A curve has parametric equations x= 2sin(t) , y= cos(2t) + 2sin(t) for -1/2 π≤t≤ 1/2π , show that dy/dx = - 2sin(t)+ 1

A parametric equation is where both x and y are expressed separately, in terms of a parameter (t). In order to differentiate them we must use the chain rule, which here would be dy/dx= dy/dt ÷ dx/dt. The differential of sin(t)= cos(t) , and the differential of cos(t)= -sin(t). To find dy/dt, we must differentiate each part of the y equation separately (so y= u + v). For du/dt we must use the chain rule again because there is a 2 in front of t and we don't want du/d(2t). du/dt = du/dk*dk/dt. Let 2t= k, so u= cos(K) , therefore du/dt= -sin(k) * 2 = -2sin(2t). The differential of 2sin(t) is simply 2cos(t). dy/dt= du/dt + dv/dt= -2sin(2t) + 2cos(t). Then, dx/dt= 2cos(t). Therefore dy/dx= (-sin(2t) + 2cos(t))/ 2cos(t). We must simplify this in order to make it match the derivative given in the question. First, re-write -2sin(2t) using the double angle formula in order to remove the 2 before the t. Sin(2t) = sin (t+t)= 2sin(t)cos(t) , so -2sin(2t)= -4sin(t)cos(t). Inserting this back into our dy/dx gives (-4sin(t)cost(t) + 2cos(t))/2cos(t). Then we can factorise the numerator, to get (2cos(t)(-2sin(t)) + 1)/ 2cos(t). This 2cos(t) terms cancel out to give dy/dx= -2sin(t)+1 , which is the answer. If we wanted to find the stationary point, we would set dy/dx = 0 and solve for t, then sub this parameter value into the original x and y parametrics to find the coordinates.