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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: hargao on December 23, 2011, 02:44:51 am

Title: CAS Techniques
Post by: hargao on December 23, 2011, 02:44:51 am: When I started my methods and spesh practice papers some time in November, these really anal questions kept on coming up ):
So I defined these functions in my brand spanking new Ti-nspire cas which made life a little bit more bearable. I’m new to posting on atarnotes forum, so please be nice haha (:

Anal question type 1 (simultaneous linear equation 0,1,infinite solution questions)

$(m+3)x + my = 7$
$(m+2)x + 4y = 13$

Find m such that there are
i) no solutions
ii) one solution
iii) infinite solutions

We can write a function in the ti calculators which automatically finds the solution set to such a question.

So for the general problem of
$ax+by=c$
$dx+ey=f$

where c,f are constants and a,b,d,e may be constants or linear factors in terms of the variable “g”

input $define \; linearsoln(a,b,c,d,e,f,g) = \{solve(a \cdot e=b\cdot d \; and \; b\cdot f\neq c\cdot e,g), solve(a\cdot e\neq b\cdot d,g),solve(a\cdot e=b\cdot d \; and \;b\cdot f=c\cdot e,g)\}$

output $\{1st \; element, 2nd \; element, 3rd \; element\}$

1.   The first element of the set in the ouput tells us the requirements for no solutions to the set of linear equations
2.   The second element of the set tells us the requirements for one solution to the set of linear equations
3.   The third element of the set tells us the requirements for infinite solutions to the set of linear equations

Example
Solve the original question above

we let:
instead of a, let a=m+3
instead of b, let b=m
instead of c, let c=7
instead of d, let d=m+2
instead of e, let e=4
instead of f, let f=13
instead of g, let g=m

input $linearsoln(m+3,m,m+2,7,m+2,4,13,m)$ instead of $linearsoln(a,b,c,d,e,f,g)$
output is $\{ m=1\pm \sqrt{13},m \neq 1\pm \sqrt(13),false\}$

1.   For there to be no solutions, $m=1+\sqrt{13} \; or \; m=1-\sqrt{13}$
2.   For there to be one solution, $m\neq 1+ \sqrt{13} \; or \; m \neq 1-\sqrt{13}$
3.   For there to be infinite solutions, there is $no \; m \in R$ that satisfies such a condition

Anal question type 2 (tedious left hand right hand area estimation thingy)

Often we get asked to find the left-hand estimate or right-hand estimate of the area bounded by a curve and the x axis (stupid pointless vcaa question that can be done on a computer with much higher accuracy)

For Left-Hand Estimate
input $define \; lhestimate(a,b,c,n)=\frac{c-b}{n} \sum_{k=0}^{n-1}(a\mid b+\frac{c-b}{n}\cdot k)$

For Right-Hand Estimate
input $define \; rhestimate(a,b,c,n)=\frac{c-b}{n} \sum_{k=1}^{n}(a\mid b+\frac{c-b}{n}\cdot k)$

Parameter “a” is where you put in the function that you want to evaluate, say $41x-sin(x)$
Parameter “b” is where you put in the lower limit of the integral you are evaluating
Parameter “c” is where you put in the upper limit of the integral you are evaluating
Parameter “n” is where you put in the number of rectangles you are splitting the integral into.

Example

Use the trapezium rule to find the area bound by the curves $y=x^2, y=0, x=0 \; and \; x=100$ with rectangle width of 0.5

So
$a=x^2$
$b=0$
$c=100$
$n=200$ as there are 100/0.5 rectangles

Therefore input $\frac{1}{2}( lhestimate(x^2,0,100,200)+rhestimate(x^2,0,100,200))$ to find the trapezoidal rule estimation of $} \int_{0}^{100}x^2 \, \mathrm{d}x$

Anal question type 3 (stationary point scavenging hunt)

Pointless max-min questions (method also suited to drawing graphs and various types of spesh applications questions)

Input $define \; statpoint(a,b)=\{solve(\frac{d}{db}(a)=0 \; and \; \frac{d^2}{db^2}(a)<0,b),solve(\frac{d}{db}(a)=0 \; and \; \frac{d^2}{db^2}(a)>0,b),solve(\frac{d}{db}(a)=0 \; and \; \frac{d^2}{db^2}(a)=0,b)\}$

Usage
Parameter “a” is the function
Parament “b” is the variable
You can also limit the domain of the function (place a restriction on b)
Eg $statpoint(x^3+x^2-x,x)\mid 0<x<4$

output $\{ 1st \; element, 2nd \; element, 3rd \; element\}$
1.   The first element gives the local maximum points
2.   The second element gives the local minimum points
3.   The third element gives the possible locations stationary points of inflection (saddle points)
note: you need to use the first derivative test to verify the nature of these points. You may change the defined function to include a higher order derivative test, ie the third derivative test, but this may cause the potential unfortunate loss of local min. and max. points depending on the function evaluated.

Example
Find the stationary points and their nature for the function $f(x)=x^4-3x^3+3x^2-x+5$

Input $statpoint(x^4-3x^3+3x^2-x+5 ,x)$
Output $\{false,x=\frac{1}{4},x=1\}$

1.   There are no local maximum points
2.   There is a local minimum point at x=0.25
3.   There may be a stationary point of inflection at x=1. Use first derivative test or third derivative test to verify that it is a stationary point of inflection.

(Not) anal (and trivial) question type 4

Finding the average rate of change of a function

$define \; avchange(a,b,c)=\frac{(a\mid x=c)-(a\mid x=b)}{c-b}$

Self-explanatory, and sort of pointless

(Not) anal (and trivial) question type 5

Finding the mean value of a function

$define \; meanval(a,b,c)=\frac{1}{c-b} \int_{b}^{c} a \, \mathrm{d}x$

Self-explanatory, and sort of pointless

Hopefully these might help you in some way, and maybe inspire you to define your own super big and awesome functions XD
Title: Re: CAS Techniques
Post by: pi on December 23, 2011, 01:18:15 pm: Nice, added to the resource list: Re: Mathematical Methods Guides and Tips :)
Title: Re: CAS Techniques
Post by: hargao on December 23, 2011, 01:40:30 pm: sorry! made an error in the statpoint(a,b) function, fixed now.