course PHY 201 Just looked at my e-mail.
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12:48:48 `q001. There are 11 questions and 7 summary questions in this assignment.
What is the area of a rectangle whose dimensions are 4 m by 3 meters.......!!!!!!!!...................................
RESPONSE --> (4X2)+(3X2)=14
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12:50:44 A 4 m by 3 m rectangle can be divided into 3 rows of 4 squares, each 1 meter on a side. This makes 3 * 4 = 12 such squares. Each 1 meter square has an area of 1 square meter, or 1 m^2. The total area of the rectangle is therefore 12 square meters, or 12 m^2.
The formula for the area of a rectangle is A = L * W, where L is the length and W the width of the rectangle. Applying this formula to the present problem we obtain area A = L * W = 4 m * 3 m = (4 * 3) ( m * m ) = 12 m^2. Note the use of the unit m, standing for meters, in the entire calculation. Note that m * m = m^2.......!!!!!!!!...................................
RESPONSE --> Ah yes, that's area. L X W
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12:53:02 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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RESPONSE --> (1/2)(L * W) 1/2(4 * 3)=6
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12:53:53 A right triangle can be joined along its hypotenuse with another identical right triangle to form a rectangle. In this case the rectangle would have dimensions 4.0 meters by 3.0 meters, and would be divided by any diagonal into two identical right triangles with legs of 4.0 meters and 3.0 meters.
The rectangle will have area A = L * W = 4.0 m * 3.0 m = 12 m^2, as explained in the preceding problem. Each of the two right triangles, since they are identical, will therefore have half this area, or 1/2 * 12 m^2 = 6.0 m^2. The formula for the area of a right triangle with base b and altitude h is A = 1/2 * b * h.......!!!!!!!!...................................
RESPONSE --> this works for a right triangle because it is esientially cutting a rectangle in have diagonaly.
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ydžxoӶT{Tv Student Name: assignment #001 001. Areas
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12:57:29 `q001. There are 11 questions and 7 summary questions in this assignment.
What is the area of a rectangle whose dimensions are 4 m by 3 meters.......!!!!!!!!...................................
RESPONSE --> W * H In this case 3 * 4 = 12
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12:57:46 A 4 m by 3 m rectangle can be divided into 3 rows of 4 squares, each 1 meter on a side. This makes 3 * 4 = 12 such squares. Each 1 meter square has an area of 1 square meter, or 1 m^2. The total area of the rectangle is therefore 12 square meters, or 12 m^2.
The formula for the area of a rectangle is A = L * W, where L is the length and W the width of the rectangle. Applying this formula to the present problem we obtain area A = L * W = 4 m * 3 m = (4 * 3) ( m * m ) = 12 m^2. Note the use of the unit m, standing for meters, in the entire calculation. Note that m * m = m^2.......!!!!!!!!...................................
RESPONSE --> A = L * W
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12:58:07 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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RESPONSE --> 1/2(L * W)
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12:58:43 A right triangle can be joined along its hypotenuse with another identical right triangle to form a rectangle. In this case the rectangle would have dimensions 4.0 meters by 3.0 meters, and would be divided by any diagonal into two identical right triangles with legs of 4.0 meters and 3.0 meters.
The rectangle will have area A = L * W = 4.0 m * 3.0 m = 12 m^2, as explained in the preceding problem. Each of the two right triangles, since they are identical, will therefore have half this area, or 1/2 * 12 m^2 = 6.0 m^2. The formula for the area of a right triangle with base b and altitude h is A = 1/2 * b * h.......!!!!!!!!...................................
RESPONSE --> it works because it is like folding a paper diagonally. Both ends are equal.
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12:59:50 `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters?
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RESPONSE --> this is like taking the area of two right triangles I would believe.
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13:00:51 A parallelogram is easily rearranged into a rectangle by 'cutting off' the protruding end, turning that portion upside down and joining it to the other end. Hopefully you are familiar with this construction. In any case the resulting rectangle has sides equal to the base and the altitude so its area is A = b * h.
The present rectangle has area A = 5.0 m * 2.0 m = 10 m^2.......!!!!!!!!...................................
RESPONSE --> Oh, well I got the right expenation, just didn't remember to solve it.
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13:02:21 `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?
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RESPONSE --> 1/2(5 * 2) = 5
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13:05:05 It is possible to join any triangle with an identical copy of itself to construct a parallelogram whose base and altitude are equal to the base and altitude of the triangle. The area of the parallelogram is A = b * h, so the area of each of the two identical triangles formed by 'cutting' the parallelogram about the approriate diagonal is A = 1/2 * b * h. The area of the present triangle is therefore A = 1/2 * 5.0 cm * 2.0 cm = 1/2 * 10 cm^2 = 5.0 cm^2.
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RESPONSE --> any triangle uses that formula because it's half of a parallelogram.
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13:06:35 `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?
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RESPONSE --> 1/2(4*5) = 10 although there are more factors it would seem that would be needed.
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13:08:35 Any trapezoid can be reconstructed to form a rectangle whose width is equal to that of the trapezoid and whose altitude is equal to the average of the two altitudes of the trapezoid. The area of the rectangle, and therefore the trapezoid, is therefore A = base * average altitude. In the present case this area is A = 4.0 km * 5.0 km = 20 km^2.
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RESPONSE --> okay, so the keyword is average altitude, which is the two altitudes.
4 * 5 = 20km^2.................................................
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13:10:40 `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?
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RESPONSE --> What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?
average altitude first (3+8)/2 = 5.5 4 * 5.5 = 22cm^2.................................................
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13:11:24 The area is equal to the product of the width and the average altitude. Average altitude is (3 cm + 8 cm) / 2 = 5.5 cm so the area of the trapezoid is A = 4 cm * 5.5 cm = 22 cm^2.
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RESPONSE --> previous question helped in this one.
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13:13:05 `q007. What is the area of a circle whose radius is 3.00 cm?
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RESPONSE --> What is the area of a circle whose radius is 3.00 cm? 2 * R^2 = A 2 * 9 = 18cm^2
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13:18:20 The area of a circle is A = pi * r^2, where r is the radius. Thus
A = pi * (3 cm)^2 = 9 pi cm^2. Note that the units are cm^2, since the cm unit is part r, which is squared. The expression 9 pi cm^2 is exact. Any decimal equivalent is an approximation. Using the 3-significant-figure approximation pi = 3.14 we find that the approximate area is A = 9 pi cm^2 = 9 * 3.14 cm^2 = 28.26 cm^2, which we round to 28.3 cm^2 to match the number of significant figures in the given radius. Be careful not to confuse the formula A = pi r^2, which gives area in square units, with the formula C = 2 pi r for the circumference. The latter gives a result which is in units of radius, rather than square units. Area is measured in square units; if you get an answer which is not in square units this tips you off to the fact that you've made an error somewhere.......!!!!!!!!...................................
RESPONSE --> I figured it had something to do with pi.
So it's pi(R)^2.................................................
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13:19:58 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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RESPONSE --> What is the circumference of a circle whose radius is exactly 3 cm?
2 pi R = 18.8.................................................
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13:20:54 The circumference of this circle is
C = 2 pi r = 2 pi * 3 cm = 6 pi cm. This is the exact area. An approximation to 3 significant figures is 6 * 3.14 cm = 18.84 cm. Note that circumference is measured in the same units as radius, in this case cm, and not in cm^2. If your calculation gives you cm^2 then you know you've done something wrong.......!!!!!!!!...................................
RESPONSE --> okay, I'll keep a note to add my units.
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13:22:13 `q009. What is the area of a circle whose diameter is exactly 12 meters?
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RESPONSE --> What is the area of a circle whose diameter is exactly 12 meters?
pi (1/2 * 12)^2 = 113.1.................................................
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13:22:34 The area of a circle is A = pi r^2, where r is the radius. The radius of this circle is half the 12 m diameter, or 6 m. So the area is
A = pi ( 6 m )^2 = 36 pi m^2. This result can be approximated to any desired accuracy by using a sufficient number of significant figures in our approximation of pi. For example using the 5-significant-figure approximation pi = 3.1416 we obtain A = 36 m^2 * 3.1416 = 113.09 m^2.......!!!!!!!!...................................
RESPONSE --> check
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13:24:17 `q010. What is the area of a circle whose circumference is 14 `pi meters?
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RESPONSE --> working backwards 14 = 2 pi R
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13:25:28 We know that A = pi r^2. We can find the area if we know the radius r. We therefore attempt to use the given information to find r.
We know that circumference and radius are related by C = 2 pi r. Solving for r we obtain r = C / (2 pi). In this case we find that r = 14 pi m / (2 pi) = (14/2) * (pi/pi) m = 7 * 1 m = 7 m. We use this to find the area A = pi * (7 m)^2 = 49 pi m^2.......!!!!!!!!...................................
RESPONSE --> perhap rounding to the first integer would have given similar results.
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13:27:09 `q011. What is the radius of circle whose area is 78 square meters?
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RESPONSE --> What is the radius of circle whose area is 78 square meters?
pi R^2 = 78 R^2 = 78/pi sqr(R^2) = sqr(78/pi) 4.98.................................................
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13:27:52 Knowing that A = pi r^2 we solve for r. We first divide both sides by pi to obtain A / pi = r^2. We then reverse the sides and take the square root of both sides, obtaining r = sqrt( A / pi ).
Note that strictly speaking the solution to r^2 = A / pi is r = +-sqrt( A / pi ), meaning + sqrt( A / pi) or - sqrt(A / pi). However knowing that r and A are both positive quantities, we can reject the negative solution. Now we substitute A = 78 m^2 to obtain r = sqrt( 78 m^2 / pi) = sqrt(78 / pi) m.{} Approximating this quantity to 2 significant figures we obtain r = 5.0 m.......!!!!!!!!...................................
RESPONSE --> check, still gota remember those units.
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13:28:40 `q012. Summary Question 1: How do we visualize the area of a rectangle?
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RESPONSE --> L * W = A Area happens to be the inside of the rectangle.
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13:29:07 We visualize the rectangle being covered by rows of 1-unit squares. We multiply the number of squares in a row by the number of rows. So the area is A = L * W.
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RESPONSE --> similar answer.
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13:29:37 `q013. Summary Question 2: How do we visualize the area of a right triangle?
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RESPONSE --> Half of a rectangles area.
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13:29:50 We visualize two identical right triangles being joined along their common hypotenuse to form a rectangle whose length is equal to the base of the triangle and whose width is equal to the altitude of the triangle. The area of the rectangle is b * h, so the area of each triangle is 1/2 * b * h.
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RESPONSE --> check
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13:31:00 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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RESPONSE --> Same as we would a rectangle, L * Altitude.
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13:31:05 The area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.
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RESPONSE --> check
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""Submitting Assignment" "MTH 164" "2168190" "2168190" "Daniel" "Drayer" "01-16-06 to 1-22-06" "ddrayer0001@email.vccs.edu" "Other than e-mail, I know of no way to submit images." "===== MTH 164 =====
L⧜t큉xq Student Name: Daniel Drayer assignment #001 001. Radian measure and the unit circle. Initials: DRD Date and Time 01-22-2006 14:37:29......!!!!!!!!...................................
14:45:41 `q002. If the first ant moves at a constant speed, moving through 1 radian every second, then approximately how long, to the nearest second, do you think it will take for the ant to move along the arc to the point where the circle meets the negative x-axis?
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RESPONSE --> 3 seconds aproximately
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14:46:31 Visual examination, perhaps accompanied by a quick sketch, shows that it takes approximately 3 arcs each of one radian to get from the positive x-axis to the negative x-axis when moving along the arc of the circle.
In figure 37 the points b, c and d lie at approximately 1, 2 and 3 radians. Remember that each radian corresponds to an arc distance equal to the radius of the circle.......!!!!!!!!...................................
RESPONSE --> check
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14:47:05 `q003. If the ant traveled at 1/2 radian per second, then after 1 second would its angular position be indicated by point a, point b, point c or point d in Figure 37?
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RESPONSE --> point a seems to be half the radian.
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14:48:31 After 1/2 second the angular position would be 1/2 radian, which would correspond to point a.
Note that after 2 seconds the angular position would be 1 radian, corresponding to point b, and after three seconds the angular position would be 3 * 1/2 radian = 3/2 radian and the ant would be at position c.......!!!!!!!!...................................
RESPONSE --> check although, 3 seconds would be one and a half radian. Wouldn't that mean, that he would be at a point not marked?
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14:49:25 `q004. How far will the ant travel in the process of completing 1 trip around the circle, starting and ending at the initial point where the circle meets the positive x-axis.
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RESPONSE --> aproximately 7 radians.
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14:50:36 The circumference of the circle is 2 pi r, where r is the radius of the circle. This is the distance traveled by the ant.
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RESPONSE --> oh, we are talking circumfrence now. 2 pi r. The unit is the same as the unit for radius.
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14:51:28 `q005. As we just saw the distance around the circle is its circumference 2 pi r, where r is the radius. Through how many radians would the ant travel from the initial point, where the circle meets the positive x-axis, if the motion was in the counterclockwise direction and ended at the original point after having completed one trip around the circle.
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RESPONSE --> 7, but this is a guess based on the figure.
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14:58:19 An arc displacement of r corresponds to an arc distance of 1 radian on the circle, so a distance of 2 `pi corresponds to 2 pi r / r = 2 pi radians.
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RESPONSE --> so, because radian and the radius are equal, you can substitute radians for R.
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15:00:27 `q006. The unit circle is a circle of radius 1 centered at the origin. What are the coordinates of the points where the unit circle meets the positive x-axis, the positive y axis, the negative x-axis and the negative y axis?
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RESPONSE --> (R,0) (0,R) (-R,0) (0,-R)
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15:00:54 The unit circle has radius 1 and is centered at the origin, so the circle meets the positive x-axis 1 unit from the origin at (x, y) = (1,0). Similarly the circle meets the positive y-axis at the 'top' of the circle, 1 unit from the origin at (x, y) = (0,1); the circle meets the negative x-axis at (-1, 0); and the circle meets the negative y-axis at (0,-1).
Figure 84 shows these points on the unit circle. Note that in this figure the small dots are located at increments of .1 unit in the x and y directions.......!!!!!!!!...................................
RESPONSE --> check
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15:06:33 `q007. Without looking at Figure 84, sketch a picture of the unit circle, complete with labeled points where the circle meets the x and y axes. Indicate the arc from the standard initial point, where the circle meets the x-axis, to the point where the circle meets the positive y axis. Describe your sketch.
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RESPONSE --> you could draw a right triangle between point A and Point B with the hypotinuse between them, the height and width would be the same.
The circle intersects the X and Y axis..................................................
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15:08:02 Your sketch should show the x and y axes and a circle of radius 1, with the points (1,0), (0, 1), (-1, 0) and (0, -1) where the circle meets the coordinate axes labeled. The arc will run along the first quadrant of the circle from (1,0) to (0,1). Your figure should match figure 84.
You should be able to quickly draw this picture any time you need it.......!!!!!!!!...................................
RESPONSE --> I'm not exactly sure, but the description matches the sketch.
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15:14:15 `q008. How many radians of angular displacement correspond to the arc displacement from the standard initial point, where the circle meets the x-axis, to the point where the circle meets the positive y axis?
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RESPONSE --> 2 pi R = Circumference. Circumference / 4 = 1/4 of the circle or (1,0) to (0,1)
1/2 pi R = distance of 1/4 the circle 1/2(pi Radian).................................................
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15:15:04 The trip around the entire circle, which corresponds to an angular displacement of 2 pi radians, corresponds to a trip from the initial point to the point where the circle meets the positive y-axis (i.e., the point (0,1)), then from this point to the point where the circle meets the negative x-axis (i.e., the point (-1,0)), then from this point to the point where the circle meets the negative y-axis (i.e., the point (0,-1)), then from this point back to the point where the circle meets the positive x-axis (i.e., the point (1,0)).
Because of the symmetry of the circle, the arc corresponding to each of these displacements is the same. The arc from (1,0) to (0,1) is 1/4 of the 2 pi radian angular displacement around the entire circle, so its angular displacement is 2 pi/4 = pi/2 radians.......!!!!!!!!...................................
RESPONSE --> check
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15:15:59 `q009. We have just seen that the angular position of the (1,0) point is 0 and the angular position of the (0,1) point is pi/2. What are the angular positions of the (-1,0) and (0,-1) points?
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RESPONSE --> it should be the same pi/2 radians.
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15:18:14 These points are reached after successive angular displacements of pi/2. The (-1,0) point is reached from the pi/2 position by an additional angular displacement of pi/2, which puts it at angular position pi.
The (0,-1) point is reached after another angular displacement of pi/2, which puts it at pi + pi/2 = 2 pi/2 + pi/2 = 3 pi/2. Note that still another angular displacement of pi/2 puts us back at the initial point, whose angular position is 0. This shows that the initial point has angular position 0, or angular position 3 pi/2 + pi/2 = 4 pi/2 = 2 pi, consistent with what we already know. You should label your picture with these angular positions pi/2, pi, 3 pi/2 and 2 pi specified at the appropriate points.......!!!!!!!!...................................
RESPONSE --> I went from (-1,0) to (0,-1)
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15:22:39 `q010. What is the angular displacement from the standard initial point of the point halfway along the arc of the circle from (1,0) to (0,1)? Note that you should begin with a sketch of the circle and of the arc specified here.
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RESPONSE --> 1/4 pi radians
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15:23:28 (1,0) is the point at which the circle meets the positive x-axis and (0,1) is the point at which the circle meets the positive y-axis. The trip along the arc of the circle from (1,0) to (0,1) will move along the first-quadrant arc from angular position 0 to angular position pi/2. Halfway along this arc, the angular position will be 1/2 * pi/2 = pi/4.
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RESPONSE --> check
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15:27:03 `q011. What will be the angular positions of the arc points halfway between the (0,1) and (-1,0) points of the circle?
What will be the angular positions of the arc points halfway between the (-1,0) and (0,-1) points of the circle? What will be the angular positions of the arc points halfway between the (0,-1) and (1,0) points of the circle?......!!!!!!!!...................................
RESPONSE --> 3/4 pi
5/4 pi 7/4 pi.................................................
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15:27:11 Halfway between the (0,1) point, which corresponds to the the the position pi/2, and the (-1,0) point, which corresponds to angular position pi, will be the point lying at angular position pi/2 + pi/4 = 2 pi / 4 + pi / 4 = (2 pi + pi)/4 = 3 pi / 4.
Halfway between the (-1,0) point, which corresponds to the the position pi,and the (0,-1) point, which corresponds to angular position 3 pi / 2, will be the point lying at angular position pi + pi/4 = 4 pi / 4 + pi / 4 = (4 pi + pi)/4 = 5 pi / 4. Halfway between the (0,-1) point, which corresponds to the the position 3 pi/2, and the (-1,0) point, which corresponds to angular position 2 pi, will be the point lying at angular position 3 pi/2 + pi/4 =62 pi / 4 + pi / 4 = (6 pi + pi)/4 = 7 pi / 4.......!!!!!!!!...................................
RESPONSE --> check
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15:27:47 `q012. What is the angular position of the point lying 1/3 of the way along the arc of the circle between the points (1,0) and (0,1)?
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RESPONSE --> 1/8 pi
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15:29:48 The arc from (1,0) to (0,1) corresponds to an angular displacement of pi/2. One-third of the arc corresponds to an angular displacement of 1/3 * pi/2 = pi/6. The angular position of the specified point is therefore pi/6.
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RESPONSE --> oh, so we take 1/3 of pi/2 radian to get 1/3 of 1/4 of the circumference.
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15:30:18
Complete Assignment 0: Complete the remainder of Assignment 0 as instructed under the Assts link on the homepage at 164.106.222.236. This includes the links to What to Include in Submitted Work, Precalculus II Week I and the optional DERIVE worksheet (requires approximately 2 hours for an orientation to a computer algebra and graphing system) When you have completed the entire assignment run the Query program. Submit SEND files from Query and q_a_.......!!!!!!!!...................................
RESPONSE --> check
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ξS؎Mzo߿ Student Name: Daniel Drayer assignment #002 002. The Fundamental Angles. Initials: DRD Date and Time 01-22-2006 16:07:55
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16:14:24 `q001. Note that this assignment has 9 activities.
If the red ant moves at an angular velocity of pi/6 radians every second, starting from the standard initial point, then what will be its angular position at the end of each of the first 12 seconds?......!!!!!!!!...................................
RESPONSE --> 2 pi radians
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16:14:46 The angular position changes by pi/6 radians every second. Starting at angular position 0, the angular positions at t = 1, 2, 3, 4, ..., 12 will be
pi/6, 2 pi/6, 3 pi/6, 4 pi/6, 5 pi/6, 6 pi/6, 7 pi/6, 8 pi/6, 9 pi/6, 10 pi/6, 11 pi/6, and 12 pi/6. You might have reduced these fractions the lowest terms, which is good. In any case this will be done in the next problem.......!!!!!!!!...................................
RESPONSE --> check
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16:17:10 `q002. Reduce the fractions pi/6, 2 pi/6, 3 pi/6, 4 pi/6, 5 pi/6, 6 pi/6, 7 pi/6, 8 pi/6, 9 pi/6, 10 pi/6, 11 pi/6, and 12 pi/6 representing the angular positions in the last problem to lowest terms.
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RESPONSE --> pi/6, 1pi/3, 1pi/2, 2pi/3, 5pi/6, 1pi, 7pi/6, 4pi/3, 3pi/2, 5pi/3, 11pi/6, 2pi.
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16:17:18 The reduced fractions are pi/6, pi/3, pi/2, 2 pi/3, 5 pi/6, pi, 7 pi/6, 4 pi/3, 3 pi/2, 5 pi/3, 11 pi/6 and 2 pi.
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RESPONSE --> check
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16:30:26 `q003. Sketch a circle centered at the origin of an x-y coordinate system, depicting the angular positions pi/6, pi/3, pi/2, 2 pi/3, 5 pi/6, pi, 7 pi/6, 4 pi/3, 3 pi/2, 5 pi/3, 11 pi/6 and 2 pi.
What are the angular positions of the following points: The point 2/3 of the way along the arc between (0,1) and (-1,0) The point 1/3 of the way along the arc from (0, 1) to (-1,0) The points 1/3 and 2/3 of the way along the arc from (-1,0) to (0,-1) The points 1/3 and 2/3 of the way along the arc from (0, -1) to (0,1)??......!!!!!!!!...................................
RESPONSE --> 5pi/3
11pi/6 1/3 from (-1,0) to (0,-1) = 7pi/6 2/3 from (-1,0) to (0,-1) = 4pi/3 (0, -1) to (0,1) Y-axis bottom to top 1/3 = 7pi/6 2/3 = 5pi/6.................................................
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16:31:10 The points lying 1/3 and 2/3 of the way along the arc between the points (0,1) and (-1,0) are at angular positions 2 pi/3 and 5 pi/6; the point 2/3 of the way between these points is at angular position 5 pi/6.
The points lying 1/3 and 2/3 of the way along the arc between the points (-1,0) and (0,1) are at angular positions 7 pi/6 and 4 pi/3. The points lying 1/3 and 2/3 of the way along the arc between the points (0,-1) and (1,0) are at angular positions 5 pi/3 and 11 pi/6. Note that you should be able to quickly sketch and label this circle, which depicts the angles which are multiples of pi/6, whenever you need it.......!!!!!!!!...................................
RESPONSE --> check
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16:35:49 `q004. If the red ant moves at an angular velocity of pi/4 radians every second then what will be its angular position at the end of each of the first 8 seconds?
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RESPONSE --> 2pi
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16:36:30 The angular position changes by pi/4 radians every second. Starting at angular position 0, the angular positions will be pi/4, 2 pi/4, 3 pi/4, 4 pi/4, 5 pi/4, 6 pi/4, 7 pi/4, and 8 pi/4. You might have reduced these fractions the lowest terms, which is good.In any case this will be done in the next problem.
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RESPONSE --> check, although I thought it just wanted 8 seconds worth.
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16:37:46 `q005. Reduce the fractions pi/4, 2 pi/4, 3 pi/4, 4 pi/4, 5 pi/4, 6 pi/4, 7 pi/4, and 8 pi/4 representing the angular positions in the last problem to lowest terms.
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RESPONSE --> pi/4, 1 pi/2, 3pi/4, pi, 5pi/4, 3pi/2, 7pi/4, 2pi
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16:38:00 The reduced fractions are pi/4, pi/2, 3 pi/4, pi, 5 pi/4, 3 pi/2, 7 pi/4, and 2 pi.
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RESPONSE --> check
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16:42:38 `q006. Sketch a circle centered at the origin of an x-y coordinate system, depicting the angular positions pi/4, pi/2, 3 pi/4, pi, 5 pi/4, 3 pi/2, 7 pi/4, and 2 pi.
What are the angular positions of the following points: The point 1/2 of the way along the arc between (0,1) and (-1,0) The point 1/2 of the way along the arc from (0, -1) to (1,0) The point 1/2 of the way along the arc from (0,-1) to (0, -1)?......!!!!!!!!...................................
RESPONSE --> 3 pi/4
7 pi/4 The point 1/2 of the way along the arc from (0,-1) to (0, -1)? [this question was copied directly] there is no distance between these two points, however between (0,1) and (0,-1) it is pi.................................................
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16:44:24 The point lying 1/2 of the way along the arc between the points (0,1) and (-1,0) (the topmost and leftmost points of the circle) is at angular position 3 pi/4.
The point lying 1/2 of the way along the arc between the points (0,-1) and (1,0) is at angular position 7 pi/4. The point lying 1/2 of the way along the arc between the points (-1,0) and (0,-1) is at angular position 5 pi/4. These angles are shown in Figure 21. Note that the degree equivalents of the angles are also given.......!!!!!!!!...................................
RESPONSE --> check
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16:51:42 `q007. If the red ant starts at angular position pi/3 and moves at an angular velocity of pi/3 radians every second then what will be its angular position at the end of each of the first 6 seconds? Reduce your fractions to lowest terms.
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RESPONSE --> 0. pi/3 1. 2 pi/3 2. pi 3. 4 pi/3 4. 5 pi/3 5. 2 pi 6. 7 pi/3 one full lap
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16:52:14 The angular position changes by pi/3 radians every second. Starting at angular position pi/3, the angular positions after successive seconds will be 2 pi/3, 3 pi/3, 4 pi/3, 5 pi/3, 6 pi/3 and 7 pi/3, which reduce to 2 pi/3, pi, 4 pi/3, 5 pi/3, 2 pi and 7 pi/3.
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RESPONSE --> check
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16:53:44 `q008. Where is the angular position 7 pi/3 located?
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RESPONSE --> it is equal to pi/3 but factors in one complete rotation. 2pi + pi/3 or 6pi/3 + pi/3 = 7pi/3
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16:54:27 If you have not done so you should refer to your figure showing the positions which are multiples of pi/6.
On your picture you will see that the sequence of angular positions 2 pi/3, pi, 4 pi/3, 5 pi/3, 2 pi, 7 pi/3 beginning in the first quadrant and moving through the second, third and fourth quadrants to the 2 pi position, then pi/3 beyond that to the 7 pi/3 position. The 7 pi/3 position is therefore identical to the pi/3 position.......!!!!!!!!...................................
RESPONSE --> check
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16:57:34 `q009. If the red ant starts at angular position pi/3 and moves at an angular velocity of pi/4 radians every second then what will be its angular position at the end of each of the first 8 seconds? Reduce your fractions to lowest terms.
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RESPONSE --> 7pi/3 because at pi/4 per second for 8 seconds is the equivalent of one full rotation, it is 7pi/3.
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16:59:33 The angular position changes by pi/4 radians every second. Starting at angular position pi/3, the angular positions after successive seconds will be
pi/3 + pi/4, pi/3 + 2 pi/4, pi/3 + 3 pi/4, pi/3 + 4 pi/4, pi/3 + 5 pi/4, pi/3 + 6 pi/4, pi/3 + 7 pi/4 and pi/3 + 8 pi/4. These fractions must be added before being reduced to lowest terms. In each case the fractions are added by changing each to the common denominator 12. This is illustrated for pi/3 + 3 pi/4: We first multiply pi/3 by 4/4 and 3 pi/4 by 3/3, obtaining the fractions 4 pi/12 and 9 pi/12. So the sum pi/3 + 3 pi/4 becomes 4 pi/12 + 9 pi/12, which is equal to 13 pi/12. The fractions add up as follows: pi/3 + pi/4 = 7 pi/12, pi/3 + 2 pi/4 = 5 pi/6, pi/3 + 3 pi/4 = 13 pi/12, pi/3 + 4 pi/4 = 4 pi/3, pi/3 + 5 pi/4 = 19 pi/12, pi/3 + 6 pi/4 = 11 pi/6, pi/3 + 7 pi/4 = 25 pi/12 and pi/3 + 8 pi/4 = 14 pi / 3.......!!!!!!!!...................................
RESPONSE --> one way versus another thinking, but for questions that are more complex, Ex. pi/4 for 6 second starting at pi/3, this would be the only way.
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17:00:09
Complete Assignment 1: Includes Class Notes #'s 1-2 (Class Notes are accessed under the Lectures button at the top of the page and are included on the CDs starting with CD #1). Introductory Experience: Pendulum modeled by Motion on a Circle (as instructed on Assts page) Sketching Exercise Graphing Vertical Position; Effects of Angular Velocity, Radius, Starting Point (as instructed on Assts page) Modeling Exercise: Circular Models (click on link on Assts page) Text Section 5.1, 'Blue' Problems (i.e., problems whose numbers are highlighted in blue) and odd multiples of 3 in text and the Web version of Ch 5 Problems Section 5.1 (use the link in the Assts page to access the problems). When you have completed the entire assignment run the Query program. Submit SEND files from Query and q_a_.......!!!!!!!!...................................
RESPONSE --> check
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