What is the length of the path along the graph of y=sqrt(9-x^2) between x=0 and x=2?
dy/dx = (1/2) (9 - x^2)^(-1/2) ( -2x) = -x / ( 9 - x^2)^(1/2)
[dy/dx]^2 = x^2 / [ 9 - x^2]
The arc length is given by
2
∫ √ ( 1 + [dy/dx]^2 ) dx =
0
2
∫ √ ( 1 + x^2 / [ 9 - x^2] ) dx =
0
2
∫ √ ( 9 / [ 9 - x^2] ) dx =
0
2
3 ∫ √ ( 1 / [ 9 - x^2] ) dx =
0
let x = 3sin (θ) → x^2 = 9sin^2( θ)
dx = 3cos (θ) dθ
Change the limits of integration
when x = 2 when x = 0
2/3 = sin (θ) 0 = sin (θ)
θ = arcsin(2/3) θ = 0
So we have
arcsin(2/3)
3 ∫ √ ( 1 / [ 9 - 9sin^2( θ) ] ) 3 cos ( θ) dθ =
0
arcsin(2/3)
3 ∫ (1/3) √ ( 1 / [ 1 - sin^2( θ) ] ) 3 cos ( θ) dθ =
0
arcsin(2/3)
3 ∫ √ ( 1 / [ 1 - sin^2( θ) ] ) cos ( θ) dθ =
0
arcsin(2/3)
3 ∫ √ ( 1 / [cos^2 ( θ) ] ) cos ( θ) dθ =
0
arcsin(2/3)
3 ∫ ( 1 / [cos ( θ) ] ) cos ( θ) dθ =
0
arcsin(2/3)
3 ∫ dθ =
0
arcsin(2/3)
3 [ θ ] = 3 [ arcsin(2/3) - 0 ] = 3 arcsin(2/3)
0
Answer "b"
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