MHB Find General Form of Z & T: Newton's Law of Cooling

[Recce]

A cup of tea is removed from the fridge and left on a desk. The temperature of the room is constantly E and T(t) represents the temperature of the tea after t minutes.
dT/dt = K(E-T) Z(t) = E-T(t)

How do I find the general form of Z and T?

 

[MarkFL]

Hello ystyegn! (Wave)

Newton's law of Cooling states that the time rate of change of the temperature $T$ of an object is proportional to the difference between the ambient temperature $M$ and the temperature of the object. Stated mathematically, this is:

$$\frac{dT}{dt}=k(M-T)$$ where $$T(0)=T_0,\,0<k\in\mathbb{R}$$.

The ODE is separable and may be written:

$$\frac{1}{T-M}\,dT=-k\,dt$$

Integrating, using the boundaries, and changing dummy variables of integration, we find:

$$\int_{T_0}^{T(t)}\frac{1}{u-M}\,du=-k\int_0^t \,dv$$

Can you proceed by applying the FTOC?

 

[Recce]

Hello ystyegn! (Wave)

Newton's law of Cooling states that the time rate of change of the temperature $T$ of an object is proportional to the difference between the ambient temperature $M$ and the temperature of the object. Stated mathematically, this is:

$$\frac{dT}{dt}=k(M-T)$$ where $$T(0)=T_0,\,0<k\in\mathbb{R}$$.

The ODE is separable and may be written:

$$\frac{1}{T-M}\,dT=-k\,dt$$

Integrating, using the boundaries, and changing dummy variables of integration, we find:

$$\int_{T_0}^{T(t)}\frac{1}{u-M}\,du=-k\int_0^t \,dv$$

Can you proceed by applying the FTOC?

Hey,

Is there a way to do this question without using integral because I haven't learn any integral yet? I don't know what this equation means $$\int_{T_0}^{T(t)}\frac{1}{u-M}\,du=-k\int_0^t \,dv$$

 

[MarkFL]

Well, we could write the given ODE in the form:

$$\d{T}{t}+kT=kM$$

Then, if we observe that multiplying though by $e^{kt}$:

$$e^{kt}\d{T}{t}+ke^{kt}T=kMe^{kt}$$

We now have on the left:

$$\frac{d}{dt}\left(e^{kt}T\right)=e^{kt}\d{T}{t}+ke^{kt}T$$

And so we may now write:

$$\frac{d}{dt}\left(e^{kt}T\right)=kMe^{kt}$$

Let's stop and think about what we have now...we have the derivative of a function of $t$ being equal to a constant times an exponential function...so let's write:

$$\frac{d}{dt}f(t)=c_1e^{kt}$$

If the derivative of a function is a constant times an exponential, then what must the function be?