A curve is defined by parametric equations: x = t^(2) + 2, and y = t(4-t^(2)). Find dy/dx in terms of t, hence, define the gradient of the curve at the point where t = 2.

dy/dx = (dy/dt)/(dx/dt) y = t(4-t² ), then using differentiation of y with respect to t, dy/dt = 4 - 3t²x = t² + 2, then using differentiation of x with respect to t, dx/dt = 2t Find dy/dx by dividing dy/dt by dx/dt (as the dt terms cancel to leave dy/dx):dy/dx = (4-3t²)/2tNow that the equation of the gradient of the parametric curve has been found (dy/dx), substitute the given value of t to establish the gradient of the curve at t = 2. dy/dx = (4-3(2)²)/2(2) = -2 (the gradient of the curve at t = 2 is dy/dx = -2)