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Chapter 6: Problem 4

The substitutions \(x=s^{2}-t^{2}\) and \(y=2\) st convert a smooth real-valued function \(f(x, y)\) into a function of \(s\) and \(t: w=f\left(s^{2}-t^{2}, 2 s t\right) .\) Find a formula for \(\|\nabla w\|=\left\|\left(\frac{\partial w}{\partial s}, \frac{\partial w}{\partial t}\right)\right\|\) in terms of \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\).

Short Answer

Expert verified
\( \|\nabla w\| = 2 \sqrt{s^2 + t^2} \cdot \sqrt{\left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \)

Step by step solution

01

Define the Partial Derivatives

First, we need to express \( \frac{\partial w}{\partial s} \) and \( \frac{\partial w}{\partial t} \) in terms of \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). Applying the chain rule to \( w = f(x, y) = f(s^2 - t^2, 2st) \), we find:\[\frac{\partial w}{\partial s} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial s}\]
02

Compute Derivatives with respect to s

Compute the derivatives of the substitutions \( x = s^2 - t^2 \) and \( y = 2st \) with respect to \( s \):\[\frac{\partial x}{\partial s} = 2s \quad \text{and} \quad \frac{\partial y}{\partial s} = 2t\]Thus, the partial derivative with respect to \( s \) becomes:\[\frac{\partial w}{\partial s} = \frac{\partial f}{\partial x} \cdot 2s + \frac{\partial f}{\partial y} \cdot 2t\]
03

Compute Derivatives with respect to t

Similarly, compute the derivatives of the substitutions \( x = s^2 - t^2 \) and \( y = 2st \) with respect to \( t \):\[\frac{\partial x}{\partial t} = -2t \quad \text{and} \quad \frac{\partial y}{\partial t} = 2s\]Thus, the partial derivative with respect to \( t \) becomes:\[\frac{\partial w}{\partial t} = \frac{\partial f}{\partial x} \cdot (-2t) + \frac{\partial f}{\partial y} \cdot 2s\]
04

Compute the Gradient Norm

Combine the results from Steps 2 and 3 to determine the gradient norm:\[\|abla w\| = \sqrt{\left(\frac{\partial w}{\partial s}\right)^2 + \left(\frac{\partial w}{\partial t}\right)^2}\]Substitute the expressions found:\[\|abla w\| = \sqrt{\left(2s \frac{\partial f}{\partial x} + 2t \frac{\partial f}{\partial y}\right)^2 + \left(-2t \frac{\partial f}{\partial x} + 2s \frac{\partial f}{\partial y}\right)^2}\]
05

Simplify the Expression

To simplify, expand both squares and combine like terms:\[(2s \frac{\partial f}{\partial x} + 2t \frac{\partial f}{\partial y})^2 = 4s^2 \left(\frac{\partial f}{\partial x}\right)^2 + 8st \frac{\partial f}{\partial x} \frac{\partial f}{\partial y} + 4t^2 \left(\frac{\partial f}{\partial y}\right)^2\]\[(-2t \frac{\partial f}{\partial x} + 2s \frac{\partial f}{\partial y})^2 = 4t^2 \left(\frac{\partial f}{\partial x}\right)^2 - 8ts \frac{\partial f}{\partial x} \frac{\partial f}{\partial y} + 4s^2 \left(\frac{\partial f}{\partial y}\right)^2\]Combining these, the coefficients for the cross terms cancel out, yielding:\[\|abla w\| = \sqrt{(4s^2 + 4t^2) \left(\frac{\partial f}{\partial x}\right)^2 + (4t^2 + 4s^2) \left(\frac{\partial f}{\partial y}\right)^2}\]\[\|abla w\| = 2 \sqrt{s^2 + t^2} \cdot \sqrt{\left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2}\]

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Key Concepts

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

Partial Derivatives
Partial derivatives are essential when dealing with multivariable functions, as they provide the rate of change of one variable while keeping others constant. When given a function with multiple inputs, like \( w = f(x, y) = f(s^2 - t^2, 2st) \), the partial derivative \( \frac{\partial w}{\partial s} \) represents how the function changes as \( s \) varies, with \( t \) held constant. Similarly, \( \frac{\partial w}{\partial t} \) shows the change in \( w \) with respect to \( t \), keeping \( s \) constant.
  • To find \( \frac{\partial w}{\partial s} \), use the chain rule, which combines the derivatives of the inner functions \( x(s, t) \) and \( y(s, t) \) with the partial derivatives of \( f \) with respect to \( x \) and \( y \).
  • The same process applies to \( \frac{\partial w}{\partial t} \).
This allows the transformation of derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) into expressions involving \( s \) and \( t \). Such expressions are crucial in calculating the gradient and understanding directional behaviors in multivariable calculus.
Gradient Norm
The gradient of a function \( abla w \) is a vector comprising partial derivatives, representing the rate and direction of the steepest ascent of a function. For a function derived from substitutes, its gradient is calculated by the vector formed by \( (\frac{\partial w}{\partial s}, \frac{\partial w}{\partial t}) \). The gradient norm, \( \| abla w \| \), is the magnitude or length of this vector and is a key concept in grasping the function鈥檚 rate of maximum increase at any point.
  • This measure is calculated using the Euclidean norm, which implies taking the square root of the sum of squares of the component functions.
  • The simplified calculation in the context of substitutions shows a final result in terms of the roots of the original function鈥檚 partial derivatives combined mathematically via transformations of \( s \) and \( t \).
The gradient norm is particularly useful in optimization problems or when locating critical points in multivariable functions.
Substitution in Calculus
Substitution is a powerful technique in calculus that helps in transforming variables to simplify complex functions or solve equations more easily. When analyzing the function \( w = f(s^2 - t^2, 2st) \), the given substitution \( x = s^2 - t^2 \) and \( y = 2st \) allow us to redefine a traditionally multivariable function in terms of new variables \( s \) and \( t \). These substitutions are significant because:
  • They simplify differentiation by converting the problem into a more manageable form, especially when combined with the chain rule.
  • The expressions for the partial derivatives of \( w \) with respect to \( s \) and \( t \), through the derivatives of substituted variables, become more straightforward.
Through these processes, substitution not only eases computation but also provides insights into the relationship between the original and transformed functions, all while maintaining the fundamental qualities of the function. These transformed expressions are then essential for evaluating other aspects like the gradient norm, which in this case is analyzed in terms of \( s \) and \( t \).

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Most popular questions from this chapter

Let \(f: \mathbb{R}^{3} \rightarrow \mathbb{R}^{2}\) be given by \(f(x, y, z)=\left(x y^{2} z^{3}+2, x \cos (y z)\right)\). (a) Find \(D f(x, y, z)\). (b) Find \(D f\left(-\frac{\pi}{2}, 1, \frac{\pi}{2}\right)\).

Let \(f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}\) be given by \(f(x, y)=\left(e^{x+y}, e^{-x} \cos y, e^{-x} \sin y\right) .\) (a) Find \(D f(x, y)\). (b) Show that \(f\) is differentiable at every point of \(\mathbb{R}^{2}\).

Let \(f: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) be the spherical coordinate transformation from \(\rho \phi \theta\) -space to \(x y z\) -space: \(f(\rho, \phi, \theta)=(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi)\) (a) Find \(D f(\rho, \phi, \theta)\). (b) Show that \(f\) is differentiable at every point of \(\mathbb{R}^{3}\). (c) Show that \(\operatorname{det} D f(\rho, \phi, \theta)=\rho^{2} \sin \phi\). (This result will be used in the next chapter.)

Let \(f: U \rightarrow \mathbb{R}\) be a smooth real-valued function of three variables defined on an open set \(U\) in \(\mathbb{R}^{3}\). Assume that the condition \(f(x, y, z)=c\) defines \(z\) implicitly as a smooth function of \(x\) and \(y, z=z(x, y),\) on some open set of points \((x, y)\) in \(\mathbb{R}^{2}\). Show that, on this open set: $$ \frac{\partial z}{\partial x}=-\frac{\partial f / \partial x}{\partial f / \partial z} \quad \text { and } \quad \frac{\partial z}{\partial y}=-\frac{\partial f / \partial y}{\partial f / \partial z} $$

(a) Suppose that \(w=f(u, v)\) is a smooth real-valued function, where \(u=x / z\) and \(v=y / z\). Show that: $$ x \frac{\partial w}{\partial x}+y \frac{\partial w}{\partial y}+z \frac{\partial w}{\partial z}=0 $$ (b) Without calculating any partial derivatives, deduce that \(w=\frac{x^{2}+x y+y^{2}}{z^{2}}\) satisfies equation (6.10)
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