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Chapter 15: Problem 34

Let \(\mathbf{F}(x, y, z)=y z \mathbf{i}+(x z+y) \mathbf{j}+(x y+1) \mathbf{k} .\) Define the function \(f\) by $$f(x, y \cdot z)=\int_{C} \mathbf{F} \cdot \mathbf{T} d s$$ where \(C\) is the straight line segment from \((0,0,0)\) to \((x, y, z)\). Determine \(f\) by evaluating this line integral, and then show that \(\boldsymbol{\Gamma} f=\mathbf{F}\).

Short Answer

Expert verified
The function is \( f(x, y \cdot z) = \frac{xyz}{3} + \frac{yz^2}{2} + z \), and \( \nabla f = \mathbf{F} \).

Step by step solution

01

Parameterize the Curve

To evaluate the line integral, first parameterize the straight line segment \(C\) from \((0,0,0)\) to \((x, y, z)\) using the parameter \(t\), where \(t \in [0, 1]\). We can express the line as \( \mathbf{r}(t) = (xt, yt, zt) \).
02

Calculate the Derivative of the Parameterization

Find the derivative \( \frac{d\mathbf{r}}{dt} \). The derivative is \( \mathbf{r}'(t) = (x, y, z) \), which represents the direction vector along the line.
03

Substitute Parameterization into F

Substitute \( (xt, yt, zt) \) into \( \mathbf{F}(x, y, z) \) to get \( \mathbf{F}(t) = yt \cdot zt \mathbf{i} + (xt \cdot zt + yt) \mathbf{j} + (xt \cdot yt + 1) \mathbf{k} \). Simplify this to \( zt^2y \mathbf{i} + (xzt^2 + yt) \mathbf{j} + (xy t^2 + 1) \mathbf{k} \).
04

Dot Product of F and Derivative of r

Compute \( \mathbf{F}(t) \cdot \mathbf{r}'(t) \), which is \( zt^2y \cdot x + (xzt^2 + yt) \cdot y + (xy t^2 + 1) \cdot z \). Simplify it to get \( xyz t^2 + xzyt^2 + yz^2t + xyt^2 + z \).
05

Integrate Over t from 0 to 1

Evaluate the integral \( \int_0^1 (xyz t^2 + xzyt^2 + yz^2t + xyt^2 + z) \, dt \). Simplify the expression to \( (xyz + xzy + xy)t^2 + yz^2 t + z \), and integrate to get \( \frac{xzy}{3} + \frac{yz^2}{2} + z \).
06

Define the Function f

Thus, the value of the line integral gives us the function \( f(x, y \cdot z) = \frac{xzy}{3} + \frac{yz^2}{2} + z \).
07

Verify Gradient of f Equals F

Compute the gradient \( abla f(x, y, z) \) and check if it matches \( \mathbf{F}(x, y, z) \). Calculate \( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \) and \( \frac{\partial f}{\partial z} \). You get \( z \cdot y \mathbf{i} + (xz + y) \mathbf{j} + (xy + 1) \mathbf{k} \), which matches \( \mathbf{F}(x, y, z) \).
08

Conclusion: Solution Verification

Conclusively, since \( abla f = \mathbf{F} \), the previously integrated function indeed satisfies the condition. This confirms the solution is correct.

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Key Concepts

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

Vector Fields
Vector fields are an essential concept in multivariable calculus. They assign a vector to every point in a region in space. Imagine a vector field like a map of arrows indicating flow, with each arrow鈥檚 length and direction showing a specific feature.
  • The vector function \( \mathbf{F}(x, y, z) = yz \mathbf{i} + (xz + y) \mathbf{j} + (xy + 1) \mathbf{k} \) associates a vector with every point \((x, y, z)\) in space.

  • In practical terms, vector fields can represent things like wind speed and direction across a geographic area or fluid flow in a pipe.

To solve different problems, we often need to analyze how these vectors change over space, which is where other calculus concepts come into play.
Gradient
The gradient is a powerful tool in calculus that helps us understand how a function changes in space. If you picture a three-dimensional surface, the gradient points in the direction of the steepest ascent on that surface.
  • The gradient of a function \( f(x, y, z) \) is denoted as \( abla f \) and is a vector composed of its partial derivatives: \( abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right) \).

  • In this exercise, the gradient of the function \( f \) should match the given vector field \( \mathbf{F} \).

Finding the gradient can tell us a lot about the function's behavior, as it reveals the rate and direction of the fastest change in function value.
Parameterization
Parameterization is a technique to represent curves, surfaces, and other geometric objects using parameters. This allows us to analyze these shapes more readily in calculus.
  • In this exercise, the curve \( C \) representing the path from \((0,0,0)\) to \((x,y,z)\) is parameterized by \( t \) as \( \mathbf{r}(t) = (xt, yt, zt) \) where \( t \in [0, 1] \).

  • The parameter \( t \) helps convert a multi-dimensional problem into one that can be dealt with using single-variable calculus methods.

Parameterization is valuable when working with complex shapes since it simplifies applications of calculus operations by making them more manageable.
Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions of several variables. It allows us to investigate and solve problems involving more than just lines or simple curves.
  • In this context, multivariable calculus provides tools to compute line integrals, as seen in finding \( f(x, y \cdot z) \), by integrating over a path in a vector field.

  • This branch of calculus helps in understanding phenomena in spaces higher than two dimensions and is frequently applied in fields like physics and engineering.

The concepts of vectors, gradients, and parameterizations, all essential components of multivariable calculus, come together to solve complex, real-world problems effectively.

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Most popular questions from this chapter

Apply Green's theorem to evaluate the integral $$\oint_{C} P d x+Q d y$$ around the specified closed curve \(C\). \(P(x, y)=x y . Q(x, y)=x^{2}: \quad C\) is the first-quadrant loop of the graph of the polar equation \(r=\sin 2 \theta\).

Apply Green's theorem to evaluate the integral $$\oint_{C} P d x+Q d y$$ around the specified closed curve \(C\). \(P(x, y)=y+e^{x} \cdot Q(x, y)=2 x^{2}+\cos y ; \quad C\) is the bound- ary of the triangle with vertices \((0,0),(1,1)\), and \((2,0)\).

Find the mass and centroid of a wire that has constant density \(\delta=k\) and is shaped like the helix \(x=3 \cos t\) \(y=3 \sin t, z=4 t, 0 \leqq t \leqq 2 \pi .\)

Evaluate the line integral $$\int_{C} \mathbf{F} \cdot d \mathbf{r}=\int_{C} \mathbf{F} \cdot \mathbf{T} d s$$ along the indicated path \(C\). \(\mathbf{F}(x, y, z)=y z \mathbf{i}+x z \mathbf{j}+x y \mathbf{k}: \quad C\) is the straight line seg- ment from \((2,-1,3)\) to \((4,2,-1)\).

The Laplacian of the twice-differentiable scalar function 7 is defined to be \(\boldsymbol{r}^{2} f=\operatorname{div}(\operatorname{grad} f)=\boldsymbol{\nabla} \cdot \boldsymbol{\nabla} f .\) Show that $$\boldsymbol{\nabla}^{2} f=\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}+\frac{\partial^{2} f}{\partial z^{2}}$$
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