Question

Line Integral/Conservative Vector Fields:Find the value of the line integral F*dr(Hint: If F is conservative, the integration may be easier on an alternative path. )

F(x,y)=<2xy,x^2>
(a) r1(t)= from 0 from 0 14 years ago

Answer 1

Here's the problem retyped
\[F(x,y)=<2xy,x^2>\]
\[r _{1}(t)=<t,t^2>\] from \[0<t<1\]
\[r _{2}(t)=<t,t^3>\] from 0<t<1

Answer 2

the "<" sign should be less than or equal to

Answer 3

I know that \[r \prime(t)=<1,2t>\]

Answer 4

^ This is for r1

Answer 5

I'm unsure of how to get F(t) though. My book says its \[F(t)=<2t^3,t^2>\]

Answer 6

Line integral Formula \[\int\limits_{}^{}F*dr\] (retyped this formula that was in my question)

Answer 7

Maybe you should use the hint.

Answer 8

http://en.wikipedia.org/wiki/Conservative_vector_field#Path_independence

Answer 9

F is conservative \[\frac {dM} {dy} = \frac {dN} {dx}=2x\]

Answer 10

F is conservative, so there is a potential: \[F=\nabla\phi\]
So you're looking for phi(x,y) such that dphi/dx= 2xy and dphi/dy=x^2

Answer 11

Can you compute phi?

Answer 12

I think so: \[\Delta \phi = <2x,2x> = v\]

Answer 13

Do you understand the notation? phi is a scalar function. \[\nabla \phi\]is the gradient of phi. so \[\nabla \phi = \left(\begin{matrix}\frac{d\phi}{dx} \\ \frac{d \phi}{dy}\end{matrix}\right)\]

Answer 14

\[\Delta\]is different from \[\nabla\]

Answer 15

^ I couldn't figure out how to make that sign on here, I didn't see an option for it.

Answer 16

\nabla

Answer 17

^ okay thanks.

Answer 18

With a good guess you can find phi pretty easily here. Do you see that \[\phi = x^2y\]

Answer 19

yes

Answer 20

and taking j's partial derivative of course would be 0 so

\[f(x,y)= x^2y\]

Answer 21

Ah wait, I meant the integral of j would be 2xy

Answer 22

What is j, and what is f?

Answer 23

we'll, we know that if the curl of that is 0, then it'll be conservative...
so we compute for
dQ/dx-dP/dy, where F=<P,Q>=<2xy,x^2> and if this is 0 then it'll be conservative..
so dQ/dx=d (x^2)/dx=2x
dP/dy=d (2xy)/dy=2x
so
dQ/dx-dP/dy=2x-2x=0.. now we have a conservative vector field. now we find the potential function f for F, such that ∇f=F=<2xy,x^2>
so we know that

\[\int\limits_{}^{}Pdx=\int\limits 2xy dx=f\]
\[\int\limits Q dy =\int\limits x^2 dy=f\]
I'll first use the 1st equation so:
\[f=\int\limits 2xy dx=x^2y+h(y)\]

I've put h(y), since in this case, since we're integrating with respect to dx, all other expressions with only y's other constants will be considered constants of integration. Now we need to find h(y)..
so since df/dy=Q=x^2 , and df/dy=d(x^2y+h(y))/dy=x^2+h'(y), so
x^2+h'(y)=x^2
h'(y)=0
integrating h'(y), we'll have h'(y)=0+c=c, so f becomes
f=x^2y+c...
now why did we want f? recall that
\[\int\limits_{t=a}^{t=b}∇f.rds=f(r(b))-f(r(a))\]
where r is a vector
recall also that if
F is a conservative vector field, then there is a function f such that ∇f=F and in this case, our f=x^2y+c. and so we'll have
a.)
\[\int\limits_{C}^{}F.dr=\int\limits_{t=a}^{t=b}∇f.dr=\int\limits_{0}^{1}∇(x^2y+c).dr=(t^2t^2)|_{0}^{1}=1 \]
b.)
\[\int\limits\limits_{C}^{}F.dr=\int\limits\limits_{t=a}^{t=b}∇f.dr=\int\limits\limits_{0}^{1}∇(x^2y+c).dr=(t^2t^3)|_{0}^{1}=1\]

Answer 24

Sorry my book has a different notation from wikipedia, allow me to make some changes:

\[v= \nabla \phi \] is \[F = \nabla f\] in my book
and from what you were doing I was assuming that you were trying tofind the potential function for F (or V according to wikipedia) so \[\nabla f(x,y)\] would be \[\nabla v(x,y)\]

Answer 25

^ that was to Thomas

Answer 26

ok, so \[f=x^2y\]

Answer 27

Now you can use the formula \[\int_{r_1} F dr\]=f(endpoint)-f(starting point)

Answer 28

Is it clear so far?

Answer 29

anonymous, where did the two t^2's come from?

Answer 30

the path is (t,t^2), so when you substitute that into f you'll get t^2t^2.

Answer 31

exactly @thomas9

Answer 32

oh okay I see now

Answer 33

I get everything that you are saying but I'm still confused as to how they are getting this for F:
\[r \prime (t) = <t,2t> \]
\[F(t)=<2t^3,t^2>\]
\[\int\limits_{C}^{}F*dr=\int\limits_{0}^{1}=4t^3dt = 1\]

Answer 34

^ Not the integration part just F(t)

Answer 35

Oh wait, I think I figured that out

Answer 36

I think they just substituted t and t^2 int F(t)
\[F(t,t^2)=2t^2+t^2\]

Answer 37

I don't understand your last line, but yes they substituted t for x and t^2 for y in F. Of course that's a completely different approach than what we did with the potential.

Answer 38

That was from my book, I think the reason why its different was because they took a different approach. even though there's seems shorter, I understood the other method. I think I can solve the second part now. Thanks.

Answer 39

@anonymoustwo44
I've put h(y), since in this case, since we're integrating with respect to dx, all other expressions with only y's other constants will be considered constants of integration. Now we need to find h(y)..
so since \[\frac{df}{dy}=Q=x^2\]
and \[\frac{df}{dy}=\frac{d(x^2y+h(y))}{dy=x^2+h^ \prime (y)}\]
so
\[x^2+h^ \prime (y) = x^2\]
\[h^\prime(y)=0\]
integrating h'(y), we'll have h'(y)=0+c=c, so f becomes
f=x^2y+c...
now why did we want f? recall that

^ I had to rewrite what you put down because it was confusing to read at first.

Answer 40

@anonymoustwo44

Its been a while since I've looked at the problem could you explain to me again how you are getting this part:



dfdy=d(x2y+h(y))dy=x2+h′(y)

Answer 41

ah ok. so I got a f(x,y) from integrating 2xy with respect to dx right? so we got f(x,y)=x^2y+h(y)
so how do I find h(y) now. Notice that if I differentiate this with respect to y, it will be equal to df/dy=Q=x^2
so d(x^2y+h(y))/dy=x^2+h'(y)
now I also know that df/dy=Q=x^2
so
x^2+h'(y)=x^2
h'(y)=0
h(y)=c
so f(x,y) becomes:
x^2y+c<answer

Answer 42

Ok thanks for the clarification.