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Author Topic: TrueTears question thread  (Read 66914 times)  Share 

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Over9000

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Re: TrueTears question thread
« Reply #270 on: March 07, 2009, 11:00:25 pm »
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thanks
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TrueTears

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Re: TrueTears question thread
« Reply #271 on: March 07, 2009, 11:03:36 pm »
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another one

Points A, B and C have position vectors andrespectively.

Find AC : BC

i keep getting 5:-3

but can you have a negative ratio? Or is that wrong lols
this is the next part to that question

Find ., where

how can you dot a magnitude with a ?

And how would you do this question

thanks!
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kamil9876

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Re: TrueTears question thread
« Reply #272 on: March 07, 2009, 11:23:42 pm »
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Generally, a dot means multiplication, it is used in certain books and in some countries in order to avoid ambiguity as 6x3(6 times 3) can be confused with 6 multiplied by some number x then multiplied by 3. You can see that . looks like the dot product without the other magnitude, so it makes sense that this question involves the dot product and hence further shows that the dot means multiplication.
Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

TrueTears

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Re: TrueTears question thread
« Reply #273 on: March 07, 2009, 11:59:08 pm »
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the asymptotes are:



from the given gradients of the asymptotes, (q=5, p=2.5)
i think you mean 2a = b :P

« Last Edit: March 08, 2009, 12:02:51 am by TrueTears »
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Re: TrueTears question thread
« Reply #274 on: March 08, 2009, 12:40:50 am »
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1.




are the 2 parametric equations for the hyperbola with equation . Determine the range of values of t which will represent the hyperbola without duplication.

Basically i've worked out the range of the hyperbola is R and domain is R\(1,4).

so range of the y = parametric equation must = range of cartesian = R

and range of the x = parametric equation must = domain of cartesian = R\(1,4)

What do i do from  here? Like if i have works for the parametric equation but then for the parametric equation is not enough to draw a whole hyperbola. But if i have then that allows the equation to work but it will have made part of the hyperbola duplicated. What do i do?
« Last Edit: March 08, 2009, 01:53:09 am by TrueTears »
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TrueTears

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Re: TrueTears question thread
« Reply #275 on: March 08, 2009, 01:50:40 am »
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another one

Points A, B and C have position vectors andrespectively.

2. this is the next part to that question

the position vector of P is . P is a point on nearest to B.

Find the position vector of D, the centre of the circle through A, B and P

« Last Edit: March 08, 2009, 03:43:30 pm by TrueTears »
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Re: TrueTears question thread
« Reply #276 on: March 08, 2009, 01:58:18 am »
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3. Engineers are trying to make a chute, they consider the equation . Express the rule for this curve in the form
, where r is a positive interger.
« Last Edit: March 08, 2009, 04:00:07 pm by TrueTears »
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Re: TrueTears question thread
« Reply #277 on: March 08, 2009, 02:08:33 am »
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4. In the triangle OAB, let M, N and P be the midpoint of the sides OB, AB and OA respectively. Let the perpendicular bisectors of sides AB and OA meet at G as shown. Let the position vector of P and M be p and m respectively and let be denoted by r as shown.



a) show that m = p + r

this i have done

b) i. Let be detnoed by v and be denoted by w, as shown. One possible expression for is = v - p + m = v+r, by a)
Use this expression to show that . p = r . p

ii. In a similar way show that , and hence that

this i have done

c)Use your results from a) and b) to show that

this i have done

d) What conclusion can you draw from a) , b) and c)?

How do you do d) ?
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/0

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Re: TrueTears question thread
« Reply #278 on: March 08, 2009, 06:21:43 am »
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another one

Points A, B and C have position vectors andrespectively.

2. this is the next part to that question

the position vector of P is . P is a point on nearest to B.

Find the position vector of D, the centre of the circle through A, B and P



If you draw a diagram you'll notice that OP must be the vector resolute of OB in the direction of OA, since that makes OA perpendicular to PB, which gives the shortest distance.

Then, since the circle goes through A, B and P, and , AB must be a diameter, with D the midpoint of AB.
« Last Edit: March 08, 2009, 06:32:26 am by /0 »

TrueTears

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Re: TrueTears question thread
« Reply #279 on: March 08, 2009, 04:00:16 pm »
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thanks /0
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Re: TrueTears question thread
« Reply #280 on: March 08, 2009, 04:02:27 pm »
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oh nvm i got Q 3
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Re: TrueTears question thread
« Reply #281 on: March 08, 2009, 05:55:24 pm »
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and another one

points A B C and D are defined by position vectors a b c and d respectively. If
a) express d in terms of a, b and c

so
and

b - a + d - c = 0

d = c + a -b

b) Show that AC and BD bisect each other
How do you do this part?

Thanks
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Mao

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Re: TrueTears question thread
« Reply #282 on: March 08, 2009, 06:04:53 pm »
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AC = c - a
BD = d - b = c + a - 2b

let X be the midpoint of AC, then, AX = XC = 0.5 (c - a)
BX = BA + AX = (a - b) + 0.5 (c - a) = 0.5 c + 0.5 a - b = 0.5 BD

implies X is also the midpoint of BD

hence, AC and BD bisect each other at X
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Re: TrueTears question thread
« Reply #283 on: March 08, 2009, 06:32:11 pm »
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Thanks mao

also got any ideas on the other 2? Q1 and Q4 d)?
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kamil9876

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Re: TrueTears question thread
« Reply #284 on: March 08, 2009, 09:58:33 pm »
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Q1.

Because you are considering the maximum domain and range of the hyperbola, then the only way to yield this is to ensure that the parametric equations have maximum range( I am pretty sure). Hence the and bits must hav their maximum ranges. The maximum ranges for and occur when is [0,pi)/{pi/2} for both functions. Now in this case so t has a domain of i.e: a domain translated units to the positive side. No duplication arises since is just one period for both of those trig functions and the values don't repeat.
« Last Edit: March 08, 2009, 10:32:36 pm by kamil9876 »
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