Compute the value of \(\int_{A}^{P} \mathbf{G} \cdot d \mathbf{L}\) for \(\mathbf{G}=2 y \mathbf{a}_{x}\) with \(A(1,-1,2)\) and \(P(2,1,2)\) using the path ( \(a\) ) straight-line segments \(A(1,-1,2)\) to \(B(1,1,2)\) to \(P(2,1,2) ;(b)\) straight-line segments \(A(1,-1,2)\) to \(C(2,-1,2)\) to \(P(2,1,2)\)

#tag_title#Step 2: Compute the line integral #tag_content#First, let's find the derivatives of the parametric paths: \(d\mathbf{r}_1 = \begin{bmatrix} 0 \\ dt \\ 0 \end{bmatrix}\), \(d\mathbf{r}_2 = \begin{bmatrix} dt \\ 0 \\ 0 \end{bmatrix}\), \(d\mathbf{r}_3 = \begin{bmatrix} dt \\ 0 \\ 0 \end{bmatrix}\), and \(d\mathbf{r}_4 = \begin{bmatrix} 0 \\ dt \\ 0 \end{bmatrix}\) Now, let's compute the line integral for both paths: For path a: \(\int_{A}^{P} \mathbf{G} \cdot d \mathbf{L} = \int_{A}^{B} \mathbf{G} \cdot d \mathbf{L} + \int_{B}^{P} \mathbf{G} \cdot d \mathbf{L}\) \(= \int_{0}^{2}(2(-1+t)\mathbf{a}_x) \cdot \left(\begin{bmatrix} 0 \\ dt \\ 0 \end{bmatrix}\right) + \int_{0}^{1}(2\mathbf{a}_x) \cdot \left(\begin{bmatrix} dt \\ 0 \\ 0 \end{bmatrix}\right)\) \(= 2\int_{0}^{2} (-1+t)dt + 2\int_{0}^{1} dt\) \(= 2[-t + \frac{1}{2}t^2]_0^2 + 2[t]_0^1\) \(= 2[-4 + 4] + 2[1] = 0 + 2 = 2\) For path b: \(\int_{A}^{P} \mathbf{G} \cdot d \mathbf{L} = \int_{A}^{C} \mathbf{G} \cdot d \mathbf{L} + \int_{C}^{P} \mathbf{G} \cdot d \mathbf{L}\) \(= \int_{0}^{1}(2y\mathbf{a}_x) \cdot \left(\begin{bmatrix} dt \\ 0 \\ 0 \end{bmatrix}\right) + \int_{0}^{2}(4\mathbf{a}_x) \cdot \left(\begin{bmatrix} 0 \\ dt \\ 0 \end{bmatrix}\right)\) \(= 2\int_{0}^{1} (-1)dt + 4\int_{0}^{2} dt\) \(= 2[-t]_0^1 + 4[t]_0^2\) \(= 2[-1] + 4[2] = -2 + 8 = 6\) So, for path a, the line integral \(\int_{A}^{P} \mathbf{G} \cdot d \mathbf{L} = 2\) and for path b, the line integral \(\int_{A}^{P} \mathbf{G} \cdot d \mathbf{L} = 6\).