Question

dt/dx= ((x^2+a^2)^(1/2))/v1+((b^2+d^2)^(1/2))/v1

the function dt/dx has a zero at a unique x on (0,d). Use this to justify your discovery that
d= (x^2+a^2)^(1/2) + (b^2+d^2)^(1/2)

Answer 1

as it stands, I think there's something wrong.

you want dt/dx = 0

but since √n is always >= 0, that means that

√(x^2+a^2) + √(b^2+d^2) = 0

since even powers are always non-negative, that means that x,a,b,d are all zero.

and 0 is not in the interval (0,d)