Silve the Newton law of cooling using the differential equations if the Temp is 70 and the ambient temperature is 45. Within 30 minutes

Newton's Law of Cooling states that the rate of change of an object's temperature is proportional to the difference between its own temperature and the ambient temperature.

$$\frac{dT}{dt} = k(T - T_{ambient})$$

Here, $T$ is the temperature of the object, $t$ is time, $T_{ambient}$ is the ambient temperature, and $k$ is the constant of proportionality.

Rearrange the equation to separate $T$ and $t$:

$$\frac{dT}{T - T_{ambient}} = k\,dt$$

Integrate both sides:

$$\int \frac{1}{T - T_{ambient}}\,dT = \int k\,dt$$ $$\ln|T - T_{ambient}| = kt + C$$

Where $C$ is the constant of integration.

Exponentiate both sides:

$$T - T_{ambient} = e^{kt + C}$$ $$T - T_{ambient} = e^{kt}e^C$$

Let $A = e^C$. Since $C$ is an arbitrary constant, $A$ is also an arbitrary positive constant.

$$T - T_{ambient} = Ae^{kt}$$ $$T(t) = T_{ambient} + Ae^{kt}$$

We are given that the initial temperature (at $t=0$) is $T_0 = 70$. The ambient temperature is $T_{ambient} = 45$.
Substitute $t=0$ and $T(0)=T_0$ into the solution:

$$T_0 = T_{ambient} + Ae^{k \cdot 0}$$ $$T_0 = T_{ambient} + A \cdot 1$$ $$A = T_0 - T_{ambient}$$

Now substitute the given values:

$$A = 70 - 45$$ $$A = 25$$

Substitute the values of $A$ and $T_{ambient}$ back into the general solution:

$$T(t) = 45 + 25e^{kt}$$

This equation describes the temperature of the object at any time $t$.

To find the temperature at a specific time, such as "within 30 minutes" (e.g., at $t=30$ minutes), the value of the constant $k$ is required. Without an additional data point (like the temperature at a different time $t > 0$), $k$ cannot be determined. The solution $T(t) = 45 + 25e^{kt}$ is valid for any time $t \ge 0$, including within the first 30 minutes.

The solution for the temperature $T(t)$ is: