(Problem 1) Compute where is the line segment from to . Note that this
is the same line segment as in Example 1, but traversed in the opposite
direction.
We compute integrals of vector-valued functions along curves.
Line Integral Let be a curve in parameterized by for and let be a vector-valued
function defined along . Then the integral of along is defined by
example 1 Compute where is the line segment from to .
We first parameterize the line segment. Since the equation of the line in the -plane is , one possible parameterization of the line segment is The derivative of is Next, and hence We are now ready to compute the contour integral: \begin{align*} \int _C \vec f(x, y) \dotp d\vec r &= \int _{-1}^1 \vec f(x(t), y(t)) \dotp \vec r\,'(t) \, dt\\ &= \int _{-1}^1 \vector{t , -\left (\frac{t}{2} + \frac 12\right )} \dotp \vector{1, \frac{1}{2}} \, dt\\[6pt] &= \int _{-1}^1 \left (t -\frac{t}{4} - \frac 14\right ) \, dt\\[6pt] &= \int _{-1}^1 \left (\frac 34 t - \frac 14\right ) \, dt\\[6pt] &= \left (\frac 38 t^2 - \frac 14 t\right )\bigg |_{-1}^1\\[6pt] &= \left (\frac 38 - \frac 14\right ) - \left (\frac 38 + \frac 14\right )\\[6 pt] &= -\frac 12 \end{align*}
We first parameterize the line segment. Since the equation of the line in the -plane is , one possible parameterization of the line segment is The derivative of is Next, and hence We are now ready to compute the contour integral: \begin{align*} \int _C \vec f(x, y) \dotp d\vec r &= \int _{-1}^1 \vec f(x(t), y(t)) \dotp \vec r\,'(t) \, dt\\ &= \int _{-1}^1 \vector{t , -\left (\frac{t}{2} + \frac 12\right )} \dotp \vector{1, \frac{1}{2}} \, dt\\[6pt] &= \int _{-1}^1 \left (t -\frac{t}{4} - \frac 14\right ) \, dt\\[6pt] &= \int _{-1}^1 \left (\frac 34 t - \frac 14\right ) \, dt\\[6pt] &= \left (\frac 38 t^2 - \frac 14 t\right )\bigg |_{-1}^1\\[6pt] &= \left (\frac 38 - \frac 14\right ) - \left (\frac 38 + \frac 14\right )\\[6 pt] &= -\frac 12 \end{align*}
An alternative method for parameterizing a line segment was discussed in chapter 1: if and are the vector representation of points, then the line segment from to is: Using and for and , respectively, we can parameterize the line segment as Thus, and the vector function is We can now compute the line integral \begin{align*} \int _C \vec f(x, y) \dotp d\vec r &= \int _0^1 \vec f(x(t), y(t)) \dotp \vec r\,'(t) \, dt\\ &= \int _0^1 \vector{2t-1, t} \dotp \vector{2, 1} \, dt\\[6pt] &= \int _0^1 \left (4t-2 -t\right ) \, dt\\[6pt] &= \int _0^1 \left (3t-2\right ) \, dt\\[6pt] &= \left (\frac 32 t^2 - 2 t\right )\bigg |_0^1\\[6pt] &= \frac 32 - 2 = -\frac 12 \end{align*}
as before.
As Example 1 and Problem 1 demonstrate, the line integral is sensitive to the
orientation of the curve . Reversing the direction gives the negative of the line
integral.
Here is a video solution of problem 1:
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0.1 Work
If the vector-valued function represents the force on an object at the point , then the line integral represents the work done by the force (or against the force, if the value of the integral is negative) on an object that travels along the curve .
Example 2 The force due to gravity by a mass at the origin of (two dimensions for
simplicity) is given by where is a positive constant that depends on the masses of
the objects involved. Find the work done by this force as an object moves in a
straight line from the point to the point .
The line segment from to can be parameterized by The work done is the value of the line integral: \begin{align*} \text{Work} &= \int _C \vec f(x,y) \dotp d\vec r\\ &= \int _0^1 \vec f(x(t),y(t)) \dotp \vec r\,'(t) \, dt\\ &= \int _0^1 \vec f(t, 1) \dotp \vec r\,'(t) \, dt\\ &= \int _0^1 \frac{k}{(t^2 + 1^2)^{3/2}} \vector{-t,-1} \dotp \vector{1,0} \, dt\\ &= \int _0^1 -\frac{kt}{(t^2 + 1^2)^{3/2}} \, dt\\ &= \frac{k}{(t^2 +1)^{1/2}} \bigg |_0^1 \\ &= k\left (\frac{1}{\sqrt 2} - 1\right ) \end{align*}
The line segment from to can be parameterized by The work done is the value of the line integral: \begin{align*} \text{Work} &= \int _C \vec f(x,y) \dotp d\vec r\\ &= \int _0^1 \vec f(x(t),y(t)) \dotp \vec r\,'(t) \, dt\\ &= \int _0^1 \vec f(t, 1) \dotp \vec r\,'(t) \, dt\\ &= \int _0^1 \frac{k}{(t^2 + 1^2)^{3/2}} \vector{-t,-1} \dotp \vector{1,0} \, dt\\ &= \int _0^1 -\frac{kt}{(t^2 + 1^2)^{3/2}} \, dt\\ &= \frac{k}{(t^2 +1)^{1/2}} \bigg |_0^1 \\ &= k\left (\frac{1}{\sqrt 2} - 1\right ) \end{align*}
Note that this answer is negative. This means that work was done against the force rather than by the force.
(Problem 2a) Use the function given in Example 2 to describe the force due to gravity.
Find the work done by this force on an object as the object moves in a straight line
from the point to the point .
Was work done by gravity or against gravity in this situation?
Was work done by gravity or against gravity in this situation?
Here is a video solution of problem 2a:
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(Problem 2b) Use the function given in Example 2 to describe the force due to
gravity. Find the work done on an object by this force as the object moves in a
straight line from the point to the point .
Was work done by gravity or against gravity in this situation?
Was work done by gravity or against gravity in this situation?
Notice that the answers in Example 2 and Problems 2a and 2b are all similar due to rotational symmetry.
Example 3 The force due to gravity by a mass at the origin of (two dimensions for
simplicity) is given by where is a positive constant that depends on the
masses of the objects involved. Find the work done on an object by this
force as the object moves in a quarter circle from the point to the point
.
The quarter circle from to can be parameterized by The work done is the value of the line integral: \begin{align*} \text{Work} &= \int _C \vec f(x,y) \dotp d\vec r\\ &= \int _0^{\pi /2} \vec f(x(t),y(t)) \dotp \vec r\,'(t) \, dt\\ &= \int _0^{\pi /2} \vec f(2\cos t, 2\sin t) \dotp \vec r\,'(t) \, dt\\ &= \int _0^{\pi /2} \frac{k}{[(2\cos t)^2 + (2\sin t)^2)^{3/2}} \vector{-2\cos t, 2 \sin t} \dotp \vector{-2\sin t, 2 \cos t} \, dt\\ &= \int _0^{\pi /2} 0 \, dt\\ &= 0 \end{align*}
The quarter circle from to can be parameterized by The work done is the value of the line integral: \begin{align*} \text{Work} &= \int _C \vec f(x,y) \dotp d\vec r\\ &= \int _0^{\pi /2} \vec f(x(t),y(t)) \dotp \vec r\,'(t) \, dt\\ &= \int _0^{\pi /2} \vec f(2\cos t, 2\sin t) \dotp \vec r\,'(t) \, dt\\ &= \int _0^{\pi /2} \frac{k}{[(2\cos t)^2 + (2\sin t)^2)^{3/2}} \vector{-2\cos t, 2 \sin t} \dotp \vector{-2\sin t, 2 \cos t} \, dt\\ &= \int _0^{\pi /2} 0 \, dt\\ &= 0 \end{align*}
Note that this answer is zero because the direction of motion is orthogonal to the force (dot product is zero) and therefore no work is done by or against the force along this path.
(Problem 3) Use the function given in Example 3 to describe the force due
to gravity. Use a line integral to determine the work done on an object by
this force as the object moves in a semicircle from the point to the point
.
Here is a video solution of problem 3:
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