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Calculate the gradient \(\nabla f\) of the following functions, \(f(x, y, z):(\mathbf{a}) f=\ln (r),\) (b) \(f=r^{n}\), (c) \(f=g(r),\) where \(r=\sqrt{x^{2}+y^{2}+z^{2}}\) and \(g(r)\) is some unspecified function of \(r .\) [Hint: Use the chain rule.]

Short Answer

Expert verified
(a) \(\nabla f = \left(\frac{x}{r^2}, \frac{y}{r^2}, \frac{z}{r^2}\right)\), (b) \(\nabla f = \left(nx r^{n-2}, ny r^{n-2}, nz r^{n-2}\right)\), (c) \(\nabla f = g'(r)\left(\frac{x}{r}, \frac{y}{r}, \frac{z}{r}\right)\).

Step by step solution

01

Understand the Gradient Concept

The gradient of a function is a vector of partial derivatives. For a function \( f(x, y, z) \), the gradient is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \). This vector points in the direction of the greatest rate of increase of the function.
02

Define \( r \) and its Partial Derivatives

Given \( r = \sqrt{x^2 + y^2 + z^2} \), recognize \( r \) as the distance from the origin to a point \( (x, y, z) \). Calculate partial derivatives: \( \frac{\partial r}{\partial x} = \frac{x}{r} \), \( \frac{\partial r}{\partial y} = \frac{y}{r} \), \( \frac{\partial r}{\partial z} = \frac{z}{r} \) using the chain rule.
03

Calculate \( \nabla f \) for \( f = \ln(r) \)

Using the chain rule, \( \frac{\partial f}{\partial x} = \frac{1}{r} \cdot \frac{\partial r}{\partial x} = \frac{x}{r^2} \), \( \frac{\partial f}{\partial y} = \frac{y}{r^2} \), \( \frac{\partial f}{\partial z} = \frac{z}{r^2} \). Thus, \( abla f = \left( \frac{x}{r^2}, \frac{y}{r^2}, \frac{z}{r^2} \right) \).
04

Calculate \( \nabla f \) for \( f = r^n \)

Use the chain rule: \( \frac{\partial f}{\partial x} = n r^{n-1} \cdot \frac{\partial r}{\partial x} = n x r^{n-2} \), \( \frac{\partial f}{\partial y} = n y r^{n-2} \), \( \frac{\partial f}{\partial z} = n z r^{n-2} \). Therefore, \( abla f = \left( n x r^{n-2}, n y r^{n-2}, n z r^{n-2} \right) \).
05

Calculate \( \nabla f \) for \( f = g(r) \)

Using the chain rule: \( \frac{\partial f}{\partial x} = g'(r) \cdot \frac{\partial r}{\partial x} = g'(r) \frac{x}{r} \), \( \frac{\partial f}{\partial y} = g'(r) \frac{y}{r} \), \( \frac{\partial f}{\partial z} = g'(r) \frac{z}{r} \). So, \( abla f = \left( g'(r) \frac{x}{r}, g'(r) \frac{y}{r}, g'(r) \frac{z}{r} \right) \).

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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 helps us find the derivative of composite functions. In simple terms, it's how we deal with functions nested within other functions, kind of like peeling an onion. When we deal with multivariable functions, such as those involving three or more variables, the chain rule provides a pathway to differentiate these functions by connecting their various layers.
For instance, consider a function that depends on another function, like how our exercise involves function values depending on the variable \( r \). Here, \( r = \sqrt{x^2 + y^2 + z^2} \), is an intermediary function that changes the outcome of the final expression \( f(r) \).
Using the chain rule, we determine how our function \( f \) changes as its inner function \( r \) changes, allowing us to find partial derivatives linked with \( r \). These derivatives connect through the chain rule providing:
  • \( \frac{\partial f}{\partial x} = \frac{df}{dr} \cdot \frac{\partial r}{\partial x} \)
  • \( \frac{\partial f}{\partial y} = \frac{df}{dr} \cdot \frac{\partial r}{\partial y} \)
  • \( \frac{\partial f}{\partial z} = \frac{df}{dr} \cdot \frac{\partial r}{\partial z} \)
Mastering the chain rule equips you to systematically tackle complex derivative problems in multivariable calculus.
Gradient Vector
The gradient vector, often denoted as \( abla f \), is an essential tool in vector calculus. It is a vector composed of all the partial derivatives of a function. In layman's terms, it tells you the direction in space where the function increases the most. This concept is crucial in fields like optimization and physics.
For example, with a function \( f(x, y, z) \) dependent on three variables, the gradient \( abla f \) forms the vector:
  • \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \)
This vector shows us how each variable affects the function's change. It's like having a compass pointing towards the steepest ascent of a hill.
When calculating the gradient in our exercise, we make use of \( r = \sqrt{x^2 + y^2 + z^2} \) to find the individual changes along the \( x, y, \) and \( z \) axes. This solution aids in determining how each coordinate direction affects the function's value.
Partial Derivatives
Partial derivatives provide a way to understand how a function changes when altering only one variable while keeping others constant. They are to multivariable calculus what regular derivatives are to single-variable calculus.
To derive these partial derivatives, we treat the function as dependent solely on one variable at a time. For example, from the function \( r = \sqrt{x^2 + y^2 + z^2} \), calculating the partial derivative with respect to \( x \) gives \( \frac{\partial r}{\partial x} = \frac{x}{r} \). The same goes for finding the derivatives with respect to \( y \) and \( z \).
  • \( \frac{\partial r}{\partial y} = \frac{y}{r} \)
  • \( \frac{\partial r}{\partial z} = \frac{z}{r} \)
By conducting this operation, you crack open the behavior of our function from various angles, allowing us to compute things like the gradient effectively.
Partial derivatives are not just abstract notions; they represent real changes in systems and are vital in understanding gradients and applying the chain rule in complex scenarios.

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Most popular questions from this chapter

If a particle's potential energy is \(U(\mathbf{r})=k\left(x^{2}+y^{2}+z^{2}\right),\) where \(k\) is a constant, what is the force on the particle?

Find the partial derivatives with respect to \(x, y,\) and \(z\) of the following functions: (a) \(f(x, y, z)=\) \(a x^{2}+b x y+c y^{2},(\mathbf{b}) g(x, y, z)=\sin \left(a x y z^{2}\right),(\mathbf{c}) h(x, y, z)=a e^{x y / z^{2}},\) where \(a, b,\) and \(c\) are constants. Remember that to evaluate \(\partial f / \partial x\) you differentiate with respect to \(x\) treating \(y\) and \(z\) as constants.

An interesting one-dimensional system is the simple pendulum, consisting of a point mass \(m\), fixed to the end of a massless rod (length \(l\) ), whose other end is pivoted from the ceiling to let it swing freely in a vertical plane, as shown in Figure \(4.26 .\) The pendulum's position can be specified by its angle \(\phi\) from the equilibrium position. (It could equally be specified by its distance \(s\) from equilibrium \(-\) indeed \(s=l \phi-\) but the angle is a little more convenient.) (a) Prove that the pendulum's potential energy (measured from the equilibrium level) is \(U(\phi)=m g l(1-\cos \phi)\). Write down the total energy \(E\) as a function of \(\phi\) and \(\dot{\phi}\). (b) Show that by differentiating your expression for \(E\) with respect to \(t\) you can get the equation of motion for \(\phi\) and that the equation of motion is just the familiar \(\Gamma=I \alpha\) (where \(\Gamma\) is the torque, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration \(\ddot{\phi}\) ). (c) Assuming that the angle \(\phi\) remains small throughout the motion, solve for \(\phi(t)\) and show that the motion is periodic with period \(\tau_{\mathrm{o}}=2 \pi \sqrt{l / g}\).

Which of the following forces is conservative? (a) \(\mathbf{F}=k(x, 2 y, 3 z)\) where \(k\) is a constant. (b) \(\mathbf{F}=k(y, x, 0) .\) (c) \(\mathbf{F}=k(-y, x, 0)\). For those which are conservative, find the corresponding potential energy \(U,\) and verify by direct differentiation that \(\mathbf{F}=-\nabla U\).

Consider a small frictionless puck perched at the top of a fixed sphere of radius \(R\). If the puck is given a tiny nudge so that it begins to slide down, through what vertical height will it descend before it leaves the surface of the sphere? [Hint: Use conservation of energy to find the puck's speed as a function of its height, then use Newton's second law to find the normal force of the sphere on the puck. At what value of this normal force does the puck leave the sphere?]

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