#$&* course Mth 164 Saturday February 20 12:21 Precalculus IIAsst # 3
......!!!!!!!!...................................
22:16:00
......!!!!!!!!...................................
query problem 5.4.72 length of ladder around corner hall widths 3 ft and 4 ft `theta relative to wall in 4' hall, ladder in contact with walls. If the angle is `theta, as indicated, then how long is the ladder?
......!!!!!!!!...................................
22:20:54 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If I draw a sketch of the ladder leaning against a 4ft wall I have the wall opposite of the ladder which is the hypotenuse. Sin =opp/hyp and cos=adj/hyp, so in knowing this the first 4ft wall would be sin=4ft/hpy. The length of the ladder is the hypot, so the equation would be hyp=4ft/sin(theta). The other hall which has the 3ft hall has a side parallel to the 4'hall, so this would be the opposite of sin which is cos. This would be the hypotnuse = 3ft/cos(theta). This will make the length = 4/sin(theta)+3/cos(theta). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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): 2 I did get the first part of the problem but I thought the way the problem read that the width of the hall was 3ft and the height was 4ft. ------------------------------------------------ Self-critique Rating: 2 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: Well we know that an isosceles triangle is really two right angles divided and it has two equal sides and angles. If A=1/2bh which is the base and height then 1/2 of the base = a cos theta and height = a sin theta. So 1/2base * height = 1/2 (a sin 'theta) (2*a cosine 'theta) = a^2 (sin theta) (cos 'theta) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
.............................................
Given Solution: STUDENT SOLUTION: A = 1/2bh Since an isosceles triangle can be separated into two right triangles . We can use right triangle math to derive an equation for the area. The triangles will have a hypoteuse of ""a "" an adjacent side (equal to 1/2 base) of a cos`theta and a Opposite side (equal to height) of a sin`theta. 1/2 base(b) = acos`theta height (h) = asin`theta A = a cos`theta * a sin`theta A = a^2 cos(`theta) sin(`theta) ** a * cos 'theta = 1/2 * base so base = 2 * a * cos(`theta). a * sin 'theta = height. So 1/2 base * height = 1/2 (a sin 'theta)(2 * a cos 'theta) = a^2 (sin 'theta)(cos 'theta) **
.........................................
22:41:28
......!!!!!!!!...................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 query problem 5.5.42 transformations to graph 3 cos x + 3 explain how you use transformations to construct the graph.
......!!!!!!!!...................................
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y=Acos(Bx-c)+D It has an amplitude of 1. You would take 1/3 and multiply 1/3(3cos(X+3))which would equal to cosx. The period doesn't cange and the y values would run 1<=0<=-1. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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 ------------------------------------------------ Self-critique Rating: 3 query problem 5.5.54 transformations to graph 4 tan(.5 x) explain how you use transformations to construct the graph.
......!!!!!!!!...................................
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Tangent does not have an amplitude and it doens't go to 2pi. It goes to pi. The period is 2pi because of the (.5x) and with A equal to 4 this means that the graph will stretch to 4<=0<=-4. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
.............................................
Given Solution: ** The period of tan x is `pi because every time x changes by `pi you get to a point on the reference circle where the values of the tangent function start repeating. The graph of tan x repeats between vertical asymptotes at x = -`pi/2 and +`pi/2. .5 x will change by `pi if x changes by `pi / .5 = 2 `pi. So the period of tan(.5x) is 2 `pi. This effective 'spreads' the graph out twice as far in the horizontal direction. The graph therefore passes thru the origin and has vertical asymptotes at -`pi and `pi (twice as far out in the horizontal direction as for tan x). 4 tan(.5x) will be just like tan(.5x) except that every point is 4 times as far from the x axis--the graph is therefore stretched vertically by factor 4. This will, among other things, make it 4 times as steep when it passes thru the x axis. **
.........................................
......!!!!!!!!...................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 describe the graph by giving the locations of its vertical asymptotes
......!!!!!!!!...................................
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The function above has a vertical asymptote of 2pi and they are -pi t0 pi. The positive asymptotes are 3pi, 5pi, 7pi etc and the same but negative on the negative side. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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: 3" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!