Sunday, March 2, 2008

Maths108 and Maths208 with TI89 Calculator

The TI-89 is a great calculator that can do calculus, matrices and vectors. Learning how to use it effectively is one of the greatest (and most rewarding) challenges.

If you can master using the TI89, you will find Maths108 and Maths208 much easier. The pretty printing of the TI89 is ideal for checking your answers and making sure you have entered the correct formulas to start with.

Normally you have to clear your calculators memory before exams and tests - so you will need to remember some of these formulas

Implicit Differenciation

If you have the TI-89 Titanium, the formula is already built into your calculator. The usage is :

impDer(equation, independent variable, dependent variable)

However, if you have an older TI-89, you can define the function by entering the following command : (Note the d is the differentiate function)

Define impDer(f, xx, yy) = -d(f, xx)/d(f, yy)

I tend to use a shorter function name and change the "impDer" to "id".

Here is an example of using this function :

To find dy/dx of
x^2 + y^2 = 1

Before you can use your new function, you must always rearrange your question so that the equation equals zero (very important - if you dont so this, the answer will be wrong), hence :

x^2 + y^2 - 1 = 0

Now use your new function (use the "VAR-LINK" button or type using the alpha keys)

impDer(x^2+y^2-1,x,y)

and you will get the answer -x/y. Make sure you dont put the x and y around the wrong way or you will end up with the answer -y/x (which is incorrect)

Important Note : If your formula contains combined variables (eg xy + 2y = 0), make sure that you expand the variables using a multiplication sign (eg x*y + 2y = 0). If you don't do this, the calculator will assume you are using a variable called xy. It does not automatically know that xy is the same as x times y.

An easier way to do Implicit Differenciation

To find dy/dx of
x^2 + y^2 = 1

Simply enter the following into your calculator

d(x^2 + y^2 = 1, x) | y = t(x)

You will notice that we did not need to reorganise the formula to equal zero. We simply told the calculator to substitute "t(x)" wherever we have used a "y".

This little trick converts "y" into a function of "x" which is easily differenciable.

The result of this calulation will be :

2.t(x) . d/dx(t(x)) + 2.x = 0

Now we can convert our t(x) back to a "y" (You can't do this on the calculator - you need to do it on paper or in your head)

First look for any "d/dx(t(x))" and change them to "dy/dx"

2.t(x) . dy/dx + 2.x = 0

And then look for any remaining "t(x)" and change them back to "y"

2.y . dy/dx + 2.x = 0

Now you can see that dy/dx appears in the middle of our equation. You can now solve this easily on paper or by using the "solve" function in the TI89.

If you use the "solve" function on the TI89, you need to substitute a single variable for the dy/dx. (in our example, we will use the variable "t" again)

solve(2 * y * t + 2 * x = 0,t)

And you will get the answer

t = -x/y (which is the same as dy/dx = -x/y)

Partial Differenciation

First Partial Derivatives : fx, fy

Second Partial Derivatives : fxx, fxy, fyy, fyx

You dont need to define any functions to do this type of problem. However, you may find it easier to store the formula in a variable to save having to type it in each time. Eg :

Define aa = ln(x^2 + y^2 + z^2)

Always use at least two characters for your variable name. You must make sure that you dont use a single character (especially x, y, z or t). If you do, it is almost guaranteed that you will mess up any future calculations using these variables.

Now that you have defined your variable, it is extremely easy.

To calculate fx

d(aa,x)

To calculate fy

d(aa,y)

To calculate fxx

d(aa,x,2) or d(d(aa,x),x)

To calculate fyy

d(aa,y,2) or d(d(aa,y),y)

To calculate fxy

d(d(aa,x),y)

To calculate fyx

d(d(aa,y),x)

In each of the examples shown above, you could have used the original formula instead of the variable "aa".

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