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Chapter 16: Problem 38

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 \(t .\) $$ \begin{array}{l}{\mathbf{F}=x^{2} \mathbf{i}+y z \mathbf{j}+y^{2} \mathbf{k}} \\\ {r(t)=3 t \mathbf{j}+4 t \mathbf{k}, \quad 0 \leq t \leq 1}\end{array} $$

Short Answer

Expert verified
The flow along the curve is 24.

Step by step solution

01

Understand the Problem

The task is to find the flow of the velocity field \( \mathbf{F} \) along the parametric curve \( \mathbf{r}(t) \). We want to compute the line integral of \( \mathbf{F} \) along the path described by \( \mathbf{r}(t) \) for \( 0 \leq t \leq 1 \).
02

Parameterize the Curve

The given curve is \( \mathbf{r}(t) = 3t \mathbf{j} + 4t \mathbf{k} \). Hence, we parameterize it as \( x(t) = 0 \), \( y(t) = 3t \), and \( z(t) = 4t \).
03

Compute \( \mathbf{F}(\mathbf{r}(t)) \)

Substitute the parameterization into \( \mathbf{F} \):\[\mathbf{F}(\mathbf{r}(t)) = x^2 \mathbf{i} + yz \mathbf{j} + y^2 \mathbf{k} = 0 \mathbf{i} + (3t)(4t) \mathbf{j} + (3t)^2 \mathbf{k} = 12t^2 \mathbf{j} + 9t^2 \mathbf{k}.\]
04

Differentiate \( \mathbf{r}(t) \)

Calculate the derivative of \( \mathbf{r}(t) \):\[ \frac{d\mathbf{r}}{dt} = \frac{d}{dt}(3t \mathbf{j} + 4t \mathbf{k}) = 3 \mathbf{j} + 4 \mathbf{k}. \]
05

Compute the Dot Product

Find the dot product \( \mathbf{F}(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt} \):\[ \mathbf{F}(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt} = (12t^2 \mathbf{j} + 9t^2 \mathbf{k}) \cdot (3 \mathbf{j} + 4 \mathbf{k}) = 12t^2 \cdot 3 + 9t^2 \cdot 4 = 36t^2 + 36t^2 = 72t^2. \]
06

Integrate Over the Interval

Integrate the dot product from \( t=0 \) to \( t=1 \) to find the total flow:\[\int_{0}^{1} 72t^2 \, dt.\]This evaluates to:\[= 72 \int_{0}^{1} t^2 \, dt = 72 \left[ \frac{t^3}{3} \right]_{0}^{1} = 72 \left( \frac{1}{3} - 0 \right) = 24.\]

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Key Concepts

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

Velocity Field
In fluid mechanics, a velocity field represents the velocity of a fluid at every point in a region of space. This concept is crucial for understanding how fluids move and interact with objects. A velocity field typically depends on the coordinates and may vary in different directions. When given a function like \( \mathbf{F} = x^2 \mathbf{i} + yz \mathbf{j} + y^2 \mathbf{k} \), \( \mathbf{F} \) describes the three-dimensional flow of fluid.
  • The \( x^2 \mathbf{i} \) component affects movement along the x-axis.
  • The \( yz \mathbf{j} \) component dictates how flow occurs in the y-direction, interacting with the z coordinate.
  • The \( y^2 \mathbf{k} \) component influences motion along the z-axis.
This composition explains what happens with the fluid when it moves throughout its space. Understanding this helps us calculate how much fluid moves along other paths, like curves.
Parametric Curve
A parametric curve is a type of curve expressed with parameter equations that allow us to specify motion along a path over an interval. Instead of traditional coordinates \( (x, y, z) \), it's defined using a parameter, often \( t \), as in \( \mathbf{r}(t) = 3t \mathbf{j} + 4t \mathbf{k} \).
  • Each coordinate (x, y, z) becomes a function of \( t \).
  • For example, this curve doesn't change in the x-direction since \( x(t) = 0 \).
  • It increases linearly in both the y and z directions; given by \( y(t) = 3t \) and \( z(t) = 4t \).
Parametric equations allow for the easy description of more complex curves and motion paths in multidimensional space, which is essential for calculating line integrals in fluid flow.
Differentiation
Differentiation is the process of finding the derivative of a function, which represents the rate of change. For parametrically-defined curves like \( \mathbf{r}(t) = 3t \mathbf{j} + 4t \mathbf{k} \), differentiation tells us how fast the curve changes as the parameter \( t \) increases.
This is needed to determine changes along the curve, especially as it pertains to line integrals.
  • The derivative \( \frac{d\mathbf{r}}{dt} = 3 \mathbf{j} + 4 \mathbf{k} \) shows constant rates because it鈥檚 linear.
  • It indicates that moving along the curve towards increasing \( t \) changes the y-direction by 3 units and z by 4 units, per unit increase in \( t \).
In line integrals, differentiating the parameterization is crucial as it is used in dot products to assess flow.
Integration
Integration is a mathematical method for calculating the area under a curve or the total accumulation of quantities. In the context of line integrals, it sums up how a function behaves over a curve. For fluid flow along a path described by \( \, \mathbf{r}(t) \, \), integration finds the total flow based on previous calculations.
  • We compute \( \int_{0}^{1} 72t^2 \, dt \) to connect the rate of fluid movement with the curve.
  • This evaluates to \( 72 \left[ \frac{t^3}{3} \right]_{0}^{1} = 24 \), which gives us the total fluid flow between \( t=0 \) and \( t=1 \).
Integration, therefore, is the final step ensuring that all dynamic changes across the curve are summed up into a single meaningful number, illustrating comprehensive flow.

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Most popular questions from this chapter

Let \(R\) be a region in the \(x y\) -plane that is bounded by a piecewise-smooth simple closed curve \(C\) and suppose that the moments of inertia of \(R\) about the \(x\) - and \(y\) -axes are known to be \(I_{x}\) and \(I_{y}\) . Evaluate the integral $$ \begin{array}{l}{\oint_{C} \nabla\left(r^{4}\right) \cdot \mathbf{n} d s} \\\ {C}\end{array} $$ where \(r=\sqrt{x^{2}+y^{2}},\) in terms of \(I_{x}\) and \(I_{y}\)

Let \(S\) be the cylinder \(x^{2}+y^{2}=a^{2}, 0 \leq z \leq h,\) together with its top, \(x^{2}+y^{2} \leq a^{2}, z=h .\) Let \(\mathbf{F}=-y \mathbf{i}+x \mathbf{j}+x^{2} \mathbf{k} .\) Use Stokes' Theorem to find the flux of \(\nabla \times \mathbf{F}\) outward through \(S .\)

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 \((\phi, \theta)=\) \((\pi / 6, \pi / 4)\)

Find the flux of the field \(\mathbf{F}(x, y, z)=4 x \mathbf{i}+4 y \mathbf{j}+2 \mathbf{k}\) outward (away from the \(z\) -axis ) through the surface cut from the bottom of the paraboloid \(z=x^{2}+y^{2}\) by the plane \(z=1\)

In Exercises \(21-26,\) find the flux of the field \(\mathbf{F}\) across the portion of the sphere \(x^{2}+y^{2}+z^{2}=a^{2}\) in the first octant in the direction away from the origin. $$ \mathbf{F}(x, y, z)=\frac{x \mathbf{i}+y \mathbf{j}+z \mathbf{k}}{\sqrt{x^{2}+y^{2}+z^{2}}} $$
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