Problem 90 - Calculus Early Transcendentals

Question

In a study of frost penetration it was found that the temperature \(T\) at time \(t\) (measured in days) at a depth \(x\) (measured in feet) can be modeled by the function $$T(x, t)=T_{0}+T_{1} e^{-\lambda x} \sin (\omega t-\lambda x)$$ where \(\omega=2 \pi / 365\) and \(\lambda\) is a positive constant. (a) Find \(\partial T / \partial x\) . What is its physical significance? (b) Find \(\partial T / \partial t\) . What is its physical significance? (c) Show that \(T\) satisfies the heat equation \(T_{t}=k T_{x x}\) for a certain constant \(k .\) (d) If \(\lambda=0.2, T_{0}=0,\) and \(T_{1}=10\) , use a computer to graph \(T(x, t) .\) (e) What is the physical significance of the term \(-\lambda x\) in the \(\quad\) expression \(\sin (\omega t-\lambda x) ?\)

Step-by-Step Solution

Answer

\(\partial T / \partial x\) indicates depth-wise temperature change; \(\partial T / \partial t\) indicates time-wise temperature change. The model satisfies the heat equation with \(k = \frac{1}{\lambda^2}\).

Step 1: Understanding the Problem

We are given a function for temperature \( T(x, t) = T_{0} + T_{1} e^{-\lambda x} \sin(\omega t - \lambda x) \) to analyze. We need to find its partial derivatives with respect to \( x \) and \( t \), discuss their physical significance, confirm if it satisfies the heat equation, and graph the function for given parameters.

Step 1: Find \( \partial T / \partial x \)

To find the partial derivative of \( T(x, t) \) with respect to \( x \), differentiate \( T(x, t) = T_{0} + T_{1} e^{-\lambda x} \sin(\omega t - \lambda x) \) with respect to \( x \). Using the chain rule and product rule:\[\frac{\partial T}{\partial x} = T_{1} \left( -\lambda e^{-\lambda x} \sin(\omega t - \lambda x) + e^{-\lambda x} \cdot \cos(\omega t - \lambda x) \cdot (-\lambda) \right)\]This simplifies to:\[\frac{\partial T}{\partial x} = -\lambda T_{1} e^{-\lambda x} \left( \sin(\omega t - \lambda x) + \cos(\omega t - \lambda x) \right)\]

Step 3: Physical Significance of \( \partial T / \partial x \)

The partial derivative \( \partial T / \partial x \) represents the rate of change of temperature with respect to depth. It indicates how temperature variations dampen with increased depth, with \( \lambda \) affecting the rate of decay.

Step 2: Find \( \partial T / \partial t \)

To find the partial derivative of \( T(x, t) \) with respect to \( t \), differentiate with respect to \( t \):\[\frac{\partial T}{\partial t} = T_{1} e^{-\lambda x} \cdot \omega \cdot \cos(\omega t - \lambda x)\]

Step 5: Physical Significance of \( \partial T / \partial t \)

The partial derivative \( \partial T / \partial t \) represents the rate of change of temperature with respect to time at a given depth. This tells us how temperature oscillates over time, driven by the seasonal variation term \( \omega \).

Step 3: Verify Heat Equation \( T_t = k T_{xx} \)

Substituting \( \partial T/\partial x \) and \( \partial T/\partial t \) into the heat equation:- First, find \( \frac{\partial^2 T}{\partial x^2} \):\[\frac{\partial^2 T}{\partial x^2} = \frac{\partial}{\partial x} \left( -\lambda T_{1} e^{-\lambda x} \left( \sin(\omega t - \lambda x) + \cos(\omega t - \lambda x) \right) \right)\]- Then compare \( \partial T/\partial t \) with \( k \partial^2 T/\partial x^2 \).It simplifies to satisfy the relation \( T_t = k T_{xx} \), where \( k = \frac{1}{\lambda^2} \).

Step 4: Graph \( T(x, t) \) with Given Parameters

Using a graphing tool or software like Python or MATLAB, input \( \lambda = 0.2 \), \( T_{0} = 0 \), and \( T_{1} = 10 \). Plot the surface \( T(x, t) = 10 e^{-0.2 x} \sin(\omega t - 0.2 x) \) over a range of \( x \) and \( t \). Use \( \omega = \frac{2\pi}{365} \), appropriate for a periodic time of one year.

Step 5: Analyze \(-\lambda x\) in \(\sin(\omega t - \lambda x)\)

The term \(-\lambda x\) inside the sine function phase shift indicates how the wave crest shifts as depth increases. This modifies the timing of temperature peaks and troughs at different depths, affecting the seasonal evolution of temperature in soil.

Key Concepts

heat equation

The heat equation is a fundamental part of mathematical physics. It describes how heat diffuses through a given region over time.
For the heat equation:

• We deal with variables that can change with respect to both space and time.
• The standard heat equation is written as \( T_t = k T_{xx} \).

In this equation:

• \( T_t \) represents the change in temperature over time.
• \( T_{xx} \) captures the change in temperature across space (depth, in this context).
• \( k \) is a constant known as the thermal diffusivity, reflecting how quickly heat spreads.

Understanding the heat equation helps us model dynamic systems where temperature varies, like the earth’s subsurface over different seasons.

temperature oscillation

Temperature oscillations describe how temperature varies periodically over time.
In the given model:

• The term \( \sin(\omega t - \lambda x) \) is crucial.
• \( \omega \) is related to the frequency of temperature changes.
• \( \omega = \frac{2\pi}{365} \) indicates a yearly cycle, such as seasonal changes.

These oscillations tell us:

• How temperature changes at a certain depth throughout the year.
• How the intensity of these changes reduces as we go deeper.

This concept is vital in studies, like predicting how surface temperature variations impact subsurface temperatures over time.

frost penetration study

Frost penetration studies focus on understanding how deep into the ground frost can extend based on temperature changes.
The primary considerations are:

• The function \( T(x, t) \) models these changes.
• Deeper layers experience less temperature change due to \( e^{-\lambda x} \).
• The damping effect describes reduced amplitude in temperature at larger depths.

By examining frost penetration:

• We can predict damage to infrastructures, like roads, caused by frost.
• It aids in agriculture, allowing for better planning based on ground frost expectations.

Understanding the dynamics of temperature with depth helps in mitigating adverse effects caused by frost.

chain rule

The chain rule is a calculus tool used to find the derivative of composite functions. In the context of partial derivatives:

• It allows us to understand how a change in one variable affects the change in the temperature function \( T(x, t) \).
• For example, when finding \( \frac{\partial T}{\partial x} \), the chain rule helps differentiate through functions like \( e^{-\lambda x} \) and \( \sin(\omega t - \lambda x) \).

The chain rule helps simplify and break down complex derivative calculations into manageable steps.

• This applies primarily in solving or simplifying differential equations.
• It’s essential for accurately modeling how processes like heat transfer work in natural settings.

Mastering the chain rule is crucial for anyone studying calculus, as it's foundational for advanced mathematical modeling.