Question

A curve is given by parametrization x=sin(2t),y=sin(3t),0≤t≤2π. Find all points (x,y) where the the tangent line is horizontal.

Answer 1

The points where the tangent line is horizontal are (√3/2, sin(π/3)) and (√3/2, sin(5π/3)).

Step-by-step explanation:

To find the points where the tangent line is horizontal, we need to find the values of t where the derivative of y with respect to t (dy/dt) is equal to 0. So, let's find dy/dt and set it equal to 0.

We have y = sin(3t), so dy/dt = 3cos(3t). Setting this equal to 0, we get 3cos(3t) = 0. Solving for t, we find t = π/6 and 5π/6.

Now, substituting these values of t into x = sin(2t), we can find the corresponding values of x. At t = π/6, x = sin(2π/6) = sin(π/3) = √3/2. At t = 5π/6, x = sin(2(5π/6)) = sin(5π/3) = √3/2. Therefore, the points (x,y) where the tangent line is horizontal are (√3/2, sin(π/3)) and (√3/2, sin(5π/3)).