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

Solution:

Given: x = 2 sin t and y = 3 cos t

We know that differentiation of parametric functions is 

dy/dx = dy/dt/ dx/dt

x = 2 sin t

dx/dt = 2 cos t

y = 3 cos t

dy/dt = - 3 sin t

Substituting the values

dy/dx = -3 sin t/ 2 cos t

= -3/2 tan t

Again by differentiation

d²y/dx² =[d/t(dy/dx)]. dt/dx

So we get

d²y/dx² = -3/2 sec ²t/ 2 cost

= -3/4 sec³t

Therefore, dy/dx = -3/2 tan t and d²y/dx² = -3/4 sec³t

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

Summary:

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