Magnitude of vector, negative in part.

AI Thread Summary
To find the magnitude of the acceleration vector given as [ bk^{2}e^{kt} - bc^{2}e^{kt} ]e_{r} + [ 2bkce^{kt} ] e_{θ}, the correct approach involves squaring each component and summing them. The initial confusion stemmed from neglecting the e_{θ} component and miscalculating the exponential term as e^{2kt} instead of e^{kt}. Once the full expression is considered, the magnitude can be calculated without negative terms complicating the result. Additionally, using proper formatting for equations can improve clarity in future discussions. The issue was resolved through collaboration, highlighting the value of seeking different perspectives.
feathermoon
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This is actually not a full problem, just a part of one I'm having trouble with:

Homework Statement



If I have a acceleration vector, say [ bk^{2}e^{kt} - bc^{2}e^{kt} ]e_{r} + [ 2bkce^{kt} ] e_{θ}. How can I find its magnitude?

Homework Equations



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

The Attempt at a Solution



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

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

I'm either doing some wrong algebra or missing something obvious I think?
 
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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.
 
Glad it worked out. FYI, I have two comments to help you in the future:

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

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

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