Problem 36
Question
Consider the gravitational potential $$ U(x, y)=\frac{-G m}{\sqrt{x^{2}+y^{2}}} $$ where \(G\) and \(m\) are constants. Show that \(U\) increases or decreases most rapidly along a line through the origin.
Step-by-Step Solution
Verified Answer
The potential \( U \) changes most rapidly along lines through the origin.
1Step 1: Understanding the Gradient
To determine the direction in which the gravitational potential \( U(x, y) \) changes most rapidly, we need to find the gradient of \( U \). The gradient is a vector pointing in the direction of greatest increase of a function.
2Step 2: Finding the Partial Derivatives
We compute the partial derivatives of \( U \) with respect to \( x \) and \( y \). This involves using the chain rule:\[\frac{\partial U}{\partial x} = \frac{\partial}{\partial x}\left(\frac{-Gm}{\sqrt{x^2 + y^2}}\right) = \frac{Gmx}{(x^2+y^2)^{3/2}}\]\[\frac{\partial U}{\partial y} = \frac{\partial}{\partial y}\left(\frac{-Gm}{\sqrt{x^2 + y^2}}\right) = \frac{Gmy}{(x^2+y^2)^{3/2}}\]
3Step 3: Forming the Gradient Vector
Now, form the gradient vector by combining the partial derivatives:\[abla U = \left( \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y} \right) = \left( \frac{Gmx}{(x^2+y^2)^{3/2}}, \frac{Gmy}{(x^2+y^2)^{3/2}} \right)\]
4Step 4: Analyzing the Gradient Direction
The gradient vector \( abla U \) points in the direction \( (x, y) \), which implies that \( U \) increases or decreases most rapidly along a line through the origin that is defined by the coordinates \( (x, y) \) themselves.
5Step 5: Conclusion
Since the gradient vector is always aligned with the vector \( (x,y) \), we conclude that the direction in which \( U \) changes most rapidly includes lines passing through the origin. Therefore, \( U \) increases or decreases most rapidly along any line through the origin.
Key Concepts
Gradient of a Potential FunctionUnderstanding Partial DerivativesDirection of Greatest Increase
Gradient of a Potential Function
The gradient is a crucial mathematical concept when talking about potential functions, like gravitational potential. It tells us the direction of steepest ascent or descent at any given point on a surface. In essence, for a function like gravitational potential \( U(x, y) \), the gradient will point towards the direction where \( U \) changes the most quickly.
For the gravitational potential given by \( U(x, y) = \frac{-Gm}{\sqrt{x^2 + y^2}} \), understanding its gradient means determining how the potential changes in different directions. The gradient vector \( abla U \) is made of the partial derivatives of \( U \) with respect to \( x \) and \( y \).
The concept of the gradient is key because it helps us visualize the direction of change and understand physical processes in better depth.
For the gravitational potential given by \( U(x, y) = \frac{-Gm}{\sqrt{x^2 + y^2}} \), understanding its gradient means determining how the potential changes in different directions. The gradient vector \( abla U \) is made of the partial derivatives of \( U \) with respect to \( x \) and \( y \).
The concept of the gradient is key because it helps us visualize the direction of change and understand physical processes in better depth.
Understanding Partial Derivatives
Partial derivatives are used to find how a function changes as one of its variables changes, while keeping other variables constant. They're like slicing through a multi-dimensional surface to examine one direction at a time.
For the gravitational potential \( U(x, y) = \frac{-Gm}{\sqrt{x^2 + y^2}} \), we calculate two key partial derivatives:
For the gravitational potential \( U(x, y) = \frac{-Gm}{\sqrt{x^2 + y^2}} \), we calculate two key partial derivatives:
- With respect to \( x \): \( \frac{\partial U}{\partial x} = \frac{Gmx}{(x^2 + y^2)^{3/2}} \)
- With respect to \( y \): \( \frac{\partial U}{\partial y} = \frac{Gmy}{(x^2 + y^2)^{3/2}} \)
Direction of Greatest Increase
The direction of greatest increase for a function is indicated by its gradient. Imagine you're standing on a hill. The direction in which you need to walk to climb up the steepest part is the direction of the gradient.
When we have the gravitational potential \( U(x, y) \), the gradient vector \( abla U \) will point in the direction of the steepest increase of \( U \). In this problem, the gradient vector is \( \left( \frac{Gmx}{(x^2+y^2)^{3/2}}, \frac{Gmy}{(x^2+y^2)^{3/2}} \right) \), which means the direction is aligned with the vector \( (x, y) \) itself.
This alignment indicates that \( U \) changes most quickly along lines passing through the origin. Hence, to encounter the most rapid change in gravitational potential, one follows the direction given by the gradient. Understanding this concept is pivotal in fields ranging from physics to machine learning, where one often seeks the path of steepest ascent or descent.
When we have the gravitational potential \( U(x, y) \), the gradient vector \( abla U \) will point in the direction of the steepest increase of \( U \). In this problem, the gradient vector is \( \left( \frac{Gmx}{(x^2+y^2)^{3/2}}, \frac{Gmy}{(x^2+y^2)^{3/2}} \right) \), which means the direction is aligned with the vector \( (x, y) \) itself.
This alignment indicates that \( U \) changes most quickly along lines passing through the origin. Hence, to encounter the most rapid change in gravitational potential, one follows the direction given by the gradient. Understanding this concept is pivotal in fields ranging from physics to machine learning, where one often seeks the path of steepest ascent or descent.
Other exercises in this chapter
Problem 36
In Problems, evaluate the given integral. $$ \int \frac{1}{1+t^{2}}\left(\mathbf{i}+t \mathbf{j}+t^{2} \mathbf{k}\right) d t $$
View solution Problem 36
Verify that the given function satisfies the wave equation: $$a^{2} \frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial^{2} u}{\partial t^{2}}$$ $$ u=\cos (x+a
View solution Problem 36
Convert the point given in cylindrical cocudinates to rectangular cocrdinates. $$ \left(2, \frac{5 \pi}{6},-3\right) $$
View solution Problem 36
Find parametric equations for the normal line at the indicated point. In Problems 35 and 36, find symmetric equations for the normal line. $$ x^{2}+y^{2}-z^{2}=
View solution