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ML-inequality, Estimation of Line Integrals

  1. Sep 9, 2013 #1
    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. jcsd
  3. Sep 9, 2013 #2
    Bring the absolute values within the integral then this |F| in the integral is smaller than [tex]||F||_\infty[/tex]. What remains is a simple line integral.
     
  4. Sep 9, 2013 #3
    Is this what you're saying?

    ##|\int\limits_C\ \vec{F} \cdot d\vec{r}| ≤ \int\limits_C\ | \vec{F} | \cdot d\vec{r} ##
     
  5. Sep 9, 2013 #4
    Any help guys?
     
  6. Sep 9, 2013 #5
    Wouldn't ## |\vec{F}|## be a scalar? In which case taking the dot product is undefined in ##\int |\vec{F}| \cdot d\vec{r} ##?
     
  7. Sep 9, 2013 #6

    statdad

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    Homework Helper

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

    and what remains when you do simplify it? (Hint: it is here where you'll use the assumption that [itex] |\mathbf{F}| \le M [/itex].)
     
  8. Sep 9, 2013 #7
    I'm not really sure how ## \int_C {\mathbf{F}} \, d{\mathbf{r}} ## can be simplified. Can you be more specific?
     
  9. Sep 9, 2013 #8

    CAF123

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    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.
     
  10. Sep 9, 2013 #9
    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.
     
  11. Sep 9, 2013 #10

    CAF123

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    ##|\underline{F}|## is a scalar. How can you reexpress ##\int |\underline{F}|\,d\underline{r}##?
     
  12. Sep 9, 2013 #11
    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?
     
  13. Sep 9, 2013 #12

    CAF123

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    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.
     
  14. Sep 9, 2013 #13
    ## \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?
     
  15. Sep 9, 2013 #14

    CAF123

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    Right. Now use the boundedness of ##|\underline{F}|## and evaluate ##\int_C d \underline{r}##
     
  16. Sep 9, 2013 #15
    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?
     
  17. Sep 9, 2013 #16

    CAF123

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    Yes, you are done.
     
  18. Sep 9, 2013 #17
    CAF123, you're the best!
     
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