Question

Find the length of the curve.

Answer 1

wat??

Answer 2

\[y=1/6x^3+1/2x\]

Answer 3

from x=1 to x=2

Answer 4

@satellite73 can you offer me any help on this one. I'd really appreciate it!!

Answer 5

Are you supposed to use the graphics calculator for this question?

Answer 6

i do not believe so why?

Answer 7

length of any curve f(x) from a to b is given by :
integral of sqrt (1 + (f'(x))^2 ) from x = a to b
where f'(x) represents derivative of f(x)
this formulla has a very simple derivation in case you'd like to know..

Answer 8

Is it 1/2x^2+ 1/2

Answer 9

yep//the very same..

Answer 10

ok

Answer 11

then what do i do.

Answer 12

i square that?

Answer 13

Well given the formula sqrt (1 + (f'(x))^2 ). Since you alrdy got your differentiation above, just substitute in.

sqrt(1 + ( 1/2x^2 +1/2)^2) like that.

Answer 14

then i put in the two numbers?

Answer 15

does the square root then go away since the derivative is squared and the square root of 1 is 1

Answer 16

nope. you have to square the derivative then add 1, then square root it. After that then integrate it from x=a to b. Then you get your answer.

Answer 17

i dont know how to square it! :(

Answer 18

Why not? in case of (a+b)^2 = a^2 + 2ab +b^2, The same principle applies to
(1/2x^2 +1/2)^2

Answer 19

so wouldnt it just become the same as the original after i integrate?

Answer 20

if i recall the length of the curve is
\[\int_a^b\sqrt{1+(f'(x))^2}dx\] is that right?

Answer 21

yes.

Answer 22

is it \(f(x)=\frac{1}{6}x^3+\frac{1}{2}x\) ?

Answer 23

thats correct!

Answer 24

wait does it matter that the 1/2x the x is on the bottom next to the 2?

Answer 25

well it sure makes a difference to me
if it is in the denominator it is \(\frac{1}{2x}\) which is a heck of a lot different than \(\frac{1}{2}x\)

Answer 26

i cannot see how it is typeset so i don't know
you have to decide

Answer 27

its on the bottom

Answer 28

ok and how about for the first piece
is it \(\frac{1}{6x^3}\) or \(\frac{1}{6}x^3\)?

Answer 29

the second way you wrote!

Answer 30

ok "one sixth x cubed" got it
so it is \(f(x)=\frac{1}{6}x^3+\frac{1}{2x}\)

Answer 31

exactly!

Answer 32

so i have the derivative as \[1/2(x^2)-1/(2x^2)\]

Answer 33

so step one is \(f'(x)=\frac{1}{2}x^2-\frac{1}{2x^2}\)

Answer 34

ok good

Answer 35

then i am lost.

Answer 36

maybe we write this as
\[\frac{x^4-1}{2x^2}\]

Answer 37

i assume i need to square it but i dont know how to do that!

Answer 38

i guess we just grind it til we find it
\[\left(\frac{x^4-1}{2x^2}\right)^2\]

Answer 39

\[\frac{(x^4-1)^2}{4x^4}\] now we add one

Answer 40

well that was easy enough i made that harder tahn needed

Answer 41

\[\frac{4x^4+(x^4-1)^2}{4x^4}\]

Answer 42

where did the 4x^4 come from in the top

Answer 43

added one in the form of \(\frac{4x^4}{4x^4}\)

Answer 44

ok

Answer 45

now comes the miracle
the denominator is obviously a perfect square, so taking the square root is easy
the miracle is the numerator is one also

Answer 46

you should expect that because these problems have to be carefully cooked up so that you can actually perform the integration

Answer 47

ok so i got 2x^2 +(x^4-1) /(2x^2)

Answer 48

careful here

Answer 49

you cannot take square roots that way
\[\sqrt{a^2+b^2}\neq a+b\]

Answer 50

oh....

Answer 51

that is clear right? you have to add first, take the square root last

Answer 52

just like in pythagoras

\[\sqrt{3^2+4^2}=5\neq 3+4=7\]

Answer 53

ok im not sure what to do then.

Answer 54

so we have to do some algebra in the top
not much
\[4x^4+(x^4-1)^2\]

Answer 55

so we get 4x^4+x^8-2x^4+1

Answer 56

\[4x^4+x^8-2x^4+1\]
\[=x^8-2x^4+1\]
\[=(x^4+1)^2\] that is the miracle, it is a perfect square
now you can take the square root

Answer 57

ok i made a typo

Answer 58

\[x^8+2x^4+1=(x^4+1)^2\] thats better

Answer 59

you got the same thing, but didn't finish by combining like terms

Answer 60

yea that was my next step.

Answer 61

now it should be a cakewalk from here
you have
\[\sqrt{\frac{(x^4+1)^2}{4x^4}}=\frac{x^4+1}{2x^2}\]

Answer 62

ok i got that too. now i integrate corect?

Answer 63

yes but i would break it up first because then it will be easy to find the anti derivative
\[\int(\frac{x^2}{2}+\frac{1}{2x^2})dx\] etc