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A bottle of white wine at room temperature \(\left(68^{\circ} \mathrm{F}\right)\) is placed in a refrigerator at \(4 \mathrm{P.M}\). Its temperature after \(t\) hr is changing at the rate of \(-18 e^{-0.6 t}{ }^{\circ} \mathrm{F} / \mathrm{hr}\). By how many degrees will the temperature of the wine have dropped by 7 P.M.? What will the temperature of the wine be at 7 P.M.?

Short Answer

Expert verified
The temperature of the wine will have dropped by approximately 34.959°F by 7 P.M., and the temperature of the wine at 7 P.M. will be approximately \(33.041^{\circ}\mathrm{F}\).

Step by step solution

01

Set up the integral for temperature change.

To find the temperature change from 4 P.M. to 7 P.M., we need to integrate the given rate, \(-18 e^{-0.6 t}\) with respect to time (\(t\)) over the interval \([0, 3]\) hours (since 7 P.M. is 3 hours after 4 P.M.). The integral is: \[ΔT = \int_0^3 -18 e^{-0.6 t} dt\]
02

Integrate the function.

Now we will integrate the function with respect to time: \[ ΔT= \int_0^3 -18 e^{-0.6 t} dt = -\frac{18}{0.6}(e^{-0.6 t})\Big|_0^3= -30(e^{-0.6 t})\Big|_0^3 \]
03

Evaluate the integral.

Now we will evaluate the integral at the given limits and subtract the values: \[ ΔT = -30(e^{-0.6(3)} - e^{-0.6(0)}) = -30(e^{-1.8} - 1) \]
04

Calculate the temperature drop by 7 P.M.

Simplify the expression of \(ΔT\): \[ ΔT = -30(e^{-1.8} - 1) ≈ -30(-0.1653 - 1) ≈ -30(-1.1653) ≈ 34.959 \] So, the temperature will have dropped by approximately 34.959°F by 7 P.M.
05

Calculate the temperature at 7 P.M.

The initial temperature of the wine is given as $68^{\circ} \mathrm{F}$. To find the temperature at 7 P.M., subtract the temperature drop from the initial temperature: \(T_{7\mathrm{P.M}} = 68 - 34.959 ≈ 33.041\) So, the temperature of the wine at 7 P.M. will be approximately \(33.041^{\circ}\mathrm{F}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
The concept of a definite integral is fundamental in calculus and is used to determine the total accumulation of quantities, such as area under a curve, total distance traveled, or in our case, the total change in temperature over a certain period. To find the definite integral, we integrate the function over a specific interval, which represents the start and end points in time or space.

In the wine cooling example, we have the rate of temperature decrease represented by the function (-18 e^{-0.6 t}). We calculate the definite integral of this rate from the time the wine was placed in the fridge (0 hours) to the time of interest (3 hours later, which is 7 P.M.). The result of this computation gives us the total change in temperature during that time span.
Exponential Functions
An exponential function is a mathematical expression where a constant base is raised to a variable exponent. In our context, the exponential function e^{-0.6 t} describes the rate of temperature change of the wine in the refrigerator. The negative exponent indicates that the temperature decrease rate diminishes over time.

Why Use Exponential Functions?

Exponential functions are ideal for modeling situations where changes occur at rates proportional to the current value—such as cooling processes, population growth, and radioactive decay.
Rate of Change
The rate of change tells us how a quantity changes with respect to another variable. In our wine example, the rate of temperature change is given in degrees Fahrenheit per hour. A negative sign in front of the rate (-18 e^{-0.6 t}) suggests that the quantity (temperature, in this case) is decreasing over time.

Understanding Rate of Change

Understanding the rate of change is crucial because it allows us to predict future values and understand how fast or slow a variable is changing. It's a core concept not only in physical scenarios like our temperature example but also in economics, biology, and almost any field involving dynamic processes.
Temperature Change Calculus
Using calculus to calculate temperature change is an exemplary application of math in real-world scenarios. By integrating a function representing the rate of temperature change over time, we can determine how much the temperature has changed after a certain period.

In the case of our wine, we integrate the rate function over a 3-hour interval to find out the total temperature drop. The beauty of using calculus in such situations is that it gives us a precise tool to deal with changing conditions over time, and it's a prime example of how differential equations are used to solve practical problems involving rates of change in various fields — not only in temperature scenarios but also in any context where quantities change over time.

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Most popular questions from this chapter

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