Find dy/dx and d²y/dx². x = 3 sin(t), y = 4 cos(t), 0 < t < 2π.

Solution:

Given, x = 3 sin(t), y = 4 cos(t)

We have to find dy/dx and d²y/dx²

dy/dx = (dy/dt)/(dx/dt)

dy/dt = d(4 cos(t))/dt

= -4 sin(t)

dx/dt = d(3 sin(t))/dt

= 3 cos(t)

dy/dx = -4sin(t)/3cos(t)

dy/dx = (-4/3)tan(t)

d²y/dx² = (d(dy/dx)/dt)/(dx/dt)

d(dy/dx)/dt = d((-4/3)tan(t))/dt

= (-4/3)sec²(t)

d²y/dx² = (-4/3)sec²(t)/3cos(t)

d²y/dx² = (-4/9)sec³t

The curve is concave upward when its second derivative is positive.

sec³t is positive for y > 0. So, -sec³t is positive for y < 0.

This implies the curve is concave upward when

y = 4 cos(t) < 0

cos(t) < 0

(π/2) + 2nπ < t < 3π/2 + 2nπ.

Where, n is any integer.

Therefore, dy/dx = (-4/3)tan(t) and d²y/dx² = (-4/9)sec³t.

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Find dy/dx and d²y/dx². x = 3 sin(t), y = 4 cos(t), 0 < t < 2π.

Summary:

x = 3 sin(t), y = 4 cos(t), 0 < t < 2π. dy/dx = (-4/3)tan(t) and d²y/dx² = (-4/9)sec³t.