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)]

eve and marks ratios are equivalent

step-by-step explanation:

answer: it's d

step-by-step explanation: