There are a host of ways to find the first derivative on the CAS calculator and to solve calculus problems quickly. Sometimes I think exam writers are well behind what these calculators can do and fail to understand how trivial some problems have become.
Let's start with simple substitution into an equation:
a) Find y at x=3 for y = 2x^2 + 2x + 2
The "|" is important between the equation y = 2x^2 + 2x + 2 and x=3
Let's now find x if we know y. This is a little harder as we have to solve the equation.
b) Find x at y=14 for y = 2x^2 + 2x + 2
The easiest way to do this is to type 14 = 2x^2 + 2x + 2, highlight it using your stylus (this is important!) and then go
interactive->advanced->solve->ok
Note that it find both possible solutions (unlike using numsolve with the incorrect range specified)
Let's find the first derivative. For this use the 2D template in the soft keyboard
Go keyboard -> 2D -> Calc ->
c) Find the 1st derivative of y=2x^2 + 2x + 2
Note that I removed the "y=" this time. I differentiated the expression on purpose as it makes the next part easier.
Finding the 1st derivative/gradient/instantaneous rate of change at a point is also easy.
d) Find the 1st derivative of y = 2x^2 + 2x + 2 at x = 3
As you can see, it is a mix between c) and a)
Last but not least we can find a point for a particular gradient.
e) Find x at y' = 14 for y = 2x^2 + 2x + 2
To find y itself we could repeat a)
It's very much a case of thinking what you need and then finding it. As you can see it can all be done with one line on a CAS calculator, things that would take multiple steps on paper. TanLine is also a useful function that can be investigated and used to quickly find tangents.
Viola!
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Showing posts with label 3A MAT. Show all posts
Showing posts with label 3A MAT. Show all posts
Tuesday, April 3, 2012
Wednesday, February 23, 2011
Drawing the first derivative
Teaching students how to visualise the first derivative in 3B MAT has been problematic over the last two years. This morning I had a bit of a breakthrough in that students weren't looking at me as if I was speaking Alien.
The major difference was that I didn't use the arrow approach. Here's what I did.
I drew a positive cubic on the board and identified the turning points. I identified clearly the x axis and the y axis and identified the coordinates for each TP. I drew their attention to (x,y)
Then I drew a second pair coordinate plane directly underneath and identified/labelled the x axis. I then deliberately (as in made a big song and dance) labelled the other axis y' asking students to think what this might mean.
I then went to the first turning point on the x,y plane and asked students what the gradient was at this point. They said zero straight away.
I then went to the second axis and said coordinates on this plane were (x,y'). Given that the TP we were examining was at (0.25) and y'(0.25) = 0, the coordinate(x,y') that we needed was at (0.25,0). We repeated this for the other turning point.
I then drew vertical dotted lines through both coordinate planes. We then looked at the slope to the left of the TP. Being a cubic (with a positive coefficient of x cubed) the slope was +ve. On the second plane I wrote +ve above the x axis to the left of the TP above the x axis. We then examined the second area and noted the slope was negative (making special note of where the point of inflection was - it wasn't mandated by the course but made sense in the context). I labelled the graph -ve underneath the x axis to the right of the TP. I then wrote +ve in the third area above the x axis.
<- It looked like this.
Once the areas were labelled it was trivial to join the dots starting where y' was positive (y' at +ve infinity), leading to where y' was negative and then changing direction midway between the x intercepts on y', back towards to the x axis until y' was +ve again (again until y' at +ve infinity). It was also a good time to discuss the type of function produced (eg a concave up quadratic) if you differentiate a cubic with a +ve coefficient of the cubed term and how that related to our y' graph.
We then repeated the process for a quartic.
yay!
The major difference was that I didn't use the arrow approach. Here's what I did.
I drew a positive cubic on the board and identified the turning points. I identified clearly the x axis and the y axis and identified the coordinates for each TP. I drew their attention to (x,y)
Then I drew a second pair coordinate plane directly underneath and identified/labelled the x axis. I then deliberately (as in made a big song and dance) labelled the other axis y' asking students to think what this might mean.
I then went to the first turning point on the x,y plane and asked students what the gradient was at this point. They said zero straight away.
I then went to the second axis and said coordinates on this plane were (x,y'). Given that the TP we were examining was at (0.25) and y'(0.25) = 0, the coordinate(x,y') that we needed was at (0.25,0). We repeated this for the other turning point.
I then drew vertical dotted lines through both coordinate planes. We then looked at the slope to the left of the TP. Being a cubic (with a positive coefficient of x cubed) the slope was +ve. On the second plane I wrote +ve above the x axis to the left of the TP above the x axis. We then examined the second area and noted the slope was negative (making special note of where the point of inflection was - it wasn't mandated by the course but made sense in the context). I labelled the graph -ve underneath the x axis to the right of the TP. I then wrote +ve in the third area above the x axis.
<- It looked like this.
Once the areas were labelled it was trivial to join the dots starting where y' was positive (y' at +ve infinity), leading to where y' was negative and then changing direction midway between the x intercepts on y', back towards to the x axis until y' was +ve again (again until y' at +ve infinity). It was also a good time to discuss the type of function produced (eg a concave up quadratic) if you differentiate a cubic with a +ve coefficient of the cubed term and how that related to our y' graph.
We then repeated the process for a quartic.
yay!
Sunday, May 17, 2009
3A MAT Ex. 8B Annuities and Amortisation
Ok.. AP's and GP's are now a thing of the past (what?? huh?? when did that happen - in 8A of course!).. we're now onto applications of growth and decay.. nope.. (we did that in 8A too.. huh?? what??)..
MAT 8B We're now onto Annuities and Amortisation - growth and decay with payments.
The calculator handles this under the Financial, Sequences or Spreadsheet.
Starting with Financial:
Once in Financial, select Compound interest.
n - represents the number of installment periods
I% - is the interest p.a.
PV - is the present value (the initial investment)
PMT - is the payment per period
FV - is the future value (the investment at period N)
P/Y - is the number of installment periods per year (how often a payment is made)
C/Y - is the number of times interest is compounded
Let's look at a simple problem say 8B q.3 in 3A MAT. Kelvin invests $620,000 into an account giving 5.8% pa. interest compounded annually from which her withdraws $50,000 at the end of every year.
a) How much is left after 10 withdrawals (N=10, FV=?).
N=10
I%=5.8
PV=-620000
PMT=50000
P/Y=1
C/Y=1
Leave the cursor on FV and press solve (at the bottom left hand corner of the window)
FV=436670
b) For how many years will Kelvin be able to withdraw 50000 per year
Find when the account is exhausted of funds (eg. N=? when FV=0)
I%=5.8
PV=-620000
I%=5.8
PMT=50000
FV=0
P/Y=1
C/Y=1
Leave the cursor on N and press solve (at the bottom left hand corner of the window)
N=22.52 therefore for 22 years.
If anyone can explain why PV is negative I would be very appreciative. I know from last year's course that it is but have no idea why.
Now Sequence:
This could also have been done through the Sequence tool using recursion
a) Tn+1=Tn*1.058-50000; T0=620000. Find T10
b) Tn+1=Tn*1.058-50000; T0=620000. Find n Where Tn=0
I'll leave the spreadsheet method for another day.
Here is a link to other CAS calculator posts.
MAT 8B We're now onto Annuities and Amortisation - growth and decay with payments.
The calculator handles this under the Financial, Sequences or Spreadsheet.
Starting with Financial:
Once in Financial, select Compound interest.
n - represents the number of installment periods
I% - is the interest p.a.
PV - is the present value (the initial investment)
PMT - is the payment per period
FV - is the future value (the investment at period N)
P/Y - is the number of installment periods per year (how often a payment is made)
C/Y - is the number of times interest is compounded
Let's look at a simple problem say 8B q.3 in 3A MAT. Kelvin invests $620,000 into an account giving 5.8% pa. interest compounded annually from which her withdraws $50,000 at the end of every year.
a) How much is left after 10 withdrawals (N=10, FV=?).
N=10
I%=5.8
PV=-620000
PMT=50000
P/Y=1
C/Y=1
Leave the cursor on FV and press solve (at the bottom left hand corner of the window)
FV=436670
b) For how many years will Kelvin be able to withdraw 50000 per year
Find when the account is exhausted of funds (eg. N=? when FV=0)
I%=5.8
PV=-620000
I%=5.8
PMT=50000
FV=0
P/Y=1
C/Y=1
Leave the cursor on N and press solve (at the bottom left hand corner of the window)
N=22.52 therefore for 22 years.
If anyone can explain why PV is negative I would be very appreciative. I know from last year's course that it is but have no idea why.
Now Sequence:
This could also have been done through the Sequence tool using recursion
a) Tn+1=Tn*1.058-50000; T0=620000. Find T10
b) Tn+1=Tn*1.058-50000; T0=620000. Find n Where Tn=0
I'll leave the spreadsheet method for another day.
Here is a link to other CAS calculator posts.
3A MAT recursive formula, AP's & GP's
Those of you attempting Exercise 8A without students with a thorough grounding in AP's & GP's in year 10 will be scratching your head at this chapter.
Q1: creating recursive formula from word descriptions
My recollection of when we did this in Discrete was that these topics were covered over multiple chapters. When attempting 8A students faced difficulties in that the calculator has some limitations. I can usually maintain 1 chapter per lesson but in this case I let it run over three lessons and found some extra resources to supplement the topic as it left many students scratching their heads. This was hard as it chewed into the revision time I had left for exams.
Things to remember for next year:
Sequence mode and Calculator usage (What not to do).
Most sequences can be done in Sequence mode. Some cannot. Here's how to get into Sequence mode.
Let's put in Tn = 2Tn-1 -5 where T1=3 (q.11 from 3a MAT). We are looking for T1 to T5. Press Type in the menu.
You will notice that the notation to the text is different in that "an" is used instead of Tn. Ignoring that, you will also notice that there is no option for "an", only for "an+1" or "an+2". (Blogger can't do subscripts so just put them in where needed!).
So now we put in T2.. Easy no? NO! If you look at Figure 1 we have options for Tn+1 where we are given a0 (the zero term) or a1(the first term). No option for a2.
The Saddler text does a pretty good job of making the calculator look clumsy and painful to use compared to paper and pen.
Q1: creating recursive formula from word descriptions
Q2 -5: creating sequences (tables of values) from recursive formula in the form Tn=...
Q6: identifying AP's, GP's or neither from sequences
Q7-9: creating sequences (tables of values) from recursive formula in the form Tn+1=...
Q10: creating sequences (tables of values) from recursive formula in the form Tn-1=...
Q11-12,15,16,17: creating sequences from multiple previous terms
Q13: recursive formula using the term counter(n) in the formula
Q14: finding unknowns in recursive formula
Q18-22: Growth and decay problems
My recollection of when we did this in Discrete was that these topics were covered over multiple chapters. When attempting 8A students faced difficulties in that the calculator has some limitations. I can usually maintain 1 chapter per lesson but in this case I let it run over three lessons and found some extra resources to supplement the topic as it left many students scratching their heads. This was hard as it chewed into the revision time I had left for exams.
Things to remember for next year:
1. Present multiple examples of recursive formula for the same sequence for Tn, Tn+1 and Tn-1.
2. Introduce the limitation that the calculator (in sequence mode) can only use up to two previous terms in its definition (eg. Tn+2=Tn+1+ Tn not Tn+3=Tn+2+Tn+1+Tn). We wasted a lot of time on this.
3. You cannot move freely between Tn-1, Tn, Tn+1 and Tn+2 representations if n itself is used in the formula. eg. Tn+1=Tn +3 is equivalent to Tn=Tn-1 +3 but Tn+1=Tn + n is not equivalent to Tn = Tn-1 + n
4. Be careful with the position when dealing with growth and decay. It is usually much easier to define T0 (Tzero) and Tn+1=.. as the initial value and formula. Thus when solving for n, n is the answer rather than n-1 (which caused no end of confusion amongst students).
5. Make students do the examples without a calculator unless it states otherwise. A lot of time can be wasted trying to make Sequence mode do things it is not intended to do.
Sequence mode and Calculator usage (What not to do).
Most sequences can be done in Sequence mode. Some cannot. Here's how to get into Sequence mode.
Let's put in Tn = 2Tn-1 -5 where T1=3 (q.11 from 3a MAT). We are looking for T1 to T5. Press Type in the menu.
Figure 1.
You will notice that the notation to the text is different in that "an" is used instead of Tn. Ignoring that, you will also notice that there is no option for "an", only for "an+1" or "an+2". (Blogger can't do subscripts so just put them in where needed!).
In this case it is not such a problem, we can just transpose our equation to Tn+1 = 2Tn + 5 as n itself is not used in the formula; The given value T1=3 now becomes T2=3, remembering that we are looking for T2 to T6 now (which is really T1-T5 of the original formula).
So now we put in T2.. Easy no? NO! If you look at Figure 1 we have options for Tn+1 where we are given a0 (the zero term) or a1(the first term). No option for a2.
The Saddler text does a pretty good job of making the calculator look clumsy and painful to use compared to paper and pen.
Here is a link to other CAS calculator posts.
Sunday, March 29, 2009
Variables on the classpad
There has been some confusion about how to define variables on the classpad in my class. Here is what we have discovered.
If we use a variable found under the mth tab -> var on the soft keyboard (the variables that are italicised) it is treated as a normal pronumeral in algebraic equations (multiplication is assumed with adjacent pronumerals). The x,y,z on the keypad is also treated this way.
eg x = 10, y=20; therefore xy=200
The multiplication sign is automatically added.
If we name a variable using the abc tab in the soft keyboard(the variables that are not italicised) then we are naming a variable that has multiple letters.
eg xy =10; x & y are undefined.
m = rise ÷ run
Potential Gotcha!
We have to be careful not to confuse functions defined under the mth tab (eg. trig ratios) and variables that we have created when using NumSolve. One of my students entered this on their calculator.
CosÎ¸=adj÷hyp
It would return the fractional value adj÷hyp rather than the value for theta. This is because the student had defined a variable "CosÎ¸" by typing Cos via the soft keyboard rather than entering the function Cos via mth->trig->Cos.
Superscripts and Subscripts
Later on students will want to use subscripted characters when creating variable names. One example is the gradient formula.
m=(Y2-Y1)÷(X2-X1)
The subscripts are found in the soft keyboard under abc->math at the bottom of the screen. Superscripts are on the line above it. Only numbers at this stage (more will be possible as more fonts are released) can be superscripted or subscripted as far as I can see.
Here is a link to other CAS calculator posts.
If we use a variable found under the mth tab -> var on the soft keyboard (the variables that are italicised) it is treated as a normal pronumeral in algebraic equations (multiplication is assumed with adjacent pronumerals). The x,y,z on the keypad is also treated this way.
eg x = 10, y=20; therefore xy=200
The multiplication sign is automatically added.
If we name a variable using the abc tab in the soft keyboard(the variables that are not italicised) then we are naming a variable that has multiple letters.
eg xy =10; x & y are undefined.
m = rise ÷ run
Potential Gotcha!
We have to be careful not to confuse functions defined under the mth tab (eg. trig ratios) and variables that we have created when using NumSolve. One of my students entered this on their calculator.
CosÎ¸=adj÷hyp
It would return the fractional value adj÷hyp rather than the value for theta. This is because the student had defined a variable "CosÎ¸" by typing Cos via the soft keyboard rather than entering the function Cos via mth->trig->Cos.
Superscripts and Subscripts
Later on students will want to use subscripted characters when creating variable names. One example is the gradient formula.
m=(Y2-Y1)÷(X2-X1)
The subscripts are found in the soft keyboard under abc->math at the bottom of the screen. Superscripts are on the line above it. Only numbers at this stage (more will be possible as more fonts are released) can be superscripted or subscripted as far as I can see.
Here is a link to other CAS calculator posts.
Location:Perth, WA, Australia
Perth WA, Australia
Thursday, February 5, 2009
Absolute value and the 3A MAT course
Ok. Absolute value - easy enough, to take the absolute value of a number, make it positive if it wasn't already. easy peasy...
I started by displaying the graph on the board using the overhead gadget for the Classpad.
I selected Graph&Tab and entered Ix-3I for y1 and I2x+4I for y2. For some reason the graph workpane doesn't allow you to use the 2d tab absolute value option - so use the abs() function under the cat tab in the soft keyboard. When you hit enter it will restore the absolute value notation.
Using the intersect function under analysis in the menu bar we know that the two lines intersect at -1/3 and -7 therefore the interval is x<=-7, x>=-1/3.
Ix-3I<=I2x+4I
I then asked students to draw a number line with the intervals marked and substitute the values back into the original inequality. We numbered the three intervals. The first interval represented x<=-7, the second -7<=x<=-1/3 and the last x>=-1/3.
Here is a link to other CAS calculator posts.
... until you start to look at IyI=IxI and ask students to graph it..
... then ask them to find the intersection of Ix-3I=I2x+4I algebraically
... then ask them to find Ix-3I<=I2x+4I
... then ask them to find Ix-3I<=I2x+4I
Students really bogged down when they reached inequalities. The approach I used was similar to that by Sadler in his 3A book for MAS. The problem was that I really wasn't sure they understood what they were doing.. they could follow the algorithm but understanding was eluding them.
I started by looking at absolute numbers and explained models for solving using number lines, graphing and algebraically. Then I used a composite approach to assist students visualise what it was they were doing with problems like:
Ix-3I<=I2x+4I
I started by displaying the graph on the board using the overhead gadget for the Classpad.
I entered Graph&Tab from the menu workpane (using the menu icon at the base of the workpane).
I selected Graph&Tab and entered Ix-3I for y1 and I2x+4I for y2. For some reason the graph workpane doesn't allow you to use the 2d tab absolute value option - so use the abs() function under the cat tab in the soft keyboard. When you hit enter it will restore the absolute value notation.
Make sure both y1 and y2 are ticked (if they are not place the cursor on the line using your stylus and hit exe). Hit the graph button in the toolbar (the first icon with the top formula pane selected). The following graph should appear:
We then looked at the original inequality again and I asked what did it really mean?
Ix-3I<=I2x+4I
One way of thinking about it was, "when is the graph y=Ix-3I less than or equal to the graph of y=I2x+4I?"
We looked at the graph and found that the part marked red on the line y=Ix-3I satisfied the inequality.
Using the intersect function under analysis in the menu bar we know that the two lines intersect at -1/3 and -7 therefore the interval is x<=-7, x>=-1/3.
We had discussed that we could also do this algebraically by using the property if IxI=IyI then x=y or x=-y to find the points of intersection.
Eg.
Ix-3I<=I2x+4I
x-3 = 2x+4
-x = 7
x=-7
x-3 = -(2x+4)
x-3=-2x-4
3x=-1
x=-1/3
I then asked students to draw a number line with the intervals marked and substitute the values back into the original inequality. We numbered the three intervals. The first interval represented x<=-7, the second -7<=x<=-1/3 and the last x>=-1/3.
We then selected a value within each of the intervals and substituted them into the inequality. If they were true then this indicated values of x that satisfied the inequality.
Ix-3I<=I2x+4I
Interval 1 (x=-8)
I-8-3I<=I2(-8)+4I
I-11I<=I-12I
11<=-12 (true)
Therefore x<=-7 is a valid interval.
Interval 2 (x=-5)
I-5-3I<=I2(-5)+4I
I-8I<=I-6I
8<=-6 (false)
Therefore -7<=x<=-1/3 is not a valid interval.
Interval 3 (x=0)
I-0-3I<=I2(-0)+4I
I-0-3I<=I2(-0)+4I
I-3I<=I6I
3<=6 (true)
Therefore x>=-1/3 is a valid interval.
The inequality Ix-3I<=I2x+4I is valid over x<=-7, x>=-1/3
Drawing students attention from the graph and back to the algebraic representation released the tension in the room, the screwed up faces and suddenly lights went back on.
Thank goodness!
Here is a link to other CAS calculator posts.
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