#$&* course Mth164 3/13 10.21 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x displacement of -1 to 6, is 6 - (-1) = 7 y displacement of 4 to 2, is 2 - 4 = -2 v = 7 i - 2 j x displacement of 1 to 6, is 6 - (1) = 5 The y displacement of 4 to 2, is 2 - 4 = -2 v = 5 i - 2 j. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If we have the initial point at (-1,4) and terminal point at (6,2). The x displacement is from -1 to 6, a displacement of 6 - (-1) = 7 The y displacement is from 4 to 2, a displacement of 2 - 4 = -2 The i and j unit vectors are in the x and y directions, respectively, so our vector is v = 7 i - 2 j. ** ** If we have the initial point at (1,4) and terminal point at (6,2). The x displacement is from 1 to 6, a displacement of 6 - (1) = 5 The y displacement is from 4 to 2, a displacement of 2 - 4 = -2 The i and j unit vectors are in the x and y directions, respectively, so our vector is v = 5 i - 2 j. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: Query problem 8.4.36 (was 8.4.38?) ||v|| + ||w|| if v = 3i-5j, w = -2i+3j.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ||v||= sqrt(3^2 + (-5)^2) = sqrt 34 ||w||= sqrt (-2^2 + 3^2 ) = sqrt 13 ||w+v|| = sqrt34+ sqrt 13 or 9.44 || w + v || = || (3 i - 5 j) + (-2 i + 3 j) || || i - 2 j || sqrt(1^2 + 2^2) sqrt(5) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** to find ||v + w|| if v = 3i-5j, w = -2i+3j, we find that ||v||= sqrt(3^2 + (-5)^2) = sqrt 34 and ||w||= sqrt (-2^2 + 3^2 ) = sqrt 13 so ||w+v|| = sqrt34+ sqrt 13 or in decimal form 9.44. To get || w + v || you have to first find w + v, then take the magnitude of this resultant. || w + v || = || (3 i - 5 j) + (-2 i + 3 j) || = || i - 2 j || = sqrt(1^2 + 2^2) = sqrt(5). This, with your work, demonstrates that || w + v || is not generally equal to || w || + || v ||. The two expressions are in fact equal if, and only if, the two vectors are parallel and in the same direction. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: Query problem 8.4.42 unit vector having same direction as v = -5 i + 12 j.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v = -5i + 12j. ||v||=sqrt(25 + 144)=sqrt 169 or 13. ||v||=13 v/||v|| = (-5i + 12j) / 13= -5/13 i + 12/13 j. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: GOOD STUDENT SOLUTION WITH AMPLIFICATION BY INSTRUCTOR We need to find a vector having magnitude 1 and the same direction as v = -5i + 12j. We first need to find ||v||: ||v||=sqrt(25 + 144)=sqrt 169 or 13. ||v||=13 v/||v|| = (-5i + 12j) / 13= -5/13 i + 12/13 j. This vector would have the same direction. ** Correct solution. The vector has the same direction because it is a multiple of the vector by a positive constant. The vector is a unit vector because when you divide a vector by a constant its new magnitude is also divided by that constant; you have divided the original vector by its original magnitude so its new magnitude is equal to its original magnitude divided by its original magnitude; the result is new magnitude 1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:
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**** Query problem 8.4.50 airplane ends up due South 200 miles with 30 mph wind from Northwest.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v + w = -200 j w = 30 cos(315 deg) i + 30 sin(315 deg) j w = 21.2 i -21.2 j v = -200 j - w 200 j - (21.2 i -21.2 j) 200 j -21.2 i +21.2 j 21.2 i -221.2 j \ sqrt( (21.2)^2 + (-221.2)^2) or 222.2 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The airplane moves as some velocity with respect to the wind, which since it is being carried along with the wind is different that its velocity with respect to the ground. If w is the direction of the wind and v the velocity of the airplane with respect to the wind. The velocity of the wind is added to the velocity of the airplane with respect to the wind. Then the actual velocity of the airplane is v + w. Since the airplane ends up 200 miles due South after 1 hour, its actual velocity is 200 mph to the South. Thus we know that v + w = -200 j, understanding that the -200 is in mph. If w is from the Northwest at 30 mph then, placing North and East in the positive y and x directions respectively we see that the direction of w is toward the Southeast, at angle 315 deg with respect to the positive x axis. Thus w = 30 cos(315 deg) i + 30 sin(315 deg) j =21.2 i -21.2 j, approx.. Now if v + w = -200 j it follows that v = -200 j - w = 200 j - (21.2 i -21.2 j) = 200 j -21.2 i +21.2 j = 21.2 i -221.2 j. The magnitude of this vector is sqrt( (21.2)^2 + (-221.2)^2) = 222.2, approx.. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Query problem 8.5.10 (was 8.5.8?) dot product of v = 3i - 4j and w = 4i - 3j.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v dot w = ||v|| * ||w|| cos(`theta) cos(`theta) = v dot w / ( ||v|| ||w|| ) `theta = cos^-1(24 / 25) 16 deg confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Using the definition of the dot product: v dot w = ||v|| * ||w|| cos(`theta) so we have cos(`theta) = v dot w / ( ||v|| ||w|| ) = 24 / 25 so `theta = cos^-1(24 / 25) = 16 deg. **
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13:40:46 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Query problem 8.5.24 resultant displacement 200 miles west then 150 miles 60 degrees north of west.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v = -200 i + 0 j w = 150 cos(120 deg) * i + 150 sin(120 deg) * j w = -75 i + 130 j v + w = -200 i + 0 j + -75 i + 130 j v + w = -275 i + 130 j sqrt( 275^2 + 130^2) 300 approx tan^-1(130/(-275)) + 180 deg 30 deg + 180 deg 210 deg confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 200 miles west is represented by the displacement vector v = -200 i + 0 j. 150 miles north of west is 150 miles at an angle of 120 deg so the displacement vector here is w = 150 cos(120 deg) * i + 150 sin(120 deg) * j = -75 i + 130 j, approx. So net displacement is v + w = -200 i + 0 j + -75 i + 130 j = -275 i + 130 j. Magnitude is `sqrt( 275^2 + 130^2) = 300 approx.. Angle is tan^-1(130/(-275)) + 180 deg = -30 deg + 180 deg = 150 deg (approximately). ** INSTRUCTIVE STUDENT ERROR: we can draw the points on the graph and we see that it travels 200 units in the negative on the x-axis and then at an angle of 60 deg it travels 150 miles north of west. we thus have two sets of points on the graph so we can say P sub 1(200,0) and P sub 2(150, 60 deg) ** correct if both of these representations are in polar coordinates ** so by using the distance formula we can say sq.rt.((200-150)^2+(0-60)^2 = 78.10 mi ** the distance formula for rectangular coordinates doesn't work with polar coordinates. ** these sections were a lot of fun to work on however there were some difficult problems.i thought it was nice that we were allowed to make use of some of the basic properties we learned in algebra one such as the commutative, associative and distributive properties to solve problems involving vectors. ** vector algebra is an entire subject in itself. Take a linear algebra course sometime if you get the chance. ** POSSIBLE ERROR IN STATEMENT OF PROBLEM: POSSIBLE CORRECTION BY RELIABLE STUDENT: The problem gives the angle 60 deg and the leg adjacent to this angle so you use the tangent to find the opposite leg, the displacement. tan 'theta = opp/adj tan 60 deg = x/150 x = 150 tan 60 deg x = about 259.81 mi. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): -30 + 180 = 150 degs ------------------------------------------------ Self-critique Rating: **** Query problem 8.5.24 resultant displacement 200 miles west then 150 miles 60 degrees north of west.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v = -200 i + 0 j w = 150 cos(120 deg) * i + 150 sin(120 deg) * j w = -75 i + 130 j v + w = -200 i + 0 j + -75 i + 130 j v + w = -275 i + 130 j sqrt( 275^2 + 130^2) 300 approx tan^-1(130/(-275)) + 180 deg 30 deg + 180 deg 210 deg confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 200 miles west is represented by the displacement vector v = -200 i + 0 j. 150 miles north of west is 150 miles at an angle of 120 deg so the displacement vector here is w = 150 cos(120 deg) * i + 150 sin(120 deg) * j = -75 i + 130 j, approx. So net displacement is v + w = -200 i + 0 j + -75 i + 130 j = -275 i + 130 j. Magnitude is `sqrt( 275^2 + 130^2) = 300 approx.. Angle is tan^-1(130/(-275)) + 180 deg = -30 deg + 180 deg = 150 deg (approximately). ** INSTRUCTIVE STUDENT ERROR: we can draw the points on the graph and we see that it travels 200 units in the negative on the x-axis and then at an angle of 60 deg it travels 150 miles north of west. we thus have two sets of points on the graph so we can say P sub 1(200,0) and P sub 2(150, 60 deg) ** correct if both of these representations are in polar coordinates ** so by using the distance formula we can say sq.rt.((200-150)^2+(0-60)^2 = 78.10 mi ** the distance formula for rectangular coordinates doesn't work with polar coordinates. ** these sections were a lot of fun to work on however there were some difficult problems.i thought it was nice that we were allowed to make use of some of the basic properties we learned in algebra one such as the commutative, associative and distributive properties to solve problems involving vectors. ** vector algebra is an entire subject in itself. Take a linear algebra course sometime if you get the chance. ** POSSIBLE ERROR IN STATEMENT OF PROBLEM: POSSIBLE CORRECTION BY RELIABLE STUDENT: The problem gives the angle 60 deg and the leg adjacent to this angle so you use the tangent to find the opposite leg, the displacement. tan 'theta = opp/adj tan 60 deg = x/150 x = 150 tan 60 deg x = about 259.81 mi. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): -30 + 180 = 150 degs ------------------------------------------------ Self-critique Rating: #*&!