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In Exercises \(7-12,\) find a potential function \(f\) for the field \(\mathbf{F}\) $$ \mathbf{F}=2 x \mathbf{i}+3 y \mathbf{j}+4 z \mathbf{k} $$

Short Answer

Expert verified
The potential function is \( f(x, y, z) = x^2 + \frac{3}{2}y^2 + 2z^2 + C \).

Step by step solution

01

Recognize the Vector Field

The vector field given is \( \mathbf{F} = 2x \mathbf{i} + 3y \mathbf{j} + 4z \mathbf{k} \). This vector field is in three-dimensional space.
02

Confirm the Field is Conservative

A vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \) is conservative if \( abla \times \mathbf{F} = \mathbf{0} \). Perform the curl operation: \[ abla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ 2x & 3y & 4z \end{vmatrix} \] Calculate each component and verify:- The \( \mathbf{i} \) component: \( \frac{\partial}{\partial y}(4z) - \frac{\partial}{\partial z}(3y) = 0 - 0 = 0 \)- The \( \mathbf{j} \) component: \( \frac{\partial}{\partial z}(2x) - \frac{\partial}{\partial x}(4z) = 0 - 0 = 0 \)- The \( \mathbf{k} \) component: \( \frac{\partial}{\partial x}(3y) - \frac{\partial}{\partial y}(2x) = 0 - 0 = 0 \)Therefore, \( abla \times \mathbf{F} = \mathbf{0} \), confirming the field is conservative.
03

Find Potential Function Components

Integrate each component of \( \mathbf{F} \) separately with respect to its variable:1. Integrate \( 2x \) with respect to \( x \): \[ f(x, y, z) = \int 2x \, dx = x^2 + C(y, z) \]2. Integrate \( 3y \) with respect to \( y \): \[ f(x, y, z) = \int 3y \, dy = y^3/2 + C(x, z) \]3. Integrate \( 4z \) with respect to \( z \): \[ f(x, y, z) = \int 4z \, dz = 2z^2 + C(x, y) \]
04

Combine Integrations for Potential Function

Combine the results of the integrations, arranging terms such that each variable's impact is represented only once. The general potential function is:\[ f(x, y, z) = x^2 + \frac{3}{2}y^2 + 2z^2 + C \]where \(C\) is an arbitrary constant (since potential functions are determined up to a constant).

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Key Concepts

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

Vector Field
A vector field is a function that assigns a vector to every point in space. In physics and mathematics, it helps us describe various quantities that have both magnitude and direction. For example, wind speed, which can vary in different locations, is best represented by a vector field. In our exercise, the given vector field is \( \mathbf{F} = 2x \mathbf{i} + 3y \mathbf{j} + 4z \mathbf{k} \). Here, \( x \), \( y \), and \( z \) are the spatial variables, and \( \mathbf{i} \), \( \mathbf{j} \), \( \mathbf{k} \) are the unit vectors in the direction of the x, y, and z axes, respectively. This vector field describes how vectors change at every point in three-dimensional space, and is crucial for understanding fields like gravitational and electric fields.
Conservative Field
A conservative field is a vector field where the line integral between any two points is independent of the path taken. This means that there is no energy loss due to a path-dependent process. For a field to be conservative, its curl must be zero: \( abla \times \mathbf{F} = \mathbf{0} \). In our step-by-step solution, we verified that \( \mathbf{F} \) is a conservative field by calculating its curl using determinants and ensuring each component simplifies to zero.

Being a conservative field allows us to define a potential function for it. This is significant in physics because potential functions enable us to calculate potential energy, which we will touch on next. Knowing a field is conservative is key in simplifying complex calculations in physics and engineering, ensuring energy conservation.
Potential Energy
Potential energy is the stored energy that can be converted into other forms of energy. It's essential in fields like mechanics and electromagnetism. For example, potential energy in gravitational fields explains why an object at a height has energy due to Earth’s gravity. Similarly, in electric fields, potential energy is stored based on the position and charge of particles.

Our task was to find a potential function \( f \), which is linked to potential energy. For a conservative vector field \( \mathbf{F} \), the potential function \( f \) satisfies \( \mathbf{F} = -abla f \). In simpler terms, this means the field can be expressed as the gradient of this function, indicating how potential energy changes across space. Through integration, we obtained the potential function in our solution, emphasizing how such functions help calculate energy changes effectively across various fields.
Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions in more than one variable; it’s vital for studying vector fields like \( \mathbf{F} \). With calculus involving several variables, we use tools like gradients, curls, and divergences to analyze behavior in fields extended over multidimensional spaces.

This level of calculus involves finding partial derivatives, which allows us to understand how a function changes with respect to each individual variable while holding others constant. As seen in our solution, integration and differentiation with respect to every variable separately helped us deduce the form of the potential function. Multivariable calculus is indispensable in engineering, physical sciences, and economics for tasks such as optimizing functions and studying rate changes in systems modeled with several variables.

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Most popular questions from this chapter

Find the outward flux of the field \(\mathbf{F}=2 x y \mathbf{i}+2 y z \mathbf{j}+2 x z \mathbf{k}\) across the surface of the cube cut from the first octant by the planes \(x=a, y=a, z=a\)

In Exercises \(13-18\) , use the surface integral in Stokes' Theorem to calculate the flux of the curl of the field \(\mathbf{F}\) across the surface \(S\) in the direction of the outward unit normal \(\mathbf{n} .\) $$ \begin{array}{l}{\mathbf{F}=(y-z) \mathbf{i}+(z-x) \mathbf{j}+(x+z) \mathbf{k}} \\ {S : \quad \mathbf{r}(r, \theta)=(r \cos \theta) \mathbf{i}+(r \sin \theta) \mathbf{j}+\left(9-r^{2}\right) \mathbf{k}} \\ {0 \leq r \leq 3, \quad 0 \leq \theta \leq 2 \pi}\end{array} $$

In Exercises \(37-40, \mathbf{F}\) is the velocity field of a fluid flowing through a region in space. Find the flow along the given curve in the direction of increasing \(t .\) $$ \begin{array}{l}{\mathbf{F}=x^{2} \mathbf{i}+y z \mathbf{j}+y^{2} \mathbf{k}} \\\ {r(t)=3 t \mathbf{j}+4 t \mathbf{k}, \quad 0 \leq t \leq 1}\end{array} $$

Integral along different paths Evaluate \(\int_{C} 2 x \cos y d x-x^{2}\) \(\sin y d y\) along the following paths \(C\) in the \(x y\) -plane. a. The parabola \(y=(x-1)^{2}\) from \((1,0)\) to \((0,1)\) b. The line segment from \((-1, \pi)\) to \((1,0)\) c. The \(x\) -axis from \((-1,0)\) to \((1,0)\) d. The astroid \(\mathbf{r}(t)=\left(\cos ^{3} t\right) \mathbf{i}+\left(\sin ^{3} t\right) \mathbf{j}, 0 \leq t \leq 2 \pi\) counterclockwise from \((1,0)\) back to \((1,0)\)

The tangent plane at a point \(P_{0}\left(f\left(u_{0}, v_{0}\right), g\left(u_{0}, v_{0}\right), h\left(u_{0}, v_{0}\right)\right)\) on a parametrized surface \(\mathbf{r}(u, v)=f(u, v) \mathbf{i}+g(u, v) \mathbf{j}+h(u, v) \mathbf{k}\) is the plane through \(P_{0}\) normal to the vector \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right) \times \mathbf{r}_{v}\left(u_{0}, v_{0}\right),\) the cross product of the tangent vectors \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right)\) and \(\mathbf{r}_{v}\left(u_{0}, v_{0}\right)\) at \(P_{0} .\) In Exercises \(49-52,\) find an equation for the plane tangent to the surface at \(P_{0} .\) Then find a Cartesian equation for the surface and sketch the surface and tangent plane together. Hemisphere The hemisphere surface \(\mathbf{r}(\phi, \theta)=(4 \sin \phi \cos \theta) \mathbf{i}\) \(+(4 \sin \phi \sin \theta) \mathbf{j}+(4 \cos \phi) \mathbf{k}, 0 \leq \phi \leq \pi / 2,0 \leq \theta \leq 2 \pi,\) at the point \(P_{0}(\sqrt{2}, \sqrt{2}, 2 \sqrt{3}) \quad\) corresponding to \((\phi, \theta)=\) \((\pi / 6, \pi / 4)\)

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