/*! 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 3 The spherical substitutions \(x=... [FREE SOLUTION] | 91Ó°ÊÓ

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The spherical substitutions \(x=\rho \sin \phi \cos \theta, y=\rho \sin \phi \sin \theta,\) and \(z=\rho \cos \phi\) convert a smooth real-valued function \(f(x, y, z)\) into a function of \(\rho, \phi,\) and \(\theta:\) $$ w=f(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) $$ (a) Find formulas for \(\frac{\partial w}{\partial \rho}, \frac{\partial w}{\partial \phi},\) and \(\frac{\partial w}{\partial \theta}\) in terms of \(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},\) and \(\frac{\partial f}{\partial z} .\) (b) If \(f(x, y, z)=x^{2}+y^{2}+z^{2},\) use your answer to part (a) to find \(\frac{\partial w}{\partial \phi},\) and verify that it is the same as the result obtained if you first write \(w\) in terms of \(\rho, \phi,\) and \(\theta\) directly, say by substituting for \(x, y,\) and \(z,\) and then differentiate that expression with respect to \(\phi .\)

Short Answer

Expert verified
\(\frac{\partial w}{\partial \rho} = \sin \phi \cos \theta \cdot \frac{\partial f}{\partial x} + \sin \phi \sin \theta \cdot \frac{\partial f}{\partial y} + \cos \phi \cdot \frac{\partial f}{\partial z}\); \(\frac{\partial w}{\partial \phi} = 0\); \(\frac{\partial w}{\partial \theta}\) simplifies as described.

Step by step solution

01

Find partial derivative \(\frac{\partial w}{\partial \rho}\)

To find \(\frac{\partial w}{\partial \rho}\), apply the chain rule: \[ \frac{\partial w}{\partial \rho} = \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial \rho} + \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial \rho} + \frac{\partial f}{\partial z} \cdot \frac{\partial z}{\partial \rho} \] \(x = \rho \sin \phi \cos \theta\), \(y = \rho \sin \phi \sin \theta\), \(z = \rho \cos \phi\). So: \[ \frac{\partial x}{\partial \rho} = \sin \phi \cos \theta, \quad \frac{\partial y}{\partial \rho} = \sin \phi \sin \theta, \quad \frac{\partial z}{\partial \rho} = \cos \phi \] Substitute these into the chain rule equation: \[ \frac{\partial w}{\partial \rho} = \sin \phi \cos \theta \cdot \frac{\partial f}{\partial x} + \sin \phi \sin \theta \cdot \frac{\partial f}{\partial y} + \cos \phi \cdot \frac{\partial f}{\partial z} \]
02

Find partial derivative \(\frac{\partial w}{\partial \phi}\)

Apply the chain rule for \(\frac{\partial w}{\partial \phi}\): \[ \frac{\partial w}{\partial \phi} = \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial \phi} + \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial \phi} + \frac{\partial f}{\partial z} \cdot \frac{\partial z}{\partial \phi} \] Using \(x = \rho \sin \phi \cos \theta\), \(y = \rho \sin \phi \sin \theta\), \(z = \rho \cos \phi\), we find: \[ \frac{\partial x}{\partial \phi} = \rho \cos \phi \cos \theta, \quad \frac{\partial y}{\partial \phi} = \rho \cos \phi \sin \theta, \quad \frac{\partial z}{\partial \phi} = -\rho \sin \phi \] Substitute these into the chain rule equation: \[ \frac{\partial w}{\partial \phi} = (\rho \cos \phi \cos \theta) \cdot \frac{\partial f}{\partial x} + (\rho \cos \phi \sin \theta) \cdot \frac{\partial f}{\partial y} - (\rho \sin \phi) \cdot \frac{\partial f}{\partial z} \]
03

Find partial derivative \(\frac{\partial w}{\partial \theta}\)

Use the chain rule to find \(\frac{\partial w}{\partial \theta}\): \[ \frac{\partial w}{\partial \theta} = \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial \theta} + \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial \theta} \] Here, we find: \[ \frac{\partial x}{\partial \theta} = -\rho \sin \phi \sin \theta, \quad \frac{\partial y}{\partial \theta} = \rho \sin \phi \cos \theta \] Substitute these into the chain rule equation: \[ \frac{\partial w}{\partial \theta} = \left(-\rho \sin \phi \sin \theta\right) \cdot \frac{\partial f}{\partial x} + \left(\rho \sin \phi \cos \theta\right) \cdot \frac{\partial f}{\partial y} \]
04

Substitute and verify function \(f(x, y, z)=x^2 + y^2 + z^2\)

For the function \(f(x, y, z) = x^2 + y^2 + z^2\), calculate the partial derivatives: \(\frac{\partial f}{\partial x} = 2x\), \(\frac{\partial f}{\partial y} = 2y\), \(\frac{\partial f}{\partial z} = 2z\). Since \(x^2 + y^2 + z^2 = \rho^2\), substitute into \(\frac{\partial w}{\partial \phi}\) from Step 2: \[ \frac{\partial w}{\partial \phi} = 2x \cdot (\rho \cos \phi \cos \theta) + 2y \cdot (\rho \cos \phi \sin \theta) - 2z \cdot (\rho \sin \phi) \] Substitute \(x = \rho \sin \phi \cos \theta\), \(y = \rho \sin \phi \sin \theta\), \(z = \rho \cos \phi\), gives: \[ 2(\rho\sin\phi\cos\theta) \cdot (\rho \cos \phi \cos \theta) + 2(\rho\sin\phi\sin\theta) \cdot (\rho \cos \phi \sin \theta) - 2(\rho \cos \phi) \cdot (\rho \sin \phi) \] Simplifying shows it equals zero, verifying results.
05

Direct computation for \(\rho^2\)

Since \(w = \rho^2\) when substituting originally, differentiate directly with respect to \(\phi\): \[ \frac{\partial w}{\partial \phi} = \frac{\partial (\rho^2)}{\partial \phi} = 0 \] This confirms \(\frac{\partial w}{\partial \phi} = 0\) by direct computation alignment.

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Key Concepts

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

Partial Derivatives
Partial derivatives represent the rate of change of a function with respect to one variable, while keeping other variables constant. In the context of spherical coordinates, they become especially useful. When you have a function like \( w = f(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) \), you can determine how \( w \) changes with each spherical coordinate \( \rho, \phi, \) and \( \theta \) independently.

To find these partial derivatives, we use the chain rule, which allows us to differentiate \( f \) concerning its variables \( x, y, z \). For instance, to get \( \frac{\partial w}{\partial \rho} \), you calculate how changes in \( \rho \) affect \( x, y, \) and \( z \), and then use those results to see how \( w \) changes based on \( f(x, y, z) \). This process helps us understand the behavior of the function when moving along the \( \rho \) direction, keeping other angles constant.
Chain Rule
The chain rule is a fundamental concept in calculus used to differentiate composite functions. In spherical coordinates, transformations from Cartesian coordinates \((x, y, z)\) require employing the chain rule to find how a function changes with respect to new variables \((\rho, \phi, \theta)\).

In practice, if \( w \) is a function of \( f \) which itself is a function of other variables, the chain rule helps break down derivatives of \( w \) into sums of products of partial derivatives. For example, to get \( \frac{\partial w}{\partial \rho} \), we differentiate \( w \) in terms of \( x, y, \) and \( z \), then multiply each by the derivative of \( x, y, \) and \( z \) with respect to \( \rho \). The chain rule simplifies complex differentiation tasks into smaller, more manageable parts, enabling us to understand and compute derivative changes through compound transformations.
Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions of several variables. It is indispensable in scenarios involving complex coordinate systems, such as spherical coordinates, which naturally appear in physical problems.

In exercises like transforming Cartesian coordinates to spherical coordinates, multivariable calculus allows us to handle multidimensional rates of change through partial derivatives. With functions such as \( f(x,y,z) \), understanding how to manage transformations using multivariable techniques like the chain rule, reveals insights into how multidimensional changes affect function behavior. This facilitates solving practical problems in fields like physics and engineering, where systems commonly depend on multiple interacting variables.
Function Transformation
Function transformation is about changing the form of a function to express it in terms of different variables. Converting Cartesian coordinates \((x,y,z)\) into spherical coordinates \((\rho, \phi, \theta)\) is a classical example of such transformations.

This process often involves a change of basis or parametric substitution, allowing us to re-express \( f(x,y,z) \) in a more insightful or computationally convenient form like \( w=f(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) \). It highlights how functions can be manipulated to suit the needs of a problem, providing a new perspective or simplifying a calculation. Mastery in function transformation unlocks the potential to tackle a wide range of applications across mathematics and science, by leveraging the elegance and symmetry of alternative coordinate systems.

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Most popular questions from this chapter

(a) Let \(f: \mathbb{R}^{3} \rightarrow \mathbb{R}^{2}\) be a smooth function. By definition, the level set \(S\) of \(f\) corresponding to the value \(\mathbf{c}=\left(c_{1}, c_{2}\right)\) is the set of all \((x, y, z)\) such that \(f(x, y, z)=\mathbf{c}\) or, equivalently, where \(f_{1}(x, y, z)=c_{1}\) and \(f_{2}(x, y, z)=c_{2}\). Typically, the last two equations define a curve in \(\mathbb{R}^{3},\) the curve of intersection of the surfaces defined by \(f_{1}=c_{1}\) and \(f_{2}=c_{2}\). Assume that the condition \(f(x, y, z)=\mathbf{c}\) defines \(x\) and \(y\) implicitly as smooth functions of \(z\) on some interval of \(z\) values in \(\mathbb{R},\) that is, \(x=x(z)\) and \(y=y(z)\) are functions that satisfy \(f(x(z), y(z), z)=\mathbf{c} .\) Then the corresponding portion of the level set is parametrized by \(\alpha(z)=(x(z), y(z), z)\) Show that \(x^{\prime}(z)\) and \(y^{\prime}(z)\) satisfy the matrix equation: $$ \left[\begin{array}{ll} \frac{\partial f_{1}}{\partial x} & \frac{\partial f_{1}}{\partial y} \\ \frac{\partial f_{2}}{\partial x} & \frac{\partial f_{2}}{\partial y} \end{array}\right]\left[\begin{array}{l} x^{\prime}(z) \\ y^{\prime}(z) \end{array}\right]=-\left[\begin{array}{l} \frac{\partial f_{1}}{\partial z} \\ \frac{\partial f_{2}}{\partial z} \end{array}\right], $$ where the partial derivatives are evaluated at the point \((x(z), y(z), z) .\) (Hint: Let \(w=\) \(\left.\left(w_{1}, w_{2}\right)=f(x(z), y(z), z) .\right)\) (b) Let \(f(x, y, z)=(x-\cos z, y-\sin z),\) and let \(C\) be the level set of \(f\) corresponding to \(\mathbf{c}=(0,0) .\) Find the functions \(x=x(z)\) and \(y=y(z)\) and the parametrization \(\alpha\) of \(C\) described in part (a), and describe \(C\) geometrically. Then, use part (a) to help find \(D \alpha(z),\) and verify that your answer makes sense.

Let \(f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}\) and \(g: \mathbb{R}^{3} \rightarrow \mathbb{R}\) be the functions: $$ f(s, t)=\left(2 s-t^{2}, s t-1,2 s^{2}+s t-t^{2}\right) \quad \text { and } \quad g(x, y, z)=(x+1)^{2} e^{y z} $$ If \(h(s, t)=g(f(s, t)),\) find \(D h(1,2)\).

Let \(f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}\) and \(g: \mathbb{R}^{3} \rightarrow \mathbb{R}^{2}\) be given by: $$ f(s, t)=\left(s-t+2, e^{2 s+3 t}, s \cos t\right) \quad \text { and } \quad g(x, y, z)=\left(x y z, x^{2}+z^{3}\right) $$ (a) Find \(D f(s, t)\) and \(D g(x, y, z)\). (b) Find \(D(g \circ f)(0,0)\). (c) Find \(D(f \circ g)(0,0,0)\).

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 \(f(x, y)\) be a smooth real-valued function defined on an open set in \(\mathbb{R}^{2} .\) The polar coordinate substitutions \(x=r \cos \theta\) and \(y=r \sin \theta\) convert \(f\) into a function of \(r\) and \(\theta\), \(w=f(r \cos \theta, r \sin \theta)\) (a) Show that: $$ \begin{array}{c} \frac{\partial^{2} w}{\partial \theta^{2}}=-r \cos \theta \frac{\partial f}{\partial x}-r \sin \theta \frac{\partial f}{\partial y}+r^{2} \sin ^{2} \theta \frac{\partial^{2} f}{\partial x^{2}}-2 r^{2} \sin \theta \cos \theta \frac{\partial^{2} f}{\partial x \partial y} \\ +r^{2} \cos ^{2} \theta \frac{\partial^{2} f}{\partial y^{2}} \end{array} $$ (b) Find an analogous expression for the mixed partial derivative \(\frac{\partial^{2} w}{\partial r \partial \theta}\). (c) Find an analogous expression for \(\frac{\partial^{2} w}{\partial r^{2}}\). (d) If \(f(x, y)=x^{2}+y^{2}+x y,\) use your answer to part (b) to find \(\frac{\partial^{2} w}{\partial r \partial \theta}\). Then, verify that it is the same as the result obtained by writing \(w\) in terms of \(r\) and \(\theta\) directly by substituting for \(x\) and \(y\) and then differentiating that expression with respect to \(r\) and \(\theta\).

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