Calculus - Arc Length?

Question

Problem: x=[(√y)(y-3)]/3

I am finding the definite integral from 1 to 4.

I believe I need to integrate with respect to y using the equation:

s=∫√{1+[g'(y)]²}

I started by taking the derivative of the original equation x=[(√y)(y-3)]/3 to find g'(y).

I came up with [(√y)/3] + [(y-3)/(2√y)]

Then I squared that quantity to equal...

[(√y)/3] + [(y-3)/3]+[(y-3)(y-3)]/4y

and after much algebra to simplify and combine...

[25y² - 60y + 81]/36y

I am stuck here, and am not even sure if my algebra is correct... any help please?

Please show all steps... Thanks in advance.

Answer 1

Following your instruction to find the definite integral:-

L = (1/3) ∫ y^(3/2) - 3y^(1/2)) dy-- (lims. 1 to 4)
L = (1/3) [ (2/5) y^(5/2) - 2y^(3/2) ]
L = (1/3) [ (2/5) (32) - 2 (8) - 2 / 5 + 2) ]
L = (1/3) [ 64 / 5 - 16 - 2 / 5 + 2 ]
L = (1/3) [62 / 5 - 14 ]
L = (1/3) [ 62 / 5 - 70 / 5 ]
L = (1/3) (- 8 / 5)
L = 8 / 15 (ignoring -ve sign)

Answer 2

To find your integral it is easier if you start off simplifying the problem. So your problem would be:

y^(1/2) * (y-3)/3 =
[y^(3/2)-3y^(1/2)]/3 =
y^(3/2)/3-y^(1/2)

The if you integrate this you would have:
2y^(5/2)/(15)-2y^(3/2)/3+C=


Which is a much simplier equation to plug your answer into.
which would be:
2(2)^5/15-2(2)^3/3+c - (2(1)^5/15-2(1)^3/3+c)=
2(32)/15-2(8)/3+c-2/15+2/3-c=
64/15-16/3-2/15+2/3=
(64-16(5)-2+2(5))/15=
(64-80-2+10)/15=8/15

Answer 3

Hmm..wish I could help but my highest level of math is College Algebra. Ha! Good luck. :)

Answer 4

i do not know