/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 A yam baking in the oven: A yam ... [FREE SOLUTION] | 91Ó°ÊÓ

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A yam baking in the oven: A yam is placed in a preheated oven to bake. An application of Newton's law of cooling gives the temperature \(Y\), in degrees, of the yam \(t\) minutes after it is placed in the oven as $$ Y=400-325 e^{-t / 50} $$ a. Make a graph of the temperature of the yam at time \(t\) over 45 minutes of baking time. b. Calculate \(\frac{d Y}{d t}\) at the time 10 minutes after the yam is placed in the oven. c. Calculate \(\frac{d Y}{d t}\) at the time 30 minutes after the yam is placed in the oven. d. Explain what your answers in parts \(b\) and \(c\) tell you about the way the yam heats over time.

Short Answer

Expert verified
The yam heats quickly initially, then the heating rate slows down over time.

Step by step solution

01

Understand the Problem

The temperature of the yam is described by the function \( Y(t) = 400 - 325e^{-t/50} \). We need to graph this function for 0 to 45 minutes and find the derivatives \( \frac{dY}{dt} \) at \( t = 10 \) and \( t = 30 \). Newton's law implies a form of exponential heating, where the temperature approaches 400 degrees over time.
02

Graph the Function

To graph \( Y(t) \) over 45 minutes, evaluate the function at various values of \( t \) (e.g., \( t = 0, 10, 20, 30, 40, 45 \)) and plot these points on a graph. The graph will show a curve that is steep at first and gradually levels off around 400 degrees.
03

Differentiate the Function

Find the derivative \( \frac{dY}{dt} \) to understand how the temperature changes over time. Start with the expression \( Y(t) = 400 - 325e^{-t/50} \). Differentiate with respect to \( t \) to get:\[\frac{dY}{dt} = 325 \cdot \frac{1}{50} \cdot e^{-t/50}\]which simplifies to:\[\frac{dY}{dt} = \frac{325}{50} e^{-t/50}\]Thus, \( \frac{dY}{dt} = 6.5e^{-t/50} \).
04

Evaluate the Derivative at t = 10

Plug in \( t = 10 \) into the derivative:\[\frac{dY}{dt}\bigg|_{t=10} = 6.5e^{-10/50}\]Calculate \( e^{-10/50} \), which is \( e^{-0.2} \approx 0.8187 \). Thus,\[\frac{dY}{dt}\bigg|_{t=10} = 6.5 \times 0.8187 \approx 5.32\]
05

Evaluate the Derivative at t = 30

Plug in \( t = 30 \) into the derivative:\[\frac{dY}{dt}\bigg|_{t=30} = 6.5e^{-30/50}\]Calculate \( e^{-30/50} \), which is \( e^{-0.6} \approx 0.5488 \). Thus,\[\frac{dY}{dt}\bigg|_{t=30} = 6.5 \times 0.5488 \approx 3.57\]
06

Analyze the Results

The derivatives \( \frac{dY}{dt} \) at \( t = 10 \) and \( t = 30 \) show how the rate of temperature increase (heating rate) slows over time. The yam heats quickly at first (rate \( \approx 5.32 \) at \( t = 10 \)) but slows as it approaches equilibrium (rate \( \approx 3.57 \) at \( t = 30 \)). This behavior is typical of exponential models like Newton's law of cooling.

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Key Concepts

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

Exponential Functions
Exponential functions are a key concept needed to understand Newton's Law of Cooling. In the case of the yam heating scenario, the temperature function given by the equation: \[ Y(t) = 400 - 325e^{-t/50} \] demonstrates an exponential approach to a limiting value. Here, the expression \( e^{-t/50} \) is the exponential component that dictates how the temperature changes over time. Unlike linear functions, which change at a constant rate, exponential functions change at rates proportional to their current value.
  • At the start (\( t = 0 \)), \( e^0 = 1 \), making the initial drop quickly change as the process begins.
  • As time \( t \) increases, \( e^{-t/50} \) becomes smaller, indicating a slower change. This is why the yam initially heats up more quickly and slows down as it approaches the oven's temperature.
  • The temperature doesn't just steadily climb; it decelerates as time progresses, creating the curved path when graphed on a chart.
This behavior is a hallmark of exponential growth or decay processes in physics and biology, making them important for modeling real-world phenomena such as cooling and heating.
Calculus Differentiation
Calculus differentiation is the process used to find the rate at which a function changes. In our yam baking example, we use differentiation to analyze how quickly the yam's temperature increases at specific moments in time. Given the temperature function\[ Y(t) = 400 - 325e^{-t/50}, \] we find its derivative with respect to time \( t \) to understand the yam's heating rate.First, you apply basic differentiation rules. The derivative of a constant is 0, and we apply the chain rule to the exponential part \( 325e^{-t/50} \). The result is \[ \frac{dY}{dt} = 325 \cdot \frac{1}{50} \cdot e^{-t/50} = 6.5e^{-t/50}. \] This derivative \( 6.5e^{-t/50} \) represents how the rate of temperature increase changes. By evaluating this at different times:
  • At \( t = 10 \), the rate is approximately 5.32, showing a quick initial temperature increase.
  • At \( t = 30 \), the rate drops to about 3.57, revealing a slower increase. This slowdown confirms the exponential nature of the heating process.
Differentiation allows us to pinpoint these changes in rate, helping visualize and predict the behavior of complex curves beyond what simple algebraic expressions can provide.
Temperature Modeling
Temperature modeling using Newton's Law of Cooling is a practical way to predict how an object like a yam changes temperature over time when placed in an environment with a different constant temperature—in this case, an oven. The function\[ Y(t) = 400 - 325e^{-t/50} \] models this process by capturing how fast or slow the yam heats up relative to the oven's constant temperature of 400 degrees.One of the fascinating aspects of temperature modeling is its predictive power:
  • The initial rapid increase in temperature represents the yam quickly absorbing the oven's heat initially.
  • Over time, the yam approaches the oven temperature more slowly, a common pattern in systems dominated by exponential functions.
  • Such models are useful not only in culinary contexts but also in engineering, climatology, and even in medical fields where body temperature changes need analysis.
By employing mathematical equations to model these processes, scientists and engineers can make predictions and adjustments to systems, ensuring reliability and efficiency in temperature-dependent procedures.

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Most popular questions from this chapter

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