A line segment AB of length 1 has endpoints A, constrained to the nonnegative y axis, and B, constrained to the nonnegative x axis. Let the region S of the xy plane be defined as follows. A point P is in S if P lies within a right triangle formed by A, B, and the origin for some A,B satisfying the condition above.
What is the area of S? What is the length of its boundary? Find an equation of each smooth segment of the boundary of S in Cartesian coordinates.
The equation for the line segment AB with endpoints A(0, y) and B(x, 0) is:
u(t) - y = (y - 0)/(0 - x) *(t - 0) = -y*t/x
u(t) = y*(1 - t/x)
However, the parameters x and y are not independent since AB has lenght 1:
x
2 + y
2 = 1 => y = √(1 - x
2), 0 x 1
So, we have a family of lines given by the parameter 0 ≤ x ≤ 1:
u(t; x) = √(1 - x
2) * (1 - t/x)
The region S that they are talking about is delimited by the positive x and y axes and the
envelope of the family of curves u(t; x). The equation of the envelope is found from the condition:
∂u/∂x = -[x/√(1 - x
2)]*(1 - t/x) + √(1 - x
2)*t/x
2 = 0 /. x
2*√(1 - x
2)
- x
2*(x - t) + (1 - x
2)*t = 0
-x
3 + x
2*t + t - x
2*t = 0
x
3 = t
x = t
1/3Substitute this value in the equation of the line and we get:
u(t; t
1/3) = √(1 - t
2/3)*(1 - t/t
1/3) = (1 - t
2/3)
3/2, 0 ≤ t ≤ 1
The area of the region S is given by the integral:
A = Int[u dt, {t, 0, 1}]
A = Int[(1 - t
2/3)
3/2 dt,{t, 0, 1}]
This integral can be reduced to Euler Beta function by the substitution:
z = t
2/3 => t = u
3/2, t = 0 -> u = 0, t = 1 -> u = 1, dt = 3/2 u
1/2 du
A = 3/2 Int[u
1/2 (1 - u)
3/2,{u, 0, 1}]
A = 3/2 B(3/2, 5/2)
Using the relation between the Euler Beta and Euler Gamma function, we get:
A = 3/2 Γ(3/2) Γ(5/2)/ Γ(4) = (3/2) * (1/2) * (3/2) * (1/2) * [Γ(1/2)]
2 / 6 = 3*π/32
For the other parts of the problem do it yourself!