Examples A

Section 1.3 Examples A

Vectors.

Example 1.3.1.

Prove that the linear velocity \(v\) of a point \(P\) in figure below of a rigid body rotating about an axis passing through a point \(O\) is given by \(\vec{v} = \vec{\omega}\times \vec{r}\text{,}\) where \(\vec{\omega} \) and \(\vec{r}\) are an angular velocity of a body and a position vector of point \(P\text{,}\) respectively.

Solution.

Let \(\vec{OP} = \vec{r}\text{,}\) \(\angle AOP = \theta,\text{,}\) and \(AP\) is perpendicular to \(OA\text{,}\) as shown in figure. Suppose the velocity of the body at point P be \(\vec{v}\) and \(\hat{n}\) be a unit vector perpendicular to \(\omega\) and \(\vec{r}\text{.}\) Then

\begin{equation*} \vec{\omega}\times\vec{r} = \left(\omega r \sin\theta\right)\hat{n} = \left(\omega AP\right)\hat{n} = \text{(speed of P)} \hat{n} \end{equation*}

= velocity of P perpendicular to \(\vec{\omega}\) and \(\vec{r} = \vec{v}\text{.}\)

Hence, \(\vec{v} = \vec{\omega}\times\vec{r}\text{.}\) proved.

Example 1.3.2.

Show that the vectors \(\vec{a}-2\vec{b}+3\vec{c}\text{,}\) \(-2\vec{a}+3\vec{b}-4\vec{c}\text{,}\) and \(-\vec{b}+2\vec{c}\) are coplanar. Where \(\vec{a}\text{,}\) \(\vec{b}\text{,}\) and \(\vec{c}\) are non-coplanar.

Solution.

Let

\begin{equation*} \vec{\alpha} = \vec{a}-2\vec{b}+3\vec{c}{;} \vec{\beta} = -2\vec{a}+3\vec{b}-4\vec{c}{;} \text{and}\quad \vec{\gamma} = -\vec{b}+2\vec{c} \end{equation*}

then,

\begin{equation*} \vec{\alpha}\cdot (\vec{\beta}\times \vec{\gamma}) = \begin{Vmatrix} 1 & -2 & 3 \\ -2 & 3 & -4 \\ 0 & -1 & 2 \end{Vmatrix} \end{equation*}

\begin{equation*} = 1(6-4)+2(-4)+3(2) =2-8+6 = 0. \end{equation*}

Hence, \(\vec{\alpha}\text{,}\) \(\vec{\beta}\text{,}\) and \(\vec{\gamma}\) are coplanar.

Example 1.3.3.

If \(\hat {a}\times{(\hat {b}\times\hat {c})} = \frac{1}{2}\hat{b}\text{,}\) find the angles that \(\hat{a}\) makes with \(\hat{b}\) and \(\hat{c}\text{.}\)

Solution.

\begin{equation*} \hat {a}\times{(\hat {b}\times\hat {c})} = \frac{1}{2}\hat{b} \end{equation*}

or,

\begin{equation*} \left(\hat{a}\cdot\hat{c}\right)\hat{b} - \left(\hat{a}\cdot\hat{b}\right)\hat{c} = \frac{1}{2}\hat{b} \end{equation*}

equating the coefficients of \(\hat{b}\) and \(\hat{c}\) , we get-

\begin{equation*} \hat{a}\cdot\hat{c} = \frac{1}{2} = 1.1.\cos\frac{\pi}{3} \end{equation*}

and

\begin{equation*} \hat{a}\cdot\hat{b} = 0 \end{equation*}

\(\Rightarrow \hat{a} \hspace{3pt} \) makes \(60^{o}\) angle with \(\hat{c}\text{,}\) and \(\hat{a}\) is \(\perp\) on \(\hat{b}\text{.}\)

Example 1.3.4.

If \(\vec{a}\text{,}\) \(\vec{b}\text{,}\) and \(\vec{c}\) are a reciprocal system to \(\vec{a'}\text{,}\) \(\vec{b'}\text{,}\) \(\vec{c'}\) then prove that \(\left[\vec{a'}\vec{b'}\vec{c'}\right] \left[\vec{a}\vec{b}\vec{c}\right] = 1\text{.}\)

Solution.

\begin{equation*} \left[\vec{a'}\vec{b'}\vec{c'}\right] = \vec{a'}\cdot\left(\vec{b'}\times\vec{c'}\right) \end{equation*}

\begin{equation} \therefore \left[\vec{a'}\vec{b'}\vec{c'}\right] = \frac{\left(\vec{b}\times\vec{c}\right)}{\left[\vec{a}\vec{b}\vec{c}\right]} \cdot\frac{\left[\left(\vec{c}\times\vec{a}\right)\times \left(\vec{a}\times\vec{b}\right)\right]}{\left[\vec{a}\vec{b}\vec{c}\right]^{2}} \tag{1.3.1} \end{equation}

Now,

\begin{align*} \left(\vec{c}\times\vec{a}\right)\times \left(\vec{a}\times\vec{b}\right) \amp = \vec{d}\times \left(\vec{a}\times\vec{b}\right), \\ \text{where}\quad \vec{d} \amp = \left(\vec{c}\times\vec{a}\right) =\left(\vec{d}\cdot\vec{b}\right)\vec{a} - \left(\vec{d}\cdot\vec{a}\right)\vec{b} \\ \amp = \lbrace\vec{c}\times\vec{a}\cdot \vec{b}\rbrace\vec{a} - \lbrace\vec{c}\times\vec{a}\cdot \vec{a}\rbrace\vec{b} \\ \amp = \left[\vec{a}\vec{b}\vec{c}\right]\vec{a} - \left[\vec{c}\vec{a}\vec{a}\right]\vec{b} = \left[\vec{a}\vec{b}\vec{c}\right]\vec{a} \end{align*}

\(\because \left[\vec{c}\vec{a}\vec{a}\right] = 0.\) From equation (1.3.1), we have -

\begin{align*} \left[\vec{a'}\vec{b'}\vec{c'}\right] \amp = \frac{\left(\vec{b}\times\vec{c}\right)\cdot \left[\vec{a}\vec{b}\vec{c}\right]\vec{a} }{\left[\vec{a}\vec{b}\vec{c}\right]^{3}} = \frac{\left[\vec{a}\vec{b}\vec{c}\right]\left(\vec{b}\times\vec{c}\right) \cdot\vec{a}}{\left[\vec{a}\vec{b}\vec{c}\right]^{3}} \\ \amp = \frac{\left[\vec{a}\vec{b}\vec{c}\right] \left[\vec{a}\vec{b}\vec{c}\right]}{\left[\vec{a}\vec{b}\vec{c}\right]^{3}} = \frac{1}{\left[\vec{a}\vec{b}\vec{c}\right]} \\ \therefore \quad \left[\vec{a'}\vec{b'}\vec{c'}\right] \left[\vec{a}\vec{b}\vec{c}\right] \amp = 1. \hspace{2cm} \text{proved.} \end{align*}

Note: The system of unit vectors \(\hat{i}\text{,}\) \(\hat{j}\text{,}\) and \(\hat{k}\) has its own reciprocal.

Example 1.3.5.

Prove that

\begin{equation*} \left[\vec{l}\vec{m}\vec{n}\right]\left[\vec{a}\vec{b}\vec{c}\right] = {\begin{Vmatrix} \vec{l}\cdot \vec{a} & \vec{l}\cdot \vec{b} & \vec{l}\cdot \vec{c} \\ \vec{m}\cdot \vec{a} & \vec{m}\cdot \vec{b} & \vec{m}\cdot \vec{c} \\ \vec{n}\cdot \vec{a} & \vec{n}\cdot \vec{b} & \vec{n}\cdot \vec{c} \end{Vmatrix}} \end{equation*}

Solution.

Let,

\begin{align*} \vec{a} \amp = a_{1}\hat{i}+a_{2}\hat{j}+ a_{3}\hat{k}; \ quad \vec{b} = b_{1}\hat{i}+b_{2}\hat{j}+ b_{3}\hat{k}; \\ and, \vec{c} \amp = c_{1}\hat{i}+c_{2}\hat{j}+ c_{3}\hat{k}\\ \vec{l} \amp = l_{1}\hat{i}+l_{2}\hat{j}+ l_{3}\hat{k}, \quad \vec{m} = m_{1}\hat{i}+m_{2}\hat{j}+ m_{3}\hat{k}, \\ and, \vec{n}\amp = n_{1}\hat{i}+n_{2}\hat{j}+ n_{3}\hat{k}, \end{align*}

then,

\begin{equation*} \vec{l}\cdot \vec{a} = \left(l_{1}\hat{i}+l_{2}\hat{j}+ l_{3}\hat{k}\right) \cdot \left(a_{1}\hat{i}+a_{2}\hat{j}+ a_{3}\hat{k}\right) = l_{1}a_{1} + l_{2}a_{2} + l_{3}a_{3} \end{equation*}

similarly, we can find the value of \(\vec{l}\cdot \vec{b}\text{,}\) \(\vec{l}\cdot \vec{c}\text{,}\) and \(\vec{m}\cdot \vec{a}\) etc. Now,

\begin{align*} \left[\vec{l}\vec{m}\vec{n}\right]\left[\vec{a}\vec{b}\vec{c}\right] \amp = {\begin{Vmatrix} l_{1} & l_{2} & l_{3} \\ m_{1} & m_{2} & m_{3} \\ n_{1} & n_{2} & n_{3} \end{Vmatrix}} {\begin{Vmatrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{Vmatrix}} \\ \amp = {\begin{Vmatrix} l_{1} & l_{2} & l_{3} \\ m_{1} & m_{2} & m_{3} \\ n_{1} & n_{2} & n_{3} \end{Vmatrix}} {\begin{Vmatrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{Vmatrix}} \end{align*}

[using properties of determinant where rows, (or, column) can be exchanged without changing the value of determinant.]

\begin{equation*} ={\begin{Vmatrix} l_{1} a_{1} + l_{2} a_{2} + l_{3} a_{3} & l_{1} b_{1} + l_{2} b_{2} + l_{3} b_{3} & l_{1} c_{1} + l_{2} c_{2} + l_{3} c_{3} \\ m_{1} a_{1} + m_{2} a_{2} + m_{3} a_{3} & m_{1} b_{1} + m_{2} b_{2} + m_{3} b_{3} & m_{1} c_{1} + m_{2} c_{2} + m_{3} c_{3} \\ n_{1} a_{1} + n_{2} a_{2} + n_{3} a_{3} & n_{1} b_{1} + nl_{2} b_{2} + n_{3} b_{3} & n_{1} c_{1} + n_{2} c_{2} + n_{3} c_{3} \end{Vmatrix}} \end{equation*}

\begin{equation*} ={\begin{Vmatrix} \vec{l}\cdot \vec{a} & \vec{l}\cdot \vec{b} & \vec{l}\cdot \vec{c} \\ \vec{m}\cdot \vec{a} & \vec{m}\cdot \vec{b} & \vec{m}\cdot \vec{c} \\ \vec{n}\cdot \vec{a} & \vec{n}\cdot \vec{b} & \vec{n}\cdot \vec{c} \end{Vmatrix}}. \end{equation*}

Field.

Example 1.3.6.

Find the velocity and acceleration of a particle which is moving along the curve \(x=a \cos t\text{,}\) \(y=a \sin t\text{,}\) and \(z= b t\) at time \(t=0\) and \(t=\frac{\pi}{2}\text{.}\)

Solution.

Let

\begin{equation*} \vec{r} = x\hat{i}+ y\hat{j} + z\hat{k} = a \cos t\hat{i}+a \sin t\hat{j}+ b t \hat{k}\text{.} \end{equation*}

Therefore,

\begin{align*} velocity \amp = \frac{\vec{\,dr}}{\,dt} = -a \sin t\hat{i}+a cost\hat{j}+ b \hat{k},\\ \text{and}, acceleration \amp = \frac{{\,d^{2}\vec{r}}}{\,dt^{2}} =\frac{\,d}{\,dt}\left(\frac{\vec{\,dr}}{\,dt} \right)= -a cost\hat{i} - a sint\hat{j}. \end{align*}

At \(t = 0\text{,}\) velocity = \(a\hat{j}+b\hat{k}\) and acceleration = \(-a\hat{i}\text{.}\)

At \(t=\frac{\pi}{2}\text{,}\) velocity = \(-a\hat{i}+b\hat{k}\) and acceleration = \(-a\hat{j}\text{.}\)

Example 1.3.7.

The position of a particle at time \(t\) is

\begin{equation*} \vec{r}=\cos(t-1)\hat{i}+\sinh(t-1)\hat{j}+\alpha t^{3}\hat{k}\text{.} \end{equation*}

Find an acceleration of the particle at time \(t=1\) when it becomes normal to the position vector.

Solution.

\begin{align*} \text{Let} \quad \vec{r} \amp =\cos(t-1)\hat{i}+\sinh (t-1)\hat{j}+\alpha t^{3}\hat{k}\\ \text{or, velocity} \amp = \frac{\vec{dr}}{dt} = - \sin(t-1)\hat{i}+\cosh (t-1)\hat{j}+3\alpha t^{2}\hat{k} \\ \text{or, acceleration } \amp = \frac{{d^{2}\vec{r}}}{dt^{2}} = - \cos(t-1)\hat{i}+\sin(t-1)\hat{j}+6\alpha t\hat{k} \end{align*}

At \(t=1\text{,}\)

\begin{equation*} \vec{a} = \frac{{d^{2}\vec{r}}}{dt^{2}} = -\hat{i}+6\alpha\hat{k}, \end{equation*}

and

\begin{equation*} \vec{r}=\hat{i}+\alpha\hat{k}. \end{equation*}

If \(\vec{r}\) and \(\vec{a}\) are normal, then \(\vec{r} \cdot \vec{a} = 0.\text{.}\)

\begin{equation*} \text{or,}\quad (\hat{i}+\alpha\hat{k})(-\hat{i}+6\alpha\hat{k}) = 0. \end{equation*}

or,

\begin{equation*} -1+6\alpha^{2} =0. \end{equation*}

\begin{equation*} \therefore \quad \alpha = \pm\frac{1}{\sqrt{6}} \hspace{1cm} Ans. \end{equation*}

Example 1.3.8.

If \(f(x,y,z)\) and \(g(x,y,z)\) are two scalar functions. Prove

\begin{equation*} \vec{\nabla} (f\pm g) = \vec{\nabla f} \pm \vec{\nabla g} \end{equation*}
\begin{equation*} \vec{\nabla} (f g) = f \vec{\nabla g} + g \vec{\nabla f} \end{equation*}

Solution.

\begin{align*} \vec{\nabla} (f\pm g) \amp = \left(\frac{\partial}{\partial x} \hat{i} + \frac{\partial}{\partial y} \hat{j} + \frac{\partial}{\partial z} \hat{k}\right) (f\pm g)\\ \amp =\left(\hat{i}\frac{\partial f}{\partial x} + \hat{j}\frac{\partial f}{\partial y} + \hat{k}\frac{\partial f}{\partial z}\right) \pm \left(\hat{i}\frac{\partial g}{\partial x} + \hat{j}\frac{\partial g}{\partial y} + \hat{k}\frac{\partial g}{\partial z}\right)\\ \amp = \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial }{\partial y} + \hat{k}\frac{\partial }{\partial z}\right)f \pm \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z}\right)g \\ \amp = \vec{\nabla f} \pm \vec{\nabla g} \end{align*}
\begin{align*} \vec{\nabla} (f g) \amp = \left(\frac{\partial}{\partial x}\hat{i} +\frac{\partial}{\partial y}\hat{j} + \frac{\partial}{\partial z}\hat{k}\right)(f g)\\ \amp = \left(\hat{i}\frac{\partial }{\partial x}(f g) + \hat{j}\frac{\partial}{\partial y}(f g) + \hat{k}\frac{\partial}{\partial z}(f g)\right) \\ \amp =\left(f \frac{\partial g}{\partial x} +g \frac{\partial f}{\partial x}\right)\hat{i} +\left(f \frac{\partial g}{\partial y} +g \frac{\partial f}{\partial y}\right)\hat{j} + \left(f \frac{\partial g}{\partial z} +g \frac{\partial f}{\partial z}\right)\hat{k}\\ \amp =f\left(\frac{\partial g}{\partial x}\hat{i} + \frac{\partial g}{\partial y}\hat{j} + \frac{\partial g}{\partial z}\hat{k}\right)+ g \left(\frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{j}+ \frac{\partial f}{\partial z}\hat{k}\right)\\ \amp =f\left(\hat{i}\frac{\partial}{\partial x} +\hat{j}\frac{\partial}{\partial y} +\hat{k}\frac{\partial}{\partial z}\right) g + g\left(\hat{i}\frac{\partial}{\partial x} +\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right) f \quad \\ \amp =f \vec{\nabla g} + g \vec{\nabla f} \end{align*}

Example 1.3.9.

Show that \(\vec{\nabla} r^{n}=n r^{n-2}\vec{r}\text{.}\)

Solution.

\begin{align*} \text{Let}\quad \phi \amp = r^{n} = \left(x^{2}+y^{2}+z^{2}\right)^{\frac{n}{2}}\\ \text{Hence, }\quad \frac{\partial \phi}{\partial x} \amp =\frac{n}{2}\left(x^{2}+y^{2} +z^{2}\right)^{\frac{n}{2}-1}\cdot 2 x \\ \amp = n\cdot\left(r^{2}\right)^{\frac{n-2}{2}}\cdot x = n\cdot r^{n-2}\cdot x\\ \text{Similarly,}\quad \frac{\partial \phi}{\partial y} \amp = n\cdot r^{n-2}\cdot y,\\ \text{and} \quad \frac{\partial \phi}{\partial z} \amp = n\cdot r^{n-2}\cdot z \\ \therefore \quad \vec{\nabla \phi} \amp = \left(\hat{i}\frac{\partial \phi}{\partial x} +\hat{j}\frac{\partial \phi}{\partial y} +\hat{k}\frac{\partial \phi}{\partial z}\right) \\ \amp = \hat{i} n \cdot r^{n-2}\cdot x + \hat{j} n \cdot r^{n-2}\cdot y + \hat{k} n \cdot r^{n-2}\cdot z\\ \amp =n \cdot r^{n-2}\left(\hat{i} x+\hat{j} y+\hat{k} z\right) = n r^{n-2}\vec{r} \end{align*}

Example 1.3.10.

If \(\phi = 3 x^{2}y-y^{3}z^{2}\text{,}\) find grad \(\phi\) at point \((1,-2,-1)\text{.}\)

Solution.

\begin{equation*} \text{grad} \phi = \vec{\nabla \phi} \end{equation*}

\begin{equation*} = \left(\hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial}{\partial y} +\hat{k}\frac{\partial}{\partial z}\right)\left(3 x^{2}y-y^{3}z^{2}\right) \end{equation*}

\begin{equation*} =\hat{i}\frac{\partial}{\partial x}\left(3 x^{2}y-y^{3}z^{2}\right) + \hat{j}\frac{\partial}{\partial y}\left(3 x^{2}y-y^{3}z^{2}\right) +\hat{k}\frac{\partial}{\partial z}\left(3 x^{2}y-y^{3}z^{2}\right) \end{equation*}

\begin{equation*} =\hat{i}(6 x y)+\hat{j}\left((3 x^{2}-3 y^{2}z^{2}\right)+\hat{k}\left(-2 y^{3}z\right) \end{equation*}

\begin{equation*} \therefore \quad \text{grad} \phi \hspace{2pt} \text{at} (1,-2,-1) \end{equation*}

\begin{equation*} = \hat{i}[6\cdot1\cdot(-2)]+\hat{j}[3.1^{2}-3(-2)^{2}(-1)^{2}]+\hat{k}[-2(-2)^{3}(-1)] \end{equation*}

\begin{equation*} = -12\hat{i}-9\hat{j}-16\hat{k}. \quad Ans. \end{equation*}

Example 1.3.11.

Find the unit vector normal to the surface \(x^{2}+3y^{2}+2z^{2}=6\) at point P (2,0,1).

Solution.

Let

\begin{align*} \phi \amp = x^{2}+3y^{2}+2z^{2}-6=0. \end{align*}

Then,

\begin{align*} grad \phi = \vec{\nabla \phi} \amp = \left(\hat{i}\frac{\partial }{\partial x} +\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right)\left(x^{2} +3y^{2}+2z^{2}-6=0\right) \\ \amp = 2x\hat{i}+6y\hat{j}+4z\hat{k} \end{align*}

\(\therefore\) grad \(\phi\) at point (2,0,1) = \(4\hat{i}+4\hat{k}\) but, grad \(\phi = \frac{\partial \phi}{\partial n}\hat{n}\text{,}\) where \(\hat{n}\) is the unit normal vector.

\begin{equation*} \therefore \hat{n}= \frac{\text{grad}\phi}{\mid \text{grad} \phi\mid} = \frac{4\hat{i}+4\hat{k}}{\sqrt{16+16}} = \frac{1}{\sqrt{2}}(\hat{i}+\hat{k}) \hspace{1cm} \text{Ans.} \end{equation*}

Example 1.3.12.

For the function \(\phi (x,y) = \frac{x}{x^{2}+y^{2}}\text{,}\) find the magnitude of the directional derivative along a line making an angle \(30^{o}\) with x- axis at position (0,2).

Solution.

Let,

\begin{align*} \vec{\nabla \phi} \amp = \left(\hat{i}\frac{\partial }{\partial x} +\hat{j}\frac{\partial}{\partial y} +\hat{k}\frac{\partial}{\partial z}\right)\left(\frac{x}{x^{2} +y^{2}}\right) \\ \amp = \hat{i}\left[\frac{1}{x^{2}+y^{2}} - \frac{x.2x}{(x^{2}+y^{2})^{2}}\right] -\hat{j}\left[\frac{x.2y}{(x^{2}+y^{2})^{2}}\right]\\ \amp= \hat{i}\left[\frac{(x^{2}+y^{2})-2x^{2}}{(x^{2}+y^{2})^{2}}\right] -\hat{j}\left[\frac{x.2y}{(x^{2}+y^{2})^{2}}\right] \\ \amp = \hat{i}\left[\frac{4-0}{(0+4)^{2}}\right] -\hat{j}\left[\frac{2.2.0}{(0+4)^{2}}\right] \end{align*}

\(\therefore\) grad \(\phi\) at (0,2) = \(\frac{\hat{i}}{4}\)

The line \(\vec{CA} \) = \(\vec{CB}+\vec{BA} = CA cos30^{o}\hat{i}+CA sin30^{o}\hat{j}\text{.}\)

Therefore, the unit vector along a line CA

\begin{equation*} = \frac{\vec{CA}}{|CA|} = \hat{i}\frac{\sqrt{3}}{2}+\hat{j}\frac{1}{2} = \hat{a} \quad \text{(say).} \end{equation*}

Hence the directional derivative at (0,2) in figure, in the direction of CA

\begin{equation*} = \vec{\nabla \phi}\cdot \hat{a} = \frac{\hat{i}}{4}\cdot\left(\hat{i}\frac{\sqrt{3}}{2}+\hat{j}\frac{1}{2}\right) = \frac{\sqrt{3}}{8} \hspace{1cm} \text{Ans.} \end{equation*}

Example 1.3.13.

Find the angle between the surfaces \(\left(x^{2}+y^{2}+z^{2}=9\right)\) and \(z= \left(x^{2}+y^{2}-3\right)\) at the point (2,-1,2).

Solution.

Let a normal to the surface \(\left(x^{2}+y^{2}+z^{2} = 9\right)\) at (2,-1,2) be

\begin{equation*} \vec{\nabla \phi_{1}} =\vec{\nabla} \left(x^{2}+y^{2}+z^{2}-9\right) = 2x\hat{i}+2y\hat{j}+2z\hat{k} = 4\hat{i}-2\hat{j}+4\hat{k} \end{equation*}

and a normal to the surface \(z= \left(x^{2}+y^{2}-3\right)\) at (2,-1,2) is

\begin{equation*} \vec{\nabla \phi_{2}} =\vec{\nabla} \left(x^{2}+y^{2}-z-3\right) = 2x\hat{i}+2y\hat{j}-\hat{k} = 4\hat{i}-2\hat{j}-\hat{k} \end{equation*}

The angle between the surfaces at any point is the angle between the normals to the surfaces at that point.

\begin{equation*} \therefore \vec{\nabla \phi_{1}}\cdot \vec{\nabla \phi_{2}} = \mid\vec{\nabla \phi_{1}}\mid\mid\vec{\nabla \phi_{2}}\mid \cos\theta, \end{equation*}

where \(\theta\) is the required angle. Now,

\begin{equation*} (4\hat{i}-2\hat{j}+4\hat{k})\cdot (4\hat{i}-2\hat{j}-\hat{k}) = \sqrt{16+4+16}\sqrt{16+4+1} \cos\theta \end{equation*}

or,

\begin{equation*} \cos\theta = \frac{16+4-4}{\sqrt{36}\sqrt{21}}=\frac{16}{6\sqrt{21}} = \frac{8\sqrt{21}}{63}=0.5891 \end{equation*}

\begin{align*} \therefore \quad \theta\amp = 54^{o}25'. \end{align*}

Divergence of a vector field.

Example 1.3.14.

Prove that

\(\displaystyle \vec{\nabla}\cdot\left(\vec{F}+\vec{G}\right)= \vec{\nabla}\cdot\vec{F} +\vec{\nabla}\cdot\vec{G}\)
\(\displaystyle \vec{\nabla}\cdot\left(\phi\vec{F}\right) = \left(\vec{\nabla \phi}\right)\cdot \vec{F} +\phi \left(\vec{\nabla} \cdot \vec{F}\right)\)

Solution.

Let \(\vec{F}=F_{1}\hat{i}+F_{2}\hat{j}+F_{3}\hat{k}\) and \(\vec{G}=G_{1}\hat{i}+G_{2}\hat{j}+G_{3}\hat{k}\text{.}\) Then,
\begin{equation*} \vec{\nabla}\cdot\left(\vec{F}+\vec{G}\right) = \left(\hat{i}\frac{\partial}{\partial x} +\hat{j}\frac{\partial}{\partial y} +\hat{k}\frac{\partial}{\partial z}\right) \end{equation*}

\begin{equation*} \cdot \left[\left(F_{1}+G_{1}\right)\hat{i}+\left(F_{2}+G_{2}\right)\hat{j} +\left(F_{3}+G_{3}\right)\hat{k}\right] \end{equation*}

\begin{equation*} =\frac{\partial}{\partial x}\left(F_{1}+G_{1}\right) + \frac{\partial}{\partial y}\left(F_{2}+G_{2}\right) +\frac{\partial}{\partial z}\left(F_{3}+G_{3}\right) \end{equation*}

\begin{equation*} =\left(\frac{\partial F_{1}}{\partial x} +\frac{\partial F_{2}}{\partial y} +\frac{\partial F_{3}}{\partial z}\right) +\left(\frac{\partial G_{1}}{\partial x} +\frac{\partial G_{2}}{\partial y} +\frac{\partial G_{3}}{\partial z}\right) \end{equation*}

\begin{equation*} =\left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y} +\hat{k}\frac{\partial}{\partial z}\right)\cdot \left(F_{1}\hat{i}+F_{2}\hat{j} +F_{3}\hat{k}\right) \end{equation*}

\begin{equation*} + \left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y} +\hat{k}\frac{\partial}{\partial z}\right) \cdot \left(G_{1}\hat{i}+G_{2}\hat{j}+G_{3}\hat{k}\right) \end{equation*}

\begin{equation*} =\vec{\nabla}\cdot\vec{F} +\vec{\nabla}\cdot\vec{G} \quad Proved. \end{equation*}
\begin{align*} \vec{\nabla}\cdot\left(\phi\vec{F}\right) \amp = \vec{\nabla}\cdot\left(\phi F_{1}\hat{i} + \phi F_{2}\hat{j}+\phi F_{3}\hat{k}\right) \\ \amp = \frac{\partial}{\partial x}\left(\phi F_{1}\right) + \frac{\partial}{\partial y}\left(\phi F_{2}\right) +\frac{\partial}{\partial z}\left(\phi F_{3}\right) \\ \amp =\frac{\partial\phi}{\partial x} F_{1} +\phi\frac{\partial F_{1}}{\partial x} +\frac{\partial\phi}{\partial y} F_{2} +\phi\frac{\partial F_{2}}{\partial y} +\frac{\partial\phi}{\partial z} F_{3} +\phi\frac{\partial F_{3}}{\partial z}\\ \amp =\frac{\partial\phi}{\partial x} F_{1} + \frac{\partial\phi}{\partial y} F_{2} +\frac{\partial\phi}{\partial z} F_{3}+ \phi\left(\frac{\partial F_{1}}{\partial x} +\frac{\partial F_{2}}{\partial y} +\frac{\partial F_{3}}{\partial z}\right)\\ \amp =\left(\hat{i}\frac{\partial}{\partial x} +\hat{j}\frac{\partial}{\partial y} +\hat{k}\frac{\partial}{\partial z}\right)\phi \cdot \left(F_{1}\hat{i}+F_{2}\hat{j}+F_{3}\hat{k}\right) \\ \amp + \phi \left(\hat{i}\frac{\partial}{\partial x} +\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right) \cdot \left(F_{1}\hat{i}+F_{2}\hat{j}+F_{3}\hat{k}\right) \\ \amp =\left(\vec{\nabla \phi}\right) \cdot \vec{F}+\phi \left(\vec{\nabla} \cdot \vec{F}\right) \quad proved. \end{align*}

Example 1.3.15.

Prove that \(\vec{\nabla}\cdot \left(\frac{\vec{r}}{r^{3}}\right)=0.\)

Solution.

Let \(\phi=r^{-3}\) and \(\vec{F}=\vec{r}\text{,}\) then

\begin{align*} \vec{\nabla}\cdot\left(\phi\vec{F}\right) \amp = \left(\vec{\nabla \phi}\right) \cdot \vec{F}+\phi \left(\vec{\nabla} \cdot \vec{F}\right) \\ \amp =\left(\vec{\nabla}r^{-3}\right) \cdot \vec{r}+r^{-3} \left(\vec{\nabla} \cdot \vec{r}\right) \\ \amp = -3r^{-3-2}\vec{r}\cdot\vec{r}+3r^{-3}=-3r^{-3}+3r^{-3} = 0 \end{align*}

\begin{equation*} \left[ \because \vec{\nabla}r^{n}= n r^{n-2}\vec{r} \right] \end{equation*}

or,

\begin{align*} \vec{\nabla}\cdot \vec{r} \amp =\left(\hat{i}\frac{\partial}{\partial x} +\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right) \cdot \left(x\hat{i}+y\hat{j}+z\hat{k}\right) =3. \end{align*}

Example 1.3.16.

Prove that \(\vec{\nabla}\cdot\vec{\nabla\phi}=\nabla^{2}\phi\text{.}\)

Solution.

\begin{align*} \vec{\nabla}\cdot\vec{\nabla\phi} \amp = \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right) \cdot \left(\frac{\partial \phi}{\partial x}\hat{i} +\frac{\partial \phi}{\partial y}\hat{j} +\frac{\partial \phi}{\partial z}\hat{k}\right)\\ \amp = \frac{\partial}{\partial x}\left(\frac{\partial \phi}{\partial x}\right) +\frac{\partial}{\partial y}\left(\frac{\partial \phi}{\partial y}\right) +\frac{\partial}{\partial z}\left(\frac{\partial \phi}{\partial z}\right) \\ \amp =\frac{\partial^{2} \phi}{\partial x^{2}} +\frac{\partial^{2} \phi}{\partial y^{2}}+\frac{\partial^{2} \phi}{\partial z^{2}} = \left(\frac{\partial^{2}}{\partial x^{2}} +\frac{\partial^{2}}{\partial y^{2}} +\frac{\partial^{2}}{\partial z^{2}}\right)\phi\\ \amp = \nabla^{2}\phi. \end{align*}

Where,

\begin{equation*} \nabla^{2}=\left(\frac{\partial^{2}}{\partial x^{2}} +\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}}\right) \end{equation*}

is called Laplacian Operator.

Example 1.3.17.

Determine the constant \(a\) if a vector \(\vec{v}=(x+3y)\hat{i}+(y-2z)\hat{j}+(x+az)\hat{k}\) is solenoidal.

Solution.

If \(\vec{\nabla}\cdot \vec{v}=0,\) then \(\vec{v}\) is solenoidal.

\begin{align*} 0 \amp =\vec{\nabla}\cdot \vec{v} \\ \amp =\frac{\partial}{\partial x}(x+3y) +\frac{\partial}{\partial y}(y-2z)+\frac{\partial}{\partial z}(x+az) \\ \amp =1+1+a = a+2 \end{align*}

Hence, \(a+2 =0 \Longrightarrow a =-2.\) Ans.

Example 1.3.18.

Prove that \(\,div\hat{r} = \frac{2}{r}\text{.}\)

Solution.

We have

\begin{align*} \hat{r} \amp =\frac{\vec{r}}{|\vec{r}|} =\frac{x\hat{i}+y\hat{j}+z\hat{k}}{r} = \frac{x}{r}\hat{i}+\frac{y}{r}\hat{j}+\frac{z}{r}\hat{k}\\ \therefore \vec{\nabla}\cdot\hat{r}\amp = \frac{\partial}{\partial x}\left(\frac{x}{r}\right) +\frac{\partial}{\partial y}\left(\frac{y}{r}\right) +\frac{\partial}{\partial z}\left(\frac{z}{r}\right) \\ \amp = \frac{1}{r^{2}}\left[3r-\left(x\frac{\partial r}{\partial x} +y\frac{\partial r}{\partial y}+z\frac{\partial r}{\partial z}\right)\right] \end{align*}

since, \(r^{2}=x^{2}+y^{2}+z^{2}\text{,}\) we have

\begin{equation*} \frac{\partial r}{\partial x}=\frac{x}{r},\quad \frac{\partial r}{\partial y}=\frac{y}{r}\quad \text{and}\quad \frac{\partial r}{\partial z}=\frac{z}{r}. \end{equation*}

Hence,

\begin{align*} \vec{\nabla}\cdot\hat{r} \amp = \frac{1}{r^{2}}\left[3r-\left(x\frac{x}{r} +y\frac{y}{r}+z\frac{z}{r}\right)\right] \\ \amp = \frac{1}{r^{2}}\left[3r-\left(\frac{x^{2}}{r}+\frac{y^{2}}{r} +\frac{z^{2}}{r}\right)\right] \\ \amp = \frac{1}{r^{2}}\left[3r-\left(\frac{r^{2}}{r}\right)\right] =\frac{1}{r^{2}}(2r)=\frac{2}{r}. \end{align*}

Curl of Vector.

Example 1.3.19.

Prove that

\begin{equation*} \vec{\nabla}\times\left(\phi\vec{F}\right) = \left(\vec{\nabla \phi}\right)\times \vec{F} +\phi \left(\vec{\nabla} \times \vec{F}\right) \end{equation*}

Solution.

We have

\begin{align*} \vec{\nabla}\times\left(\phi\vec{F}\right)\amp = \vec{\nabla}\times\left(\phi\vec{F_{1}\hat{i}} +\phi\vec{F_{2}\hat{j}}+\phi\vec{F_{3}\hat{k}}\right)\\ \amp = {det\begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ \phi F_{1} & \phi F_{2} & \phi F_{3} \end{bmatrix}}\\ \amp =\left[\frac{\partial}{\partial y}\left(\phi F_{3}\right) -\frac{\partial}{\partial z}\left(\phi F_{2}\right)\right]\hat{i}\\ \amp \quad -\left[\frac{\partial}{\partial x}\left(\phi F_{3}\right) -\frac{\partial}{\partial z}\left(\phi F_{1}\right)\right]\hat{j}\\ \amp \quad +\left[\frac{\partial}{\partial x}\left(\phi F_{2}\right) -\frac{\partial}{\partial y}\left(\phi F_{1}\right)\right]\hat{k} \end{align*}

\begin{equation*} =\left[\phi\frac{\partial F_{3}}{\partial y} +F_{3}\frac{\partial \phi}{\partial y}-\phi\frac{\partial F_{2}}{\partial z} -F_{2}\frac{\partial \phi}{\partial z}\right]\hat{i} \end{equation*}

\begin{equation*} -\left[\phi\frac{\partial F_{3}}{\partial x}+F_{3}\frac{\partial \phi}{\partial x} -\phi\frac{\partial F_{1}}{\partial z}-F_{1}\frac{\partial \phi}{\partial z}\right]\hat{j} \end{equation*}

\begin{equation*} +\left[\phi\frac{\partial F_{2}}{\partial x}+F_{2}\frac{\partial \phi}{\partial x} -\phi\frac{\partial F_{1}}{\partial y}-F_{1}\frac{\partial \phi}{\partial y}\right]\hat{k} \end{equation*}

\begin{equation*} =\phi \left[\left(\frac{\partial F_{3}}{\partial y} -\frac{\partial F_{2}}{\partial z}\right)\hat{i}+\left(\frac{\partial F_{1}}{\partial z} -\frac{\partial F_{3}}{\partial x}\right)\hat{j}+\left(\frac{\partial F_{2}}{\partial x} -\frac{\partial F_{1}}{\partial y}\right)\right]\hat{k} \end{equation*}

\begin{equation*} + \left[\left(F_{3}\frac{\partial \phi}{\partial y} -F_{2}\frac{\partial \phi}{\partial z}\right)\hat{i} +\left(F_{1}\frac{\partial \phi}{\partial z} -F_{3}\frac{\partial \phi}{\partial x}\right)\hat{j} +\left(F_{3}\frac{\partial \phi}{\partial x} -F_{3}\frac{\partial \phi}{\partial y}\right)\right]\hat{k} \end{equation*}

\begin{equation*} =\phi \left(\vec{\nabla}\times\vec{F}\right) +{\begin{Vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y} & \frac{\partial \phi}{\partial z}\\ F_{1} & F_{2} & F_{3} \end{Vmatrix}} \end{equation*}

\begin{equation*} = \left(\vec{\nabla}\phi\times\vec{F}\right) + \phi\left(\vec{\nabla}\times\vec{F}\right) \end{equation*}

Example 1.3.20.

Prove that

\(\displaystyle \vec{\nabla}\times \left(\vec{\nabla}\phi\right) = 0 \)
and \(\vec{\nabla}\cdot \left(\vec{\nabla}\times\vec{A}\right) = 0.\)

Solution.

\begin{equation*} \vec{\nabla}\times \left(\vec{\nabla}\phi\right) = \vec{\nabla}\times \left(\frac{\partial \phi}{\partial x}\hat{i} +\frac{\partial \phi}{\partial y}\hat{j}+\frac{\partial \phi}{\partial z}\hat{k}\right) \end{equation*}

\begin{equation*} = {\begin{Vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ \frac{\partial \phi}{\partial x} & \frac{\partial\phi}{\partial y} & \frac{\partial \phi}{\partial z} \end{Vmatrix}} \end{equation*}

\begin{equation*} =\left[\frac{\partial}{\partial y}\left(\frac{\partial \phi}{\partial z}\right) -\frac{\partial}{\partial z}\left(\frac{\partial \phi}{\partial y}\right)\right]\hat{i} +\left[\frac{\partial}{\partial z}\left(\frac{\partial \phi}{\partial x}\right) -\frac{\partial}{\partial z}\left(\frac{\partial \phi}{\partial x}\right)\right]\hat{j} \end{equation*}

\begin{equation*} +\left[\frac{\partial}{\partial x}\left(\frac{\partial \phi}{\partial y}\right) -\frac{\partial}{\partial y}\left(\frac{\partial \phi}{\partial x}\right)\right]\hat{k} \end{equation*}

\begin{equation*} =\left(\frac{\partial^{2} \phi}{\partial y \partial z} -\frac{\partial^{2} \phi}{\partial z \partial y}\right)\hat{i} +\left(\frac{\partial^{2} \phi}{\partial z \partial x} -\frac{\partial^{2} \phi}{\partial x \partial z}\right)\hat{j} +\left(\frac{\partial^{2} \phi}{\partial x \partial y} -\frac{\partial^{2} \phi}{\partial y \partial x}\right)\hat{k} \end{equation*}

\begin{equation*} =0 \end{equation*}
\begin{equation*} \vec{\nabla}\cdot \left(\vec{\nabla}\times\vec{A}\right) = \vec{\nabla}\cdot {det\begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ A_{1} & A_{2} & A_{3} \end{bmatrix}} \end{equation*}

\begin{equation*} = \vec{\nabla}\cdot \left\{\left[\frac{\partial A_{3}}{\partial y} -\frac{\partial A_{2}}{\partial z}\right]\hat{i}+\left[\frac{\partial A_{1}}{\partial z} -\frac{\partial A_{3}}{\partial x}\right]\hat{j}+\left[\frac{\partial A_{2}}{\partial x} -\frac{\partial A_{1}}{\partial y}\right]\hat{k}\right\} \end{equation*}

\begin{equation*} =\frac{\partial}{\partial x}\left[\frac{\partial A_{3}}{\partial y} -\frac{\partial A_{2}}{\partial z}\right] +\frac{\partial}{\partial y}\left[\frac{\partial A_{1}}{\partial z} -\frac{\partial A_{3}}{\partial x}\right] +\frac{\partial}{\partial z}\left[\frac{\partial A_{2}}{\partial x} -\frac{\partial A_{1}}{\partial y}\right] \end{equation*}

\begin{equation*} =\frac{\partial^{2}A_{3}}{\partial x\partial y} -\frac{\partial^{2}A_{2}}{\partial x\partial z} +\frac{\partial^{2}A_{1}}{\partial y\partial z} -\frac{\partial^{2}A_{3}}{\partial y\partial x} +\frac{\partial^{2}A_{2}}{\partial z\partial x} -\frac{\partial^{2}A_{1}}{\partial z\partial y}=0. \end{equation*}

Example 1.3.21.

Prove that

\begin{equation*} \vec{\nabla}\times \left(\vec{\nabla}\times\vec{A}\right) = - \nabla^{2}\vec{A}+\vec{\nabla}\left(\vec{\nabla}\cdot\vec{A}\right) \end{equation*}

Solution.

We have

\begin{equation*} \vec{a}\times\left(\vec{b}\times\vec{c}\right) = \left(\vec{a}\cdot\vec{c}\right)\vec{b}-\left(\vec{a}\cdot\vec{b}\right)\vec{c} \end{equation*}

Put, \(\vec{a}=\vec{b}=\vec{\nabla}\text{,}\) then

\begin{equation*} \vec{\nabla}\times \left(\vec{\nabla}\times\vec{A}\right) = \left(\vec{\nabla}\cdot\vec{A}\right)\vec{\nabla}-\left(\vec{\nabla}\cdot\vec{\nabla}\right)\vec{A} =\vec{\nabla}\left(\vec{\nabla}\cdot\vec{A}\right)-\nabla^{2}\vec{A}. \end{equation*}

Example 1.3.22.

If a vector \(\vec{v}=(x+y+az)\hat{i}+(bx+3y-z)\hat{j}+(3x+cy+z)\hat{k}\) is irrotational, find the values of the constants \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\)

Solution.

We have \(\,curl\vec{v}=0\text{,}\) or, \(\vec{\bigtriangledown}\times\vec{v}=0\text{,}\) or,

\begin{equation*} {\begin{Vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ x+y+az & bx+3y-z & 3x+cy+z \end{Vmatrix}} = 0 \end{equation*}

\begin{equation*} or, \quad (c+1)\hat{i}-(3-a)\hat{j}+(b-1)\hat{k} = 0.\hat{i}+0.\hat{j}+0.\hat{k} \end{equation*}

now, equating the coefficients of \(\hat{i}\text{,}\) \(\hat{j}\text{,}\) \(\hat{k}\text{,}\) we get-

\begin{equation*} c+1=0 \Rightarrow c=-1, \quad 3-a=0 \Rightarrow a=3,\quad \text{and}\quad b-1=0 \Rightarrow b=1. \end{equation*}

Example 1.3.23.

If \(r=(x^{2}+y^{2}+z^{2})^{1/2}\) and \((1/r)\) is a solution of Laplace’s equation, then show that

\(\bigtriangledown^{2}\left(\frac{1}{r}\right)=0\text{,}\) and
evaluate \(\vec{\bigtriangledown}\times\left(\frac{\vec{r}}{r^{2}}\right)\)

Solution.

\begin{equation*} \nabla^{2}\left(\frac{1}{r}\right) = \left(\frac{\partial^{2}}{\partial x^{2}} +\frac{\partial^{2}}{\partial y^{2}} +\frac{\partial^{2}}{\partial z^{2}}\right)\left[\frac{1}{(x^{2}+y^{2}+z^{2})^{1/2}}\right], \end{equation*}
here,
\begin{equation*} \nabla^{2} = \frac{\partial^{2}}{\partial x^{2}} +\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}} \end{equation*}
is a Laplacian operator. But,
\begin{equation*} \frac{\partial}{\partial x}\left[\frac{1}{(x^{2}+y^{2}+z^{2})^{1/2}}\right] = -\frac{1}{2}\left[\frac{1(2x)}{(x^{2}+y^{2}+z^{2})^{3/2}}\right] \end{equation*}
and
\begin{equation*} \frac{\partial^{2}}{\partial x^{2}}\left[\frac{1}{(x^{2}+y^{2}+z^{2})^{1/2}}\right] = \frac{\partial}{\partial x}\left[\frac{-x}{(x^{2}+y^{2}+z^{2})^{3/2}}\right] \end{equation*}

\begin{equation*} = \left[\frac{+\frac{3}{2}(x)2x}{(x^{2}+y^{2}+z^{2})^{5/2}}\right] - \left[\frac{-1}{(x^{2}+y^{2}+z^{2})^{3/2}}\right] \end{equation*}
Similarly,
\begin{equation*} \frac{\partial^{2}}{\partial y^{2}}\left[\frac{1}{(x^{2}+y^{2}+z^{2})^{1/2}}\right] \end{equation*}

\begin{equation*} = \left[\frac{+3y^{2}}{(x^{2}+y^{2}+z^{2})^{5/2}}\right] - \left[\frac{-1}{(x^{2}+y^{2}+z^{2})^{3/2}}\right] \end{equation*}
also,
\begin{equation*} \frac{\partial^{2}}{\partial z^{2}}\left[\frac{1}{(x^{2}+y^{2}+z^{2})^{1/2}}\right] \end{equation*}

\begin{equation*} = \left[\frac{3(x^{2}+y^{2}+z^{2})}{(x^{2}+y^{2}+z^{2})^{5/2}}\right] - \left[\frac{3}{(x^{2}+y^{2}+z^{2})^{3/2}}\right] = 0. \end{equation*}
Let
\begin{equation*} \vec{r}=x\hat{i}+y\hat{J}+z\hat{k}\text{,} \end{equation*}
so that
\begin{equation*} r^{2}=x^{2}+y^{2}+z^{2} \end{equation*}
also, suppose \(r^{2}=\phi\) and \(\vec{r}=\vec{F}\text{.}\) Then as
\begin{equation*} \vec{\nabla}\times\left(\phi\vec{F}\right) = \left(\vec{\nabla \phi}\right)\times \vec{F} +\phi \left(\vec{\nabla} \times \vec{F}\right) \end{equation*}
we have
\begin{equation*} \vec{\nabla}\times \left(\frac{\vec{r}}{r^{2}}\right) = \vec{\nabla}\left(\frac{1}{r^{2}}\right)\times \vec{r} + \frac{1}{r^{2}}\left(\vec{\nabla}\times \vec{r}\right) \end{equation*}

\begin{equation*} =\left[\vec{\nabla}\left(\frac{1}{x^{2}+y^{2} +z^{2}}\right)\right] \times (x\hat{i}+y\hat{J}+z\hat{k}) \end{equation*}

\begin{equation*} +\frac{1}{x^{2}+y^{2}+z^{2}}\left[\vec{\nabla} \times (x^{2}+y^{2}+z^{2})\right] \end{equation*}

\begin{equation*} =\left[\left(\hat{i}\frac{\partial}{\partial x} +\hat{j}\frac{\partial}{\partial y} +\hat{k}\frac{\partial}{\partial z}\right) \left(\frac{1}{x^{2}+y^{2}+z^{2}}\right)\right] \times (x\hat{i}+y\hat{J}+z\hat{k}) +0 \end{equation*}

\begin{equation*} = \frac{-2}{(x^{2}+y^{2}+z^{2})^{2}}(x\hat{i}+y\hat{J}+z\hat{k}) \times (x\hat{i}+y\hat{J}+z\hat{k})=0. \end{equation*}

Example 1.3.24.

Find the curl of a vector field \(\vec{v}=2xy\hat{i}+3\sin y\hat{j}\text{.}\)

Solution.

We have

\begin{equation*} \left(\vec{\nabla}\times\vec{v}\right) = {\begin{Vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ v_{1} & v_{2} & v_{3} \end{Vmatrix}} \end{equation*}

\begin{equation*} = \left[\frac{\partial v_{3}}{\partial y} -\frac{\partial v_{2}}{\partial z}\right]\hat{i} +\left[\frac{\partial v_{1}}{\partial z} -\frac{\partial v_{3}}{\partial x}\right]\hat{j} +\left[\frac{\partial v_{2}}{\partial x} -\frac{\partial v_{1}}{\partial y}\right]\hat{k} \end{equation*}

\begin{equation*} = \hat{k}(-\frac{\partial v_{1}}{\partial y})= (-2x)\hat{k} \end{equation*}

Prev Top Next