#$&* course mth 164 10-10 8 Precalculus II Asst # 3 query problem 5.4.72 length of ladder around corner hall widths 3 ft and 4 ft `theta relative to wall in 4' hall, ladder in contact with walls.
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22:20:54 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The ladder rests on the corner of the hall, where angle theta is represented on both sides of the corner with respect to the ladder. The 4ft hall has a hypotenuse of (4/sin(theta)), which is the length of the ladder in the 4ft hall. The 3ft hall has a hypotenuse of (3/cos(thets)), which is the length of the ladder in the 3ft hall. Adding the two hypotenuses of both halls gives you the total length of the ladder. (4/sin(theta))+(3/cos(theta))= the length of the ladder. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I had trouble setting up the picture to start off with; it was difficult to understand what the question wanted me to draw. I did not find this problem in section 5.4, but I understand how to find the functions needed. ------------------------------------------------ self-critique rating #$&*: query problem 5.4.78 area of isosceles triangle A = a^2 sin`theta cos`theta, a length of equal side how can we tell that the area of the triangle is a^2 sin(`theta) cos(`theta)?
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22:41:27 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The triangle is a isosceles triangle, divide it into two right triangles so we can find the corresponding height which is sin(theta). The base of the entire triangle is cos(theta), and the hypotenuse for both triangles is “a”. So by doubling the hypotenuse you get the length of the base, which is cos(theta) and multiplying those two values by the height of the triangle, sin(theta), then we can find the area with the equation- a^2sin(theta)cos(theta). confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ‘OK’ ------------------------------------------------ self-critique rating #$&*: query problem 5.5.42 transformations to graph 3 cos x + 3 explain how you use transformations to construct the graph.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph cos(x) starts off at (0,1). The 3 transforms the amplitude or the maximum and minimum range of the graph from 1 and -1, to 3 and -3; this stretch causes the graph stop increasing on the y axis at (0,3) and stop decreasing on the y axis at (0,-3). Finally, move the initial starting point of the graph from (0,0), to (0,6). So the maximum range of this graph would be (0,6) and the minimum would be (0,0). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ‘OK’ ------------------------------------------------ self-critique rating #$&*: query problem 5.5.54 transformations to graph 4 tan(.5 x) explain how you use transformations to construct the graph.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You would stretch the maximum and minimum range by 4, which would be (pi/4,4) (pi/4,-4). Pi/.5=2pi, so the graph’s horizontal length has doubled from –pi, pi; to –2pi, 2pi. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ‘OK’ ------------------------------------------------ self-critique rating #$&*: describe the graph by giving the locations of its vertical asymptotes
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: pi, 3pi, 5pi, 7pi, 9pi -pi, -3pi, -5pi, -7pi, -9pi confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ‘OK’ ------------------------------------------------ self-critique rating #$&*: "