Calculate the Laplacian of the following functions:

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

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.

(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 .

localid="1657536994674" ${\nabla }^2{T}_a=\left\{\begin{array}{cc}\frac{{\partial }^2\left(x^{2}i+2xyj+3zk+4\right)}{\partial x^2}& \frac{{\partial }^2\left(x^{2}i+2xyj+3zk+4\right)}{\partial y^2}\\ \frac{{\partial }^2\left(x^{2}i+2xyj+3zk+4\right)}{\partial z^2}& \end{array}\right\}$

localid="1657537065407" $$\begin{gathered} =\left\{\begin{array}{cc}\frac{\partial }{\partial x}\left(\frac{{\partial }^2\left(x^{2}i+2xyj+3zk+4\right)}{\partial x^2}\right)& +\frac{\partial }{\partial x}\left(\frac{\partial \left(x^{2}i+2xyj+3zk+4\right)}{\partial y}\right)\\ +\frac{\partial }{\partial x}\left(\frac{{\partial }^2\left(x^{2}i+2xyj+3zk+4\right)}{\partial z}\right)& \end{array}\right\} \\ =\left\{\begin{array}{cc}\frac{\partial }{\partial x}\left(2x+2y+0+0\right)& +\frac{\partial }{\partial x}\left(0+2x+0+0\right)\\ +\frac{\partial }{\partial x}\left(0+0+3+0\right)& \end{array}\right\} \\ =2 \end{gathered}$$

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

(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 localid="1657538369566" ${\nabla }^2{T}_b=\frac{{\partial }^2{T}_b}{\partial x^2}+\frac{{\partial }^2{T}_b}{\partial y^2}+\frac{{\partial }^2{T}_b}{\partial z^2}$.

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

localid="1657604767825" $$\begin{gathered} {\nabla }^2{T}_b=\left\{\begin{array}{cc}\frac{{\partial }^2\left(\sin x\sin y\sin z\right)}{\partial x^2}& \frac{{\partial }^2\left(\sin x\sin y\sin z\right)}{\partial y^2}\\ \frac{{\partial }^2\left(\sin x\sin y\sin z\right)}{\partial z^2}& \end{array}\right\} \\ =\left\{\begin{array}{cc}\frac{\partial }{\partial x}\left(\frac{\partial \left(\sin x\sin y\sin z\right)}{\partial x}\right)+\frac{\partial }{\partial y}& \left(\frac{\partial \left(\sin x\sin y\sin z\right)}{\partial y}\right)\\ +\frac{\partial }{\partial z}\left(\frac{\partial \left(\sin x\sin y\sin z\right)}{\partial z}\right)& \end{array}\right\} \\ =\left\{\begin{array}{cc}\frac{\partial }{\partial x}\left(\cos x\sin y\sin z\right)+\frac{\partial }{\partial x}\left(\sin x\cos y\sin z\right)& \\ +\frac{\partial }{\partial z}\left(\sin x\sin y\cos z\right)& \end{array}\right\} \\ =-\sin x\sin y\sin z-\sin y\sin x\sin z-\sin z\sin x\sin y \end{gathered}$$

Solve further as,

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

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

(c)

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

The vector${\overrightarrow{T}}_c$is defined as $e^{-5x}\sin 4y\cos 3z$. 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 role="math" localid="1657605202299" $e^{-5x}\sin 4y\cos 3z$for ${\overrightarrow{T}}_c$into ${∆}^2{T}_c=\frac{{\partial }^2{T}_c}{\partial x^2}+\frac{{\partial }^2{T}_c}{\partial y^2}+\frac{{\partial }^2{T}_c}{\partial z^2}$.

role="math" localid="1657605719160" $$\begin{gathered} {\nabla }^2{T}_b=\left\{\begin{array}{cc}\frac{{\partial }^2\left(e^{-5x}\sin 4y\cos 3z\right)}{\partial x^2}& +\frac{{\partial }^2\left(e^{-5x}\sin 4y\cos 3z\right)}{\partial y^2}\\ \frac{{\partial }^2\left(e^{-5x}\sin 4y\cos 3z\right)}{\partial z^2}& \end{array}\right\} \\ =\left\{\begin{array}{cc}\frac{\partial }{\partial x}\left(\frac{\partial \left(e^{-5x}\sin 4y\cos 3z\right)}{\partial x}\right)+\frac{\partial }{\partial y}& \left(\frac{\partial \left(e^{-5x}\sin 4y\cos 3z\right)}{\partial y}\right)\\ +\frac{\partial }{\partial z}\left(\frac{\partial \left(e^{-5x}\sin 4y\cos 3z\right)}{\partial z}\right)& \end{array}\right\} \\ =\left\{\begin{array}{cc}\frac{\partial }{\partial x}\left(-5e^{-5x}\sin 4y\cos 3z\right)+\frac{\partial }{\partial y}& \left(4e^{-5x}\cos 4y\cos 3z\right)\\ +\frac{\partial }{\partial z}\left(-3e^{-5x}\sin 4y\sin 3z\right)& \end{array}\right\} \\ =25e^{-5x}\sin 4y\cos 3z-16e^{-5x}\sin y\cos 3z-9e^{-5x}\sin 4y\cos 3z \end{gathered}$$

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.

(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 role="math" localid="1657606397182" ${\nabla }^2\overrightarrow{v}=\frac{{\partial }^{2}v}{\partial x^2}+\frac{{\partial }^{2}v}{\partial y^2}+\frac{{\partial }^{2}v}{\partial z^2}$.

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$2\stackrel{⏜}{x}+6x\stackrel{⏜}{y}$ .