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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: golden on October 26, 2011, 12:01:56 pm

Title: VCAA 2008 Exam 2 Q4e.
Post by: golden on October 26, 2011, 12:01:56 pm

Could someone please explain how to do that question?

Thank you.

Also, with question 5d, do we have to put an open dot for the Arg(z) graph?

Title: Re: VCAA 2008 Exam 2 Q4e.
Post by: nubs on October 27, 2011, 05:43:24 pm

I just briefly looked at the question so I might be missing something
There are a few ways to do this if I'm right, the easiest on to explain is:
If you look at the graph, it seems as if the minimum and maximum for the rabbits occur where the gradient is undefined
So if you look at dy/dx, which has already been given to you, you can see that the gradient is undefined where the denominator = 0
So at
25x-5xy=0
5x-xy=0
x(5-y)=0
so at x = 0 or y = 5
We can tell from the graph that it can't be at x = 0, so it must be at y = 5
So if you sub in y = 5 into the equation of the graph with value c = 27.5, and the solve for x, you should get 2 values, one being a minimum and one being a maximum
and also by observation of the graph, it seems as if y = 5 makes sense

But again, I haven't read the whole question so I can't be 100% sure
lemme know how it works out

Title: Re: VCAA 2008 Exam 2 Q4e.
Post by: nubs on October 27, 2011, 05:45:47 pm

for question 5d, not sure why you would need an open point, never really thought of it before though

Title: Re: VCAA 2008 Exam 2 Q4e.
Post by: golden on October 28, 2011, 09:02:35 am

Quote from: Nirbaan on October 27, 2011, 05:43:24 pm

I just briefly looked at the question so I might be missing something
There are a few ways to do this if I'm right, the easiest on to explain is:
If you look at the graph, it seems as if the minimum and maximum for the rabbits occur where the gradient is undefined
So if you look at dy/dx, which has already been given to you, you can see that the gradient is undefined where the denominator = 0
So at
25x-5xy=0
5x-xy=0
x(5-y)=0
so at x = 0 or y = 5
We can tell from the graph that it can't be at x = 0, so it must be at y = 5
So if you sub in y = 5 into the equation of the graph with value c = 27.5, and the solve for x, you should get 2 values, one being a minimum and one being a maximum
and also by observation of the graph, it seems as if y = 5 makes sense

But again, I haven't read the whole question so I can't be 100% sure
lemme know how it works out

All good. Thanks for replying!