Question

how to simplify the equation sqrt(1+(f'(x))^2) to remove the sqrt when f(x)= ((3x^4)/8 )+ (1/(12x^2))

Answer 1

some sort of "find the length of the curve" question?

Answer 2

yeah

Answer 3

they really work hard to cook these questions up to get something you can find the anti-deritive of when you do the algebra

Answer 4

i tried it once, it took me like half an hour to do it, maybe longer

Answer 5

first you need the derivative
did you get that?

Answer 6

im not sure if im on the right track
i am trying to leave it in terms of f'(x) until i can get rid of the sqrt

Answer 7

i got 1^(1/2) + f'(x)^(3/2)

Answer 8

you need to do it step by step
a) find the derivative

Answer 9

im really lost on this

Answer 10

yeah i see

Answer 11

lets go slow, there are no shortcuts

Answer 12

thanks, you are a life saver

Answer 13

before we begin, suppose you had to find \[\sqrt{1+6^2}\] what would you do?

Answer 14

answer: you would
a) square 6
b) add 1
c) take the square root
that is what we have to do, no shortcuts

Answer 15

a?

Answer 16

first we need \(f'(x)\)
i get \[f'(x)=\frac{9x^6-1}{6x^3}\]

Answer 17

lol no, it wasn't multiple choice, you have to do all three in that order
that is what we have to do
a) take the derivative
b) square it
c) add 1
d) take the square root

Answer 18

ah ok, mb

Answer 19

so step one, as i said, take the derivative
i cheated and used wolfram, but i am sure it is doable by hand if you have to "show your work"

Answer 20

oh before i screw up totally, it is \[f(x)=\frac{3x^4}{8}+\frac{1}{12x^2}\] right ?

Answer 21

yes it is

Answer 22

ok then the derivative is easy enough

Answer 23

its (9x^6 -1)/6x^3

Answer 24

yes
step two
square it

Answer 25

i might leave it as \[f'(x)^2=\frac{(9x^6-1)^2}{36x^6}\]

Answer 26

yeah i just got that

Answer 27

step 3
add 1

Answer 28

sorry working on limited desk space

Answer 29

no problem

Answer 30

\[1+(f')^2=1+\frac{(9x^6-1)^2}{36x^6}=\frac{(9x^6-1)^2+36x^6}{36x^6}\]

Answer 31

now the miracle occurs
ready?

Answer 32

kk

Answer 33

turns out this sucker is a perfect square ! (really imagine making this up)

Answer 34

we can remove the denominator and the 36x^6 on the numerator with 1 right?

Answer 35

if you see it, good, if not we can do the algebra

Answer 36

oh no

Answer 37

don't break it apart whatever you do !!

Answer 38

ok

Answer 39

lets do a tiny bit of algebra in the numerator

Answer 40

\[(9x^6-1)^2+36x^6=81x^{12}-18x^6+1+36x^6=81x^{12}+18x^6+1=(9x^6+1)^2\] a total miracle

Answer 41

oops that kind of bled over, hold on

Answer 42

\[(9x^6-1)^2+36x^6\\=81x^{12}-18x^6+1+36x^6\\=81x^{12}+18x^6+1\\=(9x^6+1)^2\]

Answer 43

in other words, this problem was cooked up so that the expression inside the radical is a perfect squre
that isnot easyh to do

Answer 44

you get \[\sqrt{\frac{(9x^6+1)^2}{36x^6}}\]
take the square root now is easy

Answer 45

let me know if you got stuck at the algebra

Answer 46

1/6 (9 x^3+1/x^3) right?

Answer 47

yup

Answer 48

nope

Answer 49

\[\frac{9x^6+1}{6x^3}\]

Answer 50

kk

Answer 51

don't take the square root of \(9x^6\) take the square root of \((9x^6+1)^2\) which is just \(9x^6+1\)

Answer 52

i plugged it into wolfram and it gave me the 1/6 solution

Answer 53

wasnt sure if was 9x^6+1 / 6x^3

Answer 54

that is not the problem you can put the \(\frac{1}{6}\) out front if you like
the problem you had was you changed the exponent on the \(9x^6\) term

Answer 55

ok thanks this was very helpful

Answer 56

that was wrong

Answer 57

but whatever, it is easy enough now

Answer 58

I think ghost meant
\[\frac{1}{6}(9x^3+\frac{1}{x^3}) \text{ which is the same as } \frac{9x^6+1}{6x^3}\]

Answer 59

ooh

Answer 60

the ghost of wolfram past

Answer 61

yeah, wolfram tends to just confuse me more

Answer 62

really it is amazing how they make these up
take a derivative
square it
add 1
and voila, a perfect square !

Answer 63

bet the lazy math teacher copies them out of the boook

Answer 64

in the calculus books I used to look up they would always put some kind of hint that 1+(f')^2 was a complete square

Answer 65

probably

Answer 66

@freckles go ahead an make one up without cheating
i will time you lol

Answer 67

no we were left in the dark, we were told it was possible that 1+f'(x) could have not been a perfect square

Answer 68

and honestly that was what i was afraid of

Answer 69

lol
\[f=\ln(\sec(x)) \\ f'=\frac{\sec(x) \tan(x)}{\sec(x)}=\tan(x) \\ 1+(f')^2=\sec^2(x)\]

Answer 70

ok ok

Answer 71

but the ones that involve polynomials are harder to come up with

Answer 72

or rational expressions

Answer 73

then you have to integrate secant
i think i came up with \[f(x)=6\sqrt[3]{x^2}\] or something simple

Answer 74

maybe that doesn't work dang

Answer 75

another reason why i loathe calc 2

Answer 76

I think also a good thing to remember
is that \[(f-\frac{1}{4f})^2 =f^2-2 \cdot f \frac{1}{2f}+(\frac{1}{4f})^2 \\ (f-\frac{1}{4 f})^2=f^2-\frac{1}{2}+(\frac{1}{4f})^2 \\ \\ \text{ and doing 1 plus on both sides gives } \\ (f+\frac{1}{4f})^2=f^2+\frac{1}{2}+(\frac{1}{4f})^2\]
like this is what happened on this one when expanded the (f')^2 thing you had a -1/2 for the middle term
\[f(x)=\frac{3}{8}x^4+\frac{1}{12 x^2 } \\ f'(x)=\frac{3}{2}x^3-\frac{1}{6 x^3} \\ (f')^2=(\frac{3}{2}x^3-\frac{1}{6x^3})^2 \text{ expanding this would give middle term is } \frac{-1}{2} \\ \text{ so adding one to the } (f')^2 \text{ will just mean you have } \\ \text{ to change the middle sign } \\ (f')^2+1=(\frac{3}{2}x^3+\frac{1}{6x^3})^2\]

Answer 77

oops I made a type in my first code line

Answer 78

that middle should be -2*f*1/(4 f)
but everything that seems good
and maybe I shouldn't have used f

Answer 79

like the last one
nice

Answer 80

doesn't mean i don't still hate calc 2 though

Answer 81

So to make a problem...
\[\text{ we want } f'(x)=a x^{n}-\frac{1}{b x^{n}} \\ \text{ where } \frac{a}{b}=\frac{1}{4} \\ (f'(x))^2=(ax^n)^2-2 ax^n \frac{1}{b x^n}+(\frac{1}{b x^n})^2 \\ (f'(x))^2=(ax^n)^2-\frac{1}{2}+(\frac{1}{bx^n})^2 \\ \text{ so } 1+(f'(x))^2=(ax^n)^2+\frac{1}{2}+(\frac{1}{b x^n})^2 =(ax^n+\frac{1}{b x^n})^2\]

so integrating f' we can find an f to use

Answer 82

\[f'(x)=ax^n-\frac{1}{b x^n} \text{ where } \frac{a}{b}=\frac{1}{4} \\ f'(x)=ax^n-\frac{1}{b}x^{-n} \\ f(x)=a \frac{x^{n+1}}{n+1}- \frac{1}{b} \frac{x^{-n+1}}{-n+1}+C \\ \text{ where } n \neq -1,1\]

Answer 83

ok show off!

Answer 84

guess my challenge was more than met

Answer 85

lol :p
I bet there could be other choices though for f (besides trigonometric like the one I chose earlier)