Question

i am lost and i dont know where to start, i know this is calculus 1 but this is calculus 2. the section is Applications of Definite Integrals and i am trying to find the lengths of the curves. The question is x=(y^3/3)+1/(4y) from y=1 to y=3

Answer 1

length or area under the curve?

Answer 2

Integrate the equation with respect to y and use y=3 and y=1 for upper and lower limits respectively. You'll have area under the graph

Answer 3

i understand that but i need help setting it up for the integral, after that i can solve it,

Answer 4

Arc length?

Answer 5

no it just says lengths of the curves, between 1-3

Answer 6

Yeah, that's usually solving for arc lengths, you'll have to find dx/dy first, here the "formula" if you will...
\[L = \int\limits_{a}^{b} \sqrt{1+(\frac{ dx }{ dy }})^2dy\] as in your case, see what you get.

Answer 7

well than is my problem, i dont know how to get it set up for the intrgral, to get (y^3/3)+1/(4y) to be ready to integrate,

Answer 8

Do one step at a time, as it shows, so find dx/dy first and then square it, etc.

Answer 9

so square (y^3/3+1/(4y))^2 like that?

Answer 10

But you have to find dx/dy first

Answer 11

ok then thats where i am lost now

Answer 12

so then would it be y^2-1/2y^2

Answer 13

\[\frac{ dx }{ dy } = y^2-\frac{ 1 }{ 4y^2 }\]

Answer 14

Now you have \[1+(y^2-\frac{ 1 }{ 4y^2 })^2\]

Answer 15

Work with that and squareroot it and then you can integrate it easily :P, I'll bbs, so keep working on it!

Answer 16

ok so far i am getting that

Answer 17

Actually, \[\sqrt{1+(\frac{ dx }{ dy }})^2 = y^2+\frac{ 1 }{ 4y^2 }\] so you just have to integrate this from 1 to 3

Answer 18

That's a very easy integral!

Answer 19

\[\int\limits_{1}^{3} y^2+\frac{ 1 }{ 4y^2 } dy\]

Answer 20

yup that is what i have, thank you, integral part i can do, the setting up for it throws me off

Answer 21

No problem, remember take it step by step :)