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June 28th, 2008, 03:36 PM
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| | Urgent help please How can i write a system between 2 derivates with the ti-89?
I mean
d/dx [f(x,y)=a]
where
dy/dx=c
for example if f(x,y)=a is x^2+3 =y
then the solution is 2x=c => x=c/2
I wasn't able to find anything... i've tried to do something like
solve(d(x^2+3=y,x)|(d(y,x)=1),x) -->wrong value
d(x^2+3=y,x)|(d(y,x)=1)) ------->2x=0 wrong
Thanks in advance
Marco
Last edited by swatt; June 29th, 2008 at 05:14 AM.
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June 29th, 2008, 12:14 AM
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| | Hello,
Instead of using |, separate your equations with and : Code: solve(d(x^2+3,x)=y and d(y,x)=-1,x)
The problem is that they rather consider y as a variable than a function. This is partly why there is an error.
There are huge problems in your syntax. For systems of equations, you must have : solve(...=!!! and :::=^^^,{variable1,variable2})
After thinking about it, I don't really know...
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Last edited by Moo; June 29th, 2008 at 02:05 AM.
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June 29th, 2008, 02:38 AM
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| | that equation give me a "false", maybe i was a bit unclear (i'm not english / american)..
I have this equation: x^2+3 =y ->d/dx-> 2x=dy/dx now i impose dy/dx=c (it's the same thing i have a system of 2 equations)
then the equation become 2x=c => x=c/2 I think there should be a way to tell to the ti-89 to leave d/dx when it find derivates on unknown quantity/costant...
Last edited by swa; June 29th, 2008 at 05:17 AM.
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June 29th, 2008, 03:11 AM
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| | I've find a way, but it's not right at all.. d(x^2+3=y|x=cy) -> 2x=c
but in this case i think calc. will calculate the system firstly, and after it will do the derivate... so this is wrong too
Last edited by swa; June 29th, 2008 at 05:11 AM.
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June 29th, 2008, 05:09 AM
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| | I've found the solution.
d(x^2+3=f(x),x) -> 2x=df(x)/dx
then by hands i write df(x)/dx=c so x=c/2.
Well, in this way it seems completely useless, but with giant eq. i usually resolve, it's a pretty nice way. | | Thread Tools | | | | Display Modes | Linear Mode |
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