Arc length Calculus

Question

Find the length of the following function over the given interval

x = 5y^3/2 from y = 0 to y = 1

Answer 1

can look at this way

ds²=dx²+dy²=(1+(dx/dy)²)*dy²

so can integrate with dy and y limits

s=arc length=∫ds=∫√(1+([15/2)y^1/2]²)dy ,,,evaluated y=0 to y=1

you'll need to use 1=(4/225)*(225/4),

s=(4/225)*(2/3)*(1+225y/4)^3/2 from y=0 to y=1

I think.

Answer 2

x = 5y^3/2 as y goes from 0 to 1

x goes from 0 to 5

y^3/2 = x/5

y = (x/5)^2/3

y' = f'(x) = (2/3)(x/5)^(-1/3)

one arc length formula, among others, is

integrate the square root of (1+(f'(x)^2)dx evaluated between 0 and 5 (convert the y limits to x limits)

the formula is related to the Pythagorean Theorem

f(x) = (x/5)^(2/3)

f'(x) =2/[3(x/5)^1/3]

(f'(x))^2 = 4/[9(x/5)^(2/3)]

integrate (1-4/(x/5)^(2/3))^(1/2)