#$&* course Mth 164 SOLUTIONS/COMMENTARY ON QUERY 10
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10:38:28 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The -3 means we end up to the left 3 units and the 4pi means you have completed 2 circles and are at the angle 0. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Can be write as (3, pi) or (3, -pi), or (-3, 0). ------------------------------------------------ Self-critique Rating:3 **** Query problem 8.1.42 rect coord of (-3.1, 182 deg)
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (-3.1, 182 deg) The -3.1 means you will move to the left 3.1 units and the 182 means it makes that angle at the polar axis. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: GOOD STUDENT SOLUTION: since we know the formula for converting polar coordinates to rectangular coordinates we can say that r=-3.1 and theta=182 deg. we can first find the value of x by saying x=r cos theta which gives (-3.1) cos 182 deg. = 3.1. We can then find y by saying y= r sin theta so y=(-3.1)sin 182 deg. using a calculator we get an approx. value of .1081. Thus the rectangular coordinates are (3.1, .1081), approx.. (-3.1, 182 deg)
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): To get the y use y = r sin(theta) Y = -3sinf(182) Y = 1.081 (-3.1, 182 deg) ------------------------------------------------ Self-critique Rating:3 Query problem 8.1.54 polar coordinates of (-.8, -2.1)
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18:50:13 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: r^2 = x^2 + y^2 r = sqrt( -.8^2 + -2.1^2) r = 2.24 tan(theta) = -2.1 / -.8 tan(theta) = 2.625 Theta = arctan(2.625) Theta = 69. 145 180 + 69. 145 = 249.145 (2.24 , 249.145degrees) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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).
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Query problem 8.1.60 write y^2 = 2 x using polar coordinates.
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19:05:48 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y^2 = 2 x y^2 -2x = 0 confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** We can rearrange the equation to give the form y^2 - 2 x = 0. Then since y = r sin (theta) and x = r cos (theta) we have (r^2 sin ^2 theta) - ( 2 r cos (theta) = 0. We can factor out r to get r [r sin^2(theta) - 2 cos(theta) ] = 0, which is equivalent to r = 0 or r sin^2(theta) - 2 cos(theta) = 0. The latter form can be solved for r. We get r = 2 cos(theta) / sin^2(theta). This form is convenient for graphing. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don’t understand what step to do after I get the problem set to 0. ------------------------------------------------ Self-critique Rating:2
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10:58:03 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: R = sqrt(-2^2 + (2sqrt(3)) R = sqrt(4 + 11.99) R = 3.99 Tan(theta) =2sqrt(3) / 2 Theta = 59.999 180 + 59.999 = 239.99 (3.99 , 239.99) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Would you count my answer right? ------------------------------------------------ Self-critique Rating:
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11:05:39 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4 x^2 y = 1 x=r cos(theta) y= r sin(theta) 4(r cos(theta)^2(r sin theta)=1 4 r^3 cos theta sin theta=1 2r^3(sin 2 theta)=1 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: GOOD STUDENT SOLUTION we can first say that since x=r cos theta and y= r sin theta that 4(r cos theta)^2(r sin theta)=1 we then have 4 r^3 cos theta sin theta=1 then using the double angle formula we have 2r^3(sin 2 theta)=1
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11:05:40 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Query problem 8.1.72 rect coord form of r = 3 / (3 - cos(`theta))
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: r = 3 / (3 - cos(`theta)) 3r - r cos(theta)=3 3sqrt(x^2+y^2)-x=3 3sqrt(x^2+y^2)=3 + x 3(x^2 + y^2) = x^2 + 6 x + 9 2 x^2 + y^2 - 6x - 9 = 0 2( x^2 - 3 x + 9/4 - 9/4 ) + y^2 = 9 2 ( x+3/2)^2 - 9/2 + y^2 = 9 2 ( x + 3/2)^2 + y^2 = 9/2 (x+3/2)^2 / (9/4) + y^2 / (9/2) = 1 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: **** Query problem 8.2.10 graph r = 2 sin(`theta).
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: r = 2 sin(`theta) r ^2 = 2 r sin (theta) 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 (0, 1) radius = 1 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** 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 you could use horizontal symmetry. The max and mins happen alternatively at the pi/2 and the negatives. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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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.
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11:54:30
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Query problem 8.2.36 graph r=2+4 cos `theta
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11:58:23 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: r=2+4 cos `theta when theta goes from 0 to pi/2 and cos goes from 1 to 0 then r goes from 6 to 2. When theta is equal to pi/3 we have cos(theta) equal to -1/2 so that r is 0. Then as theta goes from pi/2 to pi/3,r goes from 2 to 0. 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. This will move to the fourth coordinate. These will make heart shaped figures. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. **
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11:58:24 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!