VCE Methods Question Thread!

[Ancora_Imparo]

First, you need to find the equation of the parabolic cross-section.

Spoiler

Let the vertex be at (0,0). Thus, the parabola will be of equation . Since it's two metres deep, and 1 1/2 metres across at the top, you know that the end points of the cross-section must be at (-3/4, 2) and (3/4, 2). Substituting one of these points and solving for gives you the equation:

Next, you need to find the area of your cross-section. You need to integrate to do this.

Spoiler

When the trough is half-full, the area of the cross-section made by the water will be exactly half of the value you just found above. Let the depth of the water be . Find the width of the water at this point. Thus, you can derive an equation involving the depth of the water which you can solve.

Spoiler

Rearrange equation to find width of water when half-full. The right end-point of the water at this depth will be at .

[Smiley_]

First, you need to find the equation of the parabolic cross-section.

Spoiler

Let the vertex be at (0,0). Thus, the parabola will be of equation . Since it's two metres deep, and 1 1/2 metres across at the top, you know that the end points of the cross-section must be at (-3/4, 2) and (3/4, 2). Substituting one of these points and solving for gives you the equation:

Next, you need to find the area of your cross-section. You need to integrate to do this.

Spoiler

When the trough is half-full, the area of the cross-section made by the water will be exactly half of the value you just found above. Let the depth of the water be . Find the width of the water at this point. Thus, you can derive an equation involving the depth of the water which you can solve.

Spoiler

Rearrange equation to find width of water when half-full. The right end-point of the water at this depth will be at .

thanks


how did you ge the 9d/32

[Ancora_Imparo]

Rearranging for gives:

We need to know the exact point on the parabola that the water level touches as it will be our upper limit in our integral. We know that the y-coordinate is . Subbing it into the above equation gives .

[zhe0001]

da hello

Using symettry of the unit circle, how would you find the exact value of ?

I know the answer is 1/2 because its the equilent of cos (60)

But not sure how to do the correct working steps on paper..like that method of working


thanks!

[Professor Polonsky]

[zhe0001]

would that get full marks on a test?

[Professor Polonsky]

Yes. In fact, you could definitely skip the second line. There's nothing more that you can write, in any case.

[09Ti08]

A long trough whose cross-section is parabolic is 1 and /2

metres wide at the top and 2 metres
deep. Find the depth of water when it is half-full

I tried to solve this problem using the area and it works. I integrated along both the x and y axes and ended up with the same answer. However, when I actually look at the volume (this is the first thing that came into my mind when I read the question, yet I hesitated as this method is not taught in the course):
. This answer is very different from what I got from the other methods.

Can anyone one please explain what's wrong with this? Thank you.

[Phy124]

I tried to solve this problem using the area and it works. I integrated along both the x and y axes and ended up with the same answer. However, when I actually look at the volume (this is the first thing that came into my mind when I read the question, yet I hesitated as this method is not taught in the course):
. This answer is very different from what I got from the other methods.

Can anyone one please explain what's wrong with this? Thank you.

I believe you're missing some square roots.

[09Ti08]

Thanks, but I think the formula for calculating the volume of a paraboloid rotating around the y axis is , and x^2 in this problem is 9*y/32.

[b^3]

The volume is not a volume of revolution, rather it's a cross section that extends into the page, which is why you can do it just by areas.

[09Ti08]

Ah, now I see that word "long" in the question, thanks b^3!

[Phy124]

Thanks, but I think the formula for calculating the volume of a paraboloid rotating around the y axis is , and x^2 in this problem is 9*y/32.

Oh I see what you were trying to do, didn't even notice the 's when I edited the code which would've given away that you were trying to equate volumes -.-

As bcub3d stated it's not a volume of revolution so that isn't going to work, it's easiest to equate areas the way I showed in my post (minus the useless pi symbols )

edit: Just realised I totally skipped the sentence where you said you were attempting to do it by volumes, too. I'm doing well tonight...

[Zealous]

What do you guys do when you come across a practice exam which is really hard?

I was doing a Kilbaha Exam 2 and it completely killed my brain (and I'm severely lacking sleep hhahah).

...So do you stop doing the exam midway?
...or just do whatever you can and accept a low mark?
.. or do other companies exams until you are ready to take it on again? ideas?

[Alwin]

What do you guys do when you come across a practice exam which is really hard?

I was doing a Kilbaha Exam 2 and it completely killed my brain (and I'm severely lacking sleep hhahah).

...So do you stop doing the exam midway?
...or just do whatever you can and accept a low mark?
.. or do other companies exams until you are ready to take it on again? ideas?

Oh them kilbaha ones haha, yeah they're the hardest ones I've come across so far.

Umm, since its killing you , I would "build up to it", I think the order in approximate hardness was:
Heffernan < VCAA  < Insight < MAV < Neap < Kilbaha

well, that's my opinion of them haha, depends on your own taste for exams
Find a company that challenges you but not impossible, then work your way up to harder ones then when you're more "battle harded" smash that Kilbaha =D