Question

I NEED HELP PLOX. :3

Answer 1

Find the exact length of the curve.
#5 Por FAVOR. :3

Answer 2

Its my last problem. Yay :D

Answer 3

Are you able to find the derivative of this?

Answer 4

So.. would\[y ^{'}= \frac{ 1-2x }{ 2\sqrt{x-x^2} }+\frac{ \cos^{-1} (\sqrt{x} )}{ 2\sqrt{x} }\]

Answer 5

that's not a simple sine. That's inverse sine or arcsine

Answer 6

https://upload.wikimedia.org/math/4/8/a/48a2af001ba7003c8306e9d19aa61b9b.png

Answer 7

Oh right right.
Soo... \[y ^{'}=\frac{ 1-2x }{ 2\sqrt{x-x^2} } - \left( \frac{ -1 }{\sqrt{1-x}} \right) \left( \frac{ 1 }{ 2\sqrt{x} } \right)\]

Answer 8

yes, now let's simplify things
\[\Large y ^{'}=\frac{ 1-2x }{ 2\sqrt{x-x^2} } - \left( \frac{ -1 }{\sqrt{1-x}} \right) \left( \frac{ 1 }{ 2\sqrt{x} } \right)\]
\[\Large y ^{'}=\frac{ 1-2x }{ 2\sqrt{x-x^2} } +\frac{ 1 }{2\sqrt{x}\sqrt{1-x}} \]
\[\Large y ^{'}=\frac{ 1-2x }{ 2\sqrt{x-x^2} } +\frac{ 1 }{2\sqrt{x(1-x)}} \]
\[\Large y ^{'}=\frac{ 1-2x }{ 2\sqrt{x-x^2} } +\frac{ 1 }{2\sqrt{x-x^2}} \]
\[\Large y ^{'}=\frac{ 1-2x +1}{ 2\sqrt{x-x^2} } \]
\[\Large y ^{'}=\frac{ 2-2x}{ 2\sqrt{x-x^2} } \]
\[\Large y ^{'}=\frac{ 2(1-x)}{ 2\sqrt{x-x^2} } \]
\[\Large y ^{'}=\frac{ \cancel{2}(1-x)}{ \cancel{2}\sqrt{x-x^2} } \]
\[\Large y ^{'}=\frac{ 1-x}{ \sqrt{x-x^2} } \]
Do you agree with these steps?

Answer 9

Oh wait the 1/2ratx wouldnt be needed right?

Answer 10

Oh nvm.

Answer 11

Yes I agree

Answer 12

Now plug it into the arc length formula

\[\Large L = \int_{a}^{b}\sqrt{1+\left(y^{\prime}\right)^2}dx\]

Answer 13

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

Answer 14

yep correct

Answer 15

the values of 'a' and 'b' aren't given to you, but you can figure them out based on what the domain of f(x) is

Answer 16

Um 1?

Answer 17

0 and 1?

Answer 18

correct
a = 0
b = 1

Answer 19

Oh wait, I did that by looking at the dericative of arctan(sqrt x)

Answer 20

How do you do it correctly?

Answer 21

find the domain?

Answer 22

Yes

Answer 23

I see the 0, but not the 1

Answer 24

the first thing that comes to mind is the `sqrt(x)` inside the arcsine

the domain for that piece is x >= 0

Answer 25

Oh i plugged in a zero for f(x) and got 0.

Answer 26

The domain of arcsine is -1 <= x <= 1

but we don't have to worry about negatives (since the square root inside won't accept negatives)

So far, the domain is 0 <= x <= 1

Answer 27

The domain of sqrt(x-x^2) is 0 <= x <= 1 as well

Answer 28

So that's why a = 0 and b = 1

Answer 29

When looking for the domain, don't we look to see what numbers make the equation = 0?

Answer 30

you might be thinking of something like 1/(x-5)

there you set x-5 equal to zero and solve to find that x = 5 makes the denominator zero. So we have to kick 5 out of the domain

Answer 31

Yea

Answer 32

we don't have that issue here

Answer 33

Oh so we have to kind of visualize the graph?

Answer 34

yes that's one way to do it. Or to analyze the domain of each piece like I did above

Answer 35

Oh okay, I see. So then what would I do after I set up the arc length equation?
Do I have to factor the numerator?

Answer 36

I get \[\int\limits_{0}^{1} \sqrt{1+\left( \frac{ 1-2x+x^2 }{ x-x^2 } \right)}\]

Answer 37

\[\Large L = \int_{a}^{b}\sqrt{1+\left(y^{\prime}\right)^2}dx\]
\[\Large L = \int_{0}^{1}\sqrt{1+\left(\frac{1-x}{\sqrt{x-x^2}}\right)^2}dx\]
\[\Large L = \int_{0}^{1}\sqrt{1+\frac{(1-x)^2}{\left(\sqrt{x-x^2}\right)^2}}dx\]
\[\Large L = \int_{0}^{1}\sqrt{1+\frac{1-2x+x^2}{x-x^2}}dx\]
Do you see what to do from here?

Answer 38

ok good, we got the same thing

Answer 39

now simplify the stuff under the radical further

Answer 40

\[\int\limits_{0}^{1} \sqrt{1\left(\frac{ 1-x }{ x } \right)}\]

Answer 41

1+*

Answer 42

Then we can say this
\[\Large L = \int_{0}^{1}\sqrt{1+\frac{1-x}{x}}dx\]
\[\Large L = \int_{0}^{1}\sqrt{\frac{x}{x}+\frac{1-x}{x}}dx\]
\[\Large L = \int_{0}^{1}\sqrt{\frac{x+1-x}{x}}dx\]
\[\Large L = \int_{0}^{1}\sqrt{\frac{1}{x}}dx\]
which at this point, you can probably see how to evaluate the integral? If not, then here's a hint:
\[\Large \sqrt{x} = x^{1/2}\]

Answer 43

\[\int\limits_{0}^{1}(x)^{-1/2} = [(x)^{1/2}]_{0}^{1} = 1\]

Answer 44

Oops.

Answer 45

You also multiply 2. lmao

Answer 46

The answer = 2

Answer 47

Yay, i got it :D

Answer 48

Thanks!

Answer 49

yep the final answer is 2