• Ac3

Simplify the following expression.

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  • Ac3

Simplify the following expression.

Mathematics
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  • Ac3
\[\sqrt{(7xcosx)^2+(-7xsinx+14cosx)^2}\]
  • Ac3
the farthest i got to was\[\sqrt{49x^2-196sinxcosx+196\cos^2x}\]
  • Ac3
We want the radical to go away if possible or get the least amount of terms under the radical at least.

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Other answers:

  • Ac3
Any idea @Preetha ?
if you are lucky it is a prefect square
doesn't look like one though, maybe some other trick
  • Ac3
I factored out a 49 and pulled it out of the radical. I was thinking maybe a trig identity for cos^2x not sure though.
was that thing really the initial problem?
  • Ac3
Well if you want the initial problem I'll go ahead and give it to you.
the 49 comes out for sure not sure what else you can do here
graph if cool http://www.wolframalpha.com/input/?i=sqrt%28%287xcos%28x%29%29^2%2B%28-7xsin%28x%29%2B14cos%28x%29%29^2%29
*is
  • Ac3
r(t)=(7tsint+7cost)i + 7j + (7tcost-7sint)k, Find the Tangential, Normal, Binormal, and the Torsion.
  • Ac3
I think that's as far as it's going to go if I pull out the 49.
tangential is just ... \[|r'(t)| \text{ }\] \[r'(t)=[7 \sin(t)+7t \cos(t)-7\sin(t)]i+[0]j+[7 \cos(t)-7 t \sin(t)-7 \cos(t)]k \\ r'(t)=7 t \cos(t) i -7t \sin(t) k\]
\[|r'(t)|=\sqrt{(7 t \cos(t))^2+(-7 t \sin(t))^2}\]
this looks prettier
  • Ac3
wait hold on let me check your derivative because I checked mine like 4 times.
it looks like your k component was different
yeah seems like you wrote 7cos(t)-7 cos(t) as 7 cos(t)+7 cos(t) instead
  • Ac3
ahhhhhhhh i screwed up the last part!!!!!
  • Ac3
thanks freckles.
np
  • Ac3
derivative of -7sint = 7cost <--- lmfao i'm an idiot.
maybe for a sec you got antiderivative and derivative mixed up :p it happens to all of us especially on those crazy trig functions
  • Ac3
It happens to me all the time.
  • Ac3
I've been doing way too many integrals
anyways glad i can help
  • Ac3
see ya man thx again!
peace

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