2007-2008 Study List Discussion

[bradp]

http://www.uil.utexas.edu/academics/cal ... List08.pdf

[The3point14]

Is anyone finding this year's Study List to be much easier than last year's?

I'm needing help with #s

04A-68
04F-67
04H-80
06A-60

Any help would be appreciated.


link:


Thank you,
The3point14, TheThing, and the Nederland CA team

XD

[thething2]

Don't worry about 06A-60. The only reason I was missing it was because my calculator was in radians. That's the 2nd straight day that I've had that error. At least it's not as bad as writing diameter instead of radius or reading the calculator wrong(which I did recently also).

[bradp]

04F-67

[unparseable or potentially dangerous latex formula] k is a constant, n is the amount of gas.
in the 8 ft, 20 psi case, we'll say there is 1 unit of volume and 1 unit of gas.
in the mixed case, there is [unparseable or potentially dangerous latex formula] units of volume because there is the 8 ft tank plus the 12 ft tank which has all of its dimensions scaled by 1.5 (12/8). There is also 2 units of gas... so

20*2 times as much gas = 40 psi.... divide by [unparseable or potentially dangerous latex formula] to get 9.14

[AuSmith]

04A-68

What do you immediately know? We easily get

1) the center of the circle [unparseable or potentially dangerous latex formula]
2) two congruent right triangles
3) two vertices of each right triangle and two legs each [unparseable or potentially dangerous latex formula] and [unparseable or potentially dangerous latex formula]

What now? Well, there's the slope formula, but finding coordinates doesn't seem to be the easiest way. Nor do I see any nifty tricks like dot products or trig identities in [unparseable or potentially dangerous latex formula]. Something very conducive to this problem is realizing slope is the tangent of the angle from the positive [unparseable or potentially dangerous latex formula]-axis. Thus, with a calculator, angles give slopes much easier than coordinates, especially through angle addition. If you need to know slopes of lines through [unparseable or potentially dangerous latex formula] and angles at [unparseable or potentially dangerous latex formula] are within reach, tangent should be something to consider.

So what's the best way to find the correct angles? We know the angle bisector between the two tangent lines passes through the center of the circle. So, we quickly know the average. The difference of the angle of one of the tangent lines and the bisector is almost handed to you, as you can easily calculate two legs of a right triangle. Voila, [unparseable or potentially dangerous latex formula] where [unparseable or potentially dangerous latex formula] and [unparseable or potentially dangerous latex formula].


04H-80

Draw the radius that meets the tangent line and draw a hypotenuse from the midpoint to the center of the circle. Find the hypotenuse and multiply by [unparseable or potentially dangerous latex formula]

06A-60

I've always heard it called an isosceles trapezoid and I like that better. Regular is used for enough things already. But, that's just me.

Notice four of the segments emanate from the same point with the same length. That screams circle. Well, that angle given is an inscribed angle. That automatically says the arc it cuts is [unparseable or potentially dangerous latex formula] and it's supplement is [unparseable or potentially dangerous latex formula]. Use the triangle area formula [unparseable or potentially dangerous latex formula] three times: [unparseable or potentially dangerous latex formula]. Solve for [unparseable or potentially dangerous latex formula].

[bradp]

how about 05B-36...

the height of the flag should be [unparseable or potentially dangerous latex formula]

[unparseable or potentially dangerous latex formula] is the flag hanging down or perhaps some secret calc apps wind that is making it move?

[thething2]

The flag is drooping down. I'm not sure why you can assume this, but that's what it is. You are looking for the diagonal in the flag.

So,
30*10/19=15.8
sqrt(15.8^2 +30^2)=33.9
33.9+15=48.9

[The3point14]

[The3point14]

what about 07A-38? I thought at one point I did have a solution to it, but now I am doubting myself. Any ideas?

[bradp]

the pond is an upside down cone.
Its volume is
[unparseable or potentially dangerous latex formula]

no height is given, so using trig you can get that [unparseable or potentially dangerous latex formula]

find the original volume [unparseable or potentially dangerous latex formula]

the volume of the rain added is [unparseable or potentially dangerous latex formula]

now,
[unparseable or potentially dangerous latex formula]
solve for r, double it

[The3point14]

Alright, I am glad I asked, I was pretty sure my attempt was just a coincidence.
EDIT: I understand it now :).


I believe this will the study list for me. yay.

[indianpapu]

on 04F-67
because the question says identical shape shouldn't we assume
that the only difference in the two figures is the length 8 to 12
so shouldn't it be 40 divided by 1+1.5

[The3point14]

Think of the container as a cube. If the first cube has 8ft long sides, and you only change one side to 12ft, the shapes are no longer the same, since the 2nd is now a rectangular prism.

Not sure if this will qualify as a good explaination, but I thought I would atleast try :).

[AuSmith]

[The3point14]

of course they sway in the wind. However, the question asks for the lowest acceptable height, or some wording of that sort, so you would have to find the "worst case scenario", or when the flag is closest to the ground.

It was neither I or TheThing that found the solution to this question, all credit goes to Elaine Nguyen from our Calc Apps team.

[AuSmith]

[indianpapu]

On the flag problem, i think,
you're looking for the diagonal because that is the furthest the flag can drop down

then you just add 15
i think this is why...

[bradp]

06B-26
what's wrong here?
using [unparseable or potentially dangerous latex formula]
[unparseable or potentially dangerous latex formula]

solving for [unparseable or potentially dangerous latex formula] and multiplying by [unparseable or potentially dangerous latex formula] i get something like 600, not 84.0

06C-37

[unparseable or potentially dangerous latex formula]

is that all there is to it? doesn't make sense to me, but it gets the right answer
i guess the wording "of the original population" is the key

[The3point14]

06B-26

When the wheel spins one side will be moving in the opposite direction, and the other in the same direction, relative to the ground (right and left respectively, but it doesn't matter).And the ratio of max speed to min speed is 2.2

[unparseable or potentially dangerous latex formula]

solving for [unparseable or potentially dangerous latex formula] you get [unparseable or potentially dangerous latex formula], the roulette wheels' velocity.

then changing mph to rpm...
[unparseable or potentially dangerous latex formula]

This is the distance traveled in a minute by the wheel, divide by the circumfrence

[unparseable or potentially dangerous latex formula]

This is how I did the question, it could be wrong, and probably is, but I did get the correct answer, any corrections will be appreciated.


06C-37

That's what I have been doing as well.

[indianpapu]

05D-38

I dont get this question
I've been trying for some time
i just dont get it

[bradp]

05D-38
Clothes are an issue of area, so the height, a single dimension, has to be squared.
age to the .8 gives the height, and then squared to get area. area is equal to # of sheep times a constant

[unparseable or potentially dangerous latex formula]
[unparseable or potentially dangerous latex formula]
[unparseable or potentially dangerous latex formula]

[unparseable or potentially dangerous latex formula]
the answer is [unparseable or potentially dangerous latex formula]

[indianpapu]

thanks man

I was missing the ^2 on my (age)^.8

[AuSmith]

[The3point14]

04F-70 ?

I think TheThing answered this question by substituting some value for the base, but can somebody please explain a quicker way or just the math behind solving this geometry question?

[bradp]

[unparseable or potentially dangerous latex formula] is the fraction of the base closer to the higher side and [unparseable or potentially dangerous latex formula] is on the other side. Using similar triangle ratios, the answer comes out either way to [unparseable or potentially dangerous latex formula]

whats with the 13s ?!

(Edited by Quelloquialism: Heck if I know what this ["13 business is about, but using \over instead of \frac made them go away, at least.)