The curve C has parametric equations x = t², y = (2 - t)^1/2, for 0 ≤ t ≤2. Find d2y/dx2 in terms of t.
Can I have the full workings also please?

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)]

fist on the left goes with the third on the right. second on the left goes with the fourth on the right. third on the left goes with the second on the right. the fourth on the left goes with first on the right

maybe i can