Question

HELP PLEASE

Answer 1

#4

Answer 2

\[y = \sqrt{x-x^2}+\sin^{-1} (\sqrt{x})\]
\[y' = \frac{ 1 }{ 2 }(x-x^2)^{-\frac{ 1 }{ 2 }}(1-2x)+\frac{ 1 }{ 2x^{\frac{ 1 }{ 2 }}\sqrt{1-x} }\]

Answer 3

Finding the Arc length by the way...

But I need to square that whole y' ...

Answer 4

@agent0smith

Answer 5

Is there an easier way to do this?
I actually dont really think i even know how to square that whole thing.

Answer 6

Oh wait... It doesnt even give me the integrands. How do i solve? D:

Answer 7

@Directrix @518nad @Nnesha @Loser66

Answer 8

@pooja195 @Callisto @radar

Answer 9

#5****

Answer 10

Hey, write it as a fraction, you can see what happens

Answer 11

Oh i think i remember... Stuff cancels right?

Answer 12

\[\large y' = \frac{ 1 }{ 2 }(x-x^2)^{-\frac{ 1 }{ 2 }}(1-2x)+\frac{ 1 }{ 2x^{\frac{ 1 }{ 2 }}\sqrt{1-x} } \]
\[\large y' = \frac{1-2x}{ 2\sqrt{ x - x^2} }+\frac{ 1 }{ 2\sqrt x \sqrt{1-x} } \]
\[\large y' = \frac{1-2x}{ 2\sqrt{ x - x^2} }+\frac{ 1 }{ 2 \sqrt{x(1-x)} } \]
\[\large y' = \frac{2-2x }{ 2\sqrt{ x - x^2} }\]

Answer 13

Yup, Thanks @agent0smith

Answer 14

Oh thanks!!

Answer 15

And the 2 as well\[\large y' = \frac{1-x }{ \sqrt{ x - x^2} }\]

Answer 16

And for the Integrands, do i use -1 and 1?

Answer 17

Probably, seems reasonable. I'd just look at a graph and it'd be easy to tell for sure.

Answer 18

\[\int\limits \sqrt{1+ \left( \frac{ 1-2x+x^2 }{ x-x^2 } \right)}\]

Now I am lost. :/

Answer 19

wait...

Answer 20

\[\frac{ (x-1)(x-1) }{ -x(x-1) } = \frac{ (x-1) }{ -x }\]

Amiright?

Answer 21

Just don't even square it first\[\large \int\limits \sqrt{1+ \frac{ (1-x)^2 }{ x(1-x) }}\]

Answer 22

Oh okay I see.

Answer 23

But how would you go about it from there?

Answer 24

\[\int\limits \sqrt{1 + \left( \frac{ 1-x }{ x } \right)}\]

Answer 25

\[\large \int\limits \sqrt{\frac{ x }{ x } + \left( \frac{ 1-x }{ x } \right)}\]

Answer 26

\[\int\limits \sqrt{\frac{ 1 }{ x }} \]
\[=\frac{ 2 }{ 3 } \left( \frac{ 1 }{ x } \right)^{\frac{ 3 }{ 2 }}\]

Eh right?

Answer 27

\[\large \int\limits x^{-1/2} = 2\sqrt x \]

Answer 28

Oh right.... so then the integrands are 1 and 0? D:

Answer 29

How do you figure out the Integrals?

Answer 30

Either by finding the domain, or just looking at a graph of the original function.

Domain appears to be from 0 to 1 from eyeballing it.

Answer 31

Oh. okay, Thanks!

Answer 32

Welcome.