Mathematics 9709 · AS & A Level · Numerical solution of equations
Numerical solution of equations — practice question
A curve is defined by the parametric equations $x = \dfrac{1}{(2t + 1)^2}$, $y = \sqrt{t + 2}$. The point $P$ on the curve is associated with parameter $p$, and the gradient at $P$ is stated to be $-1$.
(i)[6]
Show that the value of $p$ satisfies $p = \sqrt{p + 2} - \dfrac{1}{2}$.
(ii)[3]
Apply an iterative method based on the equation in part (i) to determine $p$ correct to 3 decimal places. Take $0.7$ as the starting value and display the outcome of each iteration to 5 decimal places.
Worked solution & mark scheme
This 9-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Find a derivative of the form $k(2t + 1)^{-3}$” …