Question

Find the arc length of y = x^(3/2) on [1,4].

Answer 1

\[\int_C ds\to \int_{1}^{4}(\sqrt{1+(y')^2})\ dx\]

Answer 2

cannot really see the integral.. \[\sqrt{1 + y ^{2}} ?\]

Answer 3

\[\Large \int_C ds\to \int_{1}^{4}(\sqrt{1+(y')^2})\ dx\]

Answer 4

i can make it in color too :)

Answer 5

what colors are we talking about? red? green? blue?

Answer 6

so y' = 3x^(2/3)/2?

Answer 7

\[\huge \color{green}{\int_C ds}\to \int_{1}^{4}(\color{#ff23cc}{ \sqrt{1+(y')^2}})\ dx\]

Answer 8

3/2 - 2/2 = 1/2

Answer 9

\[\int\limits_{1}^{4}\sqrt{1 + 3x^(2/3)/2} dx\]

Answer 10

not 2/3 ... 1/2; which makes the ^2 nice and purrty

Answer 11

oh woops :P soo tired.

Answer 12

so then it's going to be 3/2^1/2 but the ^2 negates the 1/2... so it will be sqrt(1-3x/2)dx

Answer 13

sooo close; (3/2)^2 = 9/4

{S} sqrt(1+ 9x/4) dx

Answer 14

ah poo I give up :p

Answer 15

lol

Answer 16

not really :P

Answer 17

\[\int \sqrt{1+\frac{9}{4}x}\ dx\]

\[\int \sqrt{\frac{4+9x}{4}}\ dx\]

\[\frac{1}2\int \sqrt{4+9x}\ dx\]

Answer 18

tricky tricky

Answer 19

if we had a 9 wed be set to rock ...

\[\frac{1}2\frac 99\int \sqrt{4+9x}\ dx\]
\[\frac{1}{18}\int 9(4+9x)^{1/2}\ dx\]

Answer 20

almost got it nailed down

Answer 21

that 1/2 comes from a 3/2 so we have to catch it when it pops out with a 2/3

Answer 22

nah, i missed it; the 2/3 is afterwards

Answer 23

\[\frac{1}{18} \int9(4+9x)^{1/2}\ dx=\frac{1}{18}\frac{2}{3}\frac{(4+9x)^{3/2}}{3/2}=\frac{4}{162}(4+9x)^{3/2}\]
maybe :)

Answer 24

http://www.wolframalpha.com/input/?i=integrate+%28sqrt%284%2B9x%29%2F2%29

Answer 25

just a wee bit askewed

Answer 26

sorry I went to get a bite to eat, let me review P

Answer 27

where did this 9/9 come from? or why exactly?

Answer 28

when that function inside gets integrated; the chain rules gonna pop out a 9

Answer 29

since there is no 9 there we need to borrow one from 9/9

Answer 30

unless you wanna go thru the whole u sub thing where u = 4+9x, du = 9 dx; du/9 = dx

\[\frac{1}{2*9}\int (u)^{1/2}\ du\]

Answer 31

either way, we need to borrow a 9 to get where we want to go with it

Answer 32

that's how I learned it, u sub....

Answer 33

ooo I is tired.... I got the rest of this. Thanks very much sir/maam have a great night, thanks for all the help tonight

Answer 34

this one ..

Answer 35

damn yuou beat me to it, yues.

Answer 36

\[\int \sqrt{1+\frac{9}{4}x}\ dx\]
is the setup so its just a matter if integrating over the interval

Answer 37

is this going to be the same setup for any eq? sqrt(x'^2? + y'^2)?

Answer 38

yes, you measure the length of any curve in the same way; if its in 3diminsion you add in a z'^2 but yes; distance is distance

Answer 39

Ah okay makes sense... I always find my biggest problem with this stuff is trying to understand why I'm doing it. Professors like to give out eq but not explain them, and not explain how a problem can use multiple ones...

Answer 40

yeah, there is alot of handwaving at times since the proofs behind the intuitions can get complicated

Answer 41

its best just to remember this picture and understand that the length of the curve is adding up the total sum of /\s that we measure

Answer 42

\[\sum \Delta s^2 = \sum (\Delta x^2+\Delta y^2)\]

\[\sum \Delta s = \sum \sqrt{\Delta x^2+\Delta y^2}\]

as these values go into infinitesimals we get

\[\int ds = \int \sqrt{dx^2+dy^2}\]

Answer 43

so x' is 1?

Answer 44

if y = x^(3/2) you would need to take the ln of both sides and switch around to get x right? Or is there an easier way tro get it?

Answer 45

\[x'=\frac{dx}{d* }\]where * is the independant variable

when \[x'=\frac{dx}{dx}\to x'=1\]

Answer 46

okay so it will always be 1 then?

Answer 47

if we are doing a dx problem...

Answer 48

no, for example:\[if\ x'=\frac{dx}{dt}\]then we have no way of determing if x'=1

Answer 49

IF we are doing a problem where x IS the independant variable; then x'=dx/dx = 1

in this problem; we get dx/dx so yes

Answer 50

hmmm?

Answer 51

spose y was the INDEPENDANT variable in this problem; then:

x' = dx/dy and y' = dy/dy=1

Answer 52

so if we are given a problem with dx x would be independant, if it's dt t is the inependant, and dty would be y independant?