How would you write the function: dT/dt= -k(T-A) As an explicit mathematical expression of T(t)?
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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 $\displaystyle \frac {dT}{T - A} = -k~dt$ now integrate both sides and then solve for $\displaystyle T$ (don't forget your arbitrary constant)
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?
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 $\displaystyle \int -k~dt = -kt + C$
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?
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
Was this part even correct? or had i gone wrong before this point? [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.
Originally Posted by sazzy3 Was this part even correct? or had i gone wrong before this point? [T-A]=e^(-kt +C) you are ok up to this point. 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
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