Innovative Teaching Methods In Formation Of Professional Competencies Of Future Mathematics Teachers

Methods for visualizing the solution of problems in Geogebra software.

Methods for visualizing the solution of problems in Geogebra software is used in the examples 1-3 below [Berman, 2005].

Example 1. Calculate the area of a curvilinear trapezoid bounded by a line

$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{-\mathrm{x}}\left({\mathrm{x}}^{\mathrm{2}}+\mathrm{3}\mathrm{x}+\mathrm{1}\right)+{\mathrm{e}}^{\mathrm{2}}$, Ox axis and two lines parallel to the axes Oy, drawn through the points of the extremum of the function y.

The solution. To determine the boundaries of integration, it is necessary to find the extremum points of a given function. This can be done manually, by examining the function, which will take a lot of time. The capabilities of the GeoGebra software are used to solve the task. The required extremum points will be found by applying the tool.

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$$\mathrm{S}=\underset{-\mathrm{2}}{\overset{\mathrm{1}}{\int }}\left({\mathrm{e}}^{-\mathrm{х}}({\mathrm{x}}^{\mathrm{2}}+\mathrm{3}\mathrm{x}+\mathrm{1}\right)+{\mathrm{e}}^{\mathrm{2}})\mathrm{d}\mathrm{x}={\mathrm{3}\mathrm{e}}^{\mathrm{2}}-\mathrm{12}{\mathrm{e}}^{-\mathrm{1}}$$

Example 2. Calculate the areas, bounded by lines y $=\mathrm{l}\mathrm{n}\mathrm{x}$ and $\mathrm{y}={\mathrm{l}\mathrm{n}}^{\mathrm{2}}\mathrm{x}$.

The solution. The sought-for region bounded by given curves is constructed in the GeoGebra packet.

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The borders of integration will be found from the figure

$$\mathrm{S}=\underset{\mathrm{0}}{\overset{\mathrm{1}}{\int }}\left({\mathrm{e}}^{\sqrt{\mathrm{y}}}-{\mathrm{e}}^{\mathrm{y}}\right)\mathrm{d}\mathrm{y}=={\mathrm{3}-\mathrm{e}}^{}$$

Example 3. Find the volume of the body bounded by the conic surface ${(\mathrm{z}\mathrm{}-\mathrm{2})}^{\mathrm{2}}=\mathrm{}\frac{{\mathrm{x}}^{\mathrm{2}}}{\mathrm{3}}\mathrm{}+\mathrm{}\frac{{\mathrm{y}}^{\mathrm{2}}}{\mathrm{2}}\mathrm{}\mathrm{}$ and the plane $\mathrm{z}=\mathrm{0}$.

The solution. The GeoGebra software is used. The equations of the surfaces will be given and will be founded the section of the cone by the plane $\mathrm{}\mathrm{z}=\mathrm{0}\mathrm{}$for the construction of a solid in the input line. The following will be gotten:

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The limits of integration are found from the figure.

$$\mathrm{V}=\mathrm{4}\underset{\mathrm{0}}{\overset{\sqrt{\mathrm{12}}}{\int }}\mathrm{d}\mathrm{x}\underset{\mathrm{0}}{\overset{\sqrt{\frac{\mathrm{4}}{\mathrm{3}}(\mathrm{12}-{\mathrm{x}}^{\mathrm{2}}})}{\int }}\left(\mathrm{2}-\sqrt{\frac{{\mathrm{x}}^{\mathrm{2}}}{\mathrm{3}}+\frac{{\mathrm{y}}^{\mathrm{2}}}{\mathrm{2}}}\right)\mathrm{d}\mathrm{y}=\frac{\mathrm{8}\sqrt{\mathrm{6}}\mathrm{\pi }}{\mathrm{3}}$$

Methods for visualizing the solution of problems in Maple software.

The Maple software is used in the examples 4-7 [Romanova & Trachevskaya, 2005].

Example 4. Calculate the areas, bounded by the lines $\mathrm{}\mathrm{у}=\mathrm{1}-{\mathrm{x}}^{\mathrm{2}}$ and $\mathrm{у}={(\mathrm{x}-\mathrm{1})}^{\mathrm{2}}$.

The solution. The figure of the calculating areas is plotted:

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The command solve must be applied to find the intersection points of the curves:

$\text{solve}\left(\left\{\mathrm{y}=\mathrm{1}-x^{\mathrm{2}},\mathrm{y}=(\mathrm{x}-\mathrm{1}{)}^{\mathrm{2}}\right\},\{x,y\}\right)$

$$\{x=\mathrm{0},y=\mathrm{1}\},\{x=\mathrm{1},y=\mathrm{0}\}$$

The area of the required figure is founded by the determination the limits of integration:

$$\mathrm{S}=\underset{\mathrm{0}}{\overset{\mathrm{1}}{\int }}\left(\mathrm{1}-{\mathrm{x}}^{\mathrm{2}}-{\left(\mathrm{x}-\mathrm{1}\right)}^{\mathrm{2}}\right)\mathrm{d}\mathrm{x}=\frac{\mathrm{1}}{\mathrm{3}}$$

Example 5. Calculate the areas, bounded by the lines $\mathrm{у}=-\mathrm{x}-\mathrm{2}$, $\mathrm{у}=\mathrm{x}-\mathrm{2}$ and $\mathrm{у}=\mathrm{2}$.

The solution. The command intersection must be applied to construct the given lines and find their intersection points

[The image of the desired shape]

The areas of the sought-for figure will be calculated with the determining the limits of integration:

$$\mathrm{V}=\underset{-\mathrm{2}}{\overset{\mathrm{2}}{\int }}\left(\mathrm{y}+\mathrm{2}-\left(-\mathrm{y}-\mathrm{2}\right)\right)\mathrm{d}\mathrm{y}=\mathrm{16}$$

Example 6. Calculate the volume of a solid bounded by an elliptic paraboloid $\frac{{\mathrm{x}}^{\mathrm{2}}}{\mathrm{4}}+\frac{{\mathrm{y}}^{\mathrm{2}}}{\mathrm{2}}=\mathrm{z}$ and the plane $\mathrm{z}=\mathrm{1}$.

The solution. It will be determined what the intersection of these surfaces is.

The image of the elliptic paraboloid and the plane will be obtained using the plot3d function.

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It should be noted that the line of intersection of these surfaces is an ellipse. It is necessary to find the equation of the ellipse, solving jointly the equations of the paraboloid and the plane:

$\frac{{\mathrm{x}}^{\mathrm{2}}}{\mathrm{4}}+\frac{{\mathrm{y}}^{\mathrm{2}}}{\mathrm{2}}=\mathrm{1}$.

The implicitplot function should be used to construct a two-dimensional graph.

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The contour of the projection of the intersection of surfaces on the Oxy axis is an ellipse with semiaxes equal to 2 and $\sqrt{\mathrm{2}}$.

The volume will be found using the integral results of the construction.

The accuracy of the results should be checked using Maple.

$$\left.\begin{array}{l}{\int }_{-\mathrm{2}}^{\mathrm{2}} \left({\int }_{-\sqrt{\mathrm{2}-\frac{x^{\mathrm{2}}}{\mathrm{2}}}}^{\sqrt{\mathrm{2}-\frac{x^{\mathrm{2}}}{\mathrm{2}}}} \left(\mathrm{1}-\left(\frac{\mathrm{1}}{\mathrm{4}}\cdot x^{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{*}{y}^{\mathrm{2}}\right)\right)\mathrm{d}y\right]\mathrm{d}x\\ \end{array}\right)$$ $$\sqrt{\mathrm{2}}\pi$$

Example 7. Calculate volume of solid, bounded by these surfaces, using triple integral $\mathrm{z}=\mathrm{10}-{\mathrm{x}}^{\mathrm{2}},\mathrm{}\mathrm{}\mathrm{z}=\mathrm{0},\mathrm{}{\mathrm{x}}^{\mathrm{2}}{+\mathrm{y}}^{\mathrm{2}}=\mathrm{4}$

The solution. Plot these surfaces (Fig. 8).

restart $:$ with $($ plottools $):$ with $($ plots $):$ with $($ student $):$ $$QI:=\mathrm{c}\mathrm{y}\mathrm{l}\mathrm{i}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{r}\left(\right[\mathrm{0,0},\mathrm{0}],\mathrm{2},.\mathrm{01}):$$ $$Q\begin{array}{rr}\mathrm{2}& \mathrm{}:=\text{ cylinderplot }(\mathrm{2},\theta =\mathrm{0}..\mathrm{2}\cdot \pi ,z=\mathrm{0}..\mathrm{10},\text{ style }=\text{ patchnogrid, lightmodel }\\ & \mathrm{}=\text{ light }\mathrm{2}):\\ Q\mathrm{3}:=& \text{ plot }\mathrm{3}d\left(\mathrm{10}-x^{\mathrm{2}},x=-\sqrt{\mathrm{10}}..\sqrt{\mathrm{10}},y=-\sqrt{\mathrm{10}}-\sqrt{\mathrm{10}},\text{ style }\right.\\ & \mathrm{}=\text{ wireframe }):\\ & \text{ plots }\left[\text{ display }\right]\left(\right[Q\mathrm{1},Q\mathrm{2},Q\mathrm{3}],\text{ axes }=\text{ normal, orientation }=[\mathrm{116,63}\left]\right);\end{array}$$

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It is possible to confine to the image of a quarter of the body. From the symmetry of the figure relative to the Ox and Oy axes, the volume of which is required to be found (Fig. 9):

restart $:$ with $($ plottools $):$ with $($ plots $):$ with $($ student $):$

$QI:=$ cylinderplot $(\mathrm{2},\theta =\mathrm{0}..\pi /\mathrm{2},z=\mathrm{0}..\mathrm{0.1}):$ $$\begin{array}{rr}Q\mathrm{2}& :=\text{ cylinderplot }(\mathrm{2},\theta =\mathrm{0}..\pi /\mathrm{2},z=\mathrm{0}..\mathrm{10},\text{ style }=\text{ patchnogrid, lightmodel }\\ & =\text{ light }\mathrm{2}):\\ Q\mathrm{3}& :=\text{ plot }\mathrm{3}d\left(\mathrm{10}-{\mathrm{x}}^{\mathrm{2}},x=\mathrm{0}..\mathrm{3},y=\mathrm{0}..\mathrm{3},\text{ style }=\text{ patchnogrid }\right):\\ Q\mathrm{4}& :=\text{ implicitplot }\mathrm{3}d(x=\mathrm{0},x=\mathrm{0}..\mathrm{3},y=\mathrm{0}..\mathrm{2},z=\mathrm{0}..\mathrm{10},\text{ style }=\text{ wireframe },\\ & \text{ linestyle }=\mathrm{3},\text{ color }=\text{ BLACK }):\end{array}$$ $$\begin{array}{rr}& \text{ Q5 }:=\text{ implicitplor }\mathrm{3}d(y=\mathrm{0},x=\mathrm{0}..\mathrm{2},y=\mathrm{0}..\mathrm{3},z=\mathrm{0}..\mathrm{10},\text{ style }=\text{ wireframe, }\\ \text{ linestyle }=\mathrm{3},\text{ color }=& BLACK):\end{array}$$plots $[$ display $\left]\right([Q\mathrm{1},Q\mathrm{2},Q\mathrm{3},Q\mathrm{4},Q\mathrm{5}]$, axes $=$ normal, orientation $=[\mathrm{116},$, $$\mathrm{63}\left]\right);$$

It is necessary to go into a cylindrical coordinate system and calculate the volume:

restart : with ( plottools): with ( plots):with (student) : $$a\mathrm{s}\mathrm{i}\mathrm{g}n(x=\rho \cdot \mathrm{c}\mathrm{o}\mathrm{s}(\phi ),y=\rho \cdot \mathrm{s}\mathrm{i}\mathrm{n}(\phi ),z=zI);$$ $$z:=\mathrm{10}-x^{\mathrm{2}}$$ $$z:=-{\rho }^{\mathrm{2}}\mathrm{c}\mathrm{o}\mathrm{s}(\phi {)}^{\mathrm{2}}+\mathrm{10}$$ $\mathrm{V}=\mathrm{4}\cdot$ Tripleint $\left.(\rho ,z)=\mathrm{0}..\left(\mathrm{10}-(\rho \cdot \mathrm{c}\mathrm{o}\mathrm{s}(\phi ){)}^{\mathrm{2}}\right),\rho =\mathrm{0}\dots \mathrm{2},\phi =\mathrm{0}..\pi /\mathrm{2}\right)$ $$V=\mathrm{4}\left({\int }_{\mathrm{0}}^{\frac{\mathrm{1}}{\mathrm{2}}\pi } {\int }_{\mathrm{0}}^{\mathrm{2}} {\int }_{\mathrm{0}}^{-p^{\mathrm{2}}\mathrm{c}\mathrm{o}\mathrm{s}(\phi {)}^{\mathrm{2}}+\mathrm{10}} \rho \mathrm{d}z\mathrm{1}\mathrm{d}\rho \mathrm{d}\phi \right)$$ $V=$ value $\left(\mathrm{4}\cdot \right.$ Tripleint $\left.\left(\rho ,zl=\mathrm{0}..\left(\mathrm{10}-(\rho \cdot \mathrm{c}\mathrm{o}\mathrm{s}(\phi ){)}^{\mathrm{2}}\right),\rho =\mathrm{0}..\mathrm{2},\phi =\mathrm{0}..\pi /\mathrm{2}\right)\right)$ $$V=\mathrm{36}\pi$$