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Chapter 16: Problem 29

(a) Evaluate the line integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r},\) where \(\mathbf{F}(x, y)=e^{x-1} \mathbf{i}+x y \mathbf{j}\) and \(C\) is given by \(\mathbf{r}(t)=t^{2} \mathbf{i}+t^{3} \mathbf{j}, 0 \leqslant t \leqslant 1\) (b) Illustrate part (a) by using a graphing calculator or computer to graph \(C\) and the vectors from the vector field corresponding to \(t=0,1 / \sqrt{2},\) and 1 (as in Figure \(13 ) .\)

Short Answer

Expert verified
The line integral evaluates to \( \frac{11}{8} - \frac{1}{e} \). Visualize this using graphing software.

Step by step solution

01

Parametrize the Path

Given the path \( \mathbf{r}(t) = t^2 \mathbf{i} + t^3 \mathbf{j} \), identify that \( x = t^2 \) and \( y = t^3 \) over the interval \( 0 \leq t \leq 1 \). This is the path \( C \) over which we want to evaluate the line integral.
02

Express \( \mathbf{F} \) in terms of \( t \)

Substitute \( x = t^2 \) and \( y = t^3 \) into \( \mathbf{F}(x, y) = e^{x-1} \mathbf{i} + x y \mathbf{j} \) to get \( \mathbf{F}(t) = e^{t^2 - 1} \mathbf{i} + t^5 \mathbf{j} \).
03

Determine \( d \mathbf{r} \)

Compute the derivative of \( \mathbf{r}(t) = t^2 \mathbf{i} + t^3 \mathbf{j} \) to find \( d \mathbf{r} = (2t \mathbf{i} + 3t^2 \mathbf{j}) dt \).
04

Evaluate the Dot Product \( \mathbf{F}(t) \cdot d \mathbf{r} \)

Find the dot product: \( \mathbf{F}(t) \cdot d \mathbf{r} = (e^{t^2 - 1} \mathbf{i} + t^5 \mathbf{j}) \cdot (2t \mathbf{i} + 3t^2 \mathbf{j}) \). This simplifies to \( 2t e^{t^2 - 1} + 3t^7 \).
05

Set Up and Compute the Integral

Evaluate the integral \( \int_{0}^{1} (2t e^{t^2 - 1} + 3t^7) \, dt \). Split this as two separate integrals: \( \int_{0}^{1} 2t e^{t^2 - 1} \, dt + \int_{0}^{1} 3t^7 \, dt \).
06

Calculate \( \int_0^1 2t e^{t^2 - 1} \, dt \)

Use substitution \( u = t^2 - 1 \), \( du = 2t \, dt \). Thus, the integral becomes \( \int e^{u} \, du \) from \( u = -1 \) to \( u = 0 \), yielding \( e^{0} - e^{-1} = 1 - \frac{1}{e} \).
07

Calculate \( \int_0^1 3t^7 \, dt \)

Integrate directly: \( \int_0^1 3t^7 \, dt = \left[ \frac{3}{8} t^8 \right]_0^1 = \frac{3}{8} \).
08

Combine Results

Combine the results of the two integrals: \( \left( 1 - \frac{1}{e} \right) + \frac{3}{8} = \frac{11}{8} - \frac{1}{e} \).
09

Visualize the Path and Vectors

Use a graphing calculator or software to plot the curve given by \( \mathbf{r}(t) = t^2 \mathbf{i} + t^3 \mathbf{j} \) and plot the vector field \( \mathbf{F}(x, y) = e^{x-1} \mathbf{i} + x y \mathbf{j} \) at points corresponding to \( t = 0, \frac{1}{\sqrt{2}}, 1 \).

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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 essentially a map that assigns a vector to every point in a space. In two dimensions, this involves
  • two components, usually denoted as \( extbf{i} \) and \( extbf{j} \), which correspond to the x and y directions, respectively.
  • The exercise uses \( extbf{F}(x, y) = e^{x-1} \textbf{i} + xy \textbf{j} \) as the vector field.
This means at every point \( (x, y) \), there is a vector pointing in a direction determined by the functions \( e^{x-1} \) and \( xy \). Vector fields help us visualize phenomena like fluid flow, where each vector represents the velocity of fluid at a point. To evaluate line integrals, understanding the behavior of the vector field along a path is crucial. This is why graphing the field alongside the path is useful for visual comprehension of how vectors interact with the given path.
Parametric Equations
Parametric equations allow us to express complex curves and paths with respect to a parameter, often time \( t \).
  • Instead of defining a curve solely by \( x \) and \( y \) coordinates, it is expressed as \( x = f(t) \) and \( y = g(t) \).
  • In the exercise, the path \( C \) is given by the parametric equations \( \mathbf{r}(t) = t^2 \mathbf{i} + t^3 \mathbf{j} \).
This form is advantageous in calculating line integrals as it simplifies integral limits to values of \( t \) rather than positional coordinates. Once parametrized, both the path and the vector field become functions of \( t \), aiding in easier computation of the line integral. Detection of changing rates and how each coordinate evolves with \( t \) is facilitated and results in a more straightforward derivation of vectors and scalar values needed for further mathematical operations.
Dot Product
The dot product, also called the scalar product, is a critical operation in calculating line integrals. It takes two vectors and returns a scalar through multiplication.
  • The dot product is calculated as \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \) in 3-dimensional space.
  • For our 2D vector field, we simplify this to \( \mathbf{F}(t) \cdot d\mathbf{r} = (a_1 \cdot b_1) + (a_2 \cdot b_2) \).
In this particular exercise, it involved finding the dot product of the vectors \( \mathbf{F}(t) = e^{t^2 - 1} \mathbf{i} + t^5 \mathbf{j} \) and the differential element \( d\mathbf{r} = (2t \mathbf{i} + 3t^2 \mathbf{j}) dt \). Calculating the dot product results in a scalar that provides the infinitesimal contribution of the vector field's force along the path. This scalar is then integrated over the desired interval to find the total effect.
Calculus
Calculus, especially integration, plays a crucial role in calculating the line integral in vector fields. It is used for accumulating infinitesimal contributions along a path to find the total effect or work.
  • Definite integration computes the total accumulated quantity across an interval. Here, it determines the work done by the vector field along the path \( C \).
  • The line integral in this exercise is set up to evaluate \( \int_C \mathbf{F} \cdot d\mathbf{r} \), which is split into simpler integrals.
The exercise involves breaking the integral into separate more manageable parts: \( \int_{0}^{1} 2t e^{t^2 - 1} dt + \int_{0}^{1} 3t^7 dt \). These are then calculated using substitution and direct integration techniques. Utilizing calculus in line integrals allows us to efficiently solve for the path-dependent effects of vector fields, yielding tangible values related to physics and engineering applications.

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Most popular questions from this chapter

(a) Evaluate the line integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r},\) where \(\mathbf{F}(x, y, z)=x \mathbf{i}-z \mathbf{j}+y \mathbf{k}\) and \(C\) is given by \(\mathbf{r}(t)=2 t \mathbf{i}+3 t \mathbf{j}-t^{2} \mathbf{k},-1 \leqslant t \leqslant 1 .\) (b) Illustrate part (a) by using a computer to graph \(C\) and the vectors from the vector field corresponding to \(t=\pm 1\) and \(\pm \frac{1}{2}(\) as in Figure 13\() .\)

\(19-26\) Find a parametric representation for the surface. The part of the hyperboloid \(x^{2}+y^{2}-z^{2}=1\) that lies to the right of the \(x\) -plane

Let \(\mathbf{F}(x, y, z)=z \tan ^{-1}\left(y^{2}\right) \mathbf{i}+z^{3} \ln \left(x^{2}+1\right) \mathbf{j}+z \mathbf{k}\) Find the flux of \(\mathbf{F}\) across the part of the paraboloid \(x^{2}+y^{2}+z=2\) that lies above the plane \(z=1\) and is oriented upward.

\(19-30\) Evaluate the surface integral \(\iint_{S} \mathbf{F} \cdot d \mathbf{S}\) for the given vector field \(\mathbf{F}\) and the oriented surface \(S .\) In other words, find the flux of \(\mathbf{F}\) across \(S .\) For closed surfaces, use the positive (outward) orientation. $$\mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}+5 \mathbf{k}, \quad $$ S is the boundary of the region enclosed by the cylinder \(x^{2}+z^{2}=1\) and the planes \(y=0\) and \(x+y=2\)

\(48-49\) Find the area of the surface correct to four decimal places by expressing the area in terms of a single integral and using your calculator to estimate the integral. The part of the surface \(z=e^{-x^{2}-y^{2}}\) that lies above the disk \(x^{2}+y^{2} \leqslant 4\)
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