91Ó°ÊÓ

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 $∇=\frac{∂}{∂x}i+\frac{∂}{∂y}j+\frac{∂}{∂z}k$. The value of Laplacian${∇}^{2}v$is obtained as

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

Here,${∇}^2$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$\stackrel{â‡\P Ä}{T_a}$is defined as $x^{2}i+2xyj+3zk+4$. The Laplacian of the vector o $\stackrel{â‡\P Ä}{T_a}$is defined as ${∇}^2{T}_a=\frac{{∂}^2{T}_a}{∂x^2}+\frac{{∂}^2{T}_a}{∂y^2}+\frac{{∂}^2{T}_a}{∂z^2}$.

Substitute $x^{2}i+2xyj+3zk+4\mathrm{for}{\stackrel{â‡\P Ä}{\mathrm{T}}}_{\mathrm{a}}\mathrm{into}{∇}^2{\mathrm{T}}_{\mathrm{a}}=\frac{{∂}^2{\mathrm{T}}_{\mathrm{a}}}{∂{\mathrm{x}}^2}+\frac{{∂}^2{\mathrm{T}}_{\mathrm{a}}}{∂{\mathrm{y}}^2}+\frac{{∂}^2{\mathrm{T}}_{\mathrm{a}}}{∂{\mathrm{z}}^2}$into .

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

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

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

(b)

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

The vector$\stackrel{â‡\P Ä}{T_b}$is defined as $\sin x\sin y\sin z$. The Laplacian of the vector$\stackrel{â‡\P Ä}{T_b}$is defined as localid="1657538369566" ${∇}^2{T}_b=\frac{{∂}^2{T}_b}{∂x^2}+\frac{{∂}^2{T}_b}{∂y^2}+\frac{{∂}^2{T}_b}{∂z^2}$.

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

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

Solve further as,

${∇}^2{T}_b=-3\sin x\sin y\sin z$

Thus, the value of ${∇}^2{T}_b$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${\stackrel{→}{T}}_c$is defined as $e^{-5x}\sin 4y\cos 3z$. The Laplacian of the vector${\stackrel{→}{T}}_c$is defined as${∇}^2{T}_c=\frac{{∂}^2{T}_c}{∂x^2}+\frac{{∂}^2{T}_c}{∂y^2}+\frac{{∂}^2{T}_c}{∂z^2}$.

Substitute role="math" localid="1657605202299" $e^{-5x}\sin 4y\cos 3z$for ${\stackrel{→}{T}}_c$into ${∆}^2{T}_c=\frac{{∂}^2{T}_c}{∂x^2}+\frac{{∂}^2{T}_c}{∂y^2}+\frac{{∂}^2{T}_c}{∂z^2}$.

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

Solve further as,

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

Thus, the value of ${∇}^2{T}_c$is 0.

(d)

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

The vector $\stackrel{→}{v}$is defined as $x^{2}i+3x{z}^{2}j-2xzk$. The Laplacian of the vector$\stackrel{→}{v}$is defined as role="math" localid="1657606397182" ${∇}^2\stackrel{→}{v}=\frac{{∂}^{2}v}{∂x^2}+\frac{{∂}^{2}v}{∂y^2}+\frac{{∂}^{2}v}{∂z^2}$.

Substitute $x^{2}i+3x{z}^{2}j-2xzk$for $\stackrel{→}{v}$into ${∇}^2\stackrel{→}{v}=\frac{{∂}^{2}v}{∂x^2}+\frac{{∂}^{2}v}{∂y^2}+\frac{{∂}^{2}v}{∂z^2}$.

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

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

Solve further as,

${∇}^{2}v=2i+6xj$

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