VCE Methods Question Thread!

[Hancock]

y = 0?

[FlorianK]

solve for y

y=2y |-2y
y-2y=0
y(1-2)=0
by NFL: y=0 or 1-2=0
as 1-2 =/=0, y=0

[hurdles]

Can anyone help me with this?

[Hancock]

This is of the form:

ax + by = e
cx + dy = f

If the det(A) = ad - bc = 0, there will either be no solution to the system or infinite solutions.





Substituting into the system (to check if infinite solutions or no solutions come from m= 9):



These are not the same equation, therefore m = 9 will yield no solution.

[Stick]

Can anyone help me with this?

All it involves is proving a singular matrix, hurdles. It's important to run the solution back into the equations, however, to ensure they don't have infinite solutions (ie they are the same line).

[Homer]

This is how i would do it not sure if its right :/

For no solution the lines should have same gradients and different y intercept



 



and
solving for m, also

[hurdles]

This is of the form:

ax + by = e
cx + dy = f

If the det(A) = ad - bc = 0, there will either be no solution to the system or infinite solutions.





Substituting into the system (to check if infinite solutions or no solutions come from m= 9):



These are not the same equation, therefore m = 9 will yield no solution.

Thanks a heap

[FlorianK]

I like to take a more logical approach to those types of questions rather than the way which hancock described, even though his way is much faster for this problem, or the matrix way.

Lets say you have two linear graphs
y=ax+b and y=cx+d

For those types of questions there are 3 things they can ask for
(1) a unique solution
This simply means that the cross each other, hence the gradients need to be different so in this case

(2) no solution
This means that they are excactly parallel to each other, but they have a different y-intercept.
So in this cas and

(3) infinitly many solutions
This means for linear graphs that the graphs are exactly on top of each other, meaning that the graphs are exactly the same.
So in this case and

So for your problem you just solve for y and then equate the gradients and then look if for the value(s) you get the y-intercepts are not equal to each other.

I attached the solution using this method below.

[Hancock]

That's great Florian, it's probably best if everyone knows multiple methods of completing questions so they double check. Great explanation.

[darklight]

Can anyone help me with this question re: integration (from the Essentials book)
-Ch12,Chapter review, SA Q13 - screenshot attached

[Hancock]

I'm pretty sure this is more a Spesh topic but we can do it in methods I suppose.

Let's change the question so that it looks more familiar. I'm going to rotate the axis/picture/plane 90 degrees anticlockwise, and then reflect it in the new vertical axis. This makes our graph look like a negative parabola with the equation:



Solving for the new axial intercepts (by letting y = 0) tells us that they are at x = 0 and x = 6. So we know that point A on the picture is at (0,6). Remember that we have just changed the x and y co-ordinates for convenience.

Point B is where the line y = x intersects the parabola. To solve for this intersection, we equate the two y vales.


Solving this yields x = 0 and x = 5.

Therefore, point B is at (5,5) since point B is not the origin.


Now, let's find the area under the entire parabola. This can be done by integrating the parabola from x = 0 to x = 6.
This will yield an area of 36 units^2.



To find the area between 2 curves, we do the integration of (upper function - lower function) between the bounds. As we can see from the graph, is the LOWER function and our changed co-ordinate function of is the UPPER FOR THE DOMAIN OF x = 0 to x = 5.

Therefore,  integration of from x = 0 to x = 5 is equal to the Area Q.

Evaluating gives

Substituting back into our P+Q equation gives

[darklight]

Wow, thanks so much! 

[TrueTears]

Hancock has nailed this question, however just providing a different perspective (albeit outside the methods/spesh course but the whole point of this thread is for cool discussions )

We're gonna use something called double integrals. Basically the general form:

Area enclosed by the graphs and and the x bounds and

In this case:

we have so

[brightsky]

Hancock has nailed this question, however just providing a different perspective (albeit outside the methods/spesh course but the whole point of this thread is for cool discussions )

We're gonna use something called double integrals. Basically the general form:

Area enclosed by the graphs and and the x bounds and

In this case:

we have so

or you can use the fact that in general area between two curves = int^(right bound)_(left bound) (top curve) - (bottom curve) dx.  the two are equivalent.

[TrueTears]

ya, i just felt like being different