91Ó°ÊÓ

Differentiation is a fundamental concept in calculus that is used to measure how a function changes as its input changes. Think of it like finding the speed of a moving car; it's about understanding how quickly or slowly something is varying.
In our exercise, we're dealing with a temperature function, represented as \( T(t) = 70 + 130 e^{-2.5t} \). To find out how the temperature changes over time, we differentiate this function with respect to \( t \).
Here’s a simple walkthrough of differentiation for our problem:

• The exponential part \( 130 e^{-2.5t} \) requires using the chain rule for differentiation. The derivative of \( e^x \) is \( e^x \) itself; however, when there’s a coefficient or a constant involved, this changes.
• For the function \( e^{-2.5t} \), the derivative is \(-2.5 e^{-2.5t} \) due to the chain rule.
• When calculating it as part of \( 130 e^{-2.5t} \), it multiplies with \( -2.5 \), resulting in \(-130 \times 2.5 e^{-2.5t} = -325 e^{-2.5t} \).

Differentiation shows us how fast the coffee is cooling down at any given moment.

Exponential decay is a process where quantities decrease rapidly at first, and then gradually slow down over time. It's often represented with a formula involving the exponential function \( e^x \). Here, we have \( T(t) = 70 + 130 e^{-2.5t} \), where the term \( 130 e^{-2.5t} \) manages the decay.
Why is the exponent negative? This negative sign in \( e^{-2.5t} \) signifies that rather than growing over time, the variable's effect decreases. This is a common trait in contexts such as cooling, radioactive decay, and depreciation.
Specifically, in our scenario:

• The coffee starts at a high temperature (200 degrees with the additional 70 degrees).
• The exponential term, \( 130 e^{-2.5t} \), decreases, lowering the overall temperature value as time increases.

The concept of exponential decay illuminates how the coffee’s temperature change becomes less drastic over time. Initially rapid, it steadily declines in speed, as the temperature evens out closer to room temperature, depicted synergistically by our cooling function.

The rate of change is a measure of how one quantity changes in respect to another. It's akin to perceiving how fast or slow something shifts over a period. In the context of our coffee cup cooling down, the aim is to determine how swiftly the temperature decreases at given points in time.
Upon deriving the temperature function \( T(t) = 70 + 130 e^{-2.5t} \) into \( \frac{dT}{dt} = -325 e^{-2.5t} \), we're now able to analyze the rate at which the coffee cools:

• At \( t = 0 \), the rate is \( -325 \), suggesting a fast cooling immediately.
• When \( t = 1 \), the rate becomes approximately \(-26.6\), indicating a slower pace after an hour.

This showcases the nature of cooling—fast initially and then steadying. Differentiation thus not only helps calculate rates but also helps us interpret them for practical insights, like understanding how a cup of hot coffee will eventually settle in a cooler environment.