Question

Prove the identity, assuming that the appropriate partial derivatives exist and are continuous. If f is a scalar field and F, G are vector fields, then fF, F.G , and F X G are defined by

\(\begin{array}{l}(fF)(x,y,z) = f(x,y,z)F(x,y,z)\\(F \times G)(x,y,z) = F(x,y,z) \times G(x,y,z)\\(F \times G)(x,y,z) = F(x,y,z) \times G(x,y,z)\end{array}\)

\(24.\;curl(fF) = f\;curl\;\,F + (\nabla f) \times F\)

Short answer

The given expression is proved

Step-by-step solution

The data available

Let F and G vector fields given by,

\(F(x,y,z) = {F_1}i + {F_2}j + {F_3}k\)

We want to show

\(curl(fF) = f\;curl\;F + F \times (\nabla f)\)

Simplification

\(\begin{array}{c}curl(fF) = curl\left( {f{F_1}i + f{F_2}j + f{F_3}k} \right)\\ = \left( {\frac{{\partial f{F_3}}}{{\partial y}} - \frac{{\partial f{F_2}}}{{\partial z}}} \right)i - \left( {\frac{{\partial f{F_3}}}{{\partial x}} - \frac{{\partial f{F_1}}}{{\partial z}}} \right)j + \left( {\frac{{\partial f{F_2}}}{{\partial z}} - \frac{{\partial f{F_1}}}{{\partial y}}} \right)k\\ = \left( {\left( {f\frac{{\partial {F_3}}}{{\partial y}} + {F_3}\frac{{\partial f}}{{\partial y}}} \right) - \left( {f\frac{{\partial {F_2}}}{{\partial z}} + {F_2}\frac{{\partial f}}{{\partial z}}} \right)} \right)i - \left( {\left( {f\frac{{\partial {F_3}}}{{\partial x}} + {F_3}\frac{{\partial f}}{{\partial x}}} \right) - \left( {f\frac{{\partial {F_1}}}{{\partial z}} + {F_1}\frac{{\partial f}}{{\partial z}}} \right)} \right)j + \\\left( {\left( {f\frac{{\partial {F_2}}}{{\partial x}} + {F_2}\frac{{\partial f}}{{\partial x}}} \right) - \left( {f\frac{{\partial {F_1}}}{{\partial y}} + {F_1}\frac{{\partial f}}{{\partial y}}} \right)} \right)k\end{array}\)

Rearrangement and further Simplification

\(\begin{array}{c}curl(fF) = \left( {{F_3}\frac{{\partial f}}{{\partial y}} - {F_2}\frac{{\partial f}}{{\partial z}}} \right)i - \left( {{F_3}\frac{{\partial f}}{{\partial x}} - {F_1}\frac{{\partial f}}{{\partial z}}} \right)j + \left( {{F_2}\frac{{\partial f}}{{\partial x}} - {F_1}\frac{{\partial f}}{{\partial y}}} \right)k\\ + \left( {f\frac{{\partial {F_3}}}{{\partial y}} - f\frac{{\partial {F_2}}}{{\partial z}}} \right)i - \left( {f\frac{{\partial {F_3}}}{{\partial x}} - f\frac{{\partial {F_1}}}{{\partial z}}} \right)j + \left( {f\frac{{\partial {F_2}}}{{\partial z}} - f\frac{{\partial {F_1}}}{{\partial y}}} \right)k\end{array}\)

Let us take f common

\(\begin{array}{c}curl(fF) = \left( {{F_3}\frac{{\partial f}}{{\partial y}}i - {F_2}\frac{{\partial f}}{{\partial z}}i} \right) - \left( {{F_3}\frac{{\partial f}}{{\partial x}}j - {F_1}\frac{{\partial f}}{{\partial z}}j} \right) + \left( {{F_2}\frac{{\partial f}}{{\partial x}}k - {F_1}\frac{{\partial f}}{{\partial y}}k} \right)\\ + f\left\{ { + \left( {\frac{{\partial {F_3}}}{{\partial y}} - \frac{{\partial {F_2}}}{{\partial z}}} \right)i - \left( {\frac{{\partial {F_3}}}{{\partial x}} - \frac{{\partial {F_1}}}{{\partial z}}} \right)j + \left( {\frac{{\partial {F_2}}}{{\partial z}} - \frac{{\partial {F_1}}}{{\partial y}}} \right)k} \right\}\end{array}\)

Group the terms with same coefficients

\(curl(fF) = {F_3}\left( {\frac{{\partial f}}{{\partial y}}i - \frac{{\partial f}}{{\partial x}}j} \right) + {F_1}\left( {\frac{{\partial f}}{{\partial z}}j - \frac{{\partial f}}{{\partial y}}k} \right) + {F_2}\left( {\frac{{\partial f}}{{\partial x}}k - \frac{{\partial f}}{{\partial z}}i} \right) + f\;curl\;F\)

Equation as determinant

\(curl(fF) = {F_3}\left| {\begin{array}{*{20}{c}}i&j\\{\frac{{\partial f}}{{\partial x}}}&{\frac{{\partial f}}{{\partial y}}}\end{array}} \right| + {F_1}\left| {\begin{array}{*{20}{c}}j&k\\{\frac{{\partial f}}{{\partial y}}}&{\frac{{\partial f}}{{\partial z}}}\end{array}} \right| + {F_2}\left| {\begin{array}{*{20}{c}}k&i\\{\frac{{\partial f}}{{\partial z}}}&{\frac{{\partial f}}{{\partial z}}}\end{array}} \right| + f\;curl\;F\)

Rewrite the determinant terms so that we can group them into three into three determinant.

\( = {F_1}\left| {\begin{array}{*{20}{c}}j&k\\{\frac{{\partial f}}{{\partial y}}}&{\frac{{\partial f}}{{\partial z}}}\end{array}} \right| - {F_2}\left| {\begin{array}{*{20}{c}}k&i\\{\frac{{\partial f}}{{\partial x}}}&{\frac{{\partial f}}{{\partial z}}}\end{array}} \right| + {F_3}\left| {\begin{array}{*{20}{c}}i&j\\{\frac{{\partial f}}{{\partial x}}}&{\frac{{\partial f}}{{\partial y}}}\end{array}} \right| + f\;curl\;F\)

Determinant and it’s solution

We can write

\(curl(fF) = \left| {\begin{array}{*{20}{c}}i&j&k\\{\frac{{\partial f}}{{\partial x}}}&{\frac{{\partial f}}{{\partial y}}}&{\frac{{\partial f}}{{\partial z}}}\\{{F_1}}&{{F_2}}&{{F_3}}\end{array}} \right| + f\;curl\;F\)

\(\begin{array}{c}curl(fF) = \left( {\frac{{\partial f}}{{\partial x}}i + \frac{{\partial f}}{{\partial y}}j + \frac{{\partial f}}{{\partial z}}k} \right) \times \left( {{F_1}i + {F_2}j + {F_3}k} \right) + f\;curl\;F\\ = \nabla f \times F + f\;curl\;F\end{array}\)

Which is the required equation to be proved.