Chapter 14: Problem 37
Show that the operation of taking the gradient of a function has the given property. Assume that \(u\) and \(v\) are differentiable functions of \(x\) and \(y\) and that \(a, b\) are constants. (a) \(\nabla(a u+b v)=a \nabla u+b \nabla v\) (b) \(\nabla(u v)=u \nabla v+v \nabla u\) (c) \(\nabla\left(\frac{u}{v}\right)=\frac{v \nabla u-u \nabla v}{v^{2}}\) (d) \(\nabla u^{n}=n u^{n-1} \nabla u\)
Short Answer
Step by step solution
Understanding the Gradient
Property (a) - Linearity of Gradient
Property (b) - Product Rule
Property (c) - Quotient Rule
Property (d) - Power Rule
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient
For a function of two variables, say \( f(x, y) \), the gradient is denoted as \( abla f \) and is a vector comprising the partial derivatives of the function with respect to each variable. In formulaic terms, it is expressed as \( \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). This yields a vector pointing towards the direction of maximum increase with a magnitude indicating the rate of that increase.
- The gradient is indispensable in various applications involving optimization, where determining the direction of ascent or descent is crucial.
- Understanding the gradient for scalar fields unveils the geometry of the field, offering a clearer visualization of potential changes.
Product Rule
For functions \( u(x, y) \) and \( v(x, y) \), the gradient of their product, \( abla(uv) \), follows the rule: \( abla(uv) = u abla v + v abla u \).
What this means is:
- For the partial derivative with respect to \( x \), you take \( u \) multiplied by the partial derivative of \( v \) plus \( v \) multiplied by the partial derivative of \( u \).
- Similarly, for the partial derivative with respect to \( y \), do the same multiplication and addition.
Quotient Rule
Consider two differentiable functions \( u(x, y) \) and \( v(x, y) \). The gradient of their quotient \( \frac{u}{v} \) can be described as follows: \[ abla\left(\frac{u}{v}\right) = \left( \frac{v abla u - u abla v}{v^2} \right) \]
This gradient computation involves:
- Subtracting \( u \), differentials multiplied by the gradient of \( v \), from \( v \) differentials multiplied by the gradient of \( u \).
- Dividing the resulting vector by the square of \( v \).
Power Rule
When we consider a function \( u(x, y) \) raised to a power \( n \), the gradient is determined using the power rule: \[ abla u^n = n u^{n-1} abla u \]
In this context:
- "\( n \)" represents the exponent of the function \( u \).
- Multiply \( n \) by the function \( u \, \) reduced by one power, i.e., \( u^{n-1} \).
- Then multiply by the gradient of the function \( u \).