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)
MW
Related Maths A Level answers
All answers ▸Integrate 1/u(u-1)^2 between 4 and 2
Differentiate the equation y^2 + y = x^3 + 2x
Why is the derivative of x^2 equal to 2x?
A curve has equation y = 2x^5 + 5x^4 1 . (a) Find: (i) dy/ dx [2 marks] (ii) d^2y/ dx^2 (b) The point on the curve where x ¼ 1 is P. (i) Determine whether y is increasing or decreasing at P, giving a reason for your answer.