Second Example of Line Integral of Conservative Vector Field Using path independence of a conservative vector field to solve a line integral
Second Example of Line Integral of Conservative Vector Field
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- Let's do another problem.
- Very similar to the last one, but with a subtle difference.
- And that subtle difference will make a big difference.
- Let's say we take the line integral over some curve c--
- I'll define the curve in a second-- of x squared plus y
- squared dx plus 2xy dy-- and this might look very familiar.
- This was very similar to what we saw last time, except last
- time we had a closed line integral.
- This is not a closed line integral.
- And our curve, c, the parameterization is x is
- equal to cosine of t, y is equal to sine of t.
- So far-- it looks like sit.
- Let me write sine of t-- so far, it looks very similar to
- the closed line integral example we did in the last
- video, but instead of t going from 0 to 2 pi, we're going
- to have t go from 0 to pi.
- t is greater than or equal to 0, is less than or equal to pi.
- So now we're essentially, our path-- if I were to draw it on
- the x-y plane-- so that is my y-axis, that is my x-axis.
- So now our path isn't all the way around the unit circle.
- Our path-- our curve c now-- just starts at t is equal to 0.
- You can imagine t is almost the angle.
- t is equal to 0, and we're going to go all the way to pi.
- So that's what our path is right now, in this example.
- So it's not a curved path.
- It's not a closed path.
- So we can't just show that f, in this example-- and we're
- going to re-look at what f looks like-- hey, if that's a
- conservative vector field, if it's a closed loop that equals
- 0, this isn't a closed loop.
- So we can't apply that.
- But let's see if we can apply some of our other tools.
- So like we saw in the last video, this might look a
- little bit foreign to you.
- But if you say that f is equal to that times i, plus that
- times j, then it might look a little bit more familiar.
- If we say that f of xy-- the vector field f is equal to x
- squared plus y squared times i plus 2xy times j and dr-- I
- don't even have to look at this right now. dr, you can always
- write it as dx times i plus dy times j.
- You'll immediately see, if you take the dot product of these 2
- things, if you take f dot dr-- they're both vector valued,
- vector valued differential, vector valued field, or vector
- valued function-- if you take f dot dr, you'll get
- this right here.
- You'll get what we have inside of the interval.
- You'll get that right there, right?
- That times that-- you take the product of the i terms--
- that times that is equal to that, and add it to the
- product of the j terms.
- 2xy times dy.
- Write like that.
- So our integral, we can rewrite it as this.
- Along this curve of f dot dr, where this is our f.
- Now, we still might want to ask ourselves, is this
- a conservative field?
- Or does it have a potential?
- Is f equal to the gradient of some function, capital F?
- I guess I could write the gradient like that, because
- it creates a vector.
- This is a vector, too.
- Is this true?
- And we saw in the last video, it is.
- I'll redo it a little bit fast this time.
- Because if this is true, we can't say this is a closed
- loop and say, oh, it's just going to be equal to 0.
- But if this is true, then we know that this-- that the
- integral is path independent.
- And we'll know that this is going to be equal to capital F,
- if we say that t is going from-- well, in this case t is
- going from 0 to pi-- we could say that this is going to be
- equal to capital F of pi minus capital F of 0.
- Or if we want to write it in terms of x and y-- because f is
- going to be a function of x and y-- we could write-- and this
- right here, these are t's.
- We could also write that this is equal to f of x of pi, y of
- pi, minus f of x of 0, y of 0.
- That's what I mean when I say f of pi.
- If we were to write f purely as a function of t.
- But we know that this capital F is going to be a function.
- It's a scalar function defined on xy.
- So we could say f of x of pi, y of pi.
- These are the t's now.
- These are all equivalent things.
- So if it is path dependent, we can find our f.
- We can just evaluate this thing by just taking our
- f, evaluating it at these two points.
- At this point, and at that point right there.
- Because it would be path independent.
- If this is a conservative, if this has a potential function,
- if this is the gradient of another scalar field, then this
- is a conservative vector field, and its line integral
- is path independent.
- It's only dependent on that point and that point.
- So let's see if we can find our f.
- So I'm going to do exactly what we did in the last video.
- If you watch that last video, it might be a
- little bit monotonous.
- But I'll do it a little bit faster here.
- So we know that the partial of f with respect to x is going to
- have to be equal to this right here.
- So that's x squared plus y squared.
- Which tells us, if we take the antiderivative, with respect to
- x, then f of xy is going to have to be equal to x to the
- third over 3 plus xy squared-- right? y squared is just a
- constant in terms of x-- plus f of y.
- There might be some function of y that, when you take
- the partial with respect to x, it just disappears.
- And then we know that the partial of f with respect to
- y has got to be equal to that thing or that thing.
- We're saying that this is the gradient of f.
- So this has to be the partial with respect to y.
- And you might want to watch the other video.
- I go through this just a little bit slower in that one.
- So the antiderivative of this with respect to y-- so we get f
- of xy-- would be equal to xy squared plus some
- function of x.
- Now we did this in the last video.
- These 2 things have to be the same thing, in order for the
- gradient of capital F to be lowercase f.
- And we have xy squared, xy squared.
- We have a function of x, we have a function purely of x.
- And then we don't have a function purely of y here,
- so this thing right here must be 0.
- So we've solved.
- Our capital F of xy must be equal to x to the 3
- over 3 plus xy squared.
- So we know that lowercase f is definitely conservative.
- It is path independent.
- It has its potential.
- It is the gradient of this thing right here.
- And so to solve our integral-- this was a 0-- to solve our
- integral, we just have to figure out x of pi, y
- of pi, x of 0, y of 0.
- Evaluate the bullet points, and then subtract the 2.
- So let's do that.
- So x was cosine of t, y is sine of t.
- Let me rewrite it down here.
- So x is equal to cosine of t.
- y is equal to sine of t.
- So x of 0 is equal to cosine of 0, which is equal to 1.
- x of pi is equal to cosine of pi, which is equal to minus 1.
- y of 0 is sine of 0, which is 0.
- y of pi, which is equal to sine of pi, which is equal to 0.
- So f of x of pi, y of pi-- this is the same thing,
- so let me rewrite this.
- Our integral is simplified to-- our integral along that path of
- f dot dr-- is going to be equal to capital F of x of pi.
- x of pi is minus 1.
- y of pi is equal to 0.
- Minus capital F of x of 0 is 1, comma y of 0 is 0.
- And so what is this equal to?
- Just remember, this right here is the same thing
- is that right there.
- That is x of pi.
- That is y of pi.
- That term right there.
- You can imagine this whole f of minus 1, 0-- that's the same
- thing as f of pi, if you think in terms of just t.
- That could be a little confusing, so I want
- to make that clear.
- So this is just straightforward to evaluate.
- What is f of minus 1, 0?
- x is minus 1. y is 0.
- So it's going to be minus 1 to the third power-- right,
- that's our x-- over 3.
- So it's minus 1/3.
- It's going to be minus 1/3 plus minus 1 times 0 squared.
- So that's just going to be a 0.
- In both cases, the y is 0.
- So this term is going to disappear.
- So we can ignore that.
- And then we have minus f of 1, comma 0.
- We put a 1 here.
- 1 to the third over 3.
- That is 1/3 plus 1 times 0 squared.
- That's just 0.
- So this is going to be equal to minus 1/3.
- Minus 1/3 is equal to minus 2/3.
- And we're done.
- And once again, because this is a conservative vector field,
- and it's path independent, we really didn't have to mess with
- the cosine of t's and sines of t's when we actually took
- our antiderivative.
- We just have to find the potential function and evaluate
- it at the 2 end points to get the answer of our integral, of
- our line integral, minus 2/3.
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