Applications-of-GeoGebra/C3/Differentiation-using-GeoGebra/English - Revision history

Snehalathak at 06:05, 15 January 2019

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Madhurig at 10:25, 14 January 2019

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Madhurig at 18:36, 13 January 2019

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Madhurig at 18:13, 13 January 2019

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Vidhya at 13:09, 11 January 2019

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Vidhya at 09:37, 8 October 2018

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Vidhya: Created page with " {|border=1 | | '''Visual Cue''' | | '''Narration''' |- | | '''Slide Number 1''' '''Title Slide''' | | Welcome to this tutorial on '''Differentiation using GeoGebra'''. |- |..."

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Created page with " {|border=1 | | '''Visual Cue''' | | '''Narration''' |- | | '''Slide Number 1''' '''Title Slide''' | | Welcome to this tutorial on '''Differentiation using GeoGebra'''. |- |..."

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{|border=1
| | '''Visual Cue'''
| | '''Narration'''

|-
| | '''Slide Number 1'''

'''Title Slide'''
| | Welcome to this tutorial on '''Differentiation using GeoGebra'''.
|-
| | '''Slide Number 2'''

'''Learning Objectives'''
| | In this tutorial, we will learn how to use '''GeoGebra''' to:

Understand Differentiation

Draw graphs of derivative of functions

|-
| | '''Slide Number 3'''

'''System Requirement'''
| | Here I am using:

'''Ubuntu Linux''' OS version 16.04

'''GeoGebra''' 5.0.481.0-d

|-
| | '''Slide Number 4'''

'''Pre-requisites'''

'''www.spoken-tutorial.org'''
| | To follow this '''tutorial''', you should be familiar with:

'''GeoGebra''' interface

Differentiation

For relevant '''tutorials''', please visit our website.

|-
| | '''Slide Number 5'''

'''Differentiation: First Principles'''

[[Image:]]

'''f(x) = x2-x'''

'''f'(x)''' is derivative of '''f(x)'''

'''A (x, f(x)), B (x+j, f(x+j))'''
| |

Let us understand differentiation using '''first principles''' for the '''function f of x'''.

'''f of x''' is equal to '''x squared''' minus '''x'''

'''f prime x''' is the derivative of '''f of x'''.

Consider 2 points, '''A''' and '''B'''.

'''A''' is '''x''' comma '''f of x''' and '''B''' is '''x''' plus '''j''' comma '''f of x''' plus '''j'''
|-
| | Show the '''GeoGebra''' window.
| | I have opened the '''GeoGebra''' interface.
|-
| | Type '''f(x)=x^2-x''' in the '''input bar''' >> '''Enter'''
| | In the '''input bar''', type the following line and press '''Enter'''.

For the '''caret symbol''', hold the '''Shift''' key down and press 6.

'''f x''' in parentheses equals '''x caret 2''' minus '''x'''
|-
| | Point to the equation in '''Algebra''' view.

Point to '''parabola''' in '''Graphics''' view.
| | The equation appears in the '''Algebra''' view and the '''function f''' is graphed as a '''parabola'''.
|-
| | Point to '''parabola''' in '''Graphics''' view.
| | It opens upwards and intersects the '''x-axis''' at two points.
|-
| | Click on '''Point on Object''' tool >> click on the parabola at '''(2,2)'''.

Point to '''A''' at '''(2,2)'''.
| | Under '''Point''', click on '''Point on Object''' and click on the parabola at 2 comma 2.

This creates point '''A'''.
|-
| | Click on '''Point''' tool and click on '''(3,6)'''.
| | Create a point '''B''' at 3 comma 6.
|-
| | Click on '''Line''' tool and click on points '''B''' and '''A'''.
| | Click on '''Line''' tool and click on points '''B''' and '''A'''.
|-
| | Click on the '''Move''' tool.

Double click on the resulting '''line g''' and click on '''Object Properties'''.

Click on '''Color''' tab and select blue.

Click on '''Style''' tab and select '''dashed style'''.
| | Let us make this line '''g '''blue and dashed.
|-
| | Click on '''Tangents''' tool under '''Perpendicular Line''' tool.
| | Under '''Perpendicular Line''', click on '''Tangents'''.
|-
| | Click on '''A''' and then on the '''parabola'''.
| | Then click on '''A''' and then on the '''parabola'''.
|-
| | Point to '''tangent h''' at point '''A''' to the '''parabola'''.
| | This draws a '''tangent h''' at point '''A''' to the '''parabola'''.
|-
| | Double click on '''tangent h''' and click on '''Object Properties'''.

Under '''Color''' tab, select red.

Close the '''Preferences''' box.
| | Let us make '''tangent h''' a red line.
|-
| | Click on '''Point''' tool and click in '''Graphics''' view.
| | Click on the '''Point''' tool and click anywhere in '''Graphics''' view.
|-
| | Point to point '''C'''.
| | This creates point '''C'''.
|-
| | Double click on point '''C''' in '''Algebra''' view and change its '''coordinates''' to '''(x(B),y(A))'''.
| | In '''Algebra''' view, double-click on '''C''' and change its '''coordinates''' to the following ones.

In parentheses, '''x B''' in parentheses comma '''y A''' in parentheses.

Press '''Enter'''.
|-
| | Point to '''C'''.
| | Now C has the same '''x coordinate''' as point '''B''' and the same '''y coordinate''' as point '''A'''.
|-
| |
| | Let us draw segments '''BC''' and '''AC.'''
|-
| | Under '''Line''', click on '''Segment''' and click on '''B '''and '''C''', and then on '''A''' and '''C'''.
| | Under '''Line''', click on '''Segment''' and click on '''B''' and '''C'''.

Then, click on '''A''' and '''C''' to draw '''AC'''.

Note that '''BC''' and '''AC''' are called '''i''' and '''j''' in the order of their creation.
|-
| | Right-click on '''AC''' >> Select '''Object Properties''' >> '''Color''' tab >> Purple

Click on '''Style''' tab >> select dashed line
| | We will make '''AC''' and '''BC''' purple and dashed segments.
|-
| | Under '''Basic''' tab >> choose '''Name and Value''' >> '''Show Label''' check box.
| | Under '''Basic''' tab, choose '''Name and Value''' from the dropdown menu next to the '''Show Label''' check box.
|-
| | Close the '''Preferences''' dialog box.
| | Close the '''Preferences''' dialog box.
|-
| | Click on '''Move''' and drag '''B''' towards '''A''' on the '''parabola'''.
| | Click on '''Move''' and drag '''B''' towards '''A''' on the '''parabola'''.
|-
| | Point to the value of '''j''' (length of '''AC''') and lines '''g''' and '''h'''.
| | Observe lines '''g''' and '''h''' and the value of '''j''' (length of '''AC''').

As '''j''' approaches 0, points '''B'''and '''A''' begin to overlap.

Lines '''g''' and '''h''' also begin to overlap.
|-
| | Point to line '''g''', '''BC''' and '''AC'''.
| | Slope of line '''g''' is the ratio of length of '''BC''' to length of '''AC'''.
|-
| | Point to all the points on the parabola.
| | Derivative of the parabola is the slopes of tangents at all points on curve.
|-
| | Point to the lines '''g''' and '''h'''.
| | When '''j''' equals 0, line '''g''' through '''A''' and '''B''' coincides with the tangent line '''h''' at '''A'''.

But slope of '''g''' is undefined as the denominator of the ratio of '''i''' to '''j''' is 0.
|-
| | Point to text-box that appears in '''GeoGebra''' window.

As '''B''' approaches '''A''', slope '''AB''' approaches slope of tangent at '''A'''.
| | As '''B''' approaches '''A''' on '''f of x''', slope of '''AB''' approaches the slope of tangent at '''A'''.
|-
| |
| | Now let us look at the '''Algebra''' behind these concepts.
|-
| | '''Slide Number 7'''

'''Differentiation: First Principles, the Algebra'''

'''f'(x) = lim_(j→0) length of Segment BC/length of Segment AC'''

.......''' = lim_(j→0) [(f(x+j) – f(x)]/[(x+j) – x]'''

Remember '''f(x) = x2-x, (x+j)2 = x2+2xj+j2'''

'''f'(x) = lim_(j→0) [(x+j)2-(x+j)-(x2-x)]/[(x+j-x]'''

| | Slope of line '''AB''' equals the ratio of the lengths of '''BC''' to '''AC'''.

Line '''AB''' becomes the tangent at point '''A''' as distance '''j''' between '''A''' and '''B''' approaches 0.

'''BC''' is the difference between '''y' coordinates''', '''f of x''' plus '''j''' and '''f of x''', for '''A''' and '''B'''.

'''AC''' is the difference between the '''x-coordinates''', '''x''' plus '''j''' and '''x'''.

Let us rewrite '''f of x''' plus '''j''' and '''f of x''' in terms of '''x squared''' minus '''x'''.

We will expand the terms in the numerator.
|-
| | '''Slide 8 Differentiation: First Principles—the Algebra-Cont’d'''

'''f'(x) = lim_(j→0) [x2+2xj+j2-x-j-x2+x]/j'''

..............'''= lim_(j→0) [2xj+j2-j]/j = lim_(j→0) [j(2x+j-1)]/j'''

..................'''= lim_(j→0) [2x+j-1] = 2x-1'''

'''f'(x2-x) = 2x - 1'''
| | After expanding the terms in the numerator, we will cancel out similar terms with opposite signs.

As '''j''' is a common factor in the numerator, we will pull it out.

Now, we can cancel '''j''' from both the numerator and denominator.

Note that as '''j''' approaches 0, '''j''' can be ignored so that '''2x''' plus '''j''' minus 1 approaches '''2x''' minus 1.

As we know, derivative of '''x squared''' minus '''x''' is '''2x''' minus 1.

Thus, the derivative of a '''function''' is the slope of the tangent at a point on the function.
|-
| |
| | Let us look at derivative graphs for some '''functions'''.
|-
| | '''Slide Number 10'''

'''Differentiation of a Polynomial Function'''

Consider '''g(x)=5+12x-x3'''

'''d(5+12x-x3)/dx = d(5)/dx + d(12x)/dx - d(x3)/dx = 0 + 12 - 3x 2 = -3x 2 +12'''

For '''g(x)=5+12x-x3, g'(x) = -3x2 +12'''
| |
Consider '''g of x''' equals 5 plus '''12 x '''minus '''x cubed'''.

Derivative '''g prime x''' is the sum and difference of derivatives of the individual components.

'''g prime x''' equals 5 plus '''12 x''' minus '''x cubed''' is calculated by applying these rules.
|-
| |
| | Let us differentiate '''g of x''' in '''GeoGebra'''.
|-
| | Open a new '''GeoGebra''' window.
| | Open a new '''GeoGebra''' window.
|-
| | Type '''g(x)=5+12x-x^3''' in '''input bar''' >> '''Enter'''
| | In the '''input bar''', type the following line and press '''Enter'''.

'''g x''' in parentheses equals 5 plus '''12 x''' minus '''x caret''' 3
|-
| | Right-click in '''Graphics''' view and select '''xAxis : yAxis''' option.

Select '''1:5'''.
| | Right-click in '''Graphics''' view and select '''xAxis''' is to '''yAxis''' option.

Select 1 is to 5.
|-
| | Click on '''Point on Object''' tool and click on the curve to create point '''A'''.
| | Click on '''Point on Object''' tool and click on curve '''g of x''' to create point '''A'''.
|-
| | Click on '''Tangent''' under '''Perpendicular Line'''.

Click on point '''A''' and the curve.
| | Under '''Perpendicular Line''', click on '''Tangent'''.

Now click on point '''A''' and the curve '''g of x'''.
|-
| | Point to tangent line '''f''' to the curve at point '''A'''.
| | This draws a tangent line '''f''' to the curve at point '''A'''.
|-
| | Click on '''Slope''' tool under '''Angle''' tool and on tangent line '''f'''.
| | Under '''Angle''', click on '''Slope''' and on tangent line '''f'''.
|-
| | Point to slope of '''line f''' at '''A''' appearing as '''m''' value in '''Graphics''' view.
| | Slope of line '''f'''at '''A''' appears as '''m''' value in '''Graphics''' view.
|-
| | Click on '''Point''' tool and in '''Graphics''' view to create point '''B'''.
| | Click on '''Point''' tool and click in '''Graphics''' view to create point '''B'''.
|-
| | Double click on point '''B''' in '''Algebra''' view and change '''coordinates''' to ('''x(A), m)'''.
| | In '''Algebra''' view, double click on '''B''' and change its '''coordinates''' to '''x A''' in parentheses comma '''m'''.

Press '''Enter'''.
|-
| | Point to points '''A''' and '''B''' and slope ''' m''' of tangent line '''g'''.
| | This makes '''x coordinate''' of '''B''' the same as that of '''A'''.

And slope '''m''' of tangent line '''f''' is its '''y coordinate'''.
|-
| | Right-click on point '''B''' and select '''Trace On''' option.
| | Right-click on point '''B''' and select '''Trace On''' option
|-
| | Click on '''Move''' tool and move point '''A''' on curve.

Observe the curve traced by point '''B'''.
| | Click on '''Move''' tool and move point '''A''' on the curve.

Observe the curve traced by point '''B'''.
|-
| | Type '''Deri''' in '''input bar''' >> select '''Derivative( <Function> )''' >> Type '''g''' instead of highlighted '''<Function>''' >> press '''Enter'''
| | In the '''input bar''', type '''capital D e r i'''.

From the menu that appears, select '''Derivative Function''' option.

Type '''g''' to replace the highlighted word '''<Function>'''.

Press '''Enter'''.
|-
| | Point to the derivative graph.
| | This will confirm that you have the correct derivative graph.
|-
| | Point to the equation of '''g prime x''' in '''Algebra''' view.
| | Note the equation of '''g prime x''' in '''Algebra''' view.
|-
| | Drag the boundary to see it properly.
| | Drag the boundary to see it properly.
|-
| | Compare slide’s calculations with equation of '''g'(x)''' in '''Algebra''' view.
| | Compare the calculations in the previous slide with the equation of '''g prime x'''
|-
| |
| | Let us use differentiation to find the maxima and minima of a '''function'''.

Let us look at '''g of x'''.
|-
| | Point to derivative curve '''g'(x)''' above the '''x-axis''' and to '''g(x)'''.
| | Derivative curve '''g prime x''' remains above the '''x-axis''' (is positive) as long as '''g of x''' is increasing.
|-
| | Point to derivative curve '''g'(x)''' below the '''x-axis''' and to '''g(x)'''.
| | '''g prime x''' remains below the '''x-axis''' (is negative) as long as '''g of x''' is decreasing.
|-
| | Point to derivative curve '''g'(x)''' intersecting '''x-axis''' at '''x = -2 '''and''' x = 2'''.
| | 2 and -2 are the values of '''x''' when '''g prime x''' equals 0.
|-
| |
| | Slope of the tangents at the corresponding points on '''g of x''' is 0.

These points on '''g of x''' are maxima or a minima.
|-
| |

Point to '''(-2,-11)''' and '''(2,21)'''.
| | Hence, for '''g of x,''' -2 comma -11 is the minimum and 2 comma 21 is the maximum.
|-
| | Point to minimum of '''g(x)''' and '''x=-3''' and '''x = -1'''.
| | In '''GeoGebra''', we can see that the minimum value of '''g of x''' lies between '''x''' equals -3 and '''x''' equals -1.
|-
| | In the '''input bar''', type '''Min''' with capital M.

From the menu that appears, select '''Min Function Start x-Value End x-Value''' option.

Type '''g''' instead of the highlighted word '''Function'''.

Press '''Tab''' to move to and highlight '''Start x-Value''' and type -3.

Again, press '''Tab''' to move to and highlight '''End x-Value''' and type -1.

Press '''Enter'''.
| | In the '''input bar''', type '''Min''' with capital M.

From the menu that appears, select '''Min Function Start x-Value End x-Value''' option.

Type '''g''' instead of the highlighted word '''Function'''.

Press '''Tab''' to move to and highlight '''Start x-Value''' and type -3.

Again, press '''Tab''' to move to and highlight '''End x-Value''' and type -1.

Press '''Enter'''.
|-
| | Point to minimum '''C''' in '''Graphics '''view and its '''co-ordinates''' in '''Algebra''' view.
| | In '''Graphics''' view, we see the minimum on '''g of x'''.

Its '''co-ordinates''' are -2 comma -11 in '''Algebra''' view.
|-
| | In the '''input bar''', type '''Max''' with capital M.

From the menu that appears, select '''Max Function Start x-Value End x-Value''' option.

Type '''g''' instead of the highlighted word '''Function'''.

Press '''Tab''' to move to and highlight '''Start x-Value''' and type 1.

Again, press '''Tab''' to move to and highlight '''End x-Value''' and type 4.

Press '''Enter.'''
| | In the '''input bar''', type '''Max''' with capital M.

From the menu that appears, select '''Max Function Start x-Value End x-Value''' option.

Type '''g''' instead of the highlighted word '''Function'''.

Press '''Tab''' to move to and highlight '''Start x-Value''' and type 1.

Again, press '''Tab''' to move to and highlight '''End x-Value''' and type 4.

Press '''Enter'''.
|-
| | Point to maximum '''C''' in '''Graphics''' view and its '''co-ordinates''' in '''Algebra''' view.
| | In '''Graphics''' view, we see the maximum on '''g of x'''.

Its '''co-ordinates''' are 2 comma 21 in '''Algebra''' view.
|-
| |
| | Finally, let us take a look at a practical application of differentiation.
|-
| | '''Slide Number 16'''

'''A Practical Application of Differentiation'''

We have a 24 inches by 15 inches piece of cardboard

We have to convert it into a box

Squares have to be cut from the four corners

What size squares should we cut out to get the maximum volume of the box?
| | '''A Practical Application of Differentiation'''

We have a 24 inches by 15 inches piece of cardboard.

We have to convert it into a box.

Squares have to be cut from the four corners.

What size squares should we cut out to get the maximum volume of the box?
|-
| | '''Slide Number 17'''

'''A Sketch of the Cardboard'''

Let’s draw the cardboard:

[[Image:]]

The volume function here is '''(24-2x)*(15-2x)*x''' cubic inches.
| | '''A Sketch of the Cardboard'''

Let us draw the cardboard:

This is the volume '''function''' here.

You could expand it into a '''cubic polynomial'''<nowiki>; but we will leave it as it is. </nowiki>
|-
| | Open a new '''GeoGebra''' window.
| | Open a new '''GeoGebra''' window.
|-
| | Type '''(24-2 x) (15-2 x) x''' in the '''input bar''' >> '''Enter'''.
| | In the '''input bar''', type the following line and press '''Enter'''.

In parentheses, 24 minus 2 space '''x''' space in parentheses 15 minus 2 space '''x''' space '''x'''
|-
| | Drag the boundary to see the equation properly in '''Algebra''' view.
| | Drag the boundary to see the equation properly in '''Algebra''' view.
|-
| | Right-click in '''Graphics''' view and set '''xAxis : yAxis''' to '''1:50'''.
| | Right-click in '''Graphics''' view and set '''xAxis''' is to '''yAxis''' to 1 is to 50.
|-
| | Point to the graph for this volume '''function''' in '''Graphics''' view.

Click in and drag the background to move '''Graphics''' view to see the maximum.
| | Observe the graph that is plotted for this volume '''function''' in '''Graphics''' view.

Click in and drag the background to move '''Graphics''' view to see the maximum.
|-
| | Point to the maximum on top of the broad peak and to '''x''' = 0 and '''x''' = 7.
| | Note that the maximum is on the top of a broad peak from '''x''' equals 0 to '''x''' equals 7.
|-
| | Point to both axes.
| | The length of the square side is plotted along the '''x-axis'''.

Volume of the box is plotted along the '''y-axis'''.
|-
| | In the '''input bar''', type '''Max''' with capital M.

From the menu that appears, select '''Max Function Start x-Value End x-value'''.

Instead of highlighted '''Function''', type '''f'''.

Press '''Tab''' to move and highlight '''Start x-Value''' and type 0.

Again, press '''Tab''' to move and highlight '''End x-Value''' and type 10.

Press '''Enter'''.
| | As before, let us find the maximum of this '''function'''.
|-
| | Point to the maximum,''' A''', in '''Graphics''' view and its '''coordinates''' in '''Algebra''' view.
| | This maps the maximum, point '''A''', on the curve.

Click in and drag the background to see

Its '''coordinates''' 3 comma 486 appear in '''Algebra''' view.

Thus, we have to cut out 3 inches squares from all corners.

This will give the maximum possible volume of 486 cubic inches for the cardboard box.
|-
| |
| | Let us summarize.
|-
| | '''Slide Number 19'''

'''Summary'''
| | In this '''tutorial''', we have learnt how to use '''GeoGebra''' to:

Understand differentiation

Draw graphs of derivatives of '''functions'''
|-
| | '''Slide Number 16'''

'''Assignment'''

Draw graphs of derivatives of the following functions in '''GeoGebra''':

'''h(x)=ex'''

'''i(x)=ln(x)'''

'''j(x)=(5x3+3x-1)/(x-1)'''

Find the derivatives of these '''functions''' independently and compare with '''GeoGebra''' graphs.
| | As an assignment:

Draw graphs of derivatives of the following functions in '''GeoGebra'''.

Find the derivatives of these '''functions''' independently and compare with '''GeoGebra''' graphs.
|-
| | '''Slide Number 17'''

'''About Spoken Tutorial project'''
| | The video at the following link summarizes the '''Spoken Tutorial''' project.

Please download and watch it.
|-
| | '''Slide Number 18'''

'''Spoken Tutorial workshops'''
| | The '''Spoken Tutorial Project '''team:

Conducts workshops using spoken tutorials and
Gives certificates on passing online tests.

For more details, please write to us.
|-
| | '''Slide Number 19'''

'''Forum for specific questions:'''

Do you have questions in THIS Spoken Tutorial?

Please visit this site

Choose the minute and second where you have the question

Explain your question briefly

Someone from our team will answer them
| | Please post your timed queries on this forum.
|-
| | '''Slide Number 20'''

'''Acknowledgement'''
| | '''Spoken Tutorial Project''' is funded by NMEICT, MHRD, Government of India.

More information on this mission is available at this link.
|-
| |
| | This is '''Vidhya Iyer''' from''' IIT Bombay''', signing off.

Thank you for joining.
|-
|}

@@ Line 87: / Line 87: @@
 Click on '''Point''' tool and click on '''(3,6)'''.
-|| Clicking on the '''Point on Object''', create point A at 2 comma 2 and B at 3 comma 6.
+|| Clicking on the '''Point on Object''' tool, create point A at 2 comma 2 and B at 3 comma 6.
 |-

@@ Line 16: / Line 16: @@
 Understand Differentiation
-Draw graphs of derivative of functions
+Draw graphs of derivative of functions.
-−
-−
 |-
 || '''Slide Number 3'''
@@ Line 27: / Line 25: @@
 '''Ubuntu Linux''' Operating System version 16.04
-'''GeoGebra''' 5.0.481.0-d
+'''GeoGebra''' 5.0.481.0-d.
 |-
@@ Line 73: / Line 71: @@
 || Type '''f(x)=x^2-x''' in the '''input bar''' >> '''Enter'''
 || In the '''input bar''', type the following line.
 For the '''caret symbol''', hold the '''Shift''' key down and press 6.
@@ Line 83: / Line 82: @@
 |-
 || Click on '''Point on Object''' tool >> click on the parabola at '''(2,2)'''.
 Point to '''A''' at '''(2,2)'''.
@@ Line 92: / Line 90: @@
 |-
-|| Click on '''Line''' tool and click on points '''B''' and '''A'''.
+|| Click on '''Line''' tool >> click on points '''B''' and '''A'''.
 || Click on '''Line''' tool and click on points '''B''' and '''A''' to draw line '''g'''.
 |-
@@ Line 102: / Line 100: @@
 Click on '''Style''' tab and select '''dashed style'''.
-|| As shown earlier in this series, make this line '''g '''blue and dashed.
+|| As shown earlier in this series, make this line '''g ''' blue and dashed.
 |-
 || Click on '''Tangents''' tool under '''Perpendicular Line''' tool.
 || Under '''Perpendicular Line''', click on '''Tangents'''.
 |-
-|| Click on '''A''' and then on the '''parabola'''.
+|| Click on '''A''' >> click on the parabola.
-|| Click on '''A''' and then on the '''parabola'''.
+|| Click on '''A''' and then on the parabola.
 |-
-|| Point to '''tangent h''' at point '''A''' to the '''parabola'''.
+|| Point to '''tangent h''' at point '''A''' to the parabola.
-|| This draws a '''tangent h''' at point '''A''' to the '''parabola'''.
+|| This draws a '''tangent h''' at point '''A''' to the parabola.
 |-
 || Double click on '''tangent h''' and click on '''Object Properties'''.
@@ Line 132: / Line 130: @@
 || In '''Algebra''' view, double-click on '''C''' and change its '''coordinates''' to the following.
 |-
-||
+||Point to B and A.
 || Now C has the same '''x coordinate''' as point '''B''' and the same '''y coordinate''' as point '''A'''.
 |-
@@ Line 147: / Line 145: @@
 || We will make '''AC''' and '''BC''' purple and dashed segments.
 |-
-|| With '''Move''' highlighted, drag '''B''' towards '''A''' on the '''parabola'''.
+|| With '''Move''' highlighted, drag '''B''' towards '''A''' on the parabola.
-|| With '''Move''' highlighted, drag '''B''' towards '''A''' on the '''parabola'''.
+|| With '''Move''' highlighted, drag '''B''' towards '''A''' on the parabola.
 |-
 || Point to the value of '''j''' (length of '''AC''') and lines '''g''' and '''h'''.
@@ Line 161: / Line 159: @@
 |-
 || Point to all the points on the parabola.
-|| Derivative of the parabola is the slope of the tangents at each point on curve.
+|| Derivative of the parabola is the slope of the tangent at each point on the curve.
 |-
 || Point to text-box that appears in '''GeoGebra''' window.
@@ Line 191: / Line 189: @@
-'''BC''' is the difference between '''y' coordinates''', '''f of x''' plus '''j''' and '''f of x''', for '''A''' and '''B'''.
+'''BC''' is the difference between '''y coordinates''', '''f of x''' plus '''j''' and '''f of x''', for '''A''' and '''B'''.
@@ Line 216: / Line 214: @@
-We will pull out '''j''' from the numerator, and cancel it.
+We will pull out '''j''' from the numerator and cancel it.
-Note that as '''j''' approaches 0, '''j''' can be ignored so that '''2x''' plus '''j''' minus 1 approaches '''2x''' minus 1.
+Note that as '''j''' approaches 0, '''j''' can be ignored. So that '''2x''' plus '''j''' minus 1 approaches '''2x''' minus 1.
@@ Line 292: / Line 290: @@
 |-
 || Point to slope of '''line f''' at '''A''' appearing as '''m''' value in '''Graphics''' view.
 || Slope of tangent line '''f''' appears as '''m''' value in '''Graphics''' view.
 |-
@@ Line 303: / Line 300: @@
 || Draw point '''B''' and change its '''coordinates''' to '''x A''' in parentheses comma '''m'''.
 |-
-|| Right-click on point '''B''' and select '''Trace On''' option.
+|| Right-click on point '''B''' >> select '''Trace On''' option.
 || Right-click on '''B''' and select '''Trace On''' option
 |-
@@ Line 334: / Line 331: @@
 Drag the boundary to see it properly
 |-
-|| Compare slide’s calculations with equation of '''g'(x)''' in '''Algebra''' view.
+|| Compare slides' calculations with equation of '''g'(x)''' in '''Algebra''' view.
 || Compare the calculations in the previous slide with the equation of '''g prime x'''
 |-
@@ Line 350: / Line 347: @@
 |-
 ||Point to slope.
-|| Slope of the tangent at the corresponding points on '''g of x''' is 0.
+|| Slope of the tangent at the corresponding point on '''g of x''' is 0.
 Such points on '''g of x''' are maxima or minima.
@@ Line 384: / Line 381: @@
 |-
 || Point to minimum '''C''' in '''Graphics''' view and its '''co-ordinates''' in '''Algebra''' view.
-|| In Graphics view, we see the minimum on '''g of x'''.
+|| In '''Graphics view''', we see the minimum on '''g of x'''.
-Its '''co-ordinates''' are -2 comma -11 in '''Algebra''' view.
+Its '''co-ordinates''' are -2 comma -11 in '''Algebra''' '''view'''.
 |-
 || In the '''input bar''', type '''Max'''.
@@ Line 419: / Line 416: @@
 We have to convert it into a box
-Squares have to be cut from the four corners
+Squares have to be cut out from the four corners
 What size squares should we cut out to get the maximum volume of the box?

@@ Line 350: / Line 350: @@
 | | 2 and -2 are the values of '''x''' when '''g prime x''' equals 0.
 |-
-| |
+| | Point to the "peak" and "valley" on '''g of x'''.
 | | Slope of the tangents at the corresponding points on '''g of x''' is 0.