/*! 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 25 Show that if $$ \phi(x, t)=\... [FREE SOLUTION] | 91Ó°ÊÓ

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Show that if $$ \phi(x, t)=\frac{f(z)}{\sqrt{t}} \text { and } \quad z=\frac{x}{2 \sqrt{t}} $$ then $$ \frac{\partial \phi}{\partial t}=-\frac{z f^{\prime}(z)+f(z)}{2 t \sqrt{t}} $$ and find a similar expression for \(\partial^{2} \phi / \partial x^{2}\). Deduce that if $$ \frac{\partial^{2} \phi}{\partial x^{2}}=\frac{1}{k} \frac{\partial \phi}{\partial t} $$ then $$ k f^{\prime \prime}(z)+2 z f^{\prime}(z)+2 f(z)=0 $$

Short Answer

Expert verified
The equation is \( k f''(z) + 2z f'(z) + 2f(z) = 0 \).

Step by step solution

01

Differentiate \(\phi(x, t)\) with respect to \(t\)

To find \( \frac{\partial \phi}{\partial t} \), start by differentiating \( \phi(x, t) = \frac{f(z)}{\sqrt{t}} \) with respect to \( t \). Since \( z = \frac{x}{2 \sqrt{t}} \), use the chain rule. First, differentiate \( \frac{1}{\sqrt{t}} \):\[ \frac{d}{dt} \left( \frac{1}{\sqrt{t}} \right) = -\frac{1}{2} t^{-3/2} = -\frac{1}{2t \sqrt{t}} \]Next, use the chain rule for \( f(z) \), where \( \frac{\partial z}{\partial t} = -\frac{x}{4t^{3/2}} = -\frac{z}{2t} \), hence:\[ \frac{\partial \phi}{\partial t} = \frac{f'(z) \cdot -\frac{z}{2t}}{\sqrt{t}} - \frac{f(z)}{2t \sqrt{t}} = -\frac{z f^{\prime}(z) + f(z)}{2 t \sqrt{t}} \]
02

Differentiate \(\phi(x,t)\) with respect to \(x\) twice

To find \( \frac{\partial^2 \phi}{\partial x^2} \), first differentiate \( \phi(x, t) \) with respect to \( x \) using the chain rule. Since \( z = \frac{x}{2 \sqrt{t}} \), we have \( \frac{\partial z}{\partial x} = \frac{1}{2 \sqrt{t}} \):\[ \frac{\partial \phi}{\partial x} = \frac{f'(z)}{\sqrt{t}} \cdot \frac{1}{2 \sqrt{t}} = \frac{f'(z)}{2t} \]Then differentiate again to find \( \frac{\partial^2 \phi}{\partial x^2} \):\[ \frac{\partial^2 \phi}{\partial x^2} = \frac{1}{2t} \frac{d}{dx}(f'(z)) = \frac{1}{2t} \cdot f''(z) \cdot \frac{1}{2 \sqrt{t}} = \frac{f''(z)}{4t} \]
03

Apply the given condition to deduce the equation for \(f(z)\)

Given that \( \frac{\partial^2 \phi}{\partial x^2} = \frac{1}{k} \frac{\partial \phi}{\partial t} \), substitute the expressions:\[ \frac{f''(z)}{4t} = \frac{1}{k} \left(-\frac{z f'(z) + f(z)}{2t \sqrt{t}}\right) \]Clear the fractions by multiplying through by \(4 t \sqrt{t}\):\[ \sqrt{t} f''(z) = -\frac{2}{k}(z f'(z) + f(z)) \]Divide by \( \sqrt{t} \) to isolate terms involving \( z \):\[ k f''(z) + 2z f'(z) + 2f(z) = 0 \]
04

Conclusion

We verified the derivative of \( \phi(x, t) \) with respect to \( t \) and found the expression for \( \frac{\partial^2 \phi}{\partial x^2} \). Using the given condition \( \frac{\partial^2 \phi}{\partial x^2} = \frac{1}{k} \frac{\partial \phi}{\partial t} \), we deduced the final equation involving \( f(z) \).

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Key Concepts

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

Chain Rule
The chain rule is a fundamental concept in calculus that simplifies the process of differentiating composite functions. It helps us understand how changes in one variable affect another through a third, dependent variable. In the exercise, the chain rule was crucial for differentiating the function \( \phi(x, t) = \frac{f(z)}{\sqrt{t}} \), where \( z = \frac{x}{2\sqrt{t}} \).

Here's why the chain rule was necessary:
  • \( \phi(x, t) \) is expressed as a function of \( t \) through both direct dependence on \( \sqrt{t} \) and indirect dependence via \( z \).
  • By applying the chain rule, we first recognize that changes in \( t \) not only affect \( \sqrt{t} \) but also \( z \), as \( z \) is a function of \( t \) as well.
The step-by-step differentiation showed how to navigate these dependencies by differentiating \( f(z) \), taking into account its dependence on \( x \) and \( t \) through the intermediate variable, \( z \). This method is powerful in breaking down what initially seem like complex interactions into manageable parts.
Heat Equation
The heat equation is a partial differential equation (PDE) that describes how heat diffuses through a given region over time. It is typically written in the form: \[\frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2}\]This equation relates the time rate of change of temperature \( u \) to the spatial rate of change of temperature gradient, modulated by a constant \( k \), known as the thermal diffusivity.

In the exercise, the goal was to show consistency with the heat equation by deriving expressions for \( \frac{\partial \phi}{\partial t} \) and \( \frac{\partial^2 \phi}{\partial x^2} \), and then ensuring that they adhered to the heat equation form.
  • The problem sets up by defining \( \phi(x, t) \) in a similar vein to a typical temperature function \( u(x, t) \).
  • By comparing \( \frac{\partial^2 \phi}{\partial x^2} \) and \( \frac{1}{k} \frac{\partial \phi}{\partial t} \), a relationship is established, resembling the behavior dictated by the heat equation.
This exercise helps reinforce understanding of how solutions to PDEs like the heat equation relate to physical phenomena, in this case, heat diffusion.
Mathematical Derivation
Mathematical derivation is the process of applying rules and logical steps to transform one or more equations into another form. It provides a structured way of approaching problems and finding solutions, as demonstrated in the exercise.

In this exercise, the goal was to manipulate the given function \( \phi(x, t) \) to derive meaningful expressions that match known physical forms, such as the heat equation.
  • The task began with deriving partial derivatives using the function \( \phi(x, t) \).
  • Following this, the derived partial derivatives were compared, and algebraic manipulation helped simplify complex expressions.
  • The end result was an expression, \( k f''(z) + 2z f'(z) + 2f(z)=0 \), which embodies the underlying conditions set by the original problem.
This methodical approach not only aids in solving specific types of equations but also builds an intuition for tackling similar problems. Understanding and applying derivation leads to a deeper comprehension of the underlying mathematical models, such as those encountered in physics and engineering.

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Most popular questions from this chapter

A flat circular metal plate has a shape defined by the region \(x^{2}+y^{2} \leqslant 1\). The plate is heated so that the temperature \(T\) at any point \((x, y)\) on it is given by $$ T=x^{2}+2 y^{2}-x $$ Find the temperatures at the hottest and coldest points on the plate and the points where they occur. (Hint: Consider the level curves of \(T\).)

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Find \(\partial f / \partial x\) and \(\partial f / \partial y\) when \(f(x, y)\) is (a) \(\mathrm{e}^{x} \cos x\) (b) \(\frac{x}{x^{2}+y^{2}}\) (c) \(\frac{x+y}{x^{2}+2 y^{2}+6}\)

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