Mathematics 9709 · AS & A Level · Numerical solution of equations
Numerical solution of equations — practice question
The diagram depicts the curve given by the parametric equations $x = 3t - 6e^{-2t}$, $y = 4t^2 e^{-t}$, for $0 \le t \le 2$. At the point $P$ on this curve, the $y$-coordinate is $1$.
(i)[2]
Show that the value of $t$ at the point $P$ satisfies the equation $t = \frac{1}{2}e^t$.
(ii)[3]
Use the iterative formula $t_{n+1} = \frac{1}{2}e^{t_n}$ with $t_1 = 0.7$ to determine the value of $t$ at $P$ correct to $3$ significant figures. Give each iteration to $5$ significant figures.
(iii)[5]
Find the gradient of the curve at $P$, and give your answer correct to $2$ significant figures.
Worked solution & mark scheme
This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Set $4t^2e^{-t} = 1$, rewrite it as $t^2 = \cdots$ and hence determine $t$.” …