Mathematics 9709 · AS & A Level · Numerical solution of equations
Numerical solution of equations — practice question
The diagram represents the curve with parametric equations $x = 3t - 6e^{-2t}$, $y = 4t^2 e^{-t}$, for $0 \leq t \leq 2$. Point $P$ lies on the curve, and its $y$-coordinate is $1$.
(i)[2]
Show that the value of $t$ at point $P$ satisfies the equation $t = \dfrac{1}{2} e^t$.
(ii)[3]
Apply the iterative formula $t_{n+1} = \dfrac{1}{2} e^{t_n}$ starting from $t_1 = 0.7$ to determine the value of $t$ at $P$ correct to $3$ significant figures. Record the result of each iteration to $5$ significant figures.
(iii)[5]
Find the gradient of the curve at $P$, giving the 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 $4t^2e^{-t}$ equal to $1$, rearrange to $t^2=\dots$ and so obtain $t=\dots$” …