Q26P - Introduction to Electrodynamics

Question

Calculate the Laplacian of the following functions:

 

$$\begin{gathered} \left(a\right){\mathbf{T}}_{\mathbf{a}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{3}\mathbf{z}\mathbf{+}\mathbf{4} \\ \left(b\right)\mathbf{ }{\mathbf{T}}_{\mathbf{b}}\mathbf{=}\mathbf{sinx}\mathbf{ }\mathbf{siny}\mathbf{ }\mathbf{sinz} \\ \left(c\right)\mathbf{ }{\mathbf{T}}_{\mathbf{c}}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{5}}\mathbf{ }\mathbf{sin}\mathbf{4}\mathbf{y}\mathbf{ }\mathbf{cos}\mathbf{ }\mathbf{3}\mathbf{z}\mathbf{.} \\ \left(d\right)\mathbf{ }\mathbf{v}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{ }\hat{\mathbf{x}}\mathbf{+}\mathbf{3}{\mathbf{xz}}^{{\mathbf{2}}^{\mathbf{ }}}\hat{\mathbf{y}}\mathbf{+}\mathbf{-}\mathbf{2}\mathbf{xz}\hat{\mathbf{z}} \end{gathered}$$ 

Step-by-Step Solution

Answer

$$\begin{gathered} \left(a\right)The value of {\nabla }^2{T_a}^{ } is 2. \\ \left(b\right)The value of {\nabla }^2{T}_b is . \\ \left(c\right)The value of {\nabla }^2{T}_c is 0. \\ \left(d\right)The value of {\nabla }^2 v is 2\hat{x}+6x\hat{y} . \end{gathered}$$

Step 1: Define the laplacian

The divergence of gradient is called as Laplacian of a vector. The vector v is defined as $v=v_{x}i+v_{y}j+v_{z}k$ . the  operator is defined as $\nabla =\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k$ . The value of Laplacian${\nabla }^{2}v$ is obtained as

 

${\nabla }^{2}v={\nabla }^2{v}_{x}i+{\nabla }^2{v}_{y}j+{\nabla }^2{v}_{z}k$ 

 

Here,${\nabla }^2$ is the second ordered partial derivative of function.

Step 2: Compute ∇ 2 T a

(a)

 

To compute an expression substitute the vectors and other required expression and then simplify

The vector$\stackrel{\rightharpoonup }{T_a}$ is defined as $x^{2}i+2xyj+3zk+4$ . The Laplacian of the vector o $\stackrel{\rightharpoonup }{T_a}$ is defined as ${\nabla }^2{T}_a=\frac{{\partial }^2{T}_a}{\partial x^2}+\frac{{\partial }^2{T}_a}{\partial y^2}+\frac{{\partial }^2{T}_a}{\partial z^2}$ .

 

Substitute $x^{2}i+2xyj+3zk+4 \mathrm{for} {\stackrel{\rightharpoonup }{\mathrm{T}}}_{\mathrm{a}} \mathrm{into} {\nabla }^2{\mathrm{T}}_{\mathrm{a}} =\frac{{\partial }^2{\mathrm{T}}_{\mathrm{a}}}{\partial {\mathrm{x}}^2}+\frac{{\partial }^2{\mathrm{T}}_{\mathrm{a}}}{\partial {\mathrm{y}}^2}+\frac{{\partial }^2{\mathrm{T}}_{\mathrm{a}}}{\partial {\mathrm{z}}^2}$ into .

${\nabla }^2{T}_a=\left\{\begin{array}{cc}\frac{{\partial }^2\left(x^{2}i+2xyj+3zk+4\right)}{\partial x^2}& \frac{{\partial }^2\left(x^{2}i+2xyj+3zk+4\right)}{\partial y^2}\\ \frac{{\partial }^2\left(x^{2}i+2xyj+3zk+4\right)}{\partial z^2}& \end{array}\right\}$

            $$\begin{gathered} =\left\{\begin{array}{cc}\frac{\partial }{\partial x}\left(\frac{{\partial }^2\left(x^{2}i+2xyj+3zk+4\right)}{\partial x^2}\right)& +\frac{\partial }{\partial x}\left(\frac{\partial \left(x^{2}i+2xyj+3zk+4\right)}{\partial y}\right)\\ +\frac{\partial }{\partial x}\left(\frac{{\partial }^2\left(x^{2}i+2xyj+3zk+4\right)}{\partial z}\right)& \end{array}\right\} \\ =\left\{\begin{array}{cc}\frac{\partial }{\partial x}\left(2x+2y+0+0\right)& +\frac{\partial }{\partial x}\left(0+2x+0+0\right)\\ +\frac{\partial }{\partial x}\left(0+0+3+0\right)& \end{array}\right\} \\ =2 \end{gathered}$$

Thus the value of ${\nabla }^2{T}_{\mathrm{a}} \mathrm{is} 2$ is 2.

Step 3: Compute ∇ 2 T b

(b)

To compute an expression substitute the vectors and other required expression and then simplify.

 

The vector$\stackrel{\rightharpoonup }{T_b}$ is defined as $\sin x \sin y \sin z$ . The Laplacian of the vector$\stackrel{\rightharpoonup }{T_b}$ is defined as ${\nabla }^2{T}_b=\frac{{\partial }^2{T}_b}{\partial x^2}+\frac{{\partial }^2{T}_b}{\partial y^2}+\frac{{\partial }^2{T}_b}{\partial z^2}$ .

 

Substitute $\sin x \sin y \sin z$  for   into ${\nabla }^2{T}_b=\frac{{\partial }^2{T}_b}{\partial x^2}+\frac{{\partial }^2{T}_b}{\partial y^2}+\frac{{\partial }^2{T}_b}{\partial z^2}$

$$\begin{gathered} {\nabla }^2{T}_b=\left\{\begin{array}{cc}\frac{{\partial }^2\left(\sin x \sin y \sin z\right)}{\partial x^2}& \frac{{\partial }^2\left(\sin x \sin y \sin z\right)}{\partial y^2}\\ \frac{{\partial }^2\left(\sin x \sin y \sin z\right)}{\partial z^2}& \end{array}\right\} \\ =\left\{\begin{array}{cc}\frac{\partial }{\partial x}\left(\frac{\partial \left(\sin x \sin y \sin z\right)}{\partial x}\right)+\frac{\partial }{\partial y}& \left(\frac{\partial \left(\sin x \sin y \sin z\right)}{\partial y}\right)\\ +\frac{\partial }{\partial z}\left(\frac{\partial \left(\sin x \sin y \sin z\right)}{\partial z}\right)& \end{array}\right\} \\ =\left\{\begin{array}{cc}\frac{\partial }{\partial x}\left(\cos x \sin y \sin z\right)+\frac{\partial }{\partial x}\left(\sin x \cos y \sin z\right)& \\ +\frac{\partial }{\partial z}\left(\sin x \sin y \cos z\right)& \end{array}\right\} \\ =-\sin x \sin y \sin z-\sin y \sin x \sin z-\sin z \sin x \sin y \end{gathered}$$

Solve further as,

${\nabla }^2{T}_b=-3\sin x\sin y \sin z$ 

 

Thus, the value of ${\nabla }^2{T}_b$ is$-3\sin x\sin y \sin z$.

Step 4: Compute ∇ 2 T c

(c)

To compute an expression substitute the vectors and other required expression and then simplify.

 

The vector${\overrightarrow{T}}_c$ is defined as $e^{-5x}\sin 4 y\cos 3z$. The Laplacian of the vector ${\overrightarrow{T}}_c$ is defined as ${\nabla }^2{T}_c=\frac{{\partial }^2{T}_c}{\partial x^2}+\frac{{\partial }^2{T}_c}{\partial y^2}+\frac{{\partial }^2{T}_c}{\partial z^2}$  .

 

Substitute $e^{-5x} \sin 4y\cos 3z$ for ${\overrightarrow{T}}_c$ into ${∆}^2{T}_c=\frac{{\partial }^2{T}_c}{\partial x^2}+\frac{{\partial }^2{T}_c}{\partial y^2}+\frac{{\partial }^2{T}_c}{\partial z^2}$ .

$$\begin{gathered} {\nabla }^2{T}_b=\left\{\begin{array}{cc}\frac{{\partial }^2\left(e^{-5x} \sin 4y \cos 3z\right)}{\partial x^2}& +\frac{{\partial }^2\left(e^{-5x} \sin 4y \cos 3z\right)}{\partial y^2}\\ \frac{{\partial }^2\left(e^{-5x} \sin 4y \cos 3z\right)}{\partial z^2}& \end{array}\right\} \\ =\left\{\begin{array}{cc}\frac{\partial }{\partial x}\left(\frac{\partial \left(e^{-5x} \sin 4y \cos 3z\right)}{\partial x}\right)+\frac{\partial }{\partial y}& \left(\frac{\partial \left(e^{-5x} \sin 4y \cos 3z\right)}{\partial y}\right)\\ +\frac{\partial }{\partial z}\left(\frac{\partial \left(e^{-5x} \sin 4y \cos 3z\right)}{\partial z}\right)& \end{array}\right\} \\ =\left\{\begin{array}{cc}\frac{\partial }{\partial x}\left(-5e^{-5x} \sin 4y \cos 3z\right)+\frac{\partial }{\partial y}& \left(4e^{-5x} \cos 4y \cos 3z\right)\\ +\frac{\partial }{\partial z}\left(-3e^{-5x} \sin 4y \sin 3z\right)& \end{array}\right\} \\ =25e^{-5x} \sin 4y \cos 3z-16e^{-5x} \sin y \cos 3z-9e^{-5x} \sin 4y \cos 3z \end{gathered}$$

Solve further as,

$$\begin{gathered} {\nabla }^2{T}_b=e^{-5x}\sin 4y\cos 3z\left(25-16-9\right) \\ =0 \end{gathered}$$ 

 

Thus, the value of ${\nabla }^2{T}_c$ is 0.

Step 5: Compute ∇ 2 v

(d)

To compute an expression substitute the vectors and other required expression and then simplify.

 

The vector $\overrightarrow{v}$ is defined as $x^{2}i+3x{z}^{2}j-2xzk$. The Laplacian of the vector$\overrightarrow{v}$ is defined as ${\nabla }^2\overrightarrow{v}=\frac{{\partial }^{2}v}{\partial x^2}+\frac{{\partial }^{2}v}{\partial y^2}+\frac{{\partial }^{2}v}{\partial z^2}$.

 

Substitute $x^{2}i+3x{z}^{2}j-2xzk$ for $\overrightarrow{v}$ into ${\nabla }^2\overrightarrow{v}=\frac{{\partial }^{2}v}{\partial x^2}+\frac{{\partial }^{2}v}{\partial y^2}+\frac{{\partial }^{2}v}{\partial z^2}$.

${\nabla }^{2}v=\frac{{\partial }^2\left(x^{2}i+3x{z}^{2}j-2xzk\right)}{\partial x^2}+\frac{{\partial }^2\left(x^{2}i+3x{z}^{2}j-2xzk\right)}{\partial y^2}+\frac{{\partial }^2\left(x^{2}i+3x{z}^{2}j-2xzk\right)}{\partial z^2}$

$$\begin{gathered} =\left\{\begin{array}{cc}\frac{\partial }{\partial x}\left(\frac{\partial \left(x^{2}i+3x{z}^{2}j-2xzk\right)}{\partial x}\right)& +\frac{\partial }{\partial y}\left(\frac{\partial \left(x^{2}i+3x{z}^{2}j-2xzk\right)}{\partial y}\right)\\ \frac{\partial }{\partial x}\left(\frac{\partial \left(x^{2}i+3x{z}^{2}j-2xzk\right)}{\partial z}\right)& \end{array}\right\} \\ =\left\{\begin{array}{cc}\frac{\partial }{\partial x}\left(2xi+3z^{2}j-2zk\right)+\frac{\partial }{\partial y}& \left(0+0-0\right)\\ +\frac{\partial }{\partial z}\left(0+6xzj-2xk\right)& \end{array}\right\} \\ =\left(2i+0-0\right)+\left(0\right)+\left(0+6xj-0\right) \end{gathered}$$

Solve further as,

 ${\nabla }^{2}v=2i+6xj$

 Thus the value of ${\nabla }^{2}v$ is $2\stackrel{⏜}{x}+6x\stackrel{⏜}{y}$  .