Question

Calc Help!! find arc length

Answer 1

Will upload my current work…(computer being slow)

Answer 2

I know the next step is to get a common denominator followed by u-substitution but Im stuck |:

Answer 3

L = ∫√(1 + (dy/dx)²)

Answer 4

@sourwing yes I have my work set up already in that format

Answer 5

I don't think it's integrable. Try to do x = y^(3/2) instead

Answer 6

x = y^(3/2)
x' = (3/2)y^(1/2)

L = ∫√(1 + 9y/4) dy, from 1 to 4

Answer 7

now this looks much better XD

Answer 8

i see what you did! ok and quick question how did we get the new bounds?

Answer 9

@sourwing

Answer 10

Manipulate the expression inside the square root so that it is

1 + 1/((9/4) x^(2/3))

For awkward 1 remaining on the left side, transform it so that

((9/4) x^(2/3))/((9/4) x^(2/3)) + 1/((9/4) x^(2/3))

which leaves

(((9/4) x^(2/3)) + 1)/((9/4) x^(2/3))

since there is square root, pull the denominator out of it,

sqrt((((9/4) x^(2/3)) + 1))/((3/2) x^(1/3))

Then use u-substitution,

u = (9/4) x^(2/3) + 1

You should be able to integrate it then.

Answer 11

y = 1^(2/3) = 1
y = 8^(2/3) = 4

Answer 12

Yup, I checked and the integration takes the form of

integrating square root of u, which is something you know how to integrate.

And for the bound,

u = (9/4)*1 + 1
u = (9/4)*8^(2/3) + 1

Answer 13

In the end, either method will do. So have fun choosing one.