Question

find arc length of x=sqrt(y)-y from y= 1 to 4

Answer 1

Do you know the arc length formula?

Answer 2

yes

Answer 3

Okay so what don't you get?

Answer 4

sqrt1+[f'x]^2

Answer 5

sec

Answer 6

\[\int\limits_{1}^{4}.5y ^{-.5}\] is what i get when i do the formula

Answer 7

i integrate and get \[y^{.5}/.5\]

Answer 8

forgot the .5, but they cancel out anyways, so your left with just y^.5

Answer 9

once i plug in my values though, i dont get the answer in my book

Answer 10

What about the square root?

Answer 11

the [f'x]^2 cancels the square root in the arc length formula no?

Answer 12

Okay, first of all, what is your \(f'\)?

Answer 13

\[f'(y)=.5y ^{-.5}-1\]

Answer 14

And when you square that, what do you get?

Answer 15

not entirely sure, i dont square that since the sqrt in the arc lenth formula should cancel the square i have to do

Answer 16

so it should in the end be 1+(.5y^-.5)-1

Answer 17

1's cancel out, and i should have left the integral of .5y^-.5

Answer 18

But it's under the square root.

Answer 19

The thing is: \(\sqrt{1+[f']^2}\neq \sqrt{1+f'}\)

Answer 20

\[\sqrt{1+f'(x)^2} = 1+f'(x)\]

Answer 21

which is what i did

Answer 22

Nope, that isn't true either!

Answer 23

Whenn you square your derivative you should get 1/(4y)-1/y^(.5)+1

Answer 24

yes i get that. but then i have to take the square root of that, so why bother squaring in the first place? ._.

Answer 25

You need to add 1 to the square of the derivative before you can take the square root of it.

Answer 26

\[
\sqrt{1+x^2}\neq 1+x
\]

Answer 27

\[
(1+x)^2 = x^2+2x+ 1\neq 1+x^2
\]

Answer 28

so then how would i go about taking the square root of something that complex? is there a trick? square root of 1/(4y)-1/y^(.5)+2 yh?

Answer 29

First just expand it and simplify.

Answer 30

In most cases it creates a perfect square for you to just be able to take the square root of the whole thing. That's not always the case, however.

Answer 31

not entirely sure i have to expand something with negative exponents and decimal coiefficients

Answer 32

my prof knew some type of trick, im not entirely sure what it was though. i dont see how anyone can take the square of what we have right now

Answer 33

What do you have?

Answer 34

\[\sqrt{.25y ^{-1}-y ^{-.5}+2}\]

Answer 35

It doesn't come out to a perfect square. In most cases, it does. Do you have the problem down written correctly? There is just one part missing that keeps this from being a perfect squre.

Answer 36

\[x=\sqrt{y}-y \] evaluate from y =1 to 4 is the problem exactly how it appears in the book

Answer 37

ok...let me try something. Give me a sec.

Answer 38

ok

Answer 39

this is problem number 5 so i dont see how the it can be so complicated ._.

Answer 40

There are just some that get all twisted. In most cases, you just have to use some algebraic manipulation. I may have something. Hang on.

Answer 41

Nope, my middle term comes out to be 2/(2y)^(.5). I need a 1/y^(.5).

Answer 42

I have tried to factor it about three different ways, and each time my middle term comes out a little bit diffferent than what should be there.

Answer 43

Alright, im going to go ahead and close this. seems to be not worth the effort. thanks for the help wio and calmat

Answer 44

Yw. I will shoot you an @ if I come up with something. I am determined to figure it out.

Answer 45

lol alright

Answer 46

\[\large {x=\sqrt y - y\\x'=\frac{1}{2\sqrt y} - 1 = \frac{1-2\sqrt y}{2\sqrt y}\\(x')^2=\frac{1-4\sqrt y+4y}{4y}\\1+(x')^2=\frac{1-4\sqrt y+8y}{4y}\\\sqrt{1+(x')^2}=\frac{1}{2\sqrt y}\sqrt{8y-4\sqrt y+1}=\frac{\sqrt 8}{2\sqrt y}\sqrt{(\sqrt y)^2-(1/2)\sqrt y+(1/8)}\\\quad=\frac{\sqrt 8}{2\sqrt y}\sqrt{\left[(\sqrt y)^2-(1/2)\sqrt y+(1/16)\right]+(1/16)}\\\frac{\sqrt 8}{2\sqrt y}\sqrt{\left[\sqrt y - (1/4)\right]^2+(1/16)} }\]

Answer 47

let \[\large{\sqrt y - (1/4) = (1/4)\tan \theta\\\frac{dy}{2\sqrt y}=(1/4)\sec^2\theta d\theta\\\sqrt{\left[\sqrt y - (1/4)\right]^2 + (1/16)}=(1/4)\sec \theta}\]

Answer 48

\[\large {\sqrt{1+(x')^2}dx=\sqrt 8 \int \sqrt{(\sqrt y)^2-(1/2)\sqrt y+(1/16)}\frac{dy}{2\sqrt y}\\=\sqrt 8 \int (1/4)\tan \theta (1/4)\sec^2\theta d\theta}\]