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Load the Student :-MultivariateCalculus package.

From the Tools menu, select Load Package, then Student Multivariate Calculus.

Loading Student:-MultivariateCalculus 

Define the vector V.

$\mathbf{V}\=⟨a\,b⟩$$\stackrel{\text{assign}}{\to }$

Find the norm of V using typeset math.

$∥\mathbf{V}∥$ = $\sqrt{a^2\+b^2}$$$

Alternatively, find the norm using context menu options.

$\mathbf{V}$ = $\stackrel{\text{norm}}{\to }$$\sqrt{a^2\+b^2}$

Normalize V using the context menu.

$\mathbf{V}$ = $\stackrel{\text{normalize}}{\to }$

Use the Norm command to find the norm of V.

Type: "Norm(V)" and press Enter.

$\mathrm{Norm}\left(\mathbf{V}\right)$ = $\sqrt{a^2\+b^2}$$$

Use the Normalize command to normalize V.

Type: "Normalize(V)" and press Enter.

$\mathrm{Normalize}\left(\mathbf{V}\right)$ = $$

Step

Instructions

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Load the Student:-MultivariateCalculus package.

Loading Student:-MultivariateCalculus.

Define Vectors A, B and C.

$\mathbf{A}\=⟨a\,b\,c⟩$$\stackrel{\text{assign}}{\to }$

$\mathbf{B}\=⟨u\,v\,w⟩$$\stackrel{\text{assign}}{\to }$

$\mathbf{C}\=⟨p\,q\,r⟩$$\stackrel{\text{assign}}{\to }$

Vector Products via Typeset Math (See Common Symbols Palette).

Compute the vector products using typeset math.

Compute $\mathbf{A}·\mathbf{B}$ using typeset math.

$\mathbf{A}·\mathbf{B}$ = $au\+bv\+cw$$$

Compute $\mathbf{A}\times \mathbf{B}$ using typeset math.

$\mathbf{A}\times \mathbf{B}$ = $$

Compute $\mathbf{A}·\left(\mathbf{B}\times \mathbf{C}\right)$using typeset math.

$\mathbf{A}·\left(\mathbf{B}\times \mathbf{C}\right)$ = $a\left(-qw\+rv\right)\+b\left(pw-ru\right)\+c\left(-pv\+qu\right)$$$

Vector Products via the Context Menu

Compute   $\mathbf{A}·\mathbf{B}$ from the context menu.

$\mathbf{A}\,\mathbf{B}$ = $\stackrel{\text{dot product}}{\to }$$au\+bv\+cw$

Compute $\mathbf{A}\times \mathbf{B}$ from the context menu.

$\mathbf{A}\,\mathbf{B}$ = $\stackrel{\text{cross product}}{\to }$

Compute $\mathbf{A}·\left(\mathbf{B}\times \mathbf{C}\right)$from the context menu.

$\mathbf{A}\,\mathbf{B}\,\mathbf{C}$ = $\stackrel{\text{scalar triple product}}{\to }$$a\left(-qw\+rv\right)\+b\left(pw-ru\right)\+c\left(-pv\+qu\right)$

Vector Products via the DotProduct, CrossProduct, and TripleScalarProduct Commands

Compute   $\mathbf{A}·\mathbf{B}$ using the DotProduct command.

$\mathrm{DotProduct}\left(\mathbf{A}\,\mathbf{B}\right)$ = $au\+bv\+cw$$$

Compute $\mathbf{A}\times \mathbf{B}$ using the CrossProduct command.

$\mathrm{CrossProduct}\left(\mathbf{A}\,\mathbf{B}\right)$ = $$

Compute $\mathbf{A}·\left(\mathbf{B}\times \mathbf{C}\right)$using the TripleScalarProduct command.

$\mathrm{TripleScalarProduct}\left(\mathbf{A}\,\mathbf{B}\,\mathbf{C}\right)$ = $a\left(-qw\+rv\right)\+b\left(pw-ru\right)\+c\left(-pv\+qu\right)$$$

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Load the Student:-MultivariateCalculus package.

Loading Student:-MultivariateCalculus 

Define vector $\mathbf{V}\mathbf{\.}$

$\mathbf{V}\=⟨f\left(x\right)\,g\left(x\right)⟩$$\stackrel{\text{assign}}{\to }$

Define vector $\mathbf{v}\mathbf{\.}$

$\mathbf{v}\=⟨x\left(t\right)\,y\left(t\right)⟩$$\stackrel{\text{assign}}{\to }$

Differentiate via Typeset Math

Differentiate $\mathbf{V}\mathbf{\.}$

$\mathbf{V}\prime$ = $$

Differentiate $\mathbf{v}\mathbf{\.}$

$\stackrel{\.}{\mathbf{v}}$ = $$

Differentiate via Context Menu

Differentiate $\mathbf{V}\mathbf{\.}$

$\mathbf{V}$ = $\stackrel{\text{differentiate}}{\to }$

Differentiate $\mathbf{v}\mathbf{\.}$

$\mathbf{v}$ = $\stackrel{\text{differentiate}}{\to }$

Differentiate via the Differentiation Operator in the Calculus Palette

Differentiate $\mathbf{V}\mathbf{\.}$ 

$\frac{\ⅆ}{\ⅆ x} \mathbf{V}$ = $$

Differentiate $\mathbf{v}\mathbf{\.}$

$\frac{\ⅆ}{\ⅆ t} \mathbf{v}$ = $$

Differentiation via the Modified diff Command

Differentiate  $\mathbf{V}\mathbf{\.}$ 

$\mathrm{diff}\left(\mathbf{V}\,x\right)$ = $$

Differentiate $\mathbf{v}\mathbf{\.}$

$\mathrm{diff}\left(\mathbf{v}\,t\right)$ = $$

Integrate $\mathit{f}\left(\mathit{x}\mathbf{\,}\mathit{y}\right)\mathbf{\=}\mathit{x}\mathbf{ }\mathit{y}$ over the upper half of the circle whose center is $⟨\mathbf{1}\mathbf{\,}\mathbf{2}⟩$ and whose radius is $\mathit{R}$.

Instructions

Results

Loading Student:-MultivariateCalculus 

Use the MultiInt Command.

$\mathrm{MultiInt}\left(x y\,\left[x\,y\right]\=\mathrm{Sector}\left(\mathrm{Circle}\left(⟨1\,2⟩\,R\,\left[r\,\mathrm{\theta }\right]\right)\,0\, \mathrm{\pi }\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^R{\∫}_0^{\mathrm{\π}}\left(r\cos \left(\mathrm{\θ}\right)\+1\right)\left(r\sin \left(\mathrm{\θ}\right)\+2\right)r\ⅆ\mathrm{\θ}\ⅆr$

$\mathrm{MultiInt}\left(x y\,\left[x\,y\right]\=\mathrm{Sector}\left(\mathrm{Circle}\left(⟨1\,2⟩\,R\,\left[r\,\mathrm{\theta }\right]\right)\,0\, \mathrm{\pi }\right)\,\mathrm{output}\=\mathrm{value}\right)$ = $\mathrm{\π}{R}^2\+\frac{2}{3}{R}^3$$$

$\mathrm{MultiInt}\left(x y\,\left[x\,y\right]\=\mathrm{Sector}\left(\mathrm{Circle}\left(⟨1\,2⟩\,R\,\left[r\,\mathrm{\theta }\right]\right)\,0\, \mathrm{\pi }\right)\,\mathrm{output}\=\mathrm{steps}\right)$

$\mathrm{\π}{R}^2\+\frac{2}{3}{R}^3$

Calculate the surface area of that portion of the surface $\mathit{z}\mathbf{\=}\mathbf{3}\mathbf{-}{\mathit{x}}^{\mathbf{2}}\mathbf{\/}\mathbf{4}\mathbf{-}{\mathit{y}}^{\mathbf{2}}\mathbf{\/}\mathbf{3}$ that is defined over the triangle whose vertices are $\left(\mathbf{0}\mathbf{\,}\mathbf{0}\right)\mathbf{\,}\left(\mathbf{3}\mathbf{\,}\mathbf{0}\right)\mathbf{\,}\left(\mathbf{1}\mathbf{\,}\mathbf{2}\right)$.

Instructions

Result

Loading Student:-MultivariateCalculus.

$Z\=3-x^2\/4-y^2\/3$$\stackrel{\text{assign}}{\to }$

Use SurfaceArea Command.

$\mathrm{SurfaceArea}\left(\mathrm{Surface}\left(⟨x\,y\,Z⟩\,\left[x\,y\right]\=\mathrm{Triangle}\left(⟨0\,0⟩\,⟨3\,0⟩\,⟨1\,2⟩\right)\right)\,\mathrm{output}\=\mathrm{plot}\,\mathrm{axes}\=\mathrm{frame}\,\mathrm{labels}\=\left[x\,y\,z\right]\right)$

$\mathrm{SurfaceArea}\left(\mathrm{Surface}\left(⟨x\,y\,Z⟩\,\left[x\,y\right]\=\mathrm{Triangle}\left(⟨0\,0⟩\,⟨3\,0⟩\,⟨1\,2⟩\right)\right)\,\mathrm{output}\=\mathrm{integral}\right)$

$-\left({\∫}_0^1{\∫}_{2x}^0\frac{1}{6}\sqrt{9x^2\+16y^2\+36}\ⅆy\ⅆx\right)-\left({\∫}_1^3{\∫}_{3-x}^0\frac{1}{6}\sqrt{9x^2\+16y^2\+36}\ⅆy\ⅆx\right)$

$\mathrm{evalf}\left(\mathrm{SurfaceArea}\left(\mathrm{Surface}\left(⟨x\,y\,Z⟩\,\left[x\,y\right]\=\mathrm{Triangle}\left(⟨0\,0⟩\,⟨3\,0⟩\,⟨1\,2⟩\right)\right)\right)\right)$ = $4.028651111$$$