[tex]\vec r(t)=14t^4\,\vec\imath+t^6\,\vec\jmath[/tex]
[tex]\mathrm d\vec r=(56t^3\,\vec\imath+6t^5\,\vec\jmath)\,\mathrm dt[/tex]
[tex]\vec f(x,y)=xy\,\vec\imath+6y^2\,\vec\jmath\implies\vec f(x(t),y(t))=14t^{10}\,\vec\imath+6t^{12}\,\vec\jmath[/tex]
Then the line integral is
[tex]\displaystyle\int_C\vec f\cdot\mathrm d\vec r=\int_0^1(14t^{10}\,\vec\imath+6t^{12}\,\vec\jmath)\cdot(56t^3\,\vec\imath+6t^5\,\vec\jmath)\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^1(36t^{17}+784t^{13})\,\mathrm dt=\boxed{58}[/tex]
Applying each separate step, it is found that the result of the line integral along the vector field is given by 58.
The curve is:
[tex]f(x,y) = xyi + 6y^2j[/tex]
The vector field is:
[tex]r(t) = (x(t), y(t)) = (14t^4, t^6)[/tex]
Applying the vector field at the curve, we have that:
[tex]f(t) = (14t^{10}, 6t^{12})[/tex]
The derivative of the vector field is:
[tex]r^{\prime}(t) = (56t^3, 6t^5)[/tex]
The dot product of the vector field along the curve with the derivative is:
[tex]f(t)r^{\prime}(t) = (14t^{10}, 6t^{12})(56t^3, 6t^5) = 784t^{13} + 36t^{17}[/tex]
Hence, the line integral is:
[tex]\int_{0}^{1} (784t^{13} + 36t^{17}) dt[/tex]
[tex]56t^{14} + 2t^{18}|_{t = 0}^{t = 1}[/tex]
[tex]56 + 2 = 58[/tex]
The result of the line integral is of 58.
A similar problem is given at https://brainly.com/question/16229252
