Question

MAKE A DECISION The temperature \(T\) (in degrees Fahrenheit) during a certain day can be approximated by \(T=0.31 t^{2}+32.9,7 \leq t \leq 15\), where \(t\) represents the hour of the day, with \(t=7\) corresponding to 7 A.M. Use the model to predict the time when the temperature was \(85^{\circ} \mathrm{F}\). Can you use this model to predict the temperature at 7 P.M.? Explain.

Short answer

The appropriate time when the temperature was \(85^{\circ} F\) can be determined by solving the equation for \(t\). However, the model only applies accurately for the range \(7 \leq t \leq 15\) (between 7 A.M and 3 P.M.), making its prediction for temperatures at 7 P.M unreliable because it falls outside this range.

Step-by-step solution

Solve for time \(t\) when \(T=85^{\circ} F\)

Start by setting the temperature \(T\) equals to \(85\). Substitute \(85\) into the equation: \(85=0.31t^2 + 32.9\). Rearrange the equation to get \(t^2 = (85 - 32.9) / 0.31\). Solve for \(t\) taking the positive square root as time cannot be negative.

Round to nearest hour

Upon solving, we get two values for \(t\), a decimal and its negative pair. We discard the negative value because time can't be negative. Round off the positive decimal to the nearest whole number if necessary as \(t\) represents the hours. This would give the approximate time when the temperature reached \(85^oF\).

Apply the model to 7 P.M.

Now test if the model can predict the temperature at 7 P.M. Substituting \(t=19\) in the equation: \(T = 0.31(19^2) + 32.9\). Calculate the value of \(T\) to find the predicted temperature at 7 P.M.

Discuss the model's limits

An important interpretation point here is the range of time \(7 \leq t \leq 15\) indicating that the model only applies from 7 A.M to 3 P.M. Using this model to predict the temperature at 7 P.M (i.e., \(t=19\)) is outside of this range and thus not reliable.

Key concepts

Quadratic Models

Quadratic models are at the heart of many natural processes, where the relationship between two variables is not linear but can still be graphed as a parabola. In our temperature model, the equation provided is a function of time, represented as \( T = 0.31t^2 + 32.9 \), where the variable \( T \) represents temperature, and \( t \) represents time. Such equations are powerful because they can capture more complex behaviors like acceleration, growth, or, in our case, changes in temperature over time.

Quadratic models are useful because they can account for variables that increase at a rate that itself increases or decreases. When graphed, they produce the familiar 'U' shaped curve known as a parabola. In the context of temperature modeling, this allows us to visualize and predict the temperatures at different times of the day, provided we stay within the model's valid range.

Mathematical Prediction

Mathematical prediction enables us to forecast future events or values by using mathematical models, such as the quadratic model in our temperature prediction problem. By substituting a specific time into our temperature model equation, we can predict the temperature at that hour.

However, the quality of any prediction relies heavily on the model's accuracy and the context in which it is used. As the given model is only validated for certain hours of the day, predictions outside of this range (7 A.M. to 3 P.M.) are speculative and likely to be inaccurate, as the real-world behavior they are supposed to represent may not adhere to the model's pattern during those times.

Solving Quadratic Equations

Solving quadratic equations is a staple of algebra that involves finding the values of the variable that make the equation true. The general form of a quadratic equation is \( ax^2 + bx + c = 0 \), and it can have zero, one, or two real solutions. These solutions may be found using the quadratic formula, factoring, or, as in our example, by isolating the variable and taking the square root.

In the context of our temperature model, setting \( T = 85 \) degrees and solving for \( t \) involved rearranging the equation to isolate \( t^2 \) and then taking the square root. We only consider the positive square root since negative time isn't practical in our context. This step is crucial because it decodes the time at which the temperature reached 85 degrees Fahrenheit within the constraints of the model.

Applying Mathematical Models

Applying mathematical models involves using mathematics to simulate real-world phenomena. It's a process that translates observations into a language we can work with to make predictions, make informed decisions, or simply understand the world around us better. In our example, the model represented the relationship between time and temperature within a given range.

A crucial aspect of applying any model is knowing its limitations. Models are built on assumptions and simplifications of reality. Thus, when a particular value, like 7 P.M., falls outside the model's validated range, it would be inappropriate to make predictions because the model was not designed to accommodate such values. Understanding the range and conditions under which a mathematical model is applicable is key to using it effectively.