2.4 Connections F, F' And Trigap Calculus

  1. 2.4 Connections F F' And Trigap Calculus Worksheet
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  3. 2.4 Connections F F' And Trigap Calculus Solutions

In this section we will define the third type of line integrals we’ll be looking at: line integrals of vector fields. We will also see that this particular kind of line integral is related to special cases of the line integrals with respect to x, y and z. Name: Block: Date: AP Calculus 4.3 Connecting f ′ and f ′ ′ with the Graph of f First Derivative Test for Local Extrema Theorem 4 First Derivative Test for Local Extrema The following test applies to a continuous function f x( ). At a critical point c: 1. If f ′ changes sign from positive to negative at c ⇒ 2. Whenever lim x → a f (x) is in the domain of F and the composition F (f (x)) makes sense. The reason for which (1.2) happens is in fact a more general (at least formally).

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Section 5-4 : Line Integrals of Vector Fields

Trigap

In the previous two sections we looked at line integrals of functions. In this section we are going to evaluate line integrals of vector fields. We’ll start with the vector field,

[vec Fleft( {x,y,z} right) = Pleft( {x,y,z} right)vec i + Qleft( {x,y,z} right)vec j + Rleft( {x,y,z} right)vec k]

and the three-dimensional, smooth curve given by

[vec rleft( t right) = xleft( t right)vec i + yleft( t right)vec j + zleft( t right)vec khspace{0.25in}hspace{0.25in}a le t le b]

The line integral of (vec F) along (C) is

2.4 connections f f [intlimits_{C}{{vec Fcenterdot d,vec r}} = int_{{,a}}^{{,b}}{{vec Fleft( {vec rleft( t right)} right)centerdot vec r'left( t right),dt}}]

Note the notation in the integral on the left side. That really is a dot product of the vector field and the differential really is a vector. Also, (vec Fleft( {vec rleft( t right)} right)) is a shorthand for,

2.4 Connections F F' And Trigap Calculus Worksheet

[vec Fleft( {vec rleft( t right)} right) = vec Fleft( {xleft( t right),yleft( t right),zleft( t right)} right)]

We can also write line integrals of vector fields as a line integral with respect to arc length as follows,

[intlimits_{C}{{vec Fcenterdot d,vec r}} = intlimits_{C}{{vec Fcenterdot vec T,ds}}]

where (vec Tleft( t right)) is the unit tangent vector and is given by,

[vec Tleft( t right) = frac{{vec r'left( t right)}}{{left {vec r'left( t right)} right }}]

If we use our knowledge on how to compute line integrals with respect to arc length we can see that this second form is equivalent to the first form given above.

[begin{align*}intlimits_{C}{{vec Fcenterdot d,vec r}} & = intlimits_{C}{{vec Fcenterdot vec T,ds}} & = int_{{,a}}^{{,b}}{{vec Fleft( {vec rleft( t right)} right)centerdot frac{{vec r'left( t right)}}{{left {vec r'left( t right)} right }},left {vec r'left( t right)} right ,dt}} & = int_{{,a}}^{{,b}}{{vec Fleft( {vec rleft( t right)} right)centerdot vec r'left( t right),dt}}end{align*}]

In general, we use the first form to compute these line integral as it is usually much easier to use. Let’s take a look at a couple of examples.

Example 1 Evaluate ( displaystyle intlimits_{C}{{vec Fcenterdot d,vec r}}) where (vec Fleft( {x,y,z} right) = 8{x^2}y,z,vec i + 5z,vec j - 4x,y,vec k) and (C) is the curve given by (vec rleft( t right) = t,vec i + {t^2},vec j + {t^3},vec k), (0 le t le 1). 2.4 connections f fShow Solution

Okay, we first need the vector field evaluated along the curve.

[vec Fleft( {vec rleft( t right)} right) = 8{t^2}left( {{t^2}} right)left( {{t^3}} right)vec i + 5{t^3},vec j - 4tleft( {{t^2}} right)vec k = 8{t^7},vec i + 5{t^3},vec j - 4{t^3},vec k]

Next, we need the derivative of the parameterization.

[vec r'left( t right) = ,vec i + 2t,vec j + 3{t^2},vec k]

Finally, let’s get the dot product taken care of.

[vec Fleft( {vec rleft( t right)} right)centerdot vec r'left( t right) = 8{t^7} + 10{t^4} - 12{t^5}]

The line integral is then,

2.4 Connections F F' And Trigap Calculus Pdf

[begin{align*}intlimits_{C}{{vec Fcenterdot d,vec r}} & = int_{{,0}}^{{,1}}{{8{t^7} + 10{t^4} - 12{t^5},dt}} & = left. {left( {{t^8} + 2{t^5} - 2{t^6}} right)} right _0^1 & = 1end{align*}] Example 2 Evaluate ( displaystyle intlimits_{C}{{vec Fcenterdot d,vec r}}) where (vec Fleft( {x,y,z} right) = x,z,vec i - y,z,vec k) and (C) is the line segment from (left( { - 1,2,0} right)) to (left( {3,0,1} right)). Show Solution

We’ll first need the parameterization of the line segment. We saw how to get the parameterization of line segments in the first section on line integrals. We’ve been using the two dimensional version of this over the last couple of sections. Here is the parameterization for the line.

[begin{align*}vec rleft( t right) & = left( {1 - t} right)leftlangle { - 1,2,0} rightrangle + tleftlangle {3,0,1} rightrangle & = leftlangle {4t - 1,2 - 2t,t} rightrangle ,hspace{1.0in} 0 le t le 1end{align*}]

So, let’s get the vector field evaluated along the curve.

2.4 connections f f [begin{align*}vec Fleft( {vec rleft( t right)} right) & = left( {4t - 1} right)left( t right),vec i - left( {2 - 2t} right)left( t right),vec k & = left( {4{t^2} - t} right)vec i - left( {2t - 2{t^2}} right)vec kend{align*}]

Now we need the derivative of the parameterization.

[vec r'left( t right) = leftlangle {4, - 2,1} rightrangle ]

The dot product is then,

2.4 Connections F F' And Trigap Calculus Solutions

[vec Fleft( {vec rleft( t right)} right)centerdot vec r'left( t right) = 4left( {4{t^2} - t} right) - left( {2t - 2{t^2}} right) = 18{t^2} - 6t]

The line integral becomes,

[begin{align*}intlimits_{C}{{vec Fcenterdot d,vec r}}& = int_{{,0}}^{{,1}}{{18{t^2} - 6t,dt}} & = left. {left( {6{t^3} - 3{t^2}} right)} right _0^1 & = 3end{align*}]

Let’s close this section out by doing one of these in general to get a nice relationship between line integrals of vector fields and line integrals with respect to (x), (y), and (z).

Given the vector field (vec Fleft( {x,y,z} right) = P,vec i + Q,vec j + R,vec k) and the curve (C) parameterized by (vec rleft( t right) = xleft( t right)vec i + yleft( t right)vec j + zleft( t right)vec k), (a le t le b) the line integral is,

[begin{align*}intlimits_{C}{{vec Fcenterdot d,vec r}} & = int_{{,a}}^{{,b}}{{left( {P,vec i + Q,vec j + R,vec k} right)centerdot left( {x',vec i + y',vec j + z',vec k} right),dt}} & = int_{{,a}}^{{,b}}{{Px' + Qy' + Rz',dt}} & = int_{{,a}}^{{,b}}{{Px',dt}} + int_{{,a}}^{{,b}}{{Qy',dt}} + int_{{,a}}^{{,b}}{{Rz',dt}} & = intlimits_{C}{{P,dx}} + intlimits_{C}{{Q,dy}} + intlimits_{C}{{R,dz}} & = intlimits_{C}{{P,dx}} + Q,dy + R,dzend{align*}] 2.4

So, we see that,

[intlimits_{C}{{vec Fcenterdot d,vec r}} = intlimits_{C}{{P,dx}} + Q,dy + R,dz]

Note that this gives us another method for evaluating line integrals of vector fields.

This also allows us to say the following about reversing the direction of the path with line integrals of vector fields.

Fact

[intlimits_{{ - C}}{{vec Fcenterdot d,vec r}} = - intlimits_{C}{{vec Fcenterdot d,vec r}}]

This should make some sense given that we know that this is true for line integrals with respect to (x), (y), and/or (z) and that line integrals of vector fields can be defined in terms of line integrals with respect to (x), (y), and (z).