# ML-inequality, Estimation of Line Integrals

1. Sep 9, 2013

### wifi

Problem:

Let $\vec{F}$ be a vector function defined on a curve C. Let $|\vec{F}|$ be bounded, say, $|\vec{F}| ≤ M$ on C, where $M$ is some positive number. Show that $|\int\limits_C\ \vec{F} \cdot d\vec{r}| ≤ ML$ (L=Length of C).

Attempt at a Solution:

I honestly have no idea where to start with this one. It's not really clear to me exactly what the question is asking me to show.

I understand we have this vector function $\vec{F}=F_1 \hat{i} + F_2 \hat{j} + F_3 \hat{k}$, and that's about it... What does it mean to say that the magnitude $|\vec{F}|$ is "bounded" in such a way that $|\vec{F}| ≤ M$? Are we imposing a restriction here?

Last edited: Sep 9, 2013
2. Sep 9, 2013

### dirk_mec1

Bring the absolute values within the integral then this |F| in the integral is smaller than $$||F||_\infty$$. What remains is a simple line integral.

3. Sep 9, 2013

### wifi

Is this what you're saying?

$|\int\limits_C\ \vec{F} \cdot d\vec{r}| ≤ \int\limits_C\ | \vec{F} | \cdot d\vec{r}$

4. Sep 9, 2013

### wifi

Any help guys?

5. Sep 9, 2013

### wifi

Wouldn't $|\vec{F}|$ be a scalar? In which case taking the dot product is undefined in $\int |\vec{F}| \cdot d\vec{r}$?

6. Sep 9, 2013

yes, the norm of ${\mathbf{F}}$ is a scalar, so your comment about the dot product with $d {\mathbf{r}}$ is correct. If the norm of ${\mathbf{F}}$ is a scalar, what can you do to simplify
$$\int_C {\mathbf{F}} \, d{\mathbf{r}} \text{ ?}$$

and what remains when you do simplify it? (Hint: it is here where you'll use the assumption that $|\mathbf{F}| \le M$.)

7. Sep 9, 2013

### wifi

I'm not really sure how $\int_C {\mathbf{F}} \, d{\mathbf{r}}$ can be simplified. Can you be more specific?

8. Sep 9, 2013

### CAF123

From calculus, you can prove that for smooth $f(x)$ defined on the curve C, (which is in the domain of f(x)) then $$\left|\int f(x)\, dx \right| \leq \int \left| f(x) \right|\,dx$$
Apply this to your problem.

9. Sep 9, 2013

### wifi

I understand that $\left|\int f(x)\, dx \right| \leq \int \left| f(x) \right| dx\,$, I don't see how that can be applied to this line integral.

10. Sep 9, 2013

### CAF123

$|\underline{F}|$ is a scalar. How can you reexpress $\int |\underline{F}|\,d\underline{r}$?

11. Sep 9, 2013

### wifi

In the integral $\int |\vec{F}| \cdot d\vec{r}$, isn't $|\vec{F}| \cdot d\vec{r}$ undefined since the dot product is an operation between two vectors?

12. Sep 9, 2013

### CAF123

It is, but the integral in my last post does not include a dot product.
If $k$ is a scalar, then $\int k f(x) dx = ?$
Once you make this step, use the condition of the boundedness of f.

13. Sep 9, 2013

### wifi

$\int k f(x) dx = k\int f(x)dx$. So I think you're trying to say $\int |\vec{F}| d\vec{r} = |\vec{F}| \int d\vec{r}$. Right?

14. Sep 9, 2013

### CAF123

Right. Now use the boundedness of $|\underline{F}|$ and evaluate $\int_C d \underline{r}$

15. Sep 9, 2013

### wifi

So since we know that $|\vec{F}| \leq M$ and $\int d\vec{r}=L$ , we have $\int |\vec{F}| d\vec{r} = |\vec{F}| \int d\vec{r} \leq M \int d\vec{r} = ML$. Yes?

16. Sep 9, 2013

### CAF123

Yes, you are done.

17. Sep 9, 2013

### wifi

CAF123, you're the best!