MF26

Differentiate $x=e^t$ with respect to t.
$\frac{dx}{dt} = e^t$
Substitute $x = e^t$ into given differential equation,
$\Rightarrow e^t \frac{dy}{dt} = \mathrm{cos}(\mathrm{ln}e^t)\mathrm{sin}(\mathrm{ln}(e^t)^3)$
$e^t \frac{dy}{dt} = \mathrm{cos}t \mathrm{sin} (3t)$
Since $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$
$\frac{dy}{dt} = \frac{1}{e^t} \mathrm{cos}t \mathrm{sin} (3t) \times e^t$
$\frac{dy}{dt} = \mathrm{cos}t \mathrm{sin} (3t)$
$\int \frac{dy}{dt} dt = \int \mathrm{cos}t \mathrm{sin} (3t) dy$
Using the product to sum formula as shown here, we have
$\int \frac{dy}{dt} dt = \frac{1}{2} \int \mathrm{sin} 4t + \mathrm{sin} 2t dt$
$y = \frac{1}{2}(-\frac{\mathrm{cos}4t}{4} - \frac{\mathrm{cos}2t}{2}) + C$
$y = \frac{1}{2}(-\frac{\mathrm{cos}(4 \mathrm{ln}x)}{4} - \frac{\mathrm{cos}(2\mathrm{ln}x)}{2}) + C$

Note: $i = \begin{bmatrix}1\\ 0\\ 0\end{bmatrix}$ and $j = \begin{bmatrix}\\ 1\\ 0\end{bmatrix}$
$\pi: 2x + 3y = -6$ — (1)
$\pi \prime: x + y + z= 5$ — (2)
$\pi_1: x = 0$ — (3)
Using GC, $\vec{OA} = \begin{bmatrix}0\\ -2\\ 7\end{bmatrix}$

$\pi: 2x + 3y = -6$ — (1)
$\pi \prime: x + y + z= 5$ — (2)
$\pi_1: y = 0$ — (3)
Using GC, $\vec{OA} = \begin{bmatrix}-3\\ 0\\ 8\end{bmatrix}$

Qn11
(i)
$l: r = \begin{bmatrix}-1\\ -1\\ 0\end{bmatrix} + \alpha \begin{bmatrix}1\\ 1\\ 1\end{bmatrix}, \alpha \in \mathbb{R}$
(ii) Since l is the common line of intersection on $\pi_1$ and $\pi_2$, we need l to be on $\pi_3$ too. For that to happen,
1. l must be parallel to $\pi_3$, that is, direction of l is perpendicular to normal to $\pi_3$
$\Rightarrow \begin{bmatrix}1\\ 1\\ 1\end{bmatrix} \bullet \begin{bmatrix}5\\ {\lambda}\\ 17\end{bmatrix} = 0$
$\lambda = -22$
2. Given that l is parallel to $\pi_3$ (since $\lambda = -22$), we need l to be on $\pi_3$, so we need $\begin{bmatrix}-1\\ -1\\ 0\end{bmatrix}$ to be on $\pi_3$
$\Rightarrow \begin{bmatrix}-1\\ -1\\ 0\end{bmatrix} \bullet \begin{bmatrix}5\\ -22\\ 17\end{bmatrix} = \mu$
$\mu = 17$
(iii)
For 3 planes to have nothing in common, then l must be parallel to $\pi_3$ (Note: if l is not parallel, l will cut $\pi_3$ at a point, which means that 3 planes will cut at a point)
$\Rightarrow \lambda = -22$ from (ii)
But $\mu \neq 17, \mu \in \mathbb{R}$

12.
(i)
$\pi_1 : r = \begin{bmatrix}4\\ 2\\ 12\end{bmatrix} + \lambda \begin{bmatrix}2\\ 2\\ 1\end{bmatrix} + \mu \begin{bmatrix}5\\ -1\\ -5\end{bmatrix}, \mu , \lambda \in \mathbb{R}$
$\vec{OP} = \begin{bmatrix}4\\ 2\\ 12\end{bmatrix}$ is on $\pi_1$ when $\lambda = \mu = 0$

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