Question about line integral of F dot dr

[mesa]

Homework Statement

Evaluate the line integral of F dot dr where f(x,y)=<3x^2,2x+y> and C is a straight line segment from (1,2) to (5,4)

Homework Equations

Unfortunately I was out with family obligations when we covered line integrals and surface integrals so am stuck with the textbook for these so I will just jump to my 'attempt at a solution' since I have limited knowledge on these subjects.

The Attempt at a Solution

First I figured the equation of our line to be y=1/2x-1/2
so if x=t then,
y=1/2t-1/2
so r(t)= ti+(1/2t+3/2)j
and r'(t)= i+1/2j

∫<3t^2i+(5/2t+3/2j>dot<i+1/2j>dt
∫(3t^2+5/4t+6/2)dt

Then we evaluate the integral from 1 to 5. How bad is it?

 

[LCKurtz]

Homework Statement

Evaluate the line integral of F dot dr where f(x,y)=<3x^2,2x+y> and C is a straight line segment from (1,2) to (5,4)

Homework Equations

Unfortunately I was out with family obligations when we covered line integrals and surface integrals so am stuck with the textbook for these so I will just jump to my 'attempt at a solution' since I have limited knowledge on these subjects.

The Attempt at a Solution

First I figured the equation of our line to be y=1/2x-1/2
so if x=t then,
y=1/2t-1/2
so r(t)= ti+(1/2t+3/2)j
and r'(t)= i+1/2j

∫<3t^2i+(5/2t+3/2j>dot<i+1/2j>dt
∫(3t^2+5/4t+6/2)dt

Then we evaluate the integral from 1 to 5. How bad is it?

That should work; I didn't check your numbers. A nicer, and more standard, parameterizaton of the line would be $\vec r(t) = \langle 1,2\rangle + t\langle 4,2\rangle$ for $0\le t\le 1$. I think you will find it simpler. Try it.

 

[mesa]

Ah hah, a surprising result!

If I have an integral of F dot T ds with F(x,y)=<4x,y^2> and C is the arc of a circle of radius 2 centered at the origin then do we just take,
x=2cos t
y=2sin t
so we have the integral of <8cos t,4sin^2 t>dot<-2sin t,2cos t> with the bounds of 0 to pi/2?

 

[mesa]

That should work; I didn't check your numbers. A nicer, and more standard, parameterizaton of the line would be $\vec r(t) = \langle 1,2\rangle + t\langle 4,2\rangle$ for $0\le t\le 1$. I think you will find it simpler. Try it.

I thought I could figure this out, but...
How did you get that parameterization?

 

[LCKurtz]

A nicer, and more standard, parameterizaton of the line would be $\vec r(t) = \langle 1,2\rangle + t\langle 4,2\rangle$ for $0\le t\le 1$. I think you will find it simpler. Try it.

I thought I could figure this out, but...
How did you get that parameterization?

Given points $P_0,~P_1$ the displacement vector from $P_0$ to$P_1$ is $D=P_1-P_0$. Then $R= P_0 +tD$ is at $P_0$ when $t=0$ and $P_1$ when $t=1$.

 

[LCKurtz]

Ah hah, a surprising result!

If I have an integral of F dot T ds with F(x,y)=<4x,y^2> and C is the arc of a circle of radius 2 centered at the origin then do we just take,
x=2cos t
y=2sin t
so we have the integral of <8cos t,4sin^2 t>dot<-2sin t,2cos t> with the bounds of 0 to pi/2?

Sure, if that's the part of the arc in which you are interested. You can use any parametrization and the idea is to use "nice" ones, like that one, for a circle. Note that going from one point to another over different curves may (but might not) give different results.

 

[mesa]

Given points $P_0,~P_1$ the displacement vector from $P_0$ to$P_1$ is $D=P_1-P_0$. Then $R= P_0 +tD$ is at $P_0$ when $t=0$ and $P_1$ when $t=1$.

Okay, so would r(t)=<5,4>+t<-4,-2> also be correct?

 

[mesa]

Sure, if that's the part of the arc in which you are interested. You can use any parametrization and the idea is to use "nice" ones, like that one, for a circle. Note that going from one point to another over different curves may (but might not) give different results.

Sorry, I forgot to mention 'in the first quadrant'...
Nice to see the other part is correct.

 

[mesa]

That should work; I didn't check your numbers. A nicer, and more standard, parameterizaton of the line would be $\vec r(t) = \langle 1,2\rangle + t\langle 4,2\rangle$ for $0\le t\le 1$. I think you will find it simpler. Try it.

Another couple questions, how do we take the derivative of r(t) in this form? Is it just 4i+2j? and then what do we do with our f(x,y)=<3x^2,2x+y> to set it up for the dot product in the integral?

 

[LCKurtz]

Okay, so would r(t)=<5,4>+t<-4,-2> also be correct?

Yes and no. Yes, it's the same line segment, but no, it is in the other direction.

Another couple questions, how do we take the derivative of r(t) in this form? Is it just 4i+2j?

Yes. The derivative of $t$ is $1$ so you just get the direction vector.

and then what do we do with our f(x,y)=<3x^2,2x+y> to set it up for the dot product in the integral?

Remember $r(t) = \langle x,y\rangle$. Your parameterization gives $x$ and $y$ in terms of $t$.

 

[mesa]

Yes and no. Yes, it's the same line segment, but no, it is in the other direction.
Yes. The derivative of $t$ is $1$ so you just get the direction vector.
Remember $r(t) = \langle x,y\rangle$. Your parameterization gives $x$ and $y$ in terms of $t$.

Okay so we would have the integral of <48t^2,10t>dot<4,2> and the new bounds will be from 0 to 1 and then just calculus as usual. Is that right?

 

[LCKurtz]

Okay so we would have the integral of <48t^2,10t>dot<4,2> and the new bounds will be from 0 to 1 and then just calculus as usual. Is that right?

I haven't worked it out. But you could check by comparing the answer against your original calculation of the integral.

 

[mesa]

I haven't worked it out. But you could check by comparing the answer against your original calculation of the integral.

Very good, although I will rejoice when the semester is finished and this material can be covered properly (hopefully my copy of Anton will resurface soon as well, Stewart is a pain).

The only other question I have is if we have multiple 'segments' then we simply evaluate each one with their respective bounds and add the up, correct?

 

[LCKurtz]

Yes.

 

[mesa]

Thank you LCKurtz, that should do it.