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anonymous

  • 4 years ago

Calc II question attached:

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  1. anonymous
    • 4 years ago
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  2. campbell_st
    • 4 years ago
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    sub x = 4 so you get \[g(4) = 4 \times \sin^-1 (4/8) + \sqrt{64 - 4^2}\] \[\sin^(-1) (1/2) = \pi/6\]

  3. campbell_st
    • 4 years ago
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    i think the answer is \[2\pi/3 + 4\sqrt{?}\]

  4. campbell_st
    • 4 years ago
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    oops ? = 3

  5. Mertsj
    • 4 years ago
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    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.

  6. anonymous
    • 4 years ago
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    So the answer is pi/6, correct?

  7. Mertsj
    • 4 years ago
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    And i don't see what calculus has to do with it.

  8. anonymous
    • 4 years ago
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    Because its on my Calc II practice test?

  9. anonymous
    • 4 years ago
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    on Chapter 8

  10. Mertsj
    • 4 years ago
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    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?

  11. Mertsj
    • 4 years ago
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    What's the name of the chapter?

  12. anonymous
    • 4 years ago
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    it is the integral of the arcsin(x/8) in disguise

  13. anonymous
    • 4 years ago
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    That is just the screenshot off the webpage, unless they typed it wrong...

  14. Mertsj
    • 4 years ago
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    Ok. maybe Broken knows how to do it.

  15. anonymous
    • 4 years ago
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    maybe?

  16. anonymous
    • 4 years ago
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    Look real close, that's g'(4) :)

  17. Mertsj
    • 4 years ago
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    Oh. gee. That's a whole different animal!

  18. anonymous
    • 4 years ago
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    And the tiniest apostrophe I've ever seen...

  19. anonymous
    • 4 years ago
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    lol, its like they are trying to make me fail :P

  20. anonymous
    • 4 years ago
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    Any thoughts on the g' version?

  21. anonymous
    • 4 years ago
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    Sigh. Hold on...

  22. anonymous
    • 4 years ago
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    \[ g(x) = x\sin^{-1}(x) + \sqrt{64-x^2} \] so I don't have to flip back...

  23. anonymous
    • 4 years ago
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    arcsin x/8, not just x...

  24. anonymous
    • 4 years ago
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    \[ g'(x) = \sin^{-1}(\frac{x}{8}) + \frac{x/8}{\sqrt{1-(x/8)^2}} - \frac{x}{\sqrt{64-x^2} } = \sin^{-1}(x/8) \] so \[g'(4) = \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}\]

  25. anonymous
    • 4 years ago
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    Thanks man! And I'm sorry for asking so many question,

  26. anonymous
    • 4 years ago
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    Haha no not at all

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