# Explicit mathematical expression

• Mar 24th 2009, 01:04 AM
sazzy3
Explicit mathematical expression
How would you write the function: dT/dt= -k(T-A)

As an explicit mathematical expression of T(t)?
• Mar 24th 2009, 01:39 AM
Jhevon
Quote:

Originally Posted by sazzy3
How would you write the function: dT/dt= -k(T-A)

As an explicit mathematical expression of T(t)?

solve the differential equation, note that it is separable, so we have

$\frac {dT}{T - A} = -k~dt$

now integrate both sides and then solve for $T$ (don't forget your arbitrary constant)
• Mar 24th 2009, 01:50 AM
sazzy3
question continued
So now i have
dT/T-A=-k dt

i get integral from T to T0 1/T-A dT= integral from t to t0 -k dt

then
ln [T-A] integral from T to T0= integral from t to t0 -k dt

then
ln [T-A]= -kt

then

[T-A]= e^-kt

Is this correct?
• Mar 24th 2009, 01:56 AM
Jhevon
Quote:

Originally Posted by sazzy3
So now i have
dT/T-A=-k dt

i get integral from T to T0 1/T-A dT= integral from t to t0 -k dt

then
ln [T-A] integral from T to T0= integral from t to t0 -k dt

then
ln [T-A]= -kt

then

[T-A]= e^-kt

Is this correct?

we are not using a definite integral here, it is indefinite. and $\int -k~dt = -kt + C$
• Mar 24th 2009, 02:00 AM
sazzy3
continued
So i just needed to change it to:

ln [T-A]=-kt +c

then
[T-A]=e^-kt +c

then for just an expression of T(t)

do i go
T= Ae^-kt +c

If not can you please show me where i have gone wrong and how it should be done?
• Mar 24th 2009, 02:03 AM
Jhevon
Quote:

Originally Posted by sazzy3
So i just needed to change it to:

ln [T-A]=-kt +c

then
[T-A]=e^-kt +c

then for just an expression of T(t)

do i go
T= Ae^-kt +c

If not can you please show me where i have gone wrong and how it should be done?

if you have T - A on the left, to solve for T you would ADD A to both sides... and make sure -kt + C is the power of e, not just the -kt, which is what you typed
• Mar 24th 2009, 02:10 AM
sazzy3
continued

[T-A]=e^(-kt +C)

and i had to find an explicit mathematical expression of T(t)

so have i done that by then making the equation

T=A e^(-kt +C)??

Very confused.
• Mar 25th 2009, 09:31 AM
Jhevon
Quote:

Originally Posted by sazzy3

[T-A]=e^(-kt +C)

you are ok up to this point.

Quote:

and i had to find an explicit mathematical expression of T(t)

so have i done that by then making the equation

T=A e^(-kt +C)??

Very confused.
as i said, you have -A on the left side of the equation, you must ADD A to both sides to solve for T