# Calculus 4: P=f(t,c)

• Sep 13th 2008, 07:59 PM
Chocolatelover2
Calculus 4: P=f(t,c)
Hello everyone,

Could someone please tell me if this is correct?

P=f(t,c)

Evaluate ft and fc.

My attempt:

ft(t,c)-->ft(18,6)=

fc=(c,t)-->fc(18,6)=

Thank you very much
• Sep 13th 2008, 10:30 PM
Aryth
What do you mean by ft and fc, do you mean:

$\displaystyle f_t$ and $\displaystyle f_c$ as in, partial derivatives, or do you mean:

$\displaystyle ft$ and $\displaystyle fc$ for something else?
• Sep 14th 2008, 10:36 AM
Chocolatelover2
Sorry, I mean http://www.mathhelpforum.com/math-he...193c048a-1.gif and http://www.mathhelpforum.com/math-he...c4897d65-1.gif as in, partial derivatives.

Could you please tell me if what I said was correct?

Thank you
• Sep 14th 2008, 07:00 PM
Aryth
More questions:

What is the function P, is there any equation or anything?

And why does $\displaystyle f_t(t,c) \rightarrow f_t(18,6)$

There's really no way to verify what you have... There's nothing to compare your answer to...