Mathematics 9709 · AS & A Level · Numerical solution of equations

Numerical solution of equations — practice question

A curve is defined by parametric equations $x = \frac{1}{(2t + 1)^2}$ and $y = \sqrt{t + 2}$. The point $P$ on the curve is associated with parameter $p$, and the gradient at $P$ is stated to be $-1$.
(i)[6]

Hence show that $p = \sqrt{p + 2} - \frac{1}{2}$.

(ii)[3]

Apply an iterative method based on the equation in part (i) to determine $p$ correct to 3 decimal places. Begin from $0.7$ 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: Obtain a derivative of the form $k(2t+1)^{-3}$

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