Find the length of the graph of f(x)=(1-x^2)^1/2 on [0, b]

We'll use the formula of calculating the length of the
graph:

L = Int sqrt{1+[f'(x)]^2} from x = a to x =
b

In our case a = 0 and b =
b

The length of the graph of f(x) = sqrt(1-x^2)
is:

L = Int sqrt{1+[(sqrt(1-x^2))']^2}dx , from 0 to
b.

We'll calculate the derivative of
sqrt(1-x^2)

[sqrt(1-x^2)]' =
-2x/2sqrt(1-x^2) 

[sqrt(1-x^2)]' =
-x/sqrt(1-x^2) 

{[sqrt(1-x^2)]'}^2 =
x^2/(1-x^2)

We'll add 1 both
sides:

1 + {[sqrt(1-x^2)]'}^2 = 1 +
x^2/(1-x^2)

1 + {[sqrt(1-x^2)]'}^2 = (1 - x^2 +
x^2)/(1-x^2)

We'll eliminate like
terms:

1 + {[sqrt(1-x^2)]'}^2 =
1/(1-x^2)

L = Int dx/(1-x^2)

L
= arcsin x from x = 0 to x = b

L = arcsin b - arcsin
0

L = arcsin
b

The length of the graph of
the function f(x) = sqrt(1-x^2), over the interval [0,b], is L = arcsin
b.