H2 Math Tue 5pm

This page contains all questions and answers asked by students from this class. The most recent questions will be at the top.

MF26

Question 1b

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
$\int \frac{dy}{dt} dt = \frac{1}{2} \int \mathrm{sin} 4t + \mathrm{sin} 2t dt$
$y = \frac{1}{2}(-\frac{\mathrm{cos}4t}{4} - \frac{\mathrm{cos}2t}{2}) + C$
$y = \frac{1}{2}(-\frac{\mathrm{cos}(4 \mathrm{ln}x)}{4} - \frac{\mathrm{cos}(2\mathrm{ln}x)}{2}) + C$

Question 2(iv)

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, $\vec{OA} = \begin{bmatrix}-3\\ 0\\ 8\end{bmatrix}$

Qn 11

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 $\mu \neq 17, \mu \in \mathbb{R}$

Qn12

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 $\lambda = \mu = 0$

About

KS Teng

KS has been teaching H2/H1 Mathematics and IB mathematics for the more than 10 years. Having taught students from all various Junior Colleges, KS adapts to students' abilities and help them better understand the topics. As someone who loves to teach mathematics and sees it as a truly useful tool in life, KS seeks to enable students to appreciate math. Therefore, his tuition mission is to motivate and cultivate students to be independent and confident thinkers.

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