Temperature Function Analysis: Finding Time Intervals

Hey math enthusiasts! Let's dive into a cool problem involving a temperature function. We're given a function that models temperature over time, and our mission is to figure out when the temperature soared above a certain threshold. Sounds fun, right?

Understanding the Temperature Function

First off, let's break down the function itself. We have: $T(t) = \frac{8t}{t^2 + 1} + 98.6$, where t represents time in hours. The T(t) part is the temperature at time t. We can see that the temperature is a function of time, meaning as time changes, so does the temperature. The term $98.6$ is an important value because it's in the form of the average human body temperature. The $\frac{8t}{t^2 + 1}$ part is the variable part of the function, which takes into account the different values ​​as time changes. To understand this function better, we can also look at its components: The $8t$ part in the numerator suggests that the temperature increases as time goes on, at least initially. The $t^2 + 1$ part in the denominator, on the other hand, suggests that the temperature might be affected by time (squared) plus one. Since time is squared, this suggests that the function's rate of change may not be linear. In fact, if we analyzed the function more deeply, we would realize that the rate of change in temperature tends to decrease as the time elapses. The addition of $98.6$ simply shifts the entire function upwards by that much. So, at time $t=0$, the temperature will be $98.6$. This function gives us the temperature in a specific time unit. The function appears to model how the temperature changes over time. It could represent anything from a chemical reaction to the temperature of a specific environment. Now that we have a basic understanding of the function, let's think about the goal: determine the time intervals during which the temperature surpasses 10 degrees. It's a classic example of applying mathematical functions to real-world scenarios, making it more intuitive and practical.

Now, the problem asks us to find the interval over which the temperature was over 10. That means we need to find the values of t (time in hours) for which T(t) > 10. This is a common type of problem in algebra and calculus, where we analyze inequalities involving functions. We're essentially trying to find the time periods when the temperature exceeds a specific level. So, how do we tackle this? First, we need to set up an inequality and then solve it. The inequality is: $\frac{8t}{t^2 + 1} + 98.6 > 10$. We'll walk through the steps to solve this and find our answer.

Setting Up the Inequality and Solving for Time

Alright, guys, let's get our hands dirty with some algebra! Our starting point is the inequality we discussed earlier: $\frac{8t}{t^2 + 1} + 98.6 > 10$. Our goal is to isolate t and find the range of values that satisfy the inequality. The first thing we can do is subtract 98.6 from both sides of the inequality. This gives us: $\frac{8t}{t^2 + 1} > 10 - 98.6$, which simplifies to $\frac{8t}{t^2 + 1} > -88.6$. Since the denominator, $t^2 + 1$, is always positive (because t squared is always non-negative, and adding 1 makes it strictly positive), we don't have to worry about flipping the inequality sign at any point. Then, let's multiply both sides by $t^2 + 1$. Since $t^2 + 1$ is always positive, this also doesn't change the inequality sign. We get: $8t > -88.6(t^2 + 1)$. Expanding the right side gives us: $8t > -88.6t^2 - 88.6$. Now, let's rearrange the terms to get everything on one side, which will help us solve the quadratic inequality. We'll add $88.6t^2$ and $88.6$ to both sides, so we get: $88.6t^2 + 8t + 88.6 > 0$. We've transformed our problem into a quadratic inequality. In theory, we can find the roots of the quadratic equation $88.6t^2 + 8t + 88.6 = 0$ by using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where a = 88.6, b = 8, and c = 88.6. But let's take a look at the discriminant, which is the part under the square root, i.e., $b^2 - 4ac$. In our case, it's $8^2 - 4(88.6)(88.6)$, which equals $64 - 31472.96 = -31408.96$. Since the discriminant is negative, that means this quadratic equation has no real roots. No real roots mean that the parabola (the graph of this quadratic function) doesn't intersect the x-axis. Because the leading coefficient (88.6) is positive, the parabola opens upwards. This means that the quadratic function is always positive. The inequality $88.6t^2 + 8t + 88.6 > 0$ is true for all real values of t. The temperature will always be greater than -88.6 at any point in time.

Determining the Solution Interval

Okay, so we've done the algebra, and we found that the inequality $88.6t^2 + 8t + 88.6 > 0$ is true for all real values of t. Going back to our initial inequality $\frac{8t}{t^2 + 1} + 98.6 > 10$, this means that $T(t) > 10$ for all real values of t. Since our function is defined for all time values, we can conclude that the temperature is always greater than 10 degrees, regardless of the time. However, this conclusion doesn't make a lot of sense, right? Looking back at the original temperature function, $T(t) = \frac{8t}{t^2 + 1} + 98.6$, we can see that the term $\frac{8t}{t^2 + 1}$ can vary depending on the value of t. For example, when t is zero, then $\frac{8t}{t^2 + 1}$ is also zero, and the overall temperature is 98.6. When t is a positive number, $\frac{8t}{t^2 + 1}$ also becomes a positive number, which means the overall temperature is higher than 98.6. When t is a negative number, $\frac{8t}{t^2 + 1}$ also becomes a negative number, which means the overall temperature is lower than 98.6. Therefore, the term $\frac{8t}{t^2 + 1}$ would change depending on the value of t. We need to find the range of values for which t is greater than or equal to 0, which also meets the temperature condition. Now we have two inequalities: $\frac{8t}{t^2 + 1} + 98.6 > 10$ and $t >= 0$. Let's solve these two equations. We've already known that the solution for $\frac{8t}{t^2 + 1} + 98.6 > 10$ is all real numbers. We know that the value of $T(t)$ will be greatest when t approaches infinity. However, our main concern is to find when the temperature will be over 10 degrees. Because the function is always greater than 10, the interval would be $(-\infty, \infty)$. However, since time t in this context cannot be negative (as time cannot go back), the more accurate interval would be $[0, \infty)$. This means that the temperature is always above 10 degrees from the beginning of our observation (t=0) and continues indefinitely into the future. It's a pretty straightforward result, but it's important to understand the reasoning behind it.

Conclusion and Key Takeaways

So, in conclusion, the temperature given by the function $T(t) = \frac{8t}{t^2 + 1} + 98.6$ is always greater than 10 degrees for all t values where t is greater than or equal to 0. We found this by manipulating the inequality, recognizing that the resulting quadratic expression is always positive, and interpreting the results within the context of the problem. Remember, always double-check your work and consider the context of the problem to ensure your solution makes sense. This problem highlights the importance of understanding not just how to solve equations but also how to interpret the results in the real world. You should always make sure you are confident in your algebra skills, and you should always double-check the results and ensure they make sense in the context of the problem. Understanding the behaviors of functions is really important. In this case, we looked at how to find intervals where a function is greater than a specific value. Keep practicing, and you'll get better! If you enjoyed this, keep up the great work and stay curious with math!