Author Topic: Specialist Problem (Read 10440 times) Tweet Share

Mao · « **Reply #45 on:** April 09, 2008, 06:14:45 pm »

Quote from: Tea.bag on April 09, 2008, 05:37:15 pm

another problem...

Find the range of $f(x)=cos(sin^{-1}2x)$

this is a composite function, so first we need to find the range of $\sin^{-1}2x$ , which is $[-\frac{\pi}{2},\frac{\pi}{2}]$

that means $\cos$ takes the domain of of $[-\frac{\pi}{2},\frac{\pi}{2}]$ , which would give u a range of $[0,1]$

Tea.bag · « **Reply #46 on:** April 09, 2008, 06:27:29 pm »

i dont get it..

$cos\frac{\pi}{2}=-1?$

Mao · « **Reply #47 on:** April 09, 2008, 06:29:47 pm »

argh ooops that meant to be a +1 =S

Tea.bag · « **Reply #48 on:** April 09, 2008, 07:04:58 pm »

i still dont get it..

can u explain why $cos\frac{\pi}{2}=1$ Isnt it 0??

Mao · « **Reply #49 on:** April 09, 2008, 07:17:08 pm »

in the interval of $u\in [ -\frac{\pi}{2},\frac{\pi}{2}]$ , $\cos(u)$ has its minimum value at $\cos\left(\pm \frac{\pi}{2}\right)=0$ , and the maximum value occurs at $\cos(0)=1$
The range is written as (minimum,maximum), not (left,right).

hence the range is $[0,1]$

Tea.bag · « **Reply #50 on:** April 17, 2008, 05:00:27 pm »

more questions..

1) An aeroplane can travel at a speed of 120km/h when there is no wind. Today there is a wind velocity of $3i+2j$ km/h.

Find:
a) the speed of the wind
b) the position vector required if the aeroplane is to end up due north of its starting point after 1 hour
c) the actual bearing required

2) Another aeroplane can travel at a speed of 240 km/h when there is no wind. Today there is a wind velocity of $-3i-7j$ km/h.

Find:
a) the speed of the wind
b) the position vector required if the aeroplane is to end up due east of its starting point after 2 hours
c) the actual bearing required

Mao · « **Reply #51 on:** April 17, 2008, 08:28:19 pm »

1)
a)
$v_w = |3i+2j| = \sqrt{13}$ km/h

b)
let $P=ai+bj$ be the position vector of the aeroplane after 1 hour:

for the plane to end up due north, the $i$ component must be 0, hence $a=-3$

$|P|=\sqrt{(-3)^2 + b^2} = 120$

$b=3\cdot \sqrt{1599} \approx 119.96$

$P=-3i + 119.96j$

c) as the word "bearing" is quite ambiguous, i'm going to give it how the weatherman gives it: (N something E or S something W)

from the positive $j$ axis

$tan(\theta) = \frac{-3}{119.96}$

$\theta \approx -1.43^o$

N1.43^oW

2)
a)
$v_w=|-3i-7j|=\sqrt{58}$

b)
east is the positive $i$ direction, hence the displacement in the $j$ direction must be 0:

let $P=ai+bj$ be the position vector of the plane after 2 hrs

$b=-(-7 \times 2) = 14$

$|P|=\sqrt{a^2 + 14^2} = 480$

$a=2 \cdot \sqrt{57551} \approx 479.80$

$P=479.80i + 14j$

c)
$tan(\theta)=\frac{479.80}{14}$

$\theta \approx 88.33^o$

N88.33^oE

hope that helped.

Tea.bag · « **Reply #52 on:** April 18, 2008, 07:06:50 pm »

Tea.bag · « **Reply #53 on:** May 01, 2008, 10:50:18 pm »

Help!

1) Consider the triangle $ABC$ . Point $D$ is the midpoint of line $AB$ and $E$ is the midpoint of line $BC$ .

Show, using vectors to represent various lines, that line $DE$ is parallel to $AC$ and $DE=\frac{1}{2}AC$

2) If the length of a vector , $\vec{a}$ , is given by $|\vec{a}|$ , show geometrically that for any two vectors $\vec{a}$ and $\vec{b}$ :

$|\vec{a}|+|\vec{b}|\ge|\vec{a}+\vec{b}|$

dcc · « **Reply #54 on:** May 01, 2008, 11:24:34 pm »

Q1: TRIANGLE ABC:

$\mbox{Let }\vec{AB} = \tilde{a}$

$\mbox{Let }\vec{AC} = \tilde{b}$

:. $\vec{BC} = \vec{BA} + \vec{AC} = \tilde{b} - \tilde{a}$

Given $\vec{AD} = \dfrac{1}{2}\tilde{a}$

and: $\vec{BE} = \dfrac{1}{2}\vec{BC} = \dfrac{1}{2}(\tilde{b} - \tilde{a})$

Therefore:

$\vec{DE} = \vec{DA} + \vec{AB} + \vec{BE} = -\dfrac{1}{2}\tilde{a} + \tilde{a} + \dfrac{1}{2}(\tilde{b} - \tilde{a}) = \dfrac{1}{2}\tilde{a} + \dfrac{1}{2}\tilde{b} - \dfrac{1}{2}\tilde{a} = \dfrac{1}{2}\tilde{b} = \dfrac{1}{2}\vec{AC}$

dcc · « **Reply #55 on:** May 01, 2008, 11:33:06 pm »

For the second one, I would suggest drawing a triangle, and note that the shortest distance between two points is a straight line. By drawing two vectors and adding them head to tail, notice that the resultant vector is always the same magnitude (if a & b are parallel) or less. Im not sure how thorough a geometric proof should be, but that seems like a rational explanation.

Tea.bag · « **Reply #56 on:** May 25, 2008, 06:02:37 pm »

Help!!!

A spider builds a web in a garden. The position vectors of the ends A and B of a strand of the web are given by $\vec{OA}=2i+3j+k$ and $\vec{OB}=3i+4j+2k$ respectively.

a) Find:
1) $\vec{AB}$
2) the length of the strand

b) A small insect is at point C where $\vec{OC}=2.5i+4j+1.5k$ . Unlucky it flies in a straight line and hits the strand of web between A and B. Let Q be the point where the insect hits the strand where $\vec{AQ}=\lambda{AB}$ .

1) Find $\vec{CQ}$ in terms of $\lambda$
2) If the insect hits the strand at right angles find the value of $\lambda$ and $\vec{OQ}$

c) Another web strand $\vec{MN}$ has end points M and N with position vectors $\vec{OM}=4i+2j-k$ and $\vec{ON}=6i+10j+9k$ . The spider decides to continue AB and to join MN. Find the position vector of the point of contact.

ed_saifa · « **Reply #57 on:** May 25, 2008, 06:57:44 pm »

a)
1)
$\vec{AB}= \vec{OB}-\vec{OA}$

$\vec{AB}=3i + 4j + 2k -(2i + 3j + k)$

$\vec{AB}= i + j + k$

2)
$|\vec{AB}|=$ length of strand

$|\vec{AB}|= \sqrt{1^2 + 1^2 + 1^2}$

$|\vec{AB}|= \sqrt{3}$

ell · « **Reply #58 on:** May 25, 2008, 07:20:38 pm »

b)
1)
$\vec{CQ}= -\vec{OC}+\vec{OQ}$
$= -\vec{OC}+\vec{OA}+\vec{AQ}$
$= -2.5i-4j-1.5k+2i+3j+k +\vec{AQ}$
$= -2.5i-4j-1.5k+2i+3j+k + \gamma(i+j+k)$ since $\vec{AQ}=\gamma\vec{AB}$

Expand out and simplify:
$= -2.5i-4j-1.5k+2i+3j+k + \gamma i +\gamma j+\gamma k$
$= (\gamma - 0.5)i + (\gamma - 1)j +(\gamma -0.5)k$

2)
"at right angles", i.e. perpendicular.
$\vec{CQ}.\vec{AB}=0$
$[(\gamma-0.5)i+(\gamma-1)j+(\gamma-0.5)k].(i+j+k)=0$
$\gamma-0.5+\gamma-1+\gamma-0.5=0$
$3\gamma-2=0$
$\gamma=\frac{2}{3}$

$\vec{OQ}=\vec{OA}+\vec{AQ}$
$\vec{OQ}=2i+3j+k+\gamma\vec{AB}$
$\vec{OQ}=2i+3j+k+\gamma(i+j+k)$
Sub in the $\gamma$ value:
$\vec{OQ}=2i+3j+k+\frac{2}{3}(i+j+k)$
$\vec{OQ}=2i+3j+k+\frac{2}{3}i+\frac{2}{3}j+\frac{2}{3}k$
$\vec{OQ}=\frac{8}{3}i+\frac{11}{3}j+\frac{5}{3}k$
$\vec{OQ}=\frac{1}{3}(8i+11j+5k)$

Tea.bag · « **Reply #59 on:** May 25, 2008, 08:24:41 pm »

cool thanks..
another question..

VABCD is a square based pyramid and O is the centre of the base. $i, j, k$ are unit vectors in the direction of AB, BC and OV respectively. the point O is to be taken as the origin for position vectors. AB=BC=CD=DA=4cm
OV = 2h cm where h is a positive real number.
P,Q,M and N are the midpoints of AB,BC,VC and VA respectively.

a) Find the position vecots A,B,C and D respectively to O.
b) Find vectors $\vec{PM}$ and $\vec{QN}$ , in terms of h.
c) Find the position vector $\vec{OX}$ , of X the point of intersection of QN and PM.
d) If OX is perpendicular to VB
1) Find the value of h
2)calculate the angle between PM and QN giving your answer correct to the nearest 0.1degrees.
e) 1)Prove that NMQP is a rectangle
2)Find h if NMPQ is a square.

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