A body with an initial temperature `theta_(1)` is allowed to cool in a surrounding which is at a constant temperature of `theta_(0) (theta lt theta_(1))` Assume that Newton's law of cooling is obeyed Let `k =` constant The temperature of the body after time t is best experssed by .

Options

A. (a)`(theta_(i)-theta_(0))e^(-kt)`
B. (b)`(theta_(i)-theta_(0)) 1n(kt)`
C. (c)`theta_(0)+(theta_(i)-theta_(0))e^(-kt)`
D. (d)`theta_(i)e^(-kt)-theta_(0)`

Explanation

To solve the problem using Newton's law of cooling, we follow these steps:

### Step 1: Understand Newton's Law of Cooling
Newton's law of cooling states that the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature. Mathematically, this can be expressed as:
\[ \frac{d\theta}{dt} = -k(\theta - \theta_0) \]
where:
- \(\theta\) is the temperature of the body at time \(t\),
- \(\theta_0\) is the constant surrounding temperature,
- \(k\) is a positive constant.

### Step 2: Rearranging the Equation
We can rearrange the equation to separate variables:
\[ \frac{d\theta}{\theta - \theta_0} = -k \, dt \]

### Step 3: Integrate Both Sides
Next, we integrate both sides. The left side will be integrated with respect to \(\theta\) and the right side with respect to \(t\):
\[ \int \frac{d\theta}{\theta - \theta_0} = -k \int dt \]
This gives us:
\[ \ln|\theta - \theta_0| = -kt + C \]
where \(C\) is the constant of integration.

### Step 4: Solve for the Constant of Integration
To find \(C\), we use the initial condition. At \(t = 0\), the temperature of the body is \(\theta_1\):
\[ \ln|\theta_1 - \theta_0| = C \]
Thus, we can rewrite the equation as:
\[ \ln|\theta - \theta_0| = -kt + \ln|\theta_1 - \theta_0| \]

### Step 5: Exponentiate to Solve for \(\theta\)
Exponentiating both sides gives us:
\[ |\theta - \theta_0| = |\theta_1 - \theta_0| e^{-kt} \]

### Step 6: Remove Absolute Values
Since \(\theta < \theta_1\) and \(\theta_0 < \theta_1\), we can drop the absolute values:
\[ \theta - \theta_0 = (\theta_1 - \theta_0)e^{-kt} \]

### Step 7: Rearranging to Find \(\theta\)
Finally, we rearrange to express \(\theta\):
\[ \theta = \theta_0 + (\theta_1 - \theta_0)e^{-kt} \]

This is the expression for the temperature of the body after time \(t\).

### Final Answer
The temperature of the body after time \(t\) is given by:
\[ \theta(t) = \theta_0 + (\theta_1 - \theta_0)e^{-kt} \]