anonymous
  • anonymous
Calc II question attached:
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
1 Attachment
campbell_st
  • campbell_st
sub x = 4 so you get \[g(4) = 4 \times \sin^-1 (4/8) + \sqrt{64 - 4^2}\] \[\sin^(-1) (1/2) = \pi/6\]
campbell_st
  • campbell_st
i think the answer is \[2\pi/3 + 4\sqrt{?}\]

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campbell_st
  • campbell_st
oops ? = 3
Mertsj
  • Mertsj
Yes. That's what i got and it makes no sense in light of the answer choices. So I'm thinking that not only do I not comprehend the solution, I don't even comprehend the problem.
anonymous
  • anonymous
So the answer is pi/6, correct?
Mertsj
  • Mertsj
And i don't see what calculus has to do with it.
anonymous
  • anonymous
Because its on my Calc II practice test?
anonymous
  • anonymous
on Chapter 8
Mertsj
  • Mertsj
But what I mean is that it doesn't seem like calculus is needed to solve it. That's why I'm so suspicious of the answer. Are you positive that it is posted correctly?
Mertsj
  • Mertsj
What's the name of the chapter?
anonymous
  • anonymous
it is the integral of the arcsin(x/8) in disguise
anonymous
  • anonymous
That is just the screenshot off the webpage, unless they typed it wrong...
Mertsj
  • Mertsj
Ok. maybe Broken knows how to do it.
anonymous
  • anonymous
maybe?
anonymous
  • anonymous
Look real close, that's g'(4) :)
Mertsj
  • Mertsj
Oh. gee. That's a whole different animal!
anonymous
  • anonymous
And the tiniest apostrophe I've ever seen...
anonymous
  • anonymous
lol, its like they are trying to make me fail :P
anonymous
  • anonymous
Any thoughts on the g' version?
anonymous
  • anonymous
Sigh. Hold on...
anonymous
  • anonymous
\[ g(x) = x\sin^{-1}(x) + \sqrt{64-x^2} \] so I don't have to flip back...
anonymous
  • anonymous
arcsin x/8, not just x...
anonymous
  • anonymous
\[ g'(x) = \sin^{-1}(\frac{x}{8}) + \frac{x/8}{\sqrt{1-(x/8)^2}} - \frac{x}{\sqrt{64-x^2} } = \sin^{-1}(x/8) \] so \[g'(4) = \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}\]
anonymous
  • anonymous
Thanks man! And I'm sorry for asking so many question,
anonymous
  • anonymous
Haha no not at all

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