# Is This Line Integral Differentiable?

• Contingency
In summary: I think I understand now. Just wanted to be sure. Thank you very much for your time and patience!No problem, glad I could help!
Contingency

## Homework Statement

$\vec { F } \left( x,y \right) =u\left( x,y \right) \hat { i } +v\left( x,y \right) \hat { j }$
$u\left( x,y \right) , v\left( x,y \right)$ are continuous on ℝ²
$\Gamma$ is piecewise smooth.
Is $\psi (x,y){ =\int { \vec { F } \left( x,y \right) \cdot \vec { dr } } }$ differentiable? How can I show this if it is?
ψ is meant to be the line integral over Γ.

## Homework Equations

Definition of differentiability; definition of partial derivatives; formula for calculating line integrals under a parametrization.

## The Attempt at a Solution

I wanted to show that the partial derivatives of ψ are continuous, which would imply differentiability. I can't deal with ψ as it's defined, only with the formula, and I think differentiability must be independent of some arbitrary parametrization..

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Contingency said:

## Homework Statement

$\vec { F } \left( x,y \right) =u\left( x,y \right) \hat { i } +v\left( x,y \right) \hat { j }$
$u\left( x,y \right) , v\left( x,y \right)$ are continuous on ℝ²
$\Gamma$ is piecewise smooth.
Is $\psi (x,y){ =\int { \vec { F } \left( x,y \right) \cdot \vec { dr } } }$ differentiable? How can I show this if it is?
ψ is meant to be the line integral over Γ.

## Homework Equations

Definition of differentiability; definition of partial derivatives; formula for calculating line integrals under a parametrization.

## The Attempt at a Solution

I wanted to show that the partial derivatives of ψ are continuous, which would imply differentiability. I can't deal with ψ as it's defined, only with the formula, and I think differentiability must be independent of some arbitrary parametrization..

So what happens if you parametrize your curve? Remember that it's piecewise smooth, so you might have to break it up a bit. Plug in what you know and remember to choose the interval.

Think about the continuity of u and v.

Could you be a bit more specific?
Parametrizing my curve allows me to use the Line Integral formula, which is an integral of 't' alone (t being a parameter). I still can't see anything as far as proving continuity of partial derivatives goes..

In the original problem, I was to find a formula for ∂ψ/∂x for a given Γ. When I tried to solve it, I did exactly what you said - plugged in all I know etc - but from what I noticed, I only used continuity to guarantee integrability.. The formula I arrived at required ψ be differentiable (used the chain rule).

Contingency said:
Could you be a bit more specific?
Parametrizing my curve allows me to use the Line Integral formula, which is an integral of 't' alone (t being a parameter). I still can't see anything as far as proving continuity of partial derivatives goes..

In the original problem, I was to find a formula for ∂ψ/∂x for a given Γ. When I tried to solve it, I did exactly what you said - plugged in all I know etc - but from what I noticed, I only used continuity to guarantee integrability.. The formula I arrived at required ψ be differentiable (used the chain rule).

Could you show me your work?

Contingency said:
Could you be a bit more specific?
Parametrizing my curve allows me to use the Line Integral formula, which is an integral of 't' alone (t being a parameter). I still can't see anything as far as proving continuity of partial derivatives goes..

In the original problem, I was to find a formula for ∂ψ/∂x for a given Γ. When I tried to solve it, I did exactly what you said - plugged in all I know etc - but from what I noticed, I only used continuity to guarantee integrability.. The formula I arrived at required ψ be differentiable (used the chain rule).

If you think about the integral dt, you are integrating a piecewise continuous function. Is the result of that necessarily differentiable? Try to think of a counterexample.

Dick said:
If you think about the integral dt, you are integrating a piecewise continuous function. Is the result of that necessarily differentiable? Try to think of a counterexample.
I don't see why my integrand is piecewise continuous.. U, V are everywhere continuous..

My problem is ψ..

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Contingency said:
I don't see why my integrand is piecewise continuous.. U, V are everywhere continuous..

My problem is ψ..

$d \vec r(t)=\vec r'(t) dt$. If you are only given that r(t) is piecewise smooth then r'(t) is only piecewise continuous. Think of the one dimensional example r(t)=|t|. I'd call that piecewise smooth.

I understand that bit now.
Here's my attempt at a solution.
Hopefully you will be able to understand where my question comes from and help me.

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Contingency said:
I understand that bit now.
Here's my attempt at a solution.
Hopefully you will be able to understand where my question comes from and help me.

Yeah, that helps. I was picturing how you were differentiating the psi wrong. But I think the whole 't' thing is leading you astray. H1 and H2 aren't functions of t. t is a dummy variable. When you've got something like $\int_0^1 f(xt,y) d(xt)$ change the variable to x'=xt. Now you've got $\int_0^x f(x',y) d(x')$. What's the x derivative of that?

I don't understand a few things:
First, I don't see d(xt) anywhere, only dt. Where did the integral with d(xt) come from?
Second, can you please explain how is it that I obtain an integral with a variable limit from an integral with a constant limit? Also the bit with x and x', what are they exactly?

As far as what you asked, the derivative of that integral is f(x', t) because of the fundamental theorem.

Contingency said:
I don't understand a few things:
First, I don't see d(xt) anywhere, only dt. Where did the integral with d(xt) come from?
Second, can you please explain how is it that I obtain an integral with a variable limit from an integral with a constant limit? Also the bit with x and x', what are they exactly?

As far as what you asked, the derivative of that integral is f(x', t) because of the fundamental theorem.

Nope, the derivative can't be f(x',t). x' is a dummy variable of integration. So is t. It's f(x,y). In my example x is the constant. x'=xt. x' is just a change of variables from t. In your formula for H2(t) you have x0*v(x0*t,y) dt. That's the same as v(x0*t,y) d(x0*t) since x0 is constant. You really want to change variables to x'=x0*t.

Okay, gotcha.
So your x is my x₀, and I understand the why the derivative is f(x,y).
I'm still unsure of where the d(xt) came from. I have xU(xt,y)d(t) in my integral. Can you move the x inside the parenthesis because it's a constant? "d" is like a linear operator?

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Contingency said:
Okay, gotcha.
So your x is my x₀, and I understand the why the derivative is f(x,y).
I'm still unsure of where the d(xt) came from. I have xU(xt,y)d(t) in my integral which translates into xd(t). Can you move the x inside the parenthesis because it's a constant? "d" is like a linear operator?

Exactly. You can move x into the d part because it's a constant. d(xt)=xdt. Suggests you do the change of variables x'=xt. So dx'=xdt.

You said that in your example, x is a constant.
Doesn't that mean that $\int _{ 0 }^{ x }{ f\left( x',y \right) dx' }$ is just a number? Why can we suddenly take the derivative with respect to x which is a constant?

I think I have a fundamental problem of understanding here..

Contingency said:
You said that in your example, x is a constant.
Doesn't that mean that $\int _{ 0 }^{ x }{ f\left( x',y \right) dx' }$ is just a number? Why can we suddenly take the derivative with respect to x which is a constant?

I think I have a fundamental problem of understanding here..

x is a constant with respect to the dt or dx' integration. That's all that I mean by 'constant'.

I think I'm not being clear enough, so I will write down on paper what's bothering me in each step of the problem. Right now I have a mess in my head.
Thanks a lot for bearing with me!

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Figured everything out. Thanks a lot for all your help!

## 1. What is a line integral?

A line integral is a type of integral that is computed along a curve or a line in a vector field. It is used to calculate the work done by a force along a path or to calculate the amount of fluid flowing through a curve.

## 2. How is a line integral differentiable?

A line integral is differentiable if the integrand is continuous and the curve is smooth. This means that the curve has a well-defined tangent vector at every point and the integrand is defined at every point along the curve.

## 3. Why is it important to know if a line integral is differentiable?

Knowing if a line integral is differentiable is important because it determines if the integral can be calculated using the Fundamental Theorem of Calculus. If the integral is differentiable, it can be evaluated using the derivative of the integrand.

## 4. What happens if a line integral is not differentiable?

If a line integral is not differentiable, it means that the integrand is not continuous or the curve is not smooth. In this case, the integral cannot be evaluated using the Fundamental Theorem of Calculus. Alternative methods, such as approximations or numerical integration, may be used to calculate the integral.

## 5. Can a line integral be differentiable at some points and not others?

Yes, it is possible for a line integral to be differentiable at some points and not others. This can happen if the integrand is continuous but the curve has sharp turns or corners, which makes it not smooth at those points. In this case, the integral can still be evaluated using the Fundamental Theorem of Calculus at the points where it is differentiable.

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