anonymous
  • anonymous
Calc II question attached:
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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 \[2\pi/3 + 4\sqrt{?}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.