VCE Methods Question Thread!

[Bhootnike]

A car travels half the distance of a journey at an average speed of 80km/h and half at an average speed of x km/h
Define a function, S, which gives the average speed for the total journey as a function of x

s = (80 + x ) / d , where x= the speed for the second half of the journey
d = total disance

hope im right lol, doesnt look right since theres 2 variable ... but ahhh, fingers crossed

[Insa]

Hey,

Does anyone know how to get this hyperbola equation y=x+3/x-2 into y=a/x-b +c?

Thank you.

[b^3]

Just remember whatever you add you have to minus so that it still equal.

[rife168]

A car travels half the distance of a journey at an average speed of 80km/h and half at an average speed of x km/h
Define a function, S, which gives the average speed for the total journey as a function of x

s = (80 + x ) / d , where x= the speed for the second half of the journey
d = total disance

hope im right lol, doesnt look right since theres 2 variable ... but ahhh, fingers crossed

It should be average speed = total distance / total time
Let d= half the total distance.
Total time = t₈₀ + t_x
t₈₀ = d/80    t_x = d/x
t_total = d/80 + d/x = d(1/80 + 1/x)
S = total distance/t_total = 2d/d(1/80 + 1/x) = 2/(1/80 + 1/x)

[Insa]

May I ask what this method is called and how do you use it? This is the first time I've seen this. Thanks for the help so far.

[b^3]

May I ask what this method is called and how do you use it? This is the first time I've seen this. Thanks for the help so far.

I don't know if it has a name, its kinda just intuitive. Like you could use long division for this, but this methods is quicker (well for me anyway, I liked using this method, and seemed to make more mistakes using simple long division)

Anyway, you want to get a factor out, so firstly you factor everything so that the coefficeint of x on the top is 1, in this case it already is. Next you need to get the same thing on the top as you do on the bottom, so we are trying to get a (x-2) on the top, easiest way to achieve that is to take too form the top (-2), now if we do that, we must keep the whole thing equal so we have to add 2 to the top as well. Then you can seperate it out and cancel one of the factors down.

Or by long divison

[Insa]

Great, I get it now! Thanks for the help!

[TrueTears]

May I ask what this method is called and how do you use it? This is the first time I've seen this. Thanks for the help so far.

It's called wishful thinking, http://abagoffruit.wordpress.com/2010/05/23/wishful-thinking-in-mathematics/

I'm serious.

[kamil9876]

I've always tried to use that sort of method instead of some long division algorithm. I eventually developed similair techniques for more difficult examples and then eventually wrote out a general procedure... I then realized that this turns out to be the long division algorithm, but written in a less compact but more "reason-based" way.



Here are some other examples:

a)


b)

The general idea is that if you have you want to somehow write the numerator as a multiple of plus a polynomial of smaller degree, ie you want to write then this simplifies down to , then the procedure continues...  Now how can you find such polynomials and ? I'll save that for another day but in the case of Q(x) being linear like above it isn't so hard.

[TrueTears]

I eventually developed similair techniques for more difficult examples and then eventually wrote out a general procedure... I then realized that this turns out to be the long division algorithm

hence why you're a mathematician

[Mr. Study]

Just this question. Oil is increasing at a rate of 10 metres.hour. Find the rate when the radius is 5m.

I have no idea how to find rate when radius is 5m.

[trinh]

A triangle ABC with A(1,1) and B(-1,4). The gradients AB, AC and BC are -3m,3m and m respectively.
a) find the value of m
b) find the coordiantes of C
 c) show that AC=2AB

Solutions to this question attached.

[trinh]

Just this question. [The radius of a circular patch of] oil is increasing at a rate of 10 metres/hour. Find the rate [of increase of area of the oil patch] when the radius is 5m.

I have no idea how to find rate when radius is 5m.

Before I answer your question, is the above what you mean (with the bold parts added)?

Edit: LOL I'll just go ahead and write an answer under the assumption that this is indeed what you mean. In the case that I misunderstood you, what I've written should still be useful to you

First write down everything you know:

, where r represents the radius of the oil patch in metres, and t represents time in hours.
and
(area of circle formula) (by differentiation), where A is the area of the oil patch in square metres.

Next, determine what you need to find. That is, when r=5.

Lastly, write what you need to find, in terms of what you already know:

(chain rule)

Substituting in our 'known values', it is thus seen that <= make sure to use the correct rate; since our rate involves , we know our rate must be "square metres per hour".

Finally, substitute in the value of given to you in the question:



Thus,

Final answer: [when its radius is 5m], the area of the oil patch increases at a rate of

[cnguyen599]

Hey guys, I got a question that has troubled me the past 30 minutes:

An aircraft flies from its base at an unknown distance, x km away. It travels straight there and back, averaging 240km/h for the outward trip and 320 km/h for the return. If the plane was away 35 minutes, find the distance, x km.

The answer is 80km

[b^3]

Firstly
Rearrange for t

Let t₁ be the time taken for the journet there and t₂ be the time taken for the journey back.


Total time is t₁+t₂, which is (convert it to hours)






EDIT: I'd thought I'd done this question before, its on pg 11 of this thread, Re: Methods [3/4] Summer Holidays Question Thread!