With
[tex]\vec r(t)=4t\,\vec\imath+6t\,\vec\jmath-t^2\,\vec k[/tex]
we have
[tex]\mathrm d\vec r=(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt[/tex]
The vector field evaluated over this parameterization is
[tex]\vec f(x,y,z)=\vec f(x(t),y(t),z(t))=4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k[/tex]
so the line integral is
[tex]\displaystyle\int_{-1}^1(4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k)\cdot(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt[/tex]
[tex]=\displaystyle\int_{-1}^1(16t+6t^2-12t^2)\,\mathrm dt=-4[/tex]
Applying the steps, the result of the line integral is -4.
The curve is:
[tex]f(x,y,z) = xi - zj + yk[/tex]
The vector field is:
[tex]r(t) = (x(t), y(t), z(t)) = (4t, 6t, -t^2)[/tex]
Applying the vector field at the curve, we have that:
[tex]f(t) = (4t, t^2, 6t)[/tex]
The derivative of the vector field is:
[tex]r^{\prime}(t) = (4, 6, -2t)[/tex]
The dot product of the vector field along the curve with the derivative is:
[tex]f(t)r^{\prime}(t) = (4t, t^2, 6t)(4, 6, -2t) = 16t + 6t - 12t^2 = -6t^2 + 16t[/tex]
Hence, the line integral is:
[tex]I = \int_{-1}^{1} (-6t^2 + 16t) dt[/tex]
[tex]I = -2t^3 + 8t^2|_{t = -1}^{t = 1}[/tex]
[tex]I = -2 + 8 -2 - 8[/tex]
[tex]I = -4[/tex]
A similar problem is given at https://brainly.com/question/12666512
