## anonymous 4 years ago Calc II question attached:

1. anonymous

2. campbell_st

sub x = 4 so you get $g(4) = 4 \times \sin^-1 (4/8) + \sqrt{64 - 4^2}$ $\sin^(-1) (1/2) = \pi/6$

3. campbell_st

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

4. campbell_st

oops ? = 3

5. 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.

6. anonymous

So the answer is pi/6, correct?

7. Mertsj

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

8. anonymous

Because its on my Calc II practice test?

9. anonymous

on Chapter 8

10. 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?

11. Mertsj

What's the name of the chapter?

12. anonymous

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

13. anonymous

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

14. Mertsj

Ok. maybe Broken knows how to do it.

15. anonymous

maybe?

16. anonymous

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

17. Mertsj

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

18. anonymous

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

19. anonymous

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

20. anonymous

Any thoughts on the g' version?

21. anonymous

Sigh. Hold on...

22. anonymous

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

23. anonymous

arcsin x/8, not just x...

24. 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 $g'(4) = \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$

25. anonymous

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

26. anonymous

Haha no not at all