Question

If \({\rm{F}}\) and \({\rm{G}}\) are vector fields, then \({\rm{curl (F + G) = curl F + curl G}}\).

Short answer

The given statement is true.

Step-by-step solution

Explanation of solution. 

Let the vectors are \({\rm{F = }}{{\rm{F}}_{\rm{1}}}{\rm{i + }}{{\rm{F}}_{\rm{2}}}{\rm{j + }}{{\rm{F}}_{\rm{3}}}{\rm{k}}\) and \({\rm{G = }}{{\rm{G}}_{\rm{1}}}{\rm{i + }}{{\rm{G}}_{\rm{2}}}{\rm{j + }}{{\rm{G}}_{\rm{3}}}{\rm{k}}\).

property of determinants.

Curf is find as,

\(\begin{aligned}\rm curl(F + G) &= curl\left( {\left( {{{\rm{F}}_{\rm{1}}}{\rm{ + }}{{\rm{G}}_{\rm{1}}}} \right){\rm{i + }}\left( {{{\rm{F}}_{\rm{2}}}{\rm{ + }}{{\rm{G}}_{\rm{2}}}} \right){\rm{j + }}\left( {{{\rm{F}}_{\rm{3}}}{\rm{ + }}{{\rm{G}}_{\rm{3}}}} \right){\rm{k}}} \right)\\ &= \left| {\begin{aligned}{{}{}}{\bf{i}}&{\bf{j}}&{\bf{k}}\\{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\{\left( {{{\rm{F}}_{\rm{1}}}{\rm{ + }}{{\rm{G}}_{\rm{1}}}} \right)}&{\left( {{{\rm{F}}_{\rm{2}}}{\rm{ + }}{{\rm{G}}_{\rm{2}}}} \right)}&{\left( {{{\rm{F}}_{\rm{3}}}{\rm{ + }}{{\rm{G}}_{\rm{3}}}} \right)}\end{aligned}} \right|\end{aligned}\)

Using the determinant's property,

Distributive property.

Remember \(\text{curla=}\nabla \text{ }\!\!*\!\!\text{ a}\)

\(\text{curl(F+G)=}\nabla \text{ }\!\!*\!\!\text{ (F+G)}\)

The cross product's distributive feature allows writing

\(\begin{aligned}curl(F + G) &= \nabla * (F + G) \\ &= \nabla * F + \nabla * {\text{ }}G \\ &= curl{\text{ }}F + curl{\text{ }}G \\ \end{aligned} \)

Therefore The given statement is true