open query 3

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course MTH 164

I will have to redo this assignment, because I just could not seem to understand most of it.

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

Precalculus II

Asst # 3

02-28-2001

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22:16:00

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query problem 5.4.72 length of ladder around corner hall widths 3 ft and 4 ft `theta relative to wall in 4' hall, ladder in contact with walls.

If the angle is `theta, as indicated, then how long is the ladder?

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22:20:54

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Your solution:

length = 4/sin(theta) + 3/cos(theta)

confidence rating #$&*:

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Given Solution:

** First you need a good picture, which I hope you drew and which you should describe.

Using the picture, in the 4' hall you can construct a right triangle with angle `theta and a side of 4 ft, with the part of the ladder in that hall forming the hypotenuse. Is the 4 ft opposite to the angle, adjacent to the angle or is it the hypotenuse? Once you answer that you can find how much ladder is in the hall.

You can also construct a right triangle with the rest of the ladder as the hypotenuse and the angle `theta as one of the angles. Identifying sides and using the definitions of the trig functions you can find the length of the hypotenuse and therefore the rest of the length of the ladder.

** The 4 ft is opposite to the angle theta between the hall and the ladder, i.e., between wall and hypotenuse. So 4 ft / hypotenuse = sin(theta) and the length of the ladder section in this hall is hypotenuse = 4 ft / sin(theta).

The triangle in the 3 ft hall has the 3 ' side parallel to the 4 ' hall, so the angle between hypotenuse and the 3 ' side is theta. Thus 3 ft / hypotenuse = cos(theta) and the length of ladder in this hass is hypotenuse = 3 ft / cos(theta).

So it is true that length = 4/sin(theta) + 3/cos(theta). **

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22:20:56

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Self-critique (if necessary):

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Self-critique Rating:

query problem 5.4.78 area of isosceles triangle A = a^2 sin`theta cos`theta, a length of equal side

how can we tell that the area of the triangle is a^2 sin(`theta) cos(`theta)?

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22:41:27

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Your solution:

You can derive from the formula of A=1/2bh

confidence rating #$&*:

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Given Solution:

STUDENT SOLUTION:

A = 1/2bh

Since an isosceles triangle can be separated into two right triangles . We can use right triangle math to derive an equation for the area. The triangles will have a hypoteuse of ""a "" an adjacent side (equal to 1/2 base) of a cos`theta and a Opposite side (equal to height) of a sin`theta.

1/2 base(b) = acos`theta

height (h) = asin`theta

A = a cos`theta * a sin`theta

A = a^2 cos(`theta) sin(`theta)

** a * cos 'theta = 1/2 * base so

base = 2 * a * cos(`theta).

a * sin 'theta = height.

So 1/2 base * height = 1/2 (a sin 'theta)(2 * a cos 'theta) = a^2 (sin 'theta)(cos 'theta) **

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22:41:28

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Self-critique (if necessary):

I wasn’t too sure about this one. I think that you start out with the area formula, but how to derive from there, I am not sure.

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Self-critique Rating:

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a is the common length of the two equal sides of your triangle, and theta the common angle made by those sides with the third side.

Sketch an isosceles triangle.

Label the equal sides a and the equal angles theta.

Starting at the vertex where the equal sides meet, sketch a line segment straight across to the third side, with your segment making a right angle to that side.

Your segment will divide the triangle into two congruent right triangles.

Look at the right triangle on the left.

Its hypotenuse is a. Do you see why?

Now, one of its sides is a cos(theta) and the other is a sin(theta).

One of its sides is the segment you sketched to divide the triangle.

Is that segment the one whose length is a sin(theta) or the one with length a cos(theta)?

What are the base and altitude of your right triangle, and what therefore is its area?

What do you conclude is the area of your original isosceles triangle?

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query problem 5.5.42 transformations to graph 3 cos x + 3

explain how you use transformations to construct the graph.

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Your solution:

confidence rating #$&*:

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Given Solution:

** The graph of cos(x) is 'centered' on the x axis, has a period of 2 `pi, as you say, and an amplitude of 1. Thus it runs from y value 1 to 0 to -1 to 0 to 1 in its first cycle, and in every subsequent cycle.

The graph of y = 3 cos x has the same description except that every y value is multiplied by 3, thereby 'stretching' the graph by factor 3. Its y values run between y = 3 and y = -3. The period is not affected by the vertical stretch and remains 2 `pi.

y = 3 cos x + 3 is the same except that we now add 3 to every y value. This means that the y values will now run from -3+3 = 0 to +3+3 = 6. The period is not affected by consistent changes in the y values and remains 2 `pi. **

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Self-critique (if necessary):

This one went over my head completely. I am not sure how to do this one at all.

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&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the parts of the given solution on which your solution didn't agree, and if necessary asking specific questions (to which I will respond).

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Self-critique Rating:

query problem 5.5.54 transformations to graph 4 tan(.5 x)

explain how you use transformations to construct the graph.

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Your solution:

confidence rating #$&*:

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Given Solution:

** The period of tan x is `pi because every time x changes by `pi you get to a point on the reference circle where the values of the tangent function start repeating. The graph of tan x repeats between vertical asymptotes at x = -`pi/2 and +`pi/2.

.5 x will change by `pi if x changes by `pi / .5 = 2 `pi. So the period of tan(.5x) is 2 `pi. This effective 'spreads' the graph out twice as far in the horizontal direction.

The graph therefore passes thru the origin and has vertical asymptotes at -`pi and `pi (twice as far out in the horizontal direction as for tan x).

4 tan(.5x) will be just like tan(.5x) except that every point is 4 times as far from the x axis--the graph is therefore stretched vertically by factor 4. This will, among other things, make it 4 times as steep when it passes thru the x axis. **

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Self-critique (if necessary):

I am still trying to grasp transformations. Is there a simple way to explain how this works?

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Self-critique Rating:

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You would do best to first review graphing by transformations, after which I can try to add clarification in the context of these problems.

This refers to stretching and shifting transformations, which were covered in Precalculus I and are also covered in earlier chapters of your text. I believe the section is entitled 'graphing techniques: transformations' or something very similar.

Review that material then take another look at the techniques as they are applied here.

At that point, you're welcome to ask more specific questions and I can do my best to clarify.

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describe the graph by giving the locations of its vertical asymptotes

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Your solution:

confidence rating #$&*:

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Given Solution:

** Each period of the function happens between its vertical asymptotes. The vertical asymptotes occur at intervals of 2 `pi, since the function has period 2 `pi.

The vertical asymptotes nearest the origin are at -`pi and +`pi.

In the positive direction the next few will be at 3 `pi, 5 `pi, 7 `pi, etc..

In the negative direction the next few will be at -3 `pi, -5 `pi, -7 `pi, etc..

Thus asymptotes occur at all positive and negative odd multiples of `pi. **

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Self-critique (if necessary):

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I've inserted some specific questions.

Insert your best attempts to answer the questions I posed there into a copy of this document, with &&&& before and after each insertion. Show me all your reasoning

and all your steps, and tell me what you're thinking about each question. Then I can provide additional clarification.

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I've also referred you to review material on graphing by transformations, which is (and was) covered in Precalculus I. Check that note.

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