Question about Texas Instruments TI-Nspire Graphic Calculator
When I try to get the Derivtive of (secx) the calculator gives me the wrong answer and i really need it for an exam.
Posted by Anonymous on
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Posted on Jan 02, 2017
SOURCE: graphing trig functions
Put the calculator in radian mode and it will graph trig functions just fine. If you don't know how to change it to radians, then go to mode and cursor over to select radians. I am assuming you know where the sine, cosine, tangent, etc. keys are.
Posted on Jan 04, 2008
SOURCE: wrong trig ratio
Your calculator is set in radians. Press mode key. Arrow down to 3rd line. Highlight degree and press enter. Your calculator is now in degrees, not radians. To get out of the mode screen and back to home screen, press 2nd and then mode.
Posted on Aug 06, 2009
SOURCE: Using Tan,sin, cosin
Alot of people seem to be having this problem. Angles can be measured in two units, degrees and radians. Your calculator is currently doing everything in terms of radians. I have an 84 but I'm sure the 83 is similar, go under MODE and look for Degrees and Radians and make sure you choose degrees. if you have to convert 360[degrees] = 2* pi [radians]
(easy to remember, 360deg in a circle, 2pi radians in a circle)
Posted on Aug 12, 2009
This means that you are using a non-algebraic variable in an expression. I know, I am not even paraphrasing. Exemple: if you write 2 + johnQ, where johnQ is a table of data, a picture or a GDB, graphical data base object.
Here is an exemple where mat is a 2X2 matrix and mot is a table.
Hope it helps.
Posted on Sep 14, 2009
The derivative of the cubic root of x is, as you wll know, equal to (1/3)*x^(-2/3). Its limit as x approaches 0 is undefined. However the calculator uses approximate representations of numbers to calculate. And you can never be sure what it is going to give as results near singularities (poles of functions).
Calculators with Computer Algebra Systems can do better. To show you this, I am enclosing a screen capture showing you the correct result.
On this screen I defined a function u =f(v) where u is the derivative of cubic root of v. I stored 0 in the dummy variable v, and I evaluated the function u at v=0. You can see that the result displayed is undefined.
That this result is deemed correct seems to me just a matter of convention. We do not know, we cannot know, or we dont care to know? It will not make any of us lose sleep.
Hope it helps.
Posted on Nov 10, 2009
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