91Ó°ÊÓ

Exponential decay is a process where the quantity decreases at a rate proportional to its current value. This appears in many natural processes, such as radioactive decay, cooling of objects, and devaluation over time. It's represented mathematically by a differential equation where the rate of change of a function is proportional to the function itself.
In our specific problem, the red-hot steel rod cooling in water follows an exponential decay model. The temperature difference between the rod and the water (ewlineewline \textrm{10 - y}ewline\(\textrm{10 - y}\)) changes at a rate proportional to that difference, which is why we use the equation: ewlineewlineewlineewline \( y' = k(10 - y) \). Here, the constant \( k \) affects how quickly the temperature declines.

Initial conditions are essential in solving differential equations because they specify the system's state at a particular starting point. These conditions allow us to find the particular solution that fits the given scenario.
The problem provides the initial temperature of the rod, \( f(0) = 350^{\textrm{°}}\textrm{C} \). This means at time \( t = 0 \), the temperature is 350 \( ^{\textrm{°}} \)\text{C}. This initial condition is crucial to find the integration constant after we solve the differential equation.

Separation of variables is a technique used to solve differential equations. It involves rearranging the equation to have all terms involving one variable (e.g., \( y \)) on one side and all terms involving the other variable (e.g., \( t \)) on the other side.
In our exercise, the differential equation is: ewline \( y' = k(10 - y) \). By separating the variables, we rewrite it as: ewline \(\frac{dy}{10 - y}=k \textrm{ dt} \).
Now, each side of the equation involves a single variable, making it possible to integrate both sides independently.

Integration in calculus is the process of finding the integral of a function, which is essentially the reverse process of differentiation. Integrating a function helps determine areas under curves, volumes, and in this case, solving differential equations.
After separating the variables in our problem, we integrate both sides: ewline ewline\(\frac{dy}{10 - y}=k dt \), where \( k = 0.1 \).
The integration steps involve:

• ewline - Integrating the left side: ewline \(\frac{dy}{10 - y}= -\textrm {ln}|10 - y| + c_1\)
• ewline - Integrating the right side:ewline \( k \textrm { dt}=0.1t +c_2\)
• ewline - Combining results: ewline \( -\textrm {ln}|10 - y| = 0.1t +c \)
• ewline The constant \( C \) is computed using initial condition \( f(0)=350 \). Afterwards, exponentiate both sides to remove the natural log and solve for \( y \), yielding: ewline \( y = 10 - 340 e^{-0.1t} \)Thus, we've derived the solution to the differential equation ewline \( y = 10 - 340e^{-0.1t}\).