#$&* course Mth 164
.............................................
Given Solution: `aThe magnitudes are 10 and 8, and the angle between the vectors is the change in angle from 30 degrees to 90 degrees, or 60 degrees. The dot product is therefore dot product = product of magnitudes * cos(angle) = 10 * 8 * cos(60 deg) = 40. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*:Ok ********************************************* Question: `q002. What are the x and y components of the vector v having magnitude 10 and angle 30 degrees, and vector w having magnitude 8 and angle 90 degrees? What do you get if you add the product of the two x components to the product of the two y components? How is this result related to the answer to the preceding exercise? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Vector v has an x component of 10cos(30) and a y of 10sin(30) Vector w has an x component of 8cos(90) and a y of 8sin(90) If we try to get the product of the two we will get 0 because of the x component of the w vector, but for y we will bet 40. This relates because we come out with 40 when we add the x component product and the y component products. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aThe vector v has x component vx = 10 cos(30 deg) = 8.7, approx., and vy = 10 sin(30 deg) = 5. The vector w has x component wx = 8 cos(90 deg) = 0 and y component wy = 8 sin(90 deg) = 8. The product of the two x components is vx * wx = 8.7 * 0 = 0, and the product of the y components is vy * wy = 5 * 8 = 40. The sum of these products is 0 + 40 = 40, which is identical to the result of the preceding exercise in which we found the dot product of v and w. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*:Ok ********************************************* Question: `q003. If vector v is represented by < v1, v2 > and vector w by < w1, w2 > then if the result of the preceding exercise is valid, how do we write in symbols the dot product of the two vectors? In symbols how do we write the magnitudes of the two vectors? How then do we write the statement that the dot product of the two vectors is equal to the product of the magnitudes of the vectors, multiplied by the cosine of the angle between the vectors? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: V1 * w1 + v2 * w2 V= sqrt(v1^2 + v2^2) W=sqrt(w1^2 + w2^2) so we can say: V1*w1 + v2*w2 = sqrt(v1^2 + v2^2) * sqrt(w1^2 + w2^2) * cos(theta) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aIf the preceding exercise generalizes then the dot product is the sum of the product of the x components and the product of the y components. In this case we would therefore say that the dot product is v1 * w1 + v2 * w2. The magnitudes of the two vectors are | v | = sqrt(v1^2 + v2^2) and | w | = sqrt(w1^2 + w2^2). The statement therefore says that v1 * w1 + v2 * w2 = sqrt(v1^2 + v2^2) * sqrt(w1^2 + w2^2) * cos(theta). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*:Ok ********************************************* Question: `q004. Use the result of the preceding exercise to find the cosine of the angle between the vectors < 2, 3 > and < -7, 4 >. What therefore is the angle between these vectors? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We use the equation from the previous equation so to find theta we get: Cos(theta) = (v1 * w1 + v2 * w2)/ sqrt(v1^2+v2^2) * sqrt(w1^2+w2^2) We fill this out with the given information and we get: Cos(theta) = .09 We the get an angle from this around 85 degrees. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aThe cosine of the angle theta between the vectors is found using the fact that dot product is product of magnitudes multiplies by the cosine of the angle, expressed in detail as v1 * w1 + v2 * w2 = sqrt(v1^2 + v2^2) * sqrt(w1^2 + w2^2) * cos(theta). We easily rearrange the equation to get cos(theta) = [ v1 * w1 + v2 * w2 ] / [ sqrt(v1^2 + v2^2) * sqrt(w1^2 + w2^2)]. In this case we have cos(theta) = [ 2 * -7 + 3 * 4 ] / [ sqrt(2^3 + 3^2) * sqrt( (-7)^2 + 4^2) ] = -2 / ( sqrt(13) * sqrt(53) ] = -2 / 26.3 = .08 approx.. The angle is therefore the solution theta to the equation cos(theta) = .08, which gives us theta = 83 degrees, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*:Ok ********************************************* Question: `q005. The vectors v = < 2, 4, 5 > and w = < -3, 7, 2 > exist in 3-dimensional space. What do you think are the magnitudes of these two vectors? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Sqrt(2^2 + 4^2 + 5^2) = Sqrt(45) = 6.6 Sqrt(-3^2 + 7^2 + 2^2) = Sqrt(76) = 8.8 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aThe magnitude of a vector is found by taking the square root of the sum of the squares of its components, just as for a vector in 2-dimensional space. In this case we get magnitudes sqrt(2^2 + 4^2 + 5^2) = sqrt(45) = 6.7, approx., and sqrt((-3)^2 + 7^2 + 2^2) = sqrt(76) = 8.7 approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*:Ok ********************************************* Question: `q006. What is the dot product of the vectors v = < 2, 4, 5 > and w = < -3, 7, 2 > ? What therefore is the angle between these vectors? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2 * (-3) + 4 * 7 + 5 * 2 = 32 Cos(theta) = 32/sqrt(45) * sqrt(76) Theta = arcos(32/ sqrt(45) * sqrt(76) = 57 degrees confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aSince dot product = product of magnitudes * cos(theta) we have cos(theta) = dot product / product of magnitudes. In the preceding problem we found the magnitudes of the vectors to be sqrt(45) and sqrt(76). So if we can find the dot product we can find the cosine of the angle and therefore the angle. The dot product is again the sum of the products of the invidual components, in this case 2 * (-3) + 4 * 7 + 5 * 2 = 32. Thus we have cos(theta) = 32 / ( sqrt(45) * sqrt(76) ) and theta = arccos[ 32 / (sqrt(45) * sqrt(76) ) = 56.8 deg, approx.. Note that these vectors can actually be constructed in 3-dimensional space, and if the construction is accurate the angle will be as indicated. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*:Ok ********************************************* Question: `q007. Calculate the angle between the two vectors < 1, 7, -3, 4 > and < -3, -5, 2, 7 > YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Theta = arcos( 1 * -3 + 7 * -5 + -3 * 2 + 4 * 7) / sqrt(1^2 + 7^2 + -3^2 + 4^2) * sqrt(-3^2^2 + -5^2 + 2^2 + 7^2) = Arcos( -16/81) = 100 degrees confidence rating #$&*:Ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aAs before we find that theta = arccos ( dot product / product of magnitudes ). In this case theta = arccos [ (1 * -3 + 7 * -5 + -3 * 2 + 4 * 7) / ( sqrt(1^2 + 7^2 + (-3)^2 + 4^2 ) * sqrt( (-3)^2 + (-5)^2 + 2^2 + 7^2 ) ] = arccos[ -16 / 80.8 ] = 101 degrees, approx.. "
#$&* course Mth 164
......!!!!!!!!...................................
10:38:28 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (3, pi) , (3, -pi), (-3,0) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** (-3, 4 pi) corresponds to 2 complete revolutions, corresponding to the angle 4 pi, which directs you along the positive x axis. • However r = -3 indicates that we move 3 units in the opposite direction, so we'll end up 3 units to the left of the origin and on the x axis. This point could also be described by the polar coordinates (3, pi) or (3, -pi), or (-3, 0). **
......!!!!!!!!...................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*:Ok **** Query problem 8.1.42 rect coord of (-3.1, 182 deg)
......!!!!!!!!...................................
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: X = rcos(theta) (-3.1) * cos(182) = 3.1 Y= rsin(theta) (-3.1) * sin(182) = .12 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: GOOD STUDENT SOLUTION: since we know the formula for converting polar coordinates to rectangular coordinates we can say that r=-3.1 and theta=182 deg. we can first find the value of x by saying x=r cos theta which gives (-3.1) cos 182 deg. = 3.1. We can then find y by saying y= r sin theta so y=(-3.1)sin 182 deg. using a calculator we get an approx. value of .1081. Thus the rectangular coordinates are (3.1, .1081), approx..
.........................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn’t put the coordinates together I just found them. ------------------------------------------------ Self-critique rating #$&*:Ok Query problem 8.1.54 polar coordinates of (-.8, -2.1)
......!!!!!!!!...................................
18:50:13 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: R^2 = x^2 + y^2 R = sqrt(-.8^2 + -2.1^2) R= 2.25 Tan(theta) = y/x Tan(theta) = (-2.1/-.8) Theta = 69 degrees + 180 = 249 (2.25, 4.4) confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: STUDENT SOLUTION: For the rectangular coordinates of (-.8, -2.1) to find the polar coordinates: we know that r^2 = x^2 + y^2 so thus we know r = sqrt( (-.8) ^2 + (-2.1) ^2) so r = sqrt( .64 + 4.41) r = sqrt 5.05 r = 2.247. To find theta we use tan (theta) = y / x So tan(theta) = (-2.1)/ (-.8) tan (theta) = 2.625 and x < 0 so theta = 69.145 deg + 180 deg = 249.145 deg or about 249 degrees. Thus the polar coordinate would be as follows, ( 2.2, 249 deg). ** ** In radians we have arctan(2.625) = 1.21; adding pi radians because x < 0 we get 4.35 rad. So the coordinates are (2.2, 4.35), approx.. ** This point is in the third quadrant so the angle would be pi + 1.21 rad, or 69.1 deg + 180 deg. When the x component is negative the angle is in the second or third quadrant; the range of the arctan is the fourth and first quadarnt so when x is negative you need to add pi rad or 180 deg to arctan(y/x).
......!!!!!!!!...................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*:Ok **** Query problem 8.1.60 write y^2 = 2 x using polar coordinates.
......!!!!!!!!...................................
19:05:48 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y^2 -2x = 0 R^2 sin^2(theta) – 2rcos(theta) = 0 R(r sin^2(theta) – 2cos(theta)) = 0 Not sure where to go from here confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** We can rearrange the equation to give the form y^2 - 2 x = 0. Then since y = r sin (theta) and x = r cos (theta) we have (r^2 sin ^2 theta) - ( 2 r cos (theta) = 0. We can factor out r to get r [r sin^2(theta) - 2 cos(theta) ] = 0, which is equivalent to r = 0 or r sin^2(theta) - 2 cos(theta) = 0. The latter form can be solved for r. We get r = 2 cos(theta) / sin^2(theta). This form is convenient for graphing. **
.........................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*: **** Query problem 8.1.52 exact polar coordinates of (-2, -2`sqrt(3))
......!!!!!!!!...................................
10:58:03 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: R^2 = x^2 + y^2 R^2 = -2^2 + -2sqrt(3)^2 = 16 R = 4 Tan(theta) = y/x = (-2sqrt(3)/ -2 = Sqrt (3) Pi/3 + pi = 4pi/3 (4, 4pi/3) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: r^2=x^2+y^2 so r^2=(-2)^2+(-2 sqrt(3))^2 = 4+12 = 16 so so r=4. tan theta= y/x = (-2 sqrt(3))/(-2) = sqrt(3). This occurs for the basic angle theta = pi/3, and also for theta = pi/3 + pi = 4 pi/3. The given point is in the third quadrant, so the angle is 4 pi/3. The polar coordinates are therefore (4, 4 pi/3). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*:Ok **** Query problem 8.1.62 write 4 x^2 y = 1 using polar coordinates.
......!!!!!!!!...................................
11:05:39 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: X = r*cos(theta) Y = r * cos(theta) 4(r*cos(theta)) ^2 (r*sin(theta) = 1 4r^3 cos(theta) sin(theta) = 1 Easy to figure with the formula to get: 2r^3(sin^2(theta) = 1 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: GOOD STUDENT SOLUTION we can first say that since x=r cos theta and y= r sin theta that 4(r cos theta)^2(r sin theta)=1 we then have 4 r^3 cos theta sin theta=1 then using the double angle formula we have 2r^3(sin 2 theta)=1
.........................................
11:05:40 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*:Ok **** Query problem 8.1.72 rect coord form of r = 3 / (3 - cos(`theta))
......!!!!!!!!...................................
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: R = 3/ (3 – cos(theta)) 3r – r*cos(theta) = 3 3sqrt(x^2 + y^2) – x = 3 3sqrt(x^2 + y^2) = 3 + x 3(x^2 + y^2) = x^2 + 6x + 9 2x^2 + y^2 – 6x – 9 = 0 This is about as far as I know to go confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Multiply both sides of the equation by 3-cos(theta) to get 3r – r cos(theta)=3. Substitute sqrt(x^2+y^2) for r and x for r cos(theta) to get 3sqrt(x^2+y^2)-x=3. Add x to both sides to obtain 3sqrt(x^2+y^2)=3 + x and square both sides: 3(x^2 + y^2) = x^2 + 6 x + 9, which simplifies to 2 x^2 + y^2 - 6x - 9 = 0. Completing the square on 2 x^2 - 6x we get 2( x^2 - 3 x + 9/4 - 9/4 ) + y^2 = 9 so 2 ( x+3/2)^2 - 9/2 + y^2 = 9 so 2 ( x + 3/2)^2 + y^2 = 9/2 so (x+3/2)^2 / (9/4) + y^2 / (9/2) = 1. This is the equation of an ellipse centered at (-3/2, 0) with semi-axes 3/2 in the x direction and 3 sqrt(2) / 2 in the y direction. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I just wasn’t for sure on completeing the square. ------------------------------------------------ Self-critique rating #$&*:Ok **** Query problem 8.2.10 graph r = 2 sin(`theta).
......!!!!!!!!...................................
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Well I graphed this and it seems to be in the shape of a circle. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** When `theta = 0 or `pi, r = 0 and the graph point coincides with the origin. If `theta = `pi/2, r = 2. If `theta = `pi/4, r = `sqrt(2). So as `theta increases from 0 to `pi/2, r increases from 0 to 2 and the graph follows an arc (actually the arc of a circle) in the first quadrant from the origin to (2, `pi/2). As `theta moves from `pi/2 to `pi the graph follows an arc in the second quadrant which leads back to the origin. As `theta goes from `pi to 3 `pi/2, angles in the third quadrant, r becomes negative since sin(`theta) is negative. At 3 `pi / 2, r will be -2 and the point (-2, 3 `pi / 2) coincides with (2, `pi/2). The graph follows the same arc as before in the first quadrant. As `theta goes from 3 `pi / 2 to 2 `pi, r will remain negative, which places the graph along the same second-quadrant arc as before. Thus the graph will consist of a closed arc in the upper half-plane, tangent to the x axis at the origin. This description doesn’t prove that the graph is a circle, but it turns out to be a circle whose radius is 1. The easiest way to prove this is to convert the equation to rectangular coordinates, as follows: We can multiply both sides by r to get r ^2 = 2 r sin (theta). Substituting x^2 + y^2 for r and y for r sin(theta) we have x ^2 + y ^2 = 2 y x ^2 + ( y ^2 -2 y ) = 0 x ^ 2 + ( y ^2 - 2 y + 1 – 1 ) = 0 x ^2 + ( y -1 ) ^2 - 1 = 0 x^2 + (y-1)^2 = 1. This is the standard form of the equation of a circle with center (0, 1) and radius 1. This is the circle described above. **
......!!!!!!!!...................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*:Ok **** Was it possible to use symmetry in any way to obtain your graph, and if so how did you use it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes it was possible because of the symmetry horizontally. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
......!!!!!!!!...................................
11:54:29 yes the horizontal symmetry caused us to bound the graph at theta=pi/2 so r has a maximum height of 2. and its minimum horizontal line is drawn at -2 this maximum and minimum happens alternatley at the intervals of pi/2 and the negatives.
.........................................
11:54:30
......!!!!!!!!...................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*:Ok **** Query problem 8.2.36 graph r=2+4 cos `theta
......!!!!!!!!...................................
11:58:23 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It seems to be coming to a point at the bottom of the graph. confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** This graph is symmetric with respect to the pole, since replacing theta by -theta doesn't affect the equation (this is so because cos(-theta) = cos(theta) ). At theta goes from 0 to pi/2, cos(theta) goes from 1 to 0 so that r will go from 6 to 2. Then when theta = pi/3 we have cos(theta) = -1/2 so that r = 0. As theta goes from pi/2 to pi/3, then, r goes from 2 to 0. To this point the graph forms half of a heart-shaped figure lying above the x axis. Between theta = pi/3 and theta = 2 pi / 3 the value of cos(theta) will go from -1/2 to -1 to -1/2, so that r will go from 0 to -2 and back to 0. The corresponding points will move from the origin to the 4th quadrant as theta goes past pi / 3, reaching the point 2 units along the pole when theta = pi then moving into the first quadrant, again reaching the origin when theta = 2 pi/3. These values will form an elongated loop inside the heart-shaped figure. From theta = 2 pi / 3 thru theta = 3 pi / 2 and on to theta = 2 pi the values of r will go from 0 to 2 then to 6, forming the lower half of the heart-shaped figure. **
.........................................
11:58:24 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*: "