Question

Given that x=cos^2t and y=ln(sint) find d^2y/dx^2 at the point t=π/4

Answer 1

y = ln(sint)
dy/dt = (1/sint)(cost) = cost/sint

for x= cos^2 t = (cost)^2
dx/dt = 2cost(-sint)

(dy/dt) / (dx/dt) = (cost/sint) / (-2sintcost)
dy/dx = -1/(2 sin^2 t) = (-1/2)(sint)^-2

d(dy/dx) / dx = (-2)(-1/2)(sint)^-3 (cost)
= cost / sin^3 t
sub in t = π/4

d(dy/dx) / dx = (√2/2) / (√2/2)^3
= 1/(√2/2)^2
= 1/(2/4) = 2

Answer 2

x=cos^2t
y=ln(sint)

dy/dt = cost/sint
dx/dt = -2cost sint

dy/dx = dy/dt / dx/dt

= (cost/sint)/(-2cost sint)
= -1/2 csc^2(t)

d^2y/dx^2 = d/dx(dy/dx)
= d/dt (dy/dx) / dx/dt
= (-csct * csct cott)/(-2cost sint)
= 1/2 csc^2(t) cot(t) * sect csct
= 1/2 csc^3(t)