• Jun 28th 2008, 03:36 PM
swatt
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

Marco
• Jun 29th 2008, 12:14 AM
Moo
Hello,

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...
• Jun 29th 2008, 02:38 AM
swa
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...
• Jun 29th 2008, 03:11 AM
swa
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
• Jun 29th 2008, 05:09 AM
swa
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.