Spesh Exam Questions

[TheRajinator]

VCAA 2008 exam 2 Q22 I am really not sure how to approach this question. I also don't understand what the kilbaha solutions have done. Can someone please explain this question? Thanks.

[b^3]

First note that is a function of , that is .
So we also know that

Now since we have a function of v, we can't just integrate with respect to t, we need to integrate with respect to v, but we don't have dv on the bottom, so we need to flip it

So now we can integrate both sides with respect to v.

Now we have a but not a or a as we haven't account for our initial conditions, and on the left hand side we are left with a change in , which we need to make sure it corresponds correctly to our terminals.
So at , and at , .
So integrating from to our change in is (final take initial)
So we are left with


Which gives the answer of B.

Hope that helps

EDIT: I really shouldn't start posting on AN during lectures.....

[TheRajinator]

wow thanks that was very clear. 

[TheRajinator]

Another question: VCAA 2006 exam 2 Q5 a) ii) I didn't understand how they got this equation.

[Lasercookie]

Looking at the the stuff we plotted onto the argand diagram in Q5ai, we can see that the straight line joining and is the perpendicular bisector of the straight line joining and .

Giving a pretty rough explanation:

A perpendicular bisector is a line that cuts another line at right angles (self explanatory). Also each point on the perp. bisector is equidistant from the endpoints of the line that it's cutting (in this case the endpoints are and ). (The animation here is neat http://mathworld.wolfram.com/PerpendicularBisectorTheorem.html)

If we have something of this form, where a and b are complex numbers. The equation says that the distance from z to a is the same as the distance from z to b. So we can see that this represents the straight line that perpendicularly bisects the points a and b. You could probably explore more proofs like this http://ceemrr.com/Geometry1/RightTriangles/RightTriangles2.html. I'm getting a bit off topic now, so back to the question


So

[TheRajinator]

Looking at the the stuff we plotted onto the argand diagram in Q5ai, we can see that the straight line joining and is the perpendicular bisector of the straight line joining and .

If we have something of this form, where a and b are complex numbers. The equation says that the distance from z to a is the same as the distance from z to b. So we can see that this represents the straight line that perpendicularly bisects the points a and b. You could probably explore more proofs like this http://ceemrr.com/Geometry1/RightTriangles/RightTriangles2.html. I'm getting a bit off topic now, so back to the question


So

You lost me on this part. I get everything about perpendicular bisectors but I don't understand how we suddenly use this notation: When you say that the distance from z to a what is z? Is this just the general form of straight lines in complex form?

[Lasercookie]

You lost me on this part. I get everything about perpendicular bisectors but I don't understand how we suddenly use this notation: When you say that the distance from z to a what is z? Is this just the general form of straight lines in complex form?

z is just any complex number representing a point on the line formed, z = x+yi (like what we do with all that loci stuff).

In general, for the distance between two complex numbers, say .
So if and a is another complex number, say (x,y,c and d are real numbers)
We know using pythagoras that the distance between those two points will be .

This step might be a bit iffy, but from that, we can see that's equal to (modulus of a complex number)
Continuing on

Does that help? (I'm not too sure if I've answered your question).

[TheRajinator]

Cheers Bro for the detailed answer, I understand. Much appreciated.
Another question from the Kilbaha Exam 2 2007 Q2 (e)

I got that cos(theta)= root(130)/13
But the next part asks for the closest distance from point A to line CD.
I did it using a different approach.

My Approach:
I find theta: cos^-1(root(130)/13) hence getting theta =28.71
then we know that the shortest distance =13sin(theta) and sub in theta to get 6.245 units for the shortest distance.
However this answer is wrong in the solutions. Can you tell me if I am doing anything wrong.

[Jenny_2108]

Cheers Bro for the detailed answer, I understand. Much appreciated.
Another question from the Kilbaha Exam 2 2007 Q2 (e)

I got that cos(theta)= root(130)/13
But the next part asks for the closest distance from point A to line CD.
I did it using a different approach.

My Approach:
I find theta: cos^-1(root(130)/13) hence getting theta =28.71
then we know that the shortest distance =13sin(theta) and sub in theta to get 6.245 units for the shortest distance.
However this answer is wrong in the solutions. Can you tell me if I am doing anything wrong.

I think the problem is you should use the orginal dimensions of the triangle
I have a look solution to see why you get wrong
They wrote but in your answer, you use the other form

Thus if you wanna follow your own method, you should calculate the shortest distance is
Anyway, why don't you use pythagoras theorem? Its safer and less confusing

[TheRajinator]

Thank you thank you I see what I did wrong. I didn't use pythagoras theorem because I thought that the length we found root(130) was the length of CD, so it doesn't make sense for us to use the root(130) in pythagoras theorem as it extends beyong point A??

[Jenny_2108]

Thank you thank you I see what I did wrong. I didn't use pythagoras theorem because I thought that the length we found root(130) was the length of CD, so it doesn't make sense for us to use the root(130) in pythagoras theorem as it extends beyong point A??

You can't use the root(130) for pythagoras theorem either because and have the same ratio but they are different triangles. I think you call them as similar triangles

[TheRajinator]

Sorry I mean you can't use 2root(5) in pythagoras theorem because it represents the length of CD. But that's what the solutions have done and they got the right answer. But why can they use 2root(5) doesn't that represent length of CD not where A intersects CD at right angles? How would you use pythag?

[Jenny_2108]

Sorry I mean you can't use 2root(5) in pythagoras theorem because it represents the length of CD. But that's what the solutions have done and they got the right answer. But why can they use 2root(5) doesn't that represent length of CD not where A intersects CD at right angles? How would you use pythag?

The shortest distance is the perpendicular line so you can pythagoras theorem

[TheRajinator]

Okay here is what I meant. the 2root(5) isn't that the distance from C to D so how can we use that to find h?

[BubbleWrapMan]

I don't see a quick way to do it with pythagoras