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

### 2 Answers | Add Yours

The lngth of graph y = f(x) is given by:

L = Int sqrt{1+(dy/dx)^2} dx, from x = 0 to x = b.

Therefore first we find dy/dx.

y = f(x) = (1-x^2)(1/2).

Therefore dy/dx = d/dx (1-x^2)^(1/2)

dy/dx = (1/2) (1/1-x^2)^(1/2-1) * (-2x)

dy/dx = -2x/{2(1-x^2)^(-1/2)

dy/dx = -x/(1-x^2)^(1/2)

sqrt(1+(dy/dx)^2) = sqrt{1+x^2/(1-x^2)^1/2}

sqrt(1+(dy/dx)^2) = sqrt { 1+x^2/1-x^2)

sqrt(1+(dy/dx)^2) = sqrt{(1-x^2+x^2)/(1-x^2)}

sqrt(1+(dy/dx)^2) = 1/(sqrt(1-x^2)

Therefore F(x) = Integral sqrt{1+(dy/dx)^2} dx = Int {1/(sqrt(1-x^2)}dx

F(x) = arc sin (x)

Therefore ,

L = F(b) - F(a) = arc sin (b) arc sin (0)

L = F(b) - F(0) = arc sinb as arc sin 0 = 0.

So the require length od arc on x from x= 0 to x = b is arc sinb.

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.**