Answer:
y''=(4-3t)/[16 t^3 (2-t)^(3/2)]
Step-by-step explanation:
x = t², y = (2 - t)^1/2
dy/dx=dy/dt×dt/dx by chain rule
dy/dt=1/2 (2-t)^(1/2-1) × (-1)
dy/dt=-1/2 (2-t)^(-1/2)
dy/dt=-1/[2(2-t)^(1/2) ]
dx/dt=2t
dy/dx=-1/[2(2-t)^(1/2) ] × 1/[2t]
dy/dx=-1/[4t (2-t)^(1/2) ]
We need to find the second derivative now.
That is we calculate d/dt(dy/dx in terms of t) then divide by derivative of x in terms of t).
dy/dx=-1/[4t (2-t)^(1/2) ]
Let's find derivative of this with respect to t.
d/dt(dy/dx)=
[0[4t (2-t)^(1/2)]-(-1)(4(2-t)^(1/2)+-4t(1/2)(2-t)^(-1/2))]/ [4t (2-t)^(1/2) ]^2
Let's simplify
d/dt(dy/dx)=
[(4(2-t)^(1/2)+-4t(1/2)(2-t)^(-1/2))]/ [4t (2-t)^(1/2) ]^2
Continuing to simplify
Apply the power in the denominator
d/dt(dy/dx)=
[(4(2-t)^(1/2)+-4t(1/2)(2-t)^(-1/2))]/ [16t^2 (2-t) ]
Multiply by (2-t)^(1/2)/(2-t)^(1/2):
d/dt(dy/dx)=
[(4(2-t)+-4t(1/2)]/ [16t^2 (2-t)^(3/2)]
Distribute/multiply:
d/dt(dy/dx)=
[(8-4t+-2t)]/ [16t^2 (2-t)^(3/2)]
Combine like terms:
d/dt(dy/dx)=
[(8-6t)]/ [16t^2 (2-t)^(3/2)]
Reducing fraction by dividing top and bottom by 2:
d/dt(dy/dx)=
[(4-3t)]/ [8t^2 (2-t)^(3/2)]
Now finally the d^2 y/dx^2 in terms of t is
d/dt(dy/dx) ÷ dx/dt=
[(4-3t)]/ [8t^2 (2-t)^(3/2)] ÷ 2t
d/dt(dy/dx) ÷ dx/dt=
[(4-3t)]/ [8t^2 (2-t)^(3/2)] × 1/( 2t)
d/dt(dy/dx) ÷ dx/dt=
[(4-3t)]/ [16t^3 (2-t)^(3/2)]
Or!!!!!!
x = t², y = (2 - t)^1/2
Since t>0, then t=sqrt(x) or x^(1/2).
Make this substitution into the equation explicitly solved for y:
y = (2 - x^1/2)^1/2
Differentiate:
y' =(1/2) (2 - x^1/2)^(-1/2) × -1/2x^(-1/2)
y'=-1/4(2 - x^1/2)^(-1/2)x^(-1/2)
y'=-1/4(2x-x^3/2)^(-1/2)
Differentiate:
y''=1/8(2x-x^3/2)^(-3/2)×(2-3/2x^1/2)
y''=(2-3/2x^1/2)/[8 (2x-x^3/2)^(3/2)]
Replace x with t^2
y''=(2-3/2t)/[8 (2t^2-t^3)^(3/2)]
Multiply top and bottom by 2
y''=(4-3t)/[16 (2t^2-t^3)^(3/2)]
Factor out t^2 inside the 3/2 power factor:
y''=(4-3t)/[16 (t^2)^(3/2) (2-t)^(3/2)]
y''=(4-3t)/[16 t^3 (2-t)^(3/2)]
