Reply With QuoteMore snuggled up than me on a cold Sunday morning under my warm blanket.Originally Posted by asianese
I'm assuming that means sit snugly inside the parabola? OMG I JUST SOLVED IT!! SO HAPPY WTF OMG HAHHAHAHAHA. I feel...so smart...yet among some of these questions...
The general equation of a circle is. Here
. Solving simultaneously the parabola and circle,
. As the solution is a double root, the discriminant equals 0.
.
. The circle has centre
.
By a similar process, the centre of the circle in the parabolais
(Different k to top one)
Last edited by asianese; 28 May 2012 at 11:58 PM.
USYD: B Sc (Adv Maths) / (B A)? I
More snuggled up than me on a cold Sunday morning under my warm blanket.
Bachelor of Science (Adv. Mathematics) - University of Sydney:
This thread is practically dead, so thought I might post a question. But unlike many of the others posted so far, a 2U student can do it.
1?I didn't use any particular mathematical method to do it, though.
"It was my husband's, before he died." "Oh...I'm sorry." "Don't be, I killed him."
University of New South Wales - Computer Science
haha yeh, but you have to show how to get the answer.1?
I guessed that the first time aswell :P
This thread is practically dead, so thought I might post a question. But unlike many of the others posted so far, a 2U student can do it.
I think the conclusion is that a can indeed equal to b, but you gave us restricted boundaries, so yeah heh.
'An art, which has an aim to achieve the beauty, is called a philosophy or in the absolute sense it is named wisdom.'- Al Farabi
I think the conclusion is that a can indeed equal to b, but you gave us restricted boundaries, so yeah heh.
^I tried to do it without log laws, so that's how I did it.
The curves are arcs of circles whose centers are vertices of the square, side 6cm. Find the area of the shaded region.
Hopefully I didn't make a calculation error.
Currently studying:
PhD (Pure Mathematics) at ANU
Correct.
Hopefully I didn't make a calculation error.
Here's a question I posted a while ago but I don't remember if anybody answered it (don't think so). Best done by the interested and well-read HSC student.
I am standing at the origin. I move to the right by one unit, then up by 1/2, then left by 1/3, then down by 1/4, then right again by 1/5 etc etc.
If I continue this infinitely, is there a coordinate I will (eventually) reach? If so, what is it?
---------------------------------------
Another:
I am a 1-Dimensional person (so 1 degree of freedom). This means I can only either move forwards or backwards.
I move forward 1 unit, then backward 1/2 units, then forward 1/3 units, then backwards 1/4 units etc etc.
If I continue this infinitely, is there a coordinate I will (eventually) reach? If so, what is it?
---------------------------------------
Another:
Consider a regular n-gon. What is the total number of m-gons (not necessarily regular) that I can make such that m < n ? (careful about rotational symmetry).
Bachelor of Science (Adv. Mathematics) - University of Sydney:
Is the answer to the first one
)
Last edited by IamBread; 19 Jun 2012 at 7:05 AM.
Signature
Yep! Methinks that is a really cool result. No doubt if you could the first, then the second is trivial. Try going up 1 more dimension to 3D.
Bachelor of Science (Adv. Mathematics) - University of Sydney:
Yeah, and it was a lot of fun to do too. I am doing the first one now, so far got the z direction coordinate to be
Signature
I like this question. The hardest part is finding a closed expression for the sum that determines the number of m-gons. I think a good Extension 2 student should be able to do so if they play around a bit with Complex Numbers. I will post a solution later if nobody else does.
I know this question can be done by elementary means by partitioning various areas, but is it possible to evaluate it using calculus?
The curves are arcs of circles whose centers are vertices of the square, side 6cm. Find the area of the shaded region.
 \left | \frac{\partial (x_1,x_2)}{\partial (y_1,y_2)} \right | dy_1 dy_2)
Pretty sure it can be done using the transformation formula you have there (by perhaps mapping it out to a triangular domain) but then you would need to find the points of intersections etc, which would be quite tedious.
Bachelor of Science (Adv. Mathematics) - University of Sydney:
Found a pretty good one that most 2U students should try.
Find the area of the rectangle.
Nice!
Bachelor of Science (Adv. Mathematics) - University of Sydney:
Is it sqrt(3)?Found a pretty good one that most 2U students should try.
Find the area of the rectangle.
Interesting one for a 4U student that I saw in one of our school's past papers:
 for which:} \\ \textup{arg}(z-2)=\textup{karg}(z^2-2z) "\textup{Given }|z-2|=2\textup{ and }0<\textup{argz}<\frac{\pi}{2}, \textup{ find the value of k (real) for which:} \\ \textup{arg}(z-2)=\textup{karg}(z^2-2z)")
That's what i got, but I don't have answer lol
I remember doing this a while ago now.Interesting one for a 4U student that I saw in one of our school's past papers:
k=2/3
Here is another good one:
The lengths of the sides of the octagon are 1, 2, 3, 4, 5, 6, 7 and 8 units in some
order. Find the maximum area of the hexagon (square units).
Yeh sqrt(3) confirmed.
There are currently 1 users browsing this thread. (0 members and 1 guests)
Posting Permissions
Bookmarks