Vector Calculus 4th Edition Chapter 8 Vector Analysis in Higher Dimensions
Page 535 Problem 1 Answer
Given: 2dx+6dy−5dz;a=(1,−1,−2)
To determine the values of the following differential forms on the ordered sets of vectors.Using the determination method.
Once again, recall that the 1−form acting on a3 dimensional vector is simply a projection of the i−th coordinate.
Once again use the linearity of the form to obtain:
(2dx+6dy−5dz)(1,−1,−2)=2dx(1,−1,2)+6dy(1,−1,−2)−5dz(1,−1,−2)⇒2−6+10=6
The values of the following differential forms on the ordered set of vectors are 6.
Page 535 Problem 2 Answer
Given:4dx∧dy−7dy∧dz;a=(0,1,−1),b=(1,3,2)
To determine the values of the following differential forms on the ordered sets of vectors.Using the determination method.
Here, given a 2−form acting on3 dimensional object. First, use the linearity of the differential forms to split the form into two parts.
(4dx∧dy−7dy∧dz)((0,1,−1),(1,3,2))=4dx∧dy((0,1,−1),(1,3,2))−7dy∧dz((0,1,−1),(1,3,2))
Next, place the vectors into the columns of the matrix but cross out the terms that are not shown in the differential form operator.
For example, we cross out the last row in the matrix in the first case since there is no dx in the form and we cross out the first row in the second matrix since there is no dx.
=4[0 1 1 3]−7[1−13 2]
⇒−4−35=−39
The values of the following differential forms on the ordered sets of vectors indicated are −39.
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Page 535 Problem 3 Answer
Given:7dx∧dy∧dz;a=(1,0,3),b=(2,−1,0),c=(5,2,1)
To determine the values of the following differential forms on the ordered sets of vectors.Using the determination method.
A 3−the form has to act on 3 vectors. Additionally, since we are in 3 dimensions, the 3 form acting on 3
vectors are equal to the determinant of the matrix formed by given vectors placed into columns:
7dx∧dy∧dz((1,0,3),(2,−1,0),(5,2,1))
=[1032−10521]
⇒182
The values of the following differential forms on the ordered sets of vectors are 182.
Page 535 Problem 4 Answer
Given: The differential form dx1
∧dx2+2dx2∧dx3+3dx3∧dx4;a=(1,2,3,4),b=(4,3,2,1).To determine the values of the differential forms on the ordered sets of vectors.Using the concepts of vectors.
The2 -form acting on 4 dimensional vectors. Just like in task, use linearity and cross out the rows in the matrix corresponding to the missing coordinates.
(dx1∧dx2+2dx2∧dx3+3dx3∧dx4)((1,2,3,4),(4,3,2,1))
=∣1243∣+2∣∣2332∣+3∣∣3421∣
=−5−10−15
=−30
Therefore the value of the differential form is −30.
The value of the differential form is −30.
Page 535 Problem 5 Answer
Given:The differential form 2dx1∧dx3∧dx4+dx2∧dx3∧dx5;a=(1,0,−1,4,2),b=(0,0,9,1,−1),c=(5,0,0,0,−2).
To determine the values of the differential forms on the ordered sets of vectors indicated.
Using the concepts of vectors.
This is a 3 -form acting on 5 dimensional vector. Once again use the linearity and we place the 3 vectors into the matrices as their columns.
Then cross out the rows that are not present in the form.
For the first part of the form, cross out the second and the fifth row, for the second part cross out first and the fourth row.
Therefore the expression can be obtained as: 2∣1−14091500∣∣+∣∣0−1209−100−2∣=−370
The value of the differential form is −370.
Page 535 Problem 6 Answer
Given: Let ω be the 1 -form on R3 defined by ω=x2ydx+y2zdy+z3xdz.To find ω(3,−1.4)(a), where a=(a1,a2,a3).
Using the concepts of vector.
If the form contains some functional expression (as in this task), to specify additional vector that will be plugged into the functional expression part of the form.
In this case, replace every x
in front of (dx,dy,dz)with3 , everyy with −1 and every z with4. Therefore,ω3,−1,4(a)=(32(−1)dx+(−1)2⋅4dy+43⋅3dz)(a1,a2,a3)
Now simply calculate the expression: −9a1+4a2+192a3
The value of ω(3,−1,4)(a) is −9a1+4a2+192a3.
Page 535 Problem 7 Answer
Given: Let ω be the 2 -form on R4 given by ω=x1x3dx1∧dx3−x2x4dx2∧dx4.
To findω12,−1,−3,1)(a,b).Using the concepts of vector.
The function is xi where i∈{1,2,3,4}plug in the appropriate coordinates of the vector (2,−1,−3,1).
We have 2 -form acting on 4 dimensional vectors, therefore cross out even rows in first matrix and odd rows in second one before calculating the determinant:
(−6dx1∧dx3+dx2∧dx4)((a1,a2,a3,a4),(b1,b2,b3,b4))
=6∣a1a3b1b3∣+∣a2a4b2b4∣
=−6a1b3+6a3b1+a2b4−a4b2
Therefore ω12,−1,−3,1)(a,b) is −6a1b3+6a3b1+a2b4−a4b2.
The solution of ω12,−1,−3,1)(a,b)is −6a1b3+6a3b1+a2b4−a4b2.
Page 535 Problem 8 Answer
Given: Let ω be the 2 -form on R3 given by ω=cosxdx∧dy−sinzdy∧dz+(y2+3)dx∧dz.To findω(0,−1,π/2)
(a,b), wherea=(a1,a2,a3) and b=(b1,b2,b3).Using the concepts of vector.
Instead of x,y and z the values of (0,−1,π2) are plugged in inside the differential form.
This is a2 -form so cross the missing coordinate row inside the matrix (dx∧dy−dy∧dz+4dx∧dz)/((a1,a2,a3),(b1,b2,b3))
=∣a1a2b1b2∣−∣a2a3b2b3∣+4∣a1a3b1b3∣==a1b2−a2b1−a2b3+a3b2+4a1b3−4a3b1
Therefore ω(0,−1,π/2)(a,b)= a1b2−a2b1−a2b3+a3b2+4a1b3−4a3b1.
The solution of the ω(0,−1,π/2)(a,b)= a1b2−a2b1−a2b3+a3b2+4a1b3−4a3b1.
Page 535 Problem 9 Answer
Given function ω
To findω(x,y,z) ((2,0,−1),(1,7,5)
Method used with determinants.
Given a function ω
Now to find out ω(x,y,z) ((2,0,−1),(1,7,5)
Find out the values using determinants
Also Here, Repeat the same process as in previous task.
But this time there is no vector to plug in.
So functions in front of the forms remain.
We have ω(x,y,z) ((2,0,−1),(1,7,5)=cosx2017−sinz0−175+(y2+3)2−115
=14cosx−7sinz+11y2+33
The solution of ω(x,y,z) ((2,0,−1),(1,7,5) is14cosx−7sinz+11y2+33
Page 535 Problem 10 Answer
Given function ω the 3 form R3
To findω(0,0,0) (a,b,c)
where a=(a1,a2,a3),b=(b1,b2,b3),c=(c1,c2,c3) Method used is determinant.
Given that, Function ω the three form R3
ω=(excos y+(y2+2)e2z)dx∧dy∧dz
To find out ω(0,0,0) (a,b,c)
Where,a=(a1,a2,a3)
b=(b1,b2,b3)
c=(c1,c2,c3)
Notice that
We are dealing with the three form on three dimensional space.
Hence the resulting form is just a determinant of vectors placed in the matrix as its columns.
We only need to deal with the functional expression in front of the matrix.
But this is easy due to the fact that we plug in x=y=z=0 to obtain 3
Hence the resulting form is three times the determinant of the matrix.
ω(0,0,0)(a,b,c)=3∣a1a3b1b3c1c3∣
Or expressed as the total result equal to 3(a1b2c3+b1c2a3+c1a2b3−c1b2a3−b1a2c3−a1c2b3)
The solution of ω(0,0,0) (a,b,c) is 3(a1b2c3+b1c2a3+c1 a2b3−c1b2a3−b1a2c3−a1c2b3)
Page 535 Problem 11 Answer
Given The given expression is ω=(excosy+(y2+2)e2z)dx∧dy∧dz.
To find The value of the expression ω(x,y,z)((1,0,0),(0,2,0),(0,0,3)).
Method used The method used for this is the concept of determinant.
Given ω=(excosy+(y2+2)e2z)dx∧dy∧dz.
ω(x,y,z)((1,0,0),(0,2,0),(0,0,3)) can be written as (excosy+(y2+2)e2z)∣100020003∣
ω(x,y,z)((1,0,0),(0,2,0),(0,0,3))= (excosy+(y2+2)e2z)∣100020003∣
Now, solve the determinant.
det[adgbehcfi]=a⋅det[ehfi]−b⋅det[dgfi]+c⋅det[dgeh]
On solving.
ω(x,y,z)((1,0,0),(0,2,0),(0,0,3))= 6(excosy+(y2+2)e2z).
The value of ω(x,y,z) ((1,0,0),(0,2,0),(0,0,3)) is equal to 6(excosy+(y2+2)e2z).
Page 535 Problem 12 Answer
Given The given expressions are ω=3dx+2dy−xdz;η=x2 dx−cosydy+7dz.To find The value of the expression ω∧η.
Method used The method used in this is vector product.
The given expressions are ω=3dx+2dy−xdz
η=x2 dx−cosydy+7dz
ω∧η= (3dx+2dy−xdz)∧(x2dx−cosydy+7dz)
Now, do the product
= 3x2dx∧dx+2x2dy∧dx−x3dz∧dx−3cosydx∧dy−2cosydy∧dy+xcosydz∧dy+21dx∧dz+14dy∧dz−7xdz∧dz [dx∧dx =dy∧dy=dz∧dz= 0]
Simplifying the terms.
= (−2×2−3cosy)dx∧dy+(x3+21)dx∧dz+(14−xcosy)dy∧dz.
The value of the expression ω∧ηis equal to (−2×2−3cosy)dx∧dy+(x3+21)dx∧dz+(14−xcosy)dy∧dz.
Page 535 Problem 13 Answer
Given The given functions are On R3:ω=ydx−xdy;η=zdx∧dy+ydx∧dz+xdy∧dz.To find The value of the expressionω∧η.
Method used Do the wedge product for simplifying.
Given functions are ω=ydx−xdy
η=zdx∧dy+ydx∧dz+xdy∧dz
Thus,ω∧η= (ydx−xdy)∧(zdx∧dy+ydx∧dz+xdy∧dz)
Multiply the terms.
= yxdx∧dy∧dz−xydy∧dx∧dz+yzdx∧dx∧dy+yydx∧dx∧dz−xzdy∧dx∧dy−xxdy∧dy∧dz
[dx∧dx=dy∧dy=dz∧dz=0]
Simplifying the terms.
= 2xydx∧dy∧dz.
By using property of alternation the terms of the second term are swapped and sign is flipped.
The value of the expression ω∧η is equal to 2xydx∧dy∧dz.
Page 536 Problem 14 Answer
Given The given functions are On R4:ω=2dx1∧dx2−x3dx2∧dx4;η=2x4dx1∧dx3+(x3−x2)dx3∧dx4.
To find The value of the expression ω∧η.
Method used The method used to simplify the given expression is wedge product.
Given ω=2dx1∧dx2−x3dx2∧dx4
η=2x4dx1∧dx3+(x3−x2)dx3∧dx4
Thus, ω∧η= (2dx1∧dx2−x3dx2∧dx4)∧(2x4dx1∧dx3+(x3−x2)dx3∧dx4)
There are 4 coordinates so the term obtained in result are those which have all the four coordinates only.
Thus,ω∧η= 2x3x4dx2∧dx4∧d1∧dx3+2(x3−x2)dx1∧dx2∧dx3∧dx4
Using alternating property reorder the terms of the product.
= (2x3x4+2×3−2×2)dx1∧dx2∧dx3∧dx4.
The value of the expression ω∧η is equal to (2x3x4+2×3−2×2)dx1∧dx2∧dx3∧dx4 .