Question

Consider the 2-dimensional force field F = (4e^(-2x)+3y^3) i +9xy^2 j

a) is F conservative? find a potential function f(x,y) whose gradient is F.

b) find the work done by the force field F in moving an object from P(0,1) to Q(1,2) along the path y= sin( x pi/2) from x= 0 to x= 1.

Answer 1

Do you know what it means for F to be conservative?

Answer 2

ok I got part a) F is conservative because \[\frac{ dP }{ dy } = \frac{ dQ }{ dx } = 9y^2\]

Answer 3

yeah, but I don't know what to do in part b)

Answer 4

the potential function is \[3xy^3 -2e ^{-2x} \]

Answer 5

right? @uri

Answer 6

So the idea of F being conservative really means that the work done in a conservative field is the same regardless of the path taken.

Answer 7

So, we can take a simpler path if we want to.

Answer 8

Let me check

Answer 9

so we don't have to use \[y = 1+\sin (\frac{ \pi x }{ 2 }) \]?

Answer 10

Py=Qx => conservative is good.
For part b, you need to find f such that
\(\Delta f=F \)
Can you do that?

Answer 11

Sorry, it's not Delta f, it's grad. the triangle should be upside down.
\(<f_x(x,y),f_y(x,y)> = F(x,y)\)

Answer 12

I'm not sure @mathgirl2012 from part a we know that \[Δf=F\]

Answer 13

You're right @Ldaniel

Answer 14

so what's \[f\] the path given in part b? @mathmate

Answer 15

Correct. We could use y = x+1 if we want to because it is simpler and the field is conservative

Answer 16

Sorry, this is the second part of (a), not (b).
This is what you need to find, from the given P and Q of F, and f has to satisfy the conditions I gave.

Answer 17

The path given in part b is one possible path, but since F is conservative then the path taken is irrelevant. The Force is the same (thus the answer to the question) if you use the path y = 1 +sin(pix/2) or y = 1+x.

Answer 18

If the field is conservative, work done is just f(b)-f(a).

Answer 19

Like @uri said, independent of path.

Answer 20

But you need to find f(x,y) first.

Answer 21

ok so what f(x,y) are we talking about here?

Answer 22

so Q(1,2) - P(0,1) using y = 1+x?

Answer 23

Not really, first find f(x,y) from P, Q that satsfy
\(<f_x(x,y),f_y(x,y)> = F(x,y)=<P(x,y),Q(x,y)>\)

Answer 24

Then if we need work done from a to b, then work = f(b)-f(a).

Answer 25

what surface do i use to find \[fx(x,y),fy(x,y)\]

Answer 26

@Ldaniel Your proposed potential function is close. Can you check if it satisfies the above condition? If not, then you can make adjustments.

Answer 27

ok

Answer 28

ok so for \[f_x (x,y) = 3y^3 +4 e ^{-2x}\]

Answer 29

and \[f_(x,y) = 9y^2x\]

Answer 30

Can you check if
\(F = (4e^{2x}+3y^3)i +9xy^2 j \)
or
\(F = (4e^{-2x}+3y^3)i +9xy^2j \)

Answer 31

i'm getting the second one

Answer 32

But is the one given in the question the correct one?

Answer 33

I edit the question it forgot the negative sign sorry

Answer 34

If the first one that you need, then you need to make a slight adjustment to f(x,y).

Answer 35

Ok, so your potential function is correct, because if you derive wrt x, you get P, and wrt y, you get Q, right?

Answer 36

yeah so not what do i do?

Answer 37

Almost done!

Answer 38

Work done from A(0,1) to B(1,2).
I want to avoid P,Q which are confusing.

Answer 39

Can you find that? It is independent of path, as @uri said.

Answer 40

to what fuction do i plot those points, to the path?

Answer 41

W=f(B)-f(A)

Answer 42

I use the potential function?

Answer 43

yes.
\( \int gradf.dr = f(r(b)-f(r(a)) \)
That is why is so much easier working with a conservative field.

Answer 44

i'm getting 25.729... does that look right?

Answer 45

I don't usually evaluate powers of e. Leave them in your answer, see what you get.

Answer 46

(24-2e^(-2))- (-2)

Answer 47

26 - 2e^(-2)

Answer 48

Yes, that's what I got too! Good job!

Answer 49

thank you so much for your help :)

Answer 50

You're welcome! :)
Just for curiosity, how did you get f(x,y) at the very beginning?

Answer 51

integration

Answer 52

Good, integrate P wrt x?

Answer 53

\[\int\limits\limits4e ^{-2x}+ 3y^3 dx\]

Answer 54

yeah, well I did both P and Q

Answer 55

Did you differentiate with respect to y to make sure you get Q?
ok that's good!

Answer 56

yeah

Answer 57

Very well! congrats!

Answer 58

and i got 3xy^3

Answer 59

Yes, that is Q, so you made sure the condition grad f = F is satisfied.

Answer 60

so thats why I know that integrate P wrt x was the potential function

Answer 61

Perfect, so the only (easy) part missing was f(B)-f(A).

Answer 62

Which you just did.

Answer 63

yeah I was confuse because of the given path

Answer 64

I didn't know what to do with it, but now i know to do nothing haha

Answer 65

The path was necessary IF the field was not potential. That could be confusing sometimes.

Answer 66

yeah

Answer 67

I feel a little guilty because I seem to have pushed @uri away. Sorry about that.

Answer 68

Anyway, good luck with your other ventures!

Answer 69

thanks :)

Answer 70

lol,It's cool. :)

Answer 71

You're welcome! :)

Answer 72

@uri you're really nice!

Answer 73

thank you @uri :)