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Homework Help: Magnitude of vector, negative in part.

  1. Dec 10, 2011 #1
    This is actually not a full problem, just a part of one I'm having trouble with:

    1. The problem statement, all variables and given/known data

    If I have a acceleration vector, say [ b[itex]k^{2}[/itex][itex]e^{kt}[/itex] - b[itex]c^{2}[/itex][itex]e^{kt}[/itex] ][itex]e_{r}[/itex] + [ 2bkc[itex]e^{kt}[/itex] ] [itex]e_{θ}[/itex]. How can I find its magnitude?

    2. Relevant equations

    Mag vector |a| = (a^2)^(1/2)

    3. The attempt at a solution

    As I square [itex]_{e}r[/itex], the cross term is still negative under the radical, and doesn't subtract cleanly:

    [ [itex]b^{2}k^{4}e^{kt}[/itex] + [itex]b^{2}c^{4}e^{kt}[/itex] - 2[itex]b^{2}k^{2}c^{2}e^{kt}[/itex] ][itex]^{1/2}[/itex]

    I'm either doing some wrong algebra or missing something obvious I think?
     
    Last edited: Dec 10, 2011
  2. jcsd
  3. Dec 10, 2011 #2
    Well I'm stupid.. neglected e_theta completely out of sheer confusion from the first part... guess what it does when you add it in.

    ...Maybe this thread should be deleted. U_U Sometimes I just need a new perspective I guess.
     
  4. Dec 10, 2011 #3

    Redbelly98

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    Glad it worked out. FYI, I have two comments to help you in the future:

    1. Note that [itex](e^{kt})^2 = e^{2kt}, \text{ not } e^{kt}[/itex]

    2. You can enclose an entire equation or large expression in itex-/itex tags, you do not need to use separate itex-/itex tags for selected terms.

    So instead of

    [ [itex]bk^{2}[/itex][itex]e^{kt}[/itex] - b[itex]c^{2}[/itex][itex]e^{kt}[/itex] ][itex]e_{r}[/itex] + [ 2bkc[itex]e^{kt}[/itex] ] [itex]e_{θ}[/itex]

    you can write

    [itex] [ bk^{2}e^{kt} - bc^{2}e^{kt} ] e_{r} + [ 2bkce^{kt} ] e_{θ} [/itex]

    which gives you

    [itex] [ bk^{2}e^{kt} - bc^{2}e^{kt} ] e_{r} + [ 2bkce^{kt} ] e_{θ} [/itex]
     
  5. Dec 10, 2011 #4
    Thanks Redbelly! I did have e^kt as e^2kt in my calculation, I just forgot to transcribe that correctly.

    The itex thing on the other hand will probably save me some time on my next questions! :D
     
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