#$&* course 9/13 1 Precalculus IIAsst # 3
......!!!!!!!!...................................
22:16:00
......!!!!!!!!...................................
query problem 5.4.72 length of ladder around corner hall widths 3 ft and 4 ft `theta relative to wall in 4' hall, ladder in contact with walls. If the angle is `theta, as indicated, then how long is the ladder?
......!!!!!!!!...................................
22:20:54 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: using the pathagorean therum. we get that the hypotenuse side is 5ft. so we have a typical 3,4,5 right triangle, with two angle theta's. to find the first angle theta, relative to the 4ft side. we use arc cos(adj/hyp) = arc cos(4/5) = 36.8 degrees for the other theta. we use the 3ft side and sin. arc sin(4/5) = 53.2 degrees length now is 4/cos(36.8) + 3/sin(53.2) = 8.8 ft. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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). **
.........................................
22:20:56
......!!!!!!!!...................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ???? I mixed up my sin and cos in the final equation. I'm not sure why they are the other way around. but nonetheless, the answer should be 11.6 ft ------------------------------------------------ Self-critique Rating:3 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)?
......!!!!!!!!...................................
22:41:27 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: area of a triangle is 1/2bh. this being said, with two equal sides, find the area of the triangle by cutting the triangle in 1/2 so that one of the two equal sides is the hypotenuse. once you have the new hypotenuse, you can use pathagorean theorum to get the hypotenuse length. from there you can say a = 1/2 b h and you end up with a(unit)^2 if we set equal 1/2 b to sin(theta) and h to cos(theta) we get a = 1/2bsin(theta) cos(theta) to make it closer to ""your"" solution for area, we say. a = l * w a= a cos(theta) * a sin(theta) simplify and get a = a sin(theta) cos(theta) a = 1/2 sin(theta) cos(theta) base = 2 a* cos(theta) so area = 1/2 ( sin(theta) cos(theta) ) * 2 a cos(theta) confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: STUDENT SOLUTION: A = 1/2bh Since an isosceles triangle can be separated into two right triangles . We can use right triangle math to derive an equation for the area. The triangles will have a hypoteuse of ""a "" an adjacent side (equal to 1/2 base) of a cos`theta and a Opposite side (equal to height) of a sin`theta. 1/2 base(b) = acos`theta height (h) = asin`theta A = a cos`theta * a sin`theta A = a^2 cos(`theta) sin(`theta) ** a * cos 'theta = 1/2 * base so base = 2 * a * cos(`theta). a * sin 'theta = height. So 1/2 base * height = 1/2 (a sin 'theta)(2 * a cos 'theta) = a^2 (sin 'theta)(cos 'theta) **
.........................................
22:41:28
......!!!!!!!!...................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): whoa...I see that i didn't quite get the final solution correct. I worked through yours and it seems a bit hairy. I think i understand it though ------------------------------------------------ Self-critique Rating: query problem 5.5.42 transformations to graph 3 cos x + 3 explain how you use transformations to construct the graph.
......!!!!!!!!...................................
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: graph 3cosx + 3. here our cos is origionally centered on the origin. and we multiply that by 3. which will narrow the graph down, making it have a greater wavelength. add 3, and simply shift the graph up 3 units in the positive y direction confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. **
.........................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK?? when i graphed cos(x), i get a pretty slow looking wavelenth. but when i graph 3cos(x) it goes up and down much faster, therefore ""shrinking"" the graph in the x direction, but stretching it in the y direction. whereas you said it will stretch the graph. I did not include any information on periods either, but i now understand how you get periods ------------------------------------------------ Self-critique Rating:3
......!!!!!!!!...................................
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: when you graph tan(x) it looks to have a period of pi because it repeats much faster and in a shorter length on the x axis. multiplying it by 4 gives us a much more aggresive graph, it is stretched in the y directin and repeats faster. but when you graph the 4tan(.5x) the graph now has a period of 2pi because pi/.5 = 2pi. multiplying that by 4 just streches the graph, making it steeper confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** The period of tan x is `pi because every time x changes by `pi you get to a point on the reference circle where the values of the tangent function start repeating. The graph of tan x repeats between vertical asymptotes at x = -`pi/2 and +`pi/2. .5 x will change by `pi if x changes by `pi / .5 = 2 `pi. So the period of tan(.5x) is 2 `pi. This effective 'spreads' the graph out twice as far in the horizontal direction. The graph therefore passes thru the origin and has vertical asymptotes at -`pi and `pi (twice as far out in the horizontal direction as for tan x). 4 tan(.5x) will be just like tan(.5x) except that every point is 4 times as far from the x axis--the graph is therefore stretched vertically by factor 4. This will, among other things, make it 4 times as steep when it passes thru the x axis. **
.........................................
......!!!!!!!!...................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:3 describe the graph by giving the locations of its vertical asymptotes
......!!!!!!!!...................................
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the locations of the vertical asymptotes are at the beginning and end of every period. in this case, -pi and pi are vertical asymptotes. a difference of 2pi units. so they will occur in intervals of 2pi confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. **
.........................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:3" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!