Official Thread for 2010-2011 Study Guide

[mathmajor]

I've already seen 3 threads about the study guide, so I thought it would be better to just post on this thread any questions/explanations about the study guide. And what is the quickest way of doing 03F-69?

[sidius]

I started all 3 threads.

Anyway,
First I converted 127 degrees to 2.2165 radians.
Then found radius of the circle using:

Shaded Area=(r^2/2)(theta-sin(theta))

plugged in given info and solved for r using solver:
.411=(r^2/2)(2.2165-sin(2.2165))
where r=.76139

then plugged into Law of Cosine:
x^2=r^2+r^2-2*r*r*cos(180-127)
where x=solution=.679

[mathmajor]

Oh haha I didn't notice. And thanks! How about 04I-70

[sidius]

Start at the center of the circle and draw a line to the upper vertex of the interior triangle.
This line will represent the radius. Notice you have now split the triangle in two. The left
triangle is an isosceles. It will have two interior angles of 33 degrees, and the vertex angle will be 114 degrees.
now the lower acute angle of the triangle on the right is 66 degrees.

I'm calling the height of the right triangle "y"
where y=(1/2)*sin(66)
and I'm calling the base of the right triangle "x"
where x=(1/2)*cos(66)

we know also know that Triangle Area/Circle Area:
(1/2)*base*height / (1/2)*pi*r^2
so
[(1/2)*(.5 +.5*cos(66))*(.5*sin(66))] / [(1/2)*pi*(.5)^2]
which equals .409 which is the solution.

[mathmajor]

Thanks, how about 05D-36 on the stated problems.

[AuSmith]

The problem confuses me, but I copied it over.

"05D-36. A block is placed on a table tied with a string
extending 18 inches perpendicular to the table edge. The
block is nudged over the table edge causing it to fall.
How long after this does the block come into closest direct
approach to the end of the string?"

[calcguy]

For problem 05D-36, I tie a string to a block so my students can visualize it. Then it is easy.

Since the block is "nudged" over the table edge, it falls straight down. We must assume that the string is light compared to the block so the block accelerates downward at g = 32.17 ft/s². The two sections of string and the segment from block to string end form a right triangle with the segment as the hypotenuse. This hypotenuse is shortest when the 2 sections of string are equal (easy to prove and probably intuitive). So use d = ½at² where d is half the string length (in feet) and a = 32.17 ft/s² and you get t = 0.216 sec.

[AuSmith]

So, the string is lying along the table? That's about the last case I considered!

[mathmajor]

That makes sense! How about on the stated problems numbers 04E-77, 06I-66, 10A-38, and 10B-38.

[88bobcat]

[88bobcat]

[88bobcat]

[mathmajor]

Thanks, but on 06I-36, the answer key says its negative, but i'm not sure how it would be negative. and how about 08G-36, 09F-37, 10B-38 10D-38, and 10H-37

[AuSmith]

[88bobcat]