Question

(a) Let F = 5y i  + 2 j  −  k and C be the line from (3, 2, -2) to (6, 1, 7).
Find integral_C F · dr=


(b) Let G = 5xe^(5x^2+2y^2+z^2) i  + 2ye^(5x^2+2y^2+z^2) j  + ze^(5x^2+2y^2+z^2) k 
and C be the same as in part (a).


find integral_C G · dr =

Answer 1

@ganeshie8 can you help me with this one for part a i use the F.T. F.L.I and i get -5 but its wrong

Answer 2

@ganeshie8 I did it again and i got -11 but its also wrong. what am I doing wrong

Answer 3

gradf(5yi+2j-k) = 5yx+2y -z right??

Answer 4

you need to parameterize and setup the line integral, right ?

Answer 5

i'm not sure

Answer 6

for part a i try using the FTFLT

Answer 7

for part b,

try \(\large \frac{e^{231}-e^{57}}{2}\)

Answer 8

its right, how did you figured it out?

Answer 9

vector field in part b is conservative so you can eyeball the potential function

Answer 10

find the curl

Answer 11

you should get 0 if the field is conservative

Answer 12

what do you do after you find that the curl is 0

Answer 13

curl=0 tells you the field is conservative so a potential function exists

Answer 14

go and find it

Answer 15

\[\vec{G} = 5xe^{5x^2+2y^2+z^2} i + 2ye^{5x^2+2y^2+z^2}j + ze^{5x^2+2y^2+z^2}k\]

Answer 16

if you stare at it long enough you will see that below potential function works :
\[f(x,y,z) = \frac{1}{2}e^{5x^2+2y^2+z^2}\]

Answer 17

take partials of that function and convince yourself that you get the components of the given vector field

Answer 18

so you take the integral of f(x,y,z)

Answer 19

find the gradient of f(x,y,z)

Answer 20

find the partial derivatives

Answer 21

if you take that gradient you get G

Answer 22

Exactly! we also say G is a gradient field

Answer 23

all below are equivalent :

1) conservative field
2) gradient field
3) path independent field
4) potential function exists for a field
5) curl is 0

Answer 24

if one of them in the list is true, then eveything else in that list is also true

Answer 25

so by taking the partial derivatives of f(x,y,z) you get the answer?

Answer 26

taking partial derivatives convinces you that below is the potential function of given vector field

\[f(x,y,z) = \frac{1}{2}e^{5x^2+2y^2+z^2}\]

Answer 27

so the work done when you go from (3, 2, -2) to (6, 1, 7) equals :
\[f(6,1,7) - f(3,2,-2)\]

Answer 28

evaluate ^

Answer 29

so technically all you need is to find the potential function if the curl is 0

Answer 30

great

Answer 31

Yep because setting up line integral is more painful

Answer 32

finding the potential function is easy

Answer 33

and you can find potential function ONLY when the field is conservative

Answer 34

potential function wont exist if the field is not conservative

Answer 35

so what about part a? is using the FTFLT a good idea?

Answer 36

whats FTFLT ?

Answer 37

fundamental theorem of line integrals

Answer 38

that works only for conservative fields right ?

Answer 39

smooth curves

Answer 40

we have used fundamental theorem of line integral for part b

Answer 41

but it wont apply for part a because the field is not conservative

Answer 42

find the curl and see if you get 0 or not

Answer 43

curl is (0,0,-5)

Answer 44

its non zero

Answer 45

no

Answer 46

that tells us that the vector field in part a is not conservative

Answer 47

so you cant find a potential function

Answer 48

that means fundamental theorem of calc for line integrals is not useful here

Answer 49

is that clear ?

Answer 50

yeah

Answer 51

so what could we use?

Answer 52

parameterize the path as usual

Answer 53

try 11.5

Answer 54

well i dont have any more trys

Answer 55

so for the parametric equation you get x=3+3t, y=2-t, z-2+9t

Answer 56

?

Answer 57

how did you get 11.5? @ganeshie8

Answer 58

evaluate the line integral

Answer 59

x=3+3t, y=2-t, z = -2+9t
dx = 3 dt
dy = -1 dt
dz = 9 dt

Answer 60

\[\int_C \vec{F}\cdot d\vec{r} = \int_C 5ydx + 2dy - dz \]

Answer 61

\[= \int\limits_0^1 5(2-t)(3 dt) + 2(-dt) - 9dt\]

Answer 62

\[= \int\limits_0^1 (5(2-t)(3 ) + 2(-1) - 9)~dt\]

Answer 63

simplify and evaluate the integral ^

Answer 64

http://www.wolframalpha.com/input/?i=%5Cint_0%5E1+%285%282-t%29*3+%2B+2%28-1%29+-9%29+dt

Answer 65

yeah i got this but i have a equation how do you know the limits of integration?

Answer 66

x=3+3t, y=2-t, z = -2+9t

you are starting at (3,2,-2), so
3+3t = 3
2-t = 2
-2+9t = -2

solve t

Answer 67

that gives you lower bound of t

Answer 68

you can find the upper bound similarly

Answer 69

oh ok I understand it now thank you @ganeshie8

Answer 70

when you parameterize a line segment like this by using the direction vector the bounds will be always (0,1)
il let you figure out why..