Evaluate y and write in terms of u and v
`y = sqrt(t ) ( sqrt(t) + 1 )^2`
u = t , v = `( sqrt(t) + 1 )^2`

x=sqrt(sin 2t),y=sqrt(cos 2 t)

x=sinsqrt(t),y=e^(sqrt(t)) find dy/dx

(1)/(2)int(dt)/(t sqrt(t-1))

y=t^(2)/(sqrt(t)) then find dy/dt

Find (dx)/(dt) when x = sin^-1(t.sqrt(1-t) +sqrt(t) sqrt(1-t^2)) .

The slopee of normal to the curve x=sqrt(t) and y=t-(1)/(sqrt(t)) at t=4 is . . .

find (dy)/(dx) , when y = sin u ,, u = e^(sqrt t) and t = log x

if f(t)=t^(3)+sqrt(2)t^(2)+3(sqrt(2)t)-1, then evaluate f(2sqrt(2))

Locus of the middle points of the line segment joining P(0, sqrt(1-t^2) + t) and Q(2t, sqrt(1-t^2) - t) cuts an intercept of length a on the line x+y=1 , then a = (A) 1/sqrt(2) (B) sqrt(2) (C) 2 (D) none of these