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

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)

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

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

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

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