta = 0$, the angle $\theta$ is $\frac{\pi}{2}$.

Therefore, the angle between the tangents to the curve at $t = \pm 1$ is $\frac{\pi}{2}$.