A curve has parametric equations -> x = 2cos(2t), y = 6sin(t). Find the gradient of the curve at t = π/3.

First we need to find the derivatives of x and y in terms of t. dx/dt can be found using the chain rule. Differentiating the inside of the bracket gives us 2. Multiplying the outside gives -2sin(2t) (Derivative of cos given in the formula book). Multiplying these terms together gives us dx/dt = -4sin(2t). For dy/dt its a simple differential from the formula book. dy/dt = 6cos(t). In order to find dy/dx, we can apply the chain rule one more time -> dy/dx = dy/dt*dt/dx. dt/dx is the reciprocal of dx/dt -> dt/dx = 1/(-4sin(2t)). Therefore dy/dx = 6cos(t)/-4sin(2t) = -3cos(t)/2sin(2t). Now we can input our value of t to find the gradient at t =  π/3 (Use a calculator). dy/dx = -3cos(π/3)/2sin(2π/3) = -(3/2)/root(3). Multiplying top and bottom by root(3) gives the final answer of -root(3)/2.