Find arc length using calculus?

Question

chupacabraj · Answer

attachment

chupacabraj · Answer

I took the derivative, doubled it, added 1 to it , took the square root of that quantity and then integrated from 0 to 3 but I keep getting something ugly that's not even correct. ;/

jim_thompson5910 · Answer

You don't double it. You square it

chupacabraj · Answer

I'm sorry, I meant square it

jim_thompson5910 · Answer

` I keep getting something ugly that's not even correct`
what did you get @chupacabraj ?

chupacabraj · Answer

Everytime I try I get a different answer lol

jim_thompson5910 · Answer

what did you get for f ' (x) ?

chupacabraj · Answer

$$\frac{ 1 }{ 2 }x(x^2 +4)$$

satellite73 · Answer

what is amazing is that they can cook up the questions to get something you can actually integrate

satellite73 · Answer

take your derivative and square it 
what do you get?

chupacabraj · Answer

$$\frac{ 1 }{ 4 }x^2(x^2 +4)^2$$

satellite73 · Answer

yeah maybe best to multiply it out

chupacabraj · Answer

$$\frac{ x^6 }{ 4 }+2x^3+4x^2$$

satellite73 · Answer

hmm i thought when you squared and added one, you would get a perfect square, but it doesn't seem like it
maybe i should try with pencil and paper

satellite73 · Answer

oh derivative is wrong

satellite73 · Answer

when you use the power rule, you subtract 1 from 3/2 to get 1/2 i.e. the square root$$\frac{1}{2}x\sqrt{x^2+4}$$

satellite73 · Answer

now it should work out. square that , then add one

satellite73 · Answer

yes the usual miracle will occur
when you square and add one, you will end up with a perfect square !

chupacabraj · Answer

$$\frac{ x^2 }{ 2 }+x+1$$

?

satellite73 · Answer

no, that is not what you get when you square $$\frac{1}{2}x\sqrt{x^2+4}$$

satellite73 · Answer

the radical will go, and $(\frac{1}{2}x)^2=\frac{1}{4}x^2$ giving you $$\frac{1}{4}x^2(x^2+4)$$

satellite73 · Answer

multiply out, add one, see that you have a perfect square

chupacabraj · Answer

$$\frac{ 1 }{ 4 }[x^2+2x+1]$$

chupacabraj · Answer

I took the square root

satellite73 · Answer

i think there is still a mistake 
lets check

satellite73 · Answer

$$\frac{1}{4}x^2(x^2+4)=\frac{1}{4}x^4+x^2$$ right?  add one get $$\frac{1}{4}x^4+x^2+1$$  factor out a $\frac{1}{4}$ get $$\frac{1}{4}(x^4+4x^2+4)$$

satellite73 · Answer

and the thing inside the parentheses is a perfect square, as is $\frac{1}{4}$

satellite73 · Answer

you good from there?

chupacabraj · Answer

yes I think I can easily integrate using u subs afterwards

chupacabraj · Answer

thank you

chupacabraj · Answer

I'll get a polynomial lol no need for u sub

chupacabraj · Answer

yayy finally 15/2!

satellite73 · Answer

yay