VCE Methods Question Thread!

[lzxnl]

Someone please refresh my memory on how to do these types of questions,

"An aircraft, used for fire spotting, flies from its base to locate a fire at an unknown distance, x km away. It travels straight to the fire and back, average 240km/h for the outward trip and 320km/h for the return trip. If the plane was away for 35 minutes, find the distance, x km."

Please provide steps I just blanked out while trying to interpret these types of questions.

Let the first part of the journey take time a and let the second part of the journey take time b (times in hours). Then, the distance on the outward journey is given by 240a and the return journey takes 320b. These used distance=speed times time. We know 240a=320b=x as it's the same journey.
We also know that a+b=35/60=7/12 as this is the total time in hours.

Can you solve these equations?

[T-Infinite]

Let the first part of the journey take time a and let the second part of the journey take time b (times in hours). Then, the distance on the outward journey is given by 240a and the return journey takes 320b. These used distance=speed times time. We know 240a=320b=x as it's the same journey.
We also know that a+b=35/60=7/12 as this is the total time in hours.

Can you solve these equations?

I'm so confused, which equations are you referring to?

[lzxnl]

I've given you three equations
240a=x
320b=x
a+b=7/12

Where a and b are in hours, x is in km

[Phy124]

t = d/v, therefore time taken to fire = x/240 and time taken to return = x/320, sum these together to equal the total time of 35/60 (divide by 60 so everything is in hours)

x/240 + x/320 = 35/60

x = 80km

Ah it appears I am too slow and lzxnl has already made a response, but I'll just leave this here because you seem confused.

[BLACKCATT]

Can i please get some help on this question?

[T-Infinite]

t = d/v, therefore time taken to fire = x/240 and time taken to return = x/320, sum these together to equal the total time of 35/60 (divide by 60 so everything is in hours)

x/240 + x/320 = 35/60

x = 80km

Ah it appears I am too slow and lzxnl has already made a response, but I'll just leave this here because you seem confused.

This is less confusing! haha, I get it now

I've given you three equations
240a=x
320b=x
a+b=7/12

Where a and b are in hours, x is in km

Lol, I didn't know what to do with those equations because the a and b thing really confused me.

[M_BONG]

Can someone lend me a hand here? Thanks

"The cross section of a channel is parabolic. It is 3 metres wide at the top and 2 metres deep. Find the depth of water, correct to nearest cm, when the channel is half full"

Answer: Depth = 126cm

[Phy124]

Can someone lend me a hand here? Thanks

"The cross section of a channel is parabolic. It is 3 metres wide at the top and 2 metres deep. Find the depth of water, correct to nearest cm, when the channel is half full"

Answer: Depth = 126cm

Easiest way to do it I think is like so:

Let the bottom of the channel (the minimum of the parabola) pass through the origin.

We then know the parabola has equation .

We can then use the point (1.5,2) to find the equation of the parabola as .

So we're dealing with this:

https://www.desmos.com/calculator/znrsivtgvz


We want to found a value for d such that the area between the graphs and is half that of the area between the graphs of and .

The latter is given by:



The former is given by:



I integrated between and because these are the values of x at which y = d found by solving

We can simply equate this integral to equal half of the original area to find the value of d at which the area is half.





It should look something like this in a calculator:

Code: [Select]

solve(int(d-8x^2/9,x,-sqrt(9d/8),sqrt(9d/8))=2,d)

Another way which I previously did it is in a spoiler below:

Spoiler

*same stuff above the horizontal line above here*

One way to go about it would be to look at just the right side and convert the equation from y in terms of x to x in terms of y. This way we can integrate between 0 and d such that we find a value for d that the area under the graph between these points is half that of the area under the graph between 0 and 2 (the full height)

Rearranging we obtain:



So we're looking at this atm https://www.desmos.com/calculator/yutdassioz

Now all we have to do is form an equation to represent the area in the interval [0,d] underneath being half that of the area in the interval [0,2] underneath , this is given by;



Putting the following into your calculator should do it.

Code: [Select]

solve(int(sqrt(9y/8),y,0,d)=1/2*int(sqrt(9y/8),y,0,2),d)
This gives the depth as approximately 1.26m

It might be helpful to draw the y axis as the horizontal axis and the x axis as the vertical axis when you convert them around. I've noticed a few students can become confused understanding the situation and integrating with respect to y if they are left in the conventional manner. For example the situation would like more like this https://www.desmos.com/calculator/0j93yyfjja which is much more familiar with students.

p.s. I'm sorry for the lack of clarity in this post, feel free to ask for a better explanation if you don't understand 

[Nato]

is tried graphing , and was wondering what the asymptotes of this graph are. I have the correct shape of it, but the position of the aymptotes, particularly the horizontal one is confusing.
thanks

[psyxwar]

is tried graphing , and was wondering what the asymptotes of this graph are. I have the correct shape of it, but the position of the aymptotes, particularly the horizontal one is confusing.
thanks

There is an asymptote at y=-1 that is approached by the two ends of the graph (and not by the middleish section) if that makes sense.

[M_BONG]

is tried graphing , and was wondering what the asymptotes of this graph are. I have the correct shape of it, but the position of the aymptotes, particularly the horizontal one is confusing.
thanks

The way I would do this (although I may be wrong) is to disregard the negative sign first so we now have
Then, since we are trying to find the horizontal asymptote we let .

If the denominator is a very large number (let x become a very large value) , the fraction part of becomes very small in value. Eg is small as the graph continue and the denominator value gradually approaches infinity. Thus, the fraction part is largely irrelevant in value. This means that we can ignore the fraction part and therefore, the function gradually approaches (but never reaches) y = -1. However, there is an absolute value symbol meaning that everything becomes positive. Therefore y = 1. BUT, there is a negative sign at the front so the asymptote is y = -1


Then find x and y intercepts (if they exist) and other relevant values such as vertical asymptote (x= 0 in this case)
Hope this makes sense haha.

[psyxwar]

If I was asked to find how many apples Bob could buy with $200 and got an answer of say, 150.9, I'm supposed to round down to 150 right?

[Yacoubb]

If I was asked to find how many apples Bob could buy with $200 and got an answer of say, 150.9, I'm supposed to round down to 150 right?

Yeah! Because technically Bob couldn't buy 151 apples with $200, but 150 apples means they can purchase the apples with the given amount!

[psyxwar]

Yeah! Because technically Bob couldn't buy 151 apples with $200, but 150 apples means they can purchase the apples with the given amount!

Well yeah, but I mean how does VCAA access rounding?

I guess a better would be having to work out from say, an exponential graph, how many years it takes for a population to double. If you get 2.3 years, do you write 3 years or 2.3 years as your final answer?

[Snorlax]

This is from essentials extended response Q4

The base of a 3 m ladder leaning against a wall is x metres from the wall.
-Express the distance, d, from the top of the ladder to the ground as a function of x and
sketch the graph of the function.

Is it just simply:




Is that what the question is asking for?