VCE Methods Question Thread!

[nerdgasm]

Hmm, I believe this style of question has appeared on VCAA exam papers before - for example, multiple choice Q20 on 2010's Exam 2.

I'm not sure if one can use a calculator to solve this style of question (perhaps by defining the initial integral they give you), but it seems like it would be handy to know how to deal with this style of question (which in a way, can be thought of as integration by recognition, if you let F(x) be the parent antiderivative of f(x), and then considering what happens when you differentiate F(kx)).

[bills]

Hey guys, just got a question regarding the Casio Classpad 330:

When graphing a function, sometimes the graph window does not display the graph and I have to adjust the View Window's min/max x and y value. Most of the time it displays the graph, and I can adjust it from there.

However, sometimes I have no idea how the graph looks like so I end up playing with the min/max x and y value, which can take some time. Sometimes, it takes so long I just give up. I have tried the Zoom > Auto feature, but it only works half the time.

Any suggestions?

[cosine]

Hey guys, just got a question regarding the Casio Classpad 330:

When graphing a function, sometimes the graph window does not display the graph and I have to adjust the View Window's min/max x and y value. Most of the time it displays the graph, and I can adjust it from there.

However, sometimes I have no idea how the graph looks like so I end up playing with the min/max x and y value, which can take some time. Sometimes, it takes so long I just give up. I have tried the Zoom > Auto feature, but it only works half the time.

Any suggestions?

This is probably not a very good idea, and you should only use it for last choice when nothing else works. You could derive the function you have, work out the min/max values of the y axes, and then you will know how high or low to set the y-vlaues on the calc!

[Zealous]

Same position as nightrider, but also a) LOL!

There is an alternate method of approaching this question:

- You can think of this expression as the area under f(x) from x=0 to x=9, and that area is going to be 5.

So what is ? Well, f(3x) means that f(x) has been dilated by a factor 1/3 from the y-axis, or squished towards the y-axis. Because we've scaled down one of the linear dimensions down to a third of its original dimensions, the area it takes up will also be scaled down by a factor of 1/3.

Have a look at this image below. It doesn't really matter what type of graph we use, as long as we are scaling down one of the dimensions (in this case the dimension parallel to the x-axis), the area will adjust itself accordingly. You can try it with a few different graphs!



Therefore:  

It's essentially what the 'u' substitution is achieving but from a geometric approach. I might actually cover these types of questions in the end of year lecture since they can be tricky.

[knightrider]

There is an alternate method of approaching this question:

- You can think of this expression as the area under f(x) from x=0 to x=9, and that area is going to be 5.

So what is ? Well, f(3x) means that f(x) has been dilated by a factor 1/3 from the y-axis, or squished towards the y-axis. Because we've scaled down one of the linear dimensions down to a third of its original dimensions, the area it takes up will also be scaled down by a factor of 1/3.

Have a look at this image below. It doesn't really matter what type of graph we use, as long as we are scaling down one of the dimensions (in this case the dimension parallel to the x-axis), the area will adjust itself accordingly. You can try it with a few different graphs!

(Image removed from quote.)

Therefore:  

It's essentially what the 'u' substitution is achieving but from a geometric approach. I might actually cover these types of questions in the end of year lecture since they can be tricky.

Nice explanation Zealous 

[StupidProdigy]

For a linear approximation question asking to find the value of where the given function of f(x) is , do I only take the positive approximation because the domain of f(x) is >0?? thankss

[Springyboy]

If a piece of wire is 80cm long find the area of the largest rectangle that can be formed by the wire.

[keltingmeith]

For a linear approximation question asking to find the value of where the given function of f(x) is , do I only take the positive approximation because the domain of f(x) is >0?? thankss

I'm honestly surprised you got a negative approximation. Care to share your working?

[keltingmeith]

If a piece of wire is 80cm long find the area of the largest rectangle that can be formed by the wire.

So, consider the rectangle with length l, width w and area A. Then, we have:




We also know that we only have 80cm of wire to make this rectangle, so we know that the perimeter of the rectangle will be 80 cm:



Now, if we solve this for l, we can actually get an expression for our area in terms of one variable:



Which leaves us with a quadratic. Can you find the maximum of a quadratic?

[Springyboy]

So, consider the rectangle with length l, width w and area A. Then, we have:




We also know that we only have 80cm of wire to make this rectangle, so we know that the perimeter of the rectangle will be 80 cm:



Now, if we solve this for l, we can actually get an expression for our area in terms of one variable:



Which leaves us with a quadratic. Can you find the maximum of a quadratic?

All good at finding quadratic max thanks for help

[StupidProdigy]

I'm honestly surprised you got a negative approximation. Care to share your working?

It came from the first step. I had h=-0.1, x=4, and f'(x)=something, anyway it was the f(x) part of the approximation formula. I had f(4), so 1/sqrt(4) and this is where I got +1/2 and -1/2...stupid mistake yeah haha?

[bills]

Can someone please explain step by step how to solve this?

2cos3(x-(pi/4)) = 0, x E [-2pi, 2pi]

This is my working out so far, but I have no idea how to continue.
2cos3(x-(pi/4)) = 0
cos3(x-(pi/4)) = 0
3(x-(pi/4)) = pi/2, 3pi/2
Sorry for the lack of LaTeX.

[cosine]

Can someone please explain step by step how to solve this?

2cos3(x-(pi/4)) = 0, x E [-2pi, 2pi]

This is my working out so far, but I have no idea how to continue.
2cos3(x-(pi/4)) = 0
cos3(x-(pi/4)) = 0
3(x-(pi/4)) = pi/2, 3pi/2
Sorry for the lack of LaTeX.

Before we can solve it, we have obviously altered the domain of the function. So we must do what we have done to the domain, and apply it to the solutions we find:







And so now we solve the below equation for this new domain:



Way too tedious to do the negative solutions on here, so you can do that





I hope I have not done a mistake anywhere, hope that helps

[Apink!]

Hi guys

I was doing the Kilbaha 2004 exam when I came across something called, "Hypergeometric" formulas. I haven't seen one in my cambridge textbook. Is this in the methods course?

One thing that I noticed, was that I was able to solve the question much more quickly using that formula rather than having to do tedious calculations.

Could someone explain this for me? Is this safe to use in the methods exam (a way to double check answers or something)

Thank you!

[Apink!]

Could someone mind helping me out?

I got A but apparently the answer is meant to be B.

Any ideas why?