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By Sandy (Sandy) on Tuesday, October 07, 2003 - 06:47 pm: Edit |

Is there anyway I could do implicit differentiation on TI 89? Please help. Thank you.

By Jason817 (Jason817) on Tuesday, October 07, 2003 - 07:06 pm: Edit |

Without a program, I really doubt it.

By Sandy (Sandy) on Tuesday, October 07, 2003 - 07:12 pm: Edit |

Aight, thanks.

You are the only one who answers my questions

lol

By Jason817 (Jason817) on Tuesday, October 07, 2003 - 07:56 pm: Edit |

I am either really helpful or really bored. The latter is probably it

By Sandy (Sandy) on Wednesday, October 08, 2003 - 03:07 pm: Edit |

I would want to think that the former is true, but you just denied it outright. Your image has completely changed for me now

By Ti89 (Ti89) on Saturday, October 11, 2003 - 10:16 pm: Edit |

Its doable, I used to have a function I downloaded from ticalc.org, called implicit(), but I'm unable to locate it because ticalc.org's web search is down at the moment. Some calculus packages for the 89, such as Calc89 have this function.

(The implicit() function I had was really simple - a line or so in Keystroke.)

By Fairyofwind2 (Fairyofwind2) on Saturday, October 11, 2003 - 10:19 pm: Edit |

Implicit differentiation takes 2 seconds to do if you know the formula.

In multivariable, partial derivatives are taken with respect to a variable so that all other variables are treated as constants. And guess what? The derivatives given by the TI-89 just so happen to be partial.

The formula for implicit differentiation of a function f[x,y]=0 is given by dy/dx=-(partial f/partial x)/(partial f/partial y).

In TI-89, all you do is do -(d(f[x,y],x]))/(d(f[x,y],y)). And that's dy/dx.

For example the derivative of x^2+y^2=a^2 is -(2x)/(2y)=-x/y since d(f[x,y],x)=2x. d(f[x,y],y)=2y. Remember you treat all other variables you are not differentiating with respect to as CONSTANTS. Done.

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