Question

How would you plug this function:

3(x square root) , whose upper limit is 10 and lower one is 0,

and this one:

^3(x "square root sign) - 1/4(^3(x "square root sign"))

into the arc length formula?

There derivatives are originially from 2(x^3/2), and 3/4(x^4/3) - 3/8(x^2/3)+7.. @Calculus1

Answer 1

can you use the equation feature please? Trying to read this is hurting my eyes.

Answer 2

\[Turing-Test.\]

Answer 3

http://www.youtube.com/watch?v=J6ZWlDks0nQ

Answer 4

SAIFOO-SAN-KHAN

Answer 5

San?? o_O
from where did that came from?

Answer 6

like master
saifoo khan san
turing test san
you know...

Answer 7

like sensei, you know

Answer 8

Ohh!! now i got it.. ^_^
thanks. =)

Answer 9

is this the music you do math to? pretty mellow vibe

Answer 10

Nope, i go for random music. currently this once's on top! =D

Answer 11

oh, cuz this is what I listen to
http://www.youtube.com/watch?v=OFSrINhfNsQ

Answer 12

LOL.

Answer 13

:)

Answer 14

alright hold on

Answer 15

is that your pic Turing?

Answer 16

i guess so.

Answer 17

yup that's me

Answer 18

that's you saifoo?

Answer 19

Yes, me and my camera! =D

Answer 20

Nice hairs! :D

Answer 21

Agree^

Answer 22

Heh, I was at a hippie commune in that pic. Only hippie physics major there ;)
what ethinicity/nationality are you?

Answer 23

\[y=2x ^{3/2}\] is the given curve where 0 =a and 10=b

and

\[y = 3/4x ^{4/3}-3/8x ^{2/3}+7\]

where 1 is a and 8 is b.

Answer 24

Pakistani.
you?

Answer 25

I am an Indian .. and you ?

Answer 26

All American ,but I live in mexico

Answer 27

ok let's do our buddy's problem, now that it's posted correctly

Answer 28

I'm trying to find the arc lengths of both functions, I already found the derivatives, I'm just unsure of how to plug that back into the function.

Answer 29

OH mostly Mexican want to leave in America.. any felony records in states ? :P

Answer 30

ok thanks.

Answer 31

what is a and b ?

Answer 32

integration ?

Answer 33

a stand for the lower limit and b stands for the upper limit.

Answer 34

^^Yes

Answer 35

so this are two separate problems of finding the area?

Answer 36

*these

Answer 37

these are after you've taken the derivative, correct?

Answer 38

yes, that is correct.

Answer 39

\[ \int_0^{10} 2x ^{3/2} dx \]

Answer 40

so why do you have y= and not dy/dx=
???

Answer 41

No actually I was wrong, both functions are before doing the derivative

Answer 42

ok let's take the derivative then...

Answer 43

whatever you are just flustered yourself.....

Answer 44

\[dy/dx=3x^{1/2}\]\[dy/dx=x^{1/3}-{x^{-1/3}\over4}\]are the derivatives then. do we agree?

Answer 45

arc length formula:\[\int\limits_{a}^{b} \sqrt{1+(dy/dx)^2}\]for the first prob\[\int\limits\limits_{0}^{10} \sqrt{1+(3x^{1/2})^2}dx=\int\limits\limits_{0}^{10} \sqrt{1+9x}dx\]can you integrate that yourself?

Answer 46

^ Yes that is what I got, except you have to bring each of the exponents down and square them.

For your other question, I got \[\sqrt{91}\]

Answer 47

the above integral\[\int\limits_{0}^{10}(1+9x)dx={2\over27}(1+9x)^{3/2}\]from 0 to 10 is\[{2\over27}(91^{3/2}-1)\]

Answer 48

sorry that should be (1+9x)^(1/2) in the integral.

Answer 49

Yes this is what I got and the decimal approximation to it would be 64.228

Answer 50

Okay, now I think I can figure out the second one. I'll let you know if I have any issues. Thanks alot.

Answer 51

the other becomes\[dy/dx=x^{1/3}-{1\over4}x^{-1/3}\]\[(dy/dx)^2=x^{2/3}+x^{-2/3}-1/2\]so arc length becomes\[\int\limits_{1}^{8}\sqrt{x^{2/3}+x^{-2/3}+{1\over2}}dx=\int\limits_{1}^{8}\sqrt{x^{2/3}+x^{-2/3}+{1\over2}}dx\]or at least that's what I got, but I can't seem to integrate it, so maybe you can help me with this one when you figure it out

Answer 52

alright, I'll look at it later and get back to you.

Answer 53

actually (dy/dx)^2 should be \[x^{2/3}+{1\over16}x^{-2/3}-{1\over2}\]my mistake

Answer 54

so\[\int\limits_{1}^{8}\sqrt{x^{2/3}+{1\over16}x^{-2/3}+{1\over2}}dx\]should be the formula...

Answer 55

I want to factor what's in the radical into (a+b)^2 but that doesn't seem possible...

Answer 56

I think I got it...

Answer 57

\[\int\limits_{1}^{8} \sqrt{x^{2/3}+{1\over2}+{1\over16}x^{-2/3}}dx=\int\limits_{1}^{8} \sqrt{x^{-2/3}(x^{4/3}+{1\over2}x^{2/3}+{1\over16})}dx\]\[=\int\limits_{1}^{8}\sqrt{x^{-2/3}(x^{2/3}+{1\over4})^2}dx=\int\limits_{1}^{8}x^{-1/3}(x^{2/3}+{1\over4})dx\]\[=\int\limits\limits_{1}^{8}x^{1/3}+{1\over4}x^{-1/3}dx={3\over4}x^{4/3}+{3\over8}x^{2/3}\]I'm prettty sure you can evaluate that from 1 to 8 yourself, right?

Answer 58

I got 93/8