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 now integrate both sides and then solve for (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
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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