Calc II question attached:

Question

anonymous · Answer

attachment

campbell_st · Answer

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 · Answer

i think the answer is $$2\pi/3 + 4\sqrt{?}$$

campbell_st · Answer

oops ? = 3

mertsj · Answer

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 · Answer

So the answer is pi/6, correct?

mertsj · Answer

And i don't see what calculus has to do with it.

anonymous · Answer

Because its on my Calc II practice test?

anonymous · Answer

on Chapter 8

mertsj · Answer

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 · Answer

What's the name of the chapter?

anonymous · Answer

it is the integral of the arcsin(x/8) in disguise

anonymous · Answer

That is just the screenshot off the webpage, unless they typed it wrong...

mertsj · Answer

Ok.  maybe Broken knows how to do it.

anonymous · Answer

maybe?

anonymous · Answer

Look real close, that's g'(4) :)

mertsj · Answer

Oh. gee.  That's a whole different animal!

anonymous · Answer

And the tiniest apostrophe I've ever seen...

anonymous · Answer

lol, its like they are trying to make me fail :P

anonymous · Answer

Any thoughts on the g' version?

anonymous · Answer

Sigh.  Hold on...

anonymous · Answer

$$ g(x) = x\sin^{-1}(x) + \sqrt{64-x^2} $$ so I don't have to flip back...

anonymous · Answer

arcsin x/8, not just x...

anonymous · Answer

$$ g'(x) = \sin^{-1}(\frac{x}{8}) + \frac{x/8}{\sqrt{1-(x/8)^2}} - \frac{x}{\sqrt{64-x^2} } = \sin^{-1}(x/8) $$
so 
$$g'(4) = \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$$

anonymous · Answer

Thanks man! And I'm sorry for asking so many question,

anonymous · Answer

Haha no not at all