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Find, if possible, a function \(f\) such that $$\nabla f=\left\langle\sin (y z), x z \cos (y z)+2 y, x y \cos (y z)+\frac{5}{z}\right\rangle .$$ If not possible, explain why.

Short Answer

Expert verified
The potential function that satisfies the given gradient is \(f(x, y, z) = x\sin(yz) + y^2 + 5\ln(z)\).

Step by step solution

01

Integrate the first component w.r.t. x

Integrating the first component with respect to \(x\), we have: $$\int \sin(yz) \, dx = \sin(yz) \cdot x + g(y, z)$$ Here, \(g(y, z)\) is an arbitrary function of \(y\) and \(z\).
02

Integrate the second component w.r.t. y

Integrating the second component with respect to \(y\), we have: $$\int (xz\cos (yz) + 2y) \, dy = xz\sin(yz) + y^2 + h(x, z)$$ Here, \(h(x, z)\) is an arbitrary function of \(x\) and \(z\).
03

Integrate the third component w.r.t. z

Integrating the third component with respect to \(z\), we have: $$\int (xy\cos (yz) + \frac{5}{z}) \, dz = xy\sin(yz) - 5 \ln(z) + k(x, y)$$ Here, \(k(x, y)\) is an arbitrary function of \(x\) and \(y\).
04

Compare the integrated functions

From Steps 1, 2, and 3, we got that: $$f(x,y,z) = \sin(yz) \cdot x + g(y, z) = xz\sin(yz) + y^2 + h(x, z) = xy\sin(yz) - 5 \ln(z) + k(x, y)$$ Now, we have to find \(g(y,z)\), \(h(x,z)\), and \(k(x,y)\).
05

Solve for the arbitrary functions

From the equations above, 1. \(xz\sin(yz) = \sin(yz)x \implies g(y,z) = h(x,z) = 0\) 2. \(y^2 = -5 \ln(z) + k(x,y) \implies k(x, y) = y^2 + 5\ln(z)\) Therefore, the potential function satisfies: $$f(x, y, z) = x\sin(yz) + y^2 + 5\ln(z)$$ As we have found a function that satisfies the gradient, our final answer is: $$f(x, y, z) = x\sin(yz) + y^2 + 5\ln(z)$$

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Key Concepts

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

Integration in Multivariable Calculus
Understanding integration in multivariable calculus is essential. It's the process of finding a function when given its gradient or rate of change. In single-variable calculus, you integrate with respect to a single variable, but with multivariable calculus, integration involves more than one variable. This means that you need to take into account the effect of each variable independently.

To find a potential function from a gradient vector, you often perform integration over each component of the vector field. For example, if you have a vector field given by three components (as in the exercise), you integrate each component concerning its respective variable.

  • Start with one variable, integrating along one direction, while treating other variables as constants.
  • Repeat the process for each direction or variable, adding arbitrary functions which stand for unknown influences of the other variables not considered in the current integration step.
  • Combine results carefully to ensure consistency across all integrations, sometimes by identifying values for the arbitrary functions that keep the identities compatible.
This method helps in constructing potential functions wherein the mixed partial derivatives equate over all relevant directions.
Gradient Fields
Gradient fields, also known as vector fields, are essential tools in vector calculus. A gradient field is a vector field that represents the gradients of a scalar function. These fields are helpful because they illustrate how a scalar quantity changes at every point in space.

The notation \( abla f \) denotes the gradient of a function \( f \). It combines all the first partial derivatives of \( f \), resulting in a vector.
  • The first component is the derivative of \( f \) concerning \( x \).
  • The second component is the derivative of \( f \) concerning \( y \).
  • The third component is the derivative of \( f \) concerning \( z \).
When these derivatives are known, finding the original scalar function \( f \) involves reversing the differentiation through integration. Gradient fields provide directions of steepest ascent, displaying how to increase the function’s value most rapidly in space.

These fields are conservative if a potential function exists, meaning that the integral around any closed loop is zero, indicating that the work done is path-independent.
Potential Functions
A potential function is a scalar field whose gradient returns a given vector field. Discovering potential functions is like decoding a complex pattern; it involves determining a function whose slope at every point matches the given gradient vector.

For a vector field to have a potential function, certain conditions like being conservative must be met. In simple terms, this means that the vector field's curl (a measure of rotation) should be zero, and the path taken to compute integrals along vector lines shouldn't matter.
  • To find a potential function, start by integrating the vector's first component with respect to \( x \), yielding a partial potential function.
  • Proceed to the remaining components, integrating each while ensuring that your potential function remains consistent.
  • Identify and solve for arbitrary functions using compatibility conditions, ensuring your potential function aligns with all components.
Successfully identifying the potential function means that you unravel the intrinsic structure of a vector field, portraying it as the gradient of a higher-dimensional surface. It's like piecing together the surface that the gradient would naturally shape. In our solved exercise, such a function was defined as \( f(x, y, z) = x\sin(yz) + y^2 + 5\ln(z) \), complying precisely with the given gradient.

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Most popular questions from this chapter

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