91Ó°ÊÓ

In the context of this exercise, the 'rate of work' refers to the number of pages you can study per hour. It is an important concept as it helps us understand the efficiency of studying under varying conditions. Here, the rate of work decreases as the temperature increases. At 2 p.m., you were able to cover 20 pages in one hour. However, as the temperature increased, your studying efficiency dropped, and by the second hour, you could only manage 10 pages. This declining productivity showcases how the 'rate of work' is affected by temperature, illustrating an inverse proportionality relationship with temperature excesses over a certain baseline (in this case, 75°F). To put it simply:

• Rate of work = How many pages per hour you can study


• Higher temperature = Lower rate of work


• Inverse relationship = As one goes up, the other goes down

Temperature dependence means that the rate at which you study is closely linked to the ambient temperature. As specified, your study rate is inversely proportional to how much the temperature exceeds 75°F. Let's break it down:

• When the temperature is exactly 75°F, it acts like a neutral point with no impact on your work rate.


• As the temperature rises above 75°F, it starts to negatively affect your study rate. In other words, the higher the temperature, the fewer pages you can cover.

This relationship is captured by the formula: \( R(t) = \frac{k}{T(t) - 75} \), where \( T(t) \) is the temperature at time \( t \) and \( k \) is a proportionality constant. This formula lets us calculate how temperature affects study rate at different times.

The proportionality constant \( k \) is a crucial factor in determining the rate at which you can study relative to the excess temperature. Mathematically, it helps us establish the inverse relationship between study rate and temperature. Suppose the rate of work at specific times is known:

• At 2 p.m.: \( R(0) = 20 \) pages per hour


• At 3 p.m.: \( R(1) = 10 \) pages per hour

Using the inverse proportionality formula: \( R(0) = \frac{k}{T(0) - 75} = 20 \) and \( R(1) = \frac{k}{T(1) - 75} = 10 \), we can solve for \( k \). This proportionality constant, once determined, allows us to predict the rate of studying at any temperature above 75°F efficiently.

Time evaluation in this problem involves determining the trend of temperature changes and how they affect study rate over time. To do this, we consider the temperature as rising at a constant rate. Assuming \( T(t) = T_0 + at \), where \( T_0 \) is the initial temperature at 2 p.m. and \( a \) is the rate of temperature increase per hour. From the given study rates:

• At 2 p.m.: 20 pages/hour implies a baseline temperature \( T_0 \)


• At 3 p.m.: 10 pages/hour implies an increased temperature \( T_1 = T_0 + a \)

We find that \( T_1 - 75 = 2(T_0 - 75) \) and solve for \( T_0 \). Knowing that the temperature rises steadily helps us determine the exact time when it was 75°F. By considering the constant increase, we can backtrack to find the exact time when the baseline temperature was established.