Eliminate t
If x=2at/(1+t²)
, Y=b(1-t²)/(1+t²)​

x=2at/(1+t²)

dx/dt =[(1+t²)(2at)'-(2at)(1+t²)']/(1+t²)²

dx/dt=[(1+t²)(2a)-(2at)(2t)]/(1+t²)²

dx/dt=[2a+2at²-4at²]/(1+t²)²

dx/dt=[2a-2at²]/(1+t²)²

y=[b(1-t²)]/(1+t²)

dy/dt=[(1+t²)(b(1-t²))'-b(1-t²)(1+t²)']/(1+t²)²

dy/dt=[((1+t²)(-2bt)-b(1-t²)(2t)]/(1+t²)²

dy/dt=(-2bt-2bt³-2bt+2bt³)/(1+t²)²

dy/dt=(-4bt)/(1+t²)²

dy/dx=(dy/dt)/(dx/dt)

dy/dx=[(-4bt)/(1+t²)²]/[(2a-2at²)/(1+t²)²]

Cancelling common denominator

dy/dx=(-4bt)/(2a-2at²)

dy/dx=(-2bt)/(a-at²)

dy/dx=(2bt)/(at²-a)

Differentiating the above equation

y''=d²y/dx²=D[(2bt)/(at²-a)]

y''=[(at²-a)D(2bt)-(2bt)D(at²-a)]/(at²-a)²

y''=[(at²-a)(2b)(dt/dx)-(2bt)(2at)(dt/dx)]/(at²-a)²

dt/dx=(1+t²)²/(2a-2at²)

y''=[(at²-a)(2b)[(1+t²)²/(2a(1-t²))]-

(4abt²)[(1+t²)²/(2a-2at²)]/(at²-a)²

y''=[(-b)(1+t²)²-(2bt²)[(1+t²)²/(1-t²)]/(at²-a)²

y''=[(-b)(1+t²)²(1-t²)-(2bt²)(1+t²)²]/[(1-t²)(at²-a)²]

y''=[b(1+t²)²(t²-1)-2bt²(1+t²)²]/[(1-t²)(at²-a)²]

Step-by-step explanation:

it would take 7/10 of an hour too fill the pool.

step-by-step explanation: