Question

Show that the operation of taking the gradient of a function has the given property. Assume that \(u\) and \(v\) are differentiable functions of \(x\) and \(y\) and that \(a,b\) are constants.

(a)\(\nabla (au + bv) = a\nabla u + b\nabla v\)

(b) \(\nabla (uv) = u\nabla v + v\nabla u\)

(c) \(\nabla \left( {\frac{u}{v}} \right) = \frac{{v\nabla u - u\nabla v}}{{{v^2}}}\)

(d) \(\nabla {u^n} = n{u^{n - 1}}\nabla u\)

Short answer

(a) It is proved that \(\nabla (au + bv) = a\nabla u + b\nabla v\).

(b) \(\nabla (uv) = u\nabla v + v\nabla u\) is proved.

(c) \(\nabla \left( {\frac{u}{v}} \right) = \frac{{v\nabla u - u\nabla v}}{{{v^2}}}\) is proved.

(d) \(\nabla \left( {{u^n}} \right) = n{u^{n - 1}}\nabla u\) is proved.

Step-by-step solution

Definition of gradient

Gradient is change in the value of a quantity (as temperature, pressure, or concentration) with change in a given variable and especially per unit on a linear scale.

Compute the value of \(\nabla (au + bv)\)

\(u\)and \(v\) are the functions of \(x\) and \(y\).

\(a\)and \(b\) are constants.

Consider the equation \(\nabla (au + bv)\).

Compute the value of \(\nabla (au + bv)\) as follows,

\(\)\(\begin{aligned}{l}\nabla (au + bv) = \left\langle {\frac{\partial }{{\partial x}}(au + bv),\frac{\partial }{{\partial y}}(au + bv)} \right\rangle \\ = \left\langle {\frac{\partial }{{\partial x}}(au) + \frac{\partial }{{\partial x}}(bv),\frac{\partial }{{\partial y}}(au) + \frac{\partial }{{\partial y}}(bv)} \right\rangle \\ = \left\langle {a\frac{{\partial u}}{{\partial x}} + b\frac{{\partial v}}{{\partial x}},a\frac{{\partial u}}{{\partial y}} + b\frac{{\partial v}}{{\partial y}}} \right\rangle \\ = a\left\langle {\frac{{\partial u}}{{\partial x}},\frac{{\partial u}}{{\partial y}}} \right\rangle + b\left\langle {\frac{{\partial v}}{{\partial x}},\frac{{\partial v}}{{\partial y}}} \right\rangle \\ = a\nabla u + b\nabla v\end{aligned}\)

Thus, the value of \(\nabla (au + bv)\) is \(a\nabla u + b\nabla v\).

Hence, \(\nabla (au + bv) = a\nabla u + b\nabla v\) is proved.

Compute the value of \(\nabla (uv)\)

(b)

\(u\)and \(v\) are the functions of \(x\) and \(y\).

\(a\)and \(b\) are constants.

Consider the equation \(\nabla (uv)\).

Compute the value of \(\nabla (uv)\) as follows,

\(\begin{aligned}{l}\nabla (uv) = \left\langle {\frac{\partial }{{\partial x}}(uv),\frac{\partial }{{\partial y}}(uv)} \right\rangle \\ = \left\langle {v\frac{{\partial u}}{{\partial x}} + u\frac{{\partial v}}{{\partial x}},v\frac{{\partial u}}{{\partial y}} + u\frac{{\partial v}}{{\partial y}}} \right\rangle \\ = v\left\langle {\frac{{\partial u}}{{\partial x}},\frac{{\partial u}}{{\partial y}}} \right\rangle + u\left\langle {\frac{{\partial v}}{{\partial x}},\frac{{\partial v}}{{\partial y}}} \right\rangle \\ = v\nabla u + u\nabla v\end{aligned}\)

Thus, the value of \(\nabla (uv)\) is \(u\nabla v + v\nabla u\).

Hence, \(\nabla (uv) = u\nabla v + v\nabla u\) is proved.

Compute the value of \(\nabla \left( {\frac{u}{v}} \right)\)

(c)

\(u\)and \(v\) are the functions of \(x\) and \(y\).

\(a\)and \(b\) are constants.

Calculation:

Consider the equation \(\nabla \left( {\frac{u}{v}} \right)\).

Compute the value of \(\nabla \left( {\frac{u}{v}} \right)\) as follows,

\(\begin{aligned}{l}\nabla \left( {\frac{u}{v}} \right) = \left\langle {\frac{\partial }{{\partial x}}\left( {\frac{u}{v}} \right),\frac{\partial }{{\partial y}}\left( {\frac{u}{v}} \right)} \right\rangle \\ = \left\langle {\frac{{v\frac{{du}}{{dx}} - u\frac{d}{d}}}{{{v^2}}},\frac{{v\frac{{du}}{{\partial y}} - u\frac{{\partial v}}{{dy}}}}{{{v^2}}}} \right\rangle \\ = \frac{{v\left\langle {\frac{{du}}{{dx}},\frac{{du}}{{dy}}} \right\rangle - u\left\langle {\frac{{dx}}{{dx}},\frac{{du}}{{dy}}} \right\rangle }}{{{v^2}}}\\ = \frac{{v\nabla u - u\nabla v}}{{{v^2}}}\end{aligned}\)

Thus, the value of \(\nabla \left( {\frac{u}{v}} \right)\) is \(\frac{{v\nabla u - u\nabla v}}{{{v^2}}}\).

Hence, \(\nabla \left( {\frac{u}{v}} \right) = \frac{{v\nabla u - u\nabla v}}{{{v^2}}}\) is proved.

Compute the value of \(\nabla \left( {{u^n}} \right)\) 

(d)

\(u\)and \(v\) are the functions of \(x\) and \(y\).

\(a\)and \(b\) are constants.

Calculation:

Consider the equation \(\nabla \left( {{u^n}} \right)\).

Compute the value of \(\nabla \left( {{u^n}} \right)\) as follows,

\(\begin{aligned}{l}\nabla \left( {{u^n}} \right) = \left\langle {\frac{\partial }{{\partial x}}\left( {{u^n}} \right),\frac{\partial }{{\partial y}}\left( {{u^n}} \right)} \right\rangle \\ = \left\langle {n{u^{n - 1}}\frac{{\partial u}}{{\partial x}},n{u^{n - 1}}\frac{{\partial u}}{{\partial y}}} \right\rangle \\ = n{u^{n - 1}}\left\langle {\frac{{\partial u}}{{\partial x}},\frac{{\partial u}}{{\partial y}}} \right\rangle \\ = n{u^{n - 1}}\nabla u\end{aligned}\)

Thus, the value of \(\nabla \left( {{u^n}} \right)\) is \(n{u^{n - 1}}\nabla u\).

Hence, \(\nabla \left( {{u^n}} \right) = n{u^{n - 1}}\nabla u\) is proved.