Applications-of-GeoGebra/C3/Differentiation-using-GeoGebra/English-timed
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Time | Narration |
00:01 | Welcome to this tutorial on Differentiation using GeoGebra. |
00:06 | In this tutorial, we will learn how to use GeoGebra to: |
00:11 | Understand Differentiation
Draw graphs of derivative of functions. |
00:18 | Here I am using:
Ubuntu Linux Operating System version 16.04 |
00:25 | GeoGebra 5.0.481.0 hyphen d. |
00:31 | To follow this tutorial, you should be familiar with: |
00:34 | GeoGebra interface
Differentiation |
00:39 | For relevant tutorials, please visit our website. |
00:43 | Differentiation: First Principles |
00:47 | Let us understand differentiation using first principles for the function f of x. |
00:54 | f of x is equal to x squared minus x |
00:58 | f prime x is the derivative of f of x. |
01:04 | Consider 2 points, A and B. |
01:08 | A is x comma f of x and B is x plus j comma f of x plus j |
01:18 | I have opened the GeoGebra interface. |
01:22 | In the input bar, type the following line. |
01:26 | For the caret symbol, hold the Shift key down and press 6.
Press Enter. |
01:33 | Observe the equation and the parabolic graph of function f. |
01:40 | Clicking on the Point on Object tool, create point A at 2 comma 2 and B at 3 comma 6. |
01:53 | Click on Line tool and click on points B and A to draw line g. |
02:04 | As shown earlier in this series, make this line g blue and dashed. |
02:11 | Under Perpendicular Line, click on Tangents. |
02:16 | Click on A and then on the parabola. |
02:21 | This draws a tangent h at point A to the parabola. |
02:27 | Let us make tangent h a red line. |
02:31 | Click on the Point tool and click anywhere in Graphics view.
This creates point C. |
02:41 | In Algebra view, double-click on C and change its coordinates to the following. |
02:49 | Now C has the same x coordinate as point B and the same y coordinate as point A. |
02:58 | Let us use the Segment tool to draw segments B C and A C. |
03:08 | We will make AC and BC purple and dashed segments. |
03:14 | With Move highlighted, drag B towards A on the parabola. |
03:22 | Observe lines g and h and the value of j (length of AC). |
03:29 | As j approaches 0, points B and A begin to overlap. |
03:37 | Lines g and h also begin to overlap. |
03:42 | Slope of line g is the ratio of length of BC to length of AC. |
03:50 | Derivative of the parabola is the slope of the tangent at each point on the curve. |
03:58 | As B approaches A on f of x, slope of AB approaches the slope of tangent at A. |
04:08 | Now let us look at the Algebra behind these concepts. |
04:14 | Differentiation: First Principles, the Algebra |
04:18 | Slope of line AB equals the ratio of the lengths of BC to AC. |
04:25 | Line AB becomes the tangent at point A as distance j between A and B approaches 0. |
04:35 | BC is the difference between y coordinates, f of x plus j and f of x, for A and B. |
04:43 | AC is the difference between the x-coordinates, x plus j and x. |
04:50 | Let us rewrite f of x plus j and f of x in terms of x squared minus x. |
04:58 | We will expand the terms in the numerator. |
05:02 | After expanding the terms in the numerator, we will cancel out similar terms with opposite signs. |
05:10 | We will pull out j from the numerator and cancel it. |
05:15 | Note that as j approaches 0, j can be ignored. So that 2x plus j minus 1 approaches 2x minus 1. |
05:25 | As we know, derivative of x squared minus x is 2x minus 1. |
05:32 | Let us look at derivative graphs for some functions. |
05:37 | Differentiation of a Polynomial Function |
05:41 | Consider g of x. |
05:44 | Derivative g prime x is the sum and difference of derivatives of the individual components. |
05:53 | g prime x' is calculated by applying these rules. |
05:59 | Let us differentiate g of x in GeoGebra. |
06:04 | Open a new GeoGebra window. |
06:07 | In the input bar, type the following line and press Enter. |
06:13 | As shown earlier in the series, zoom out to see function g properly. |
06:24 | Right-click in Graphics view and select xAxis is to yAxis option.
Select 1 is to 5. |
06:35 | I will zoom out again. |
06:42 | As shown earlier, draw point A on curve g and a tangent f at this point. |
06:50 | Under Angle, click on Slope and on tangent line f. |
06:58 | Slope of tangent line f appears as m value in Graphics view. |
07:04 | Draw point B and change its coordinates to x A in parentheses comma m. |
07:13 | Right-click on B and select Trace On option |
07:20 | With Move tool highlighted, move point A on curve. |
07:31 | Observe the curve traced by point B. |
07:35 | Let us check whether we have the correct derivative graph. |
07:39 | In the input bar, type d e r i.
From the menu that appears, select Derivative Function option. |
07:49 | Type g' to replace the highlighted word Function.
Press Enter. |
07:55 | Note the equation of g prime x in Algebra view.
Drag the boundary to see it properly |
08:04 | Compare the calculations in the previous slide with the equation of g prime x |
08:11 | Let us find the maxima and minima of the function g of x. |
08:16 | Derivative curve g prime x remains above the x axis (is positive) as long as g of x is increasing. |
08:27 | g prime x remains below the x axis is negative as long as g of x is decreasing. |
08:37 | 2 and -2 are the values of x when g prime x equals 0. |
08:45 | Slope of the tangent at the corresponding point on g of x is 0. |
08:52 | Such points on g of x are maxima or minima. |
08:58 | Hence, for g of x, -2 comma -11 is the minimum and 2 comma 21 is the maximum. |
09:08 | In GeoGebra, we can see that the minimum value of g of x lies between x equals -3 and x equals -1. |
09:20 | In the input bar, type M i n. |
09:24 | From the menu that appears, select Min Function Start x Value, End x Value option. |
09:31 | Type g for Function. |
09:35 | Press Tab to go to the next argument. |
09:38 | Type -4 and -1 as Start and End x-Values.
Press Enter. |
09:47 | In Graphics view, we see the minimum on g of x. |
09:52 | Its co-ordinates are -2 comma -11 in Algebra view. |
09:58 | In the input bar, type Max. |
10:02 | From the menu that appears, select Max Function Start x Value, End x Value option. |
10:09 | Type g, 1 and 4 as the arguments.
Press Enter. |
10:17 | We see the maximum on g of x, 2 comma 21. |
10:24 | Finally, let us take a look at a practical application of differentiation. |
10:31 | We have a 24 inches by 15 inches piece of cardboard. |
10:36 | We have to convert it into a box. |
10:39 | Squares have to be cut from the four corners. |
10:43 | What size squares should we cut out to get the maximum volume of the box? |
10:49 | A Sketch of the Cardboard |
10:51 | Let us draw the cardboard: |
10:54 | This is the volume function here. |
10:58 | You could expand it into a cubic polynomial but we will leave it as it is. |
11:05 | Open a new GeoGebra window. |
11:08 | In the input bar, type the following line and press Enter. |
11:13 | Drag the boundary to see the equation properly in Algebra view. |
11:19 | Right click in Graphics view and set xAxis is to yAxis to 1 is to 50. |
11:29 | Now, zoom out to see the function properly. |
11:38 | Observe the graph that is plotted for the volume function in Graphics view. |
11:44 | Drag the background to see the maximum. |
11:48 | Note that the maximum is on the top of this broad peak. |
11:53 | The length of the square side is plotted along the x axis. |
11:58 | Volume of the box is plotted along the y axis. |
12:02 | As before, let us find the maximum of this function. |
12:13 | This maps the maximum, point A, on the curve. |
12:18 | Its coordinates 3 comma 486 appear in Algebra view. |
12:24 | Thus, we have to cut out 3 inch squares from all corners. |
12:30 | This will give the maximum possible volume of 486 cubic inches for the cardboard box. |
12:39 | Let us summarize. |
12:41 | In this tutorial, we have learnt how to use GeoGebra to:
Understand differentiation, Draw graphs of derivatives of functions |
12:53 | As an assignment:
Draw graphs of derivatives of the following functions in GeoGebra. |
13:00 | Find the derivatives of these functions independently and compare with GeoGebra graphs. |
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13:44 | This is Vidhya Iyer from IIT Bombay, signing off.
Thank you for joining. |