#$&* course Mth 164 3/22 4:30 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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**** Query problem 7.3.14 b = 4, c = 1, alpha = 120 deg
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00:36:10
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Given Solution: a= sqrt(4^2 + 1^2 -2(4)(1)cos120) a = sqrt(16+1-8cos120) a = sqrt(17-8cos120) a = 4.58 b^2 = a^2 + c^2 -2accos`beta 4^2 = 4.58^2 + 1^2 - 2(4.58)(1)cos`beta 9.16cos`beta = 4.58^2 + 1^2 - 4^2 9.16cos`beta = 5.98 beta = cos^-1(5.98/9.16) beta = 49.2deg c^2 = a^2 + b^2 -2abcos`gamma 1^2 = 4.58^2 + 4^2-2(4.58)(4)cos`gamma 36.64cos`gamma = 35.98 gamma = cos^-1(35.98/36.64) gamma = 10.89
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK **** Query problem 7.3.32 500 ft tower 15 deg slope 2 guy wires 100 ft on either side of base on slope
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00:45:52 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We have two triangles formed with this question. Triangle 1 faces up the hill from the tower and has sides a1 = 100 ft, b1 = ?, c1 = 500 ft and has angles A1 = ?, B1 = 90 deg - 15 deg = 75 deg, C1 = ?. Triangle 2 faces down the hill from the tower and has sides a2 = 100 ft, b2 = ?, c2 = 500 ft and has angles A1 = ?, B1 = 90 deg + 15 deg = 105 deg, C1 = ?. We have two sides and the included angle of each triangle so we can use Law of Cosines to find b1 and b2 which are how long the two guy wires should be as follows: length of guy wire uphill = b1 b1^2 = 100^2 + 500^2 -2(100)(500)cos 75 deg b1^2 = 10000 + 250000 - 100000cos 75 deg b1^2 = 260000 - 100000cos 75deg b1 = `sqrt(260000 - 100000cos 75deg) or approx. 483.86 ft length of guy wire downhill = b2 b2^2 = 100^2 + 500^2 -2(100)(500)cos 105 deg b2^2 = 10000 + 250000 - 100000cos 105 deg b2^2 = 260000 - 100000cos 105deg b2 = `sqrt(260000 - 100000cos 105deg) or approx. 534.68 ft confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** GOOD STUDENT SOLUTION: A triangle containing the tower side and the side of the slope up the hill will have angle 75 deg between tower and hill. The triangle with the side down the slope has angle 105 deg between tower and hill. The guy wire will form the hypotenuse c on each side. If c1 = length of guy wire on the right hand side of the tower and c2 = length of guy wire on the left hand side of the tower then the Law of Cosines gives us (c1)^2 = a^2 + b^2 - 2ab cos `gamma = 500ft^2 + 100ft^2 - 2(500)(100)(cos75 deg) = 260,000 - 25,881.90451 (c1)^2 = 234118.0955 c = approx. 483.8575 feet (c2)^2 = a^2 + b^2 - 2ab cos`gamma = 500ft^2 + 100ft^2 -2(500)(100)(cos105 deg) = 260,000 - (-25881.90451) (c2)^2 = 285,881.9045 c = approx. 534.68 feet ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK **** query problem 7.5.6 coiled spring 10 cm displ period 3 sec moving downward at equil when t = 0. What is the equation of motion of the object?
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20:29:55 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Amplitude = A= |10| = 10 cm Period = T = 3 sec therefore omega = 2pi/3 sec The object moving down so the reference point at t = 0 would be 0 so we have: d = -10 cm * sin(2pi/3 sec * t)
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Given Solution: ** Modeling with reference-circle point moving counterclockwise at angular velocity omega on a circle of radius A, and using the sine function, displacement will be y(t) = A sin (`omega * t + theta0), where theta0 is the initial angular position on the reference circle. We know that the spring is at equilibrium when t = 0, so theta0 is either 0 or pi. Since the spring is initially moving downward, we conclude that theta0 is pi. The amplitude of motion will be 10 cm so A = 10 cm. The period of motion is 2`pi/`omega = 3 sec, so `omega = 2`pi/(3 sec), so the equation of motion is y = 10 cm sin( 2 pi / (3 sec) * t + pi). Alternatively we could model using the cosine function, though this is less intuitive for an object moving up and down: x(t) = A cos(omega * t + theta0), in which case since the spring is at equilibrium when t = 0, theta0 is pi/2 (motion at t = 0 therefore being in the negative direction) and we get x(t) = 10 cm cos(2 pi / (3 sec) * t + pi/2). STUDENT COMMENT: I don’t understand why + pi is necessary on the end INSTRUCTOR RESPONSE: One stated condition is that the object is moving downward when t = 0. The equation you gave would have had it moving upward. In order for the object to be at equilibrium, with a sine model, the angle on the reference circle must be either 0 or pi. If the angle is 0, then since the positive direction is counterclockwise (e.g., from the positive x axis into the first quadrant), the sine function will immediately move into positive values, indicating upward motion. If the angle is pi, then continued positive motion around the reference circle will take you into the third quadrant, where the sine takes negative values; this indicates downward motion.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I’m not sure that I am rapping my head around these concepts correctly. Am I correct in saying that we are creating an equation or a function of displacement in this case of the y coordinate which is cm with respect to time which is the x coordinate which is also seconds? I am certain that the equation I supplied is correct. Am I correct in explaining that the movement is simple harmonic with a maximum displacement of 10 cm and that we are moving in a clockwise direction from y displacement 0 pulled down to -10 then when released it oscillates first to 10 then to -10 discounting any friction at a frequency of 3/2pi per sec and that the time to complete one oscillation or period is 3sec ? I guess that I am confused about when to start at pi moving clockwise or 0 moving counterclockwise, although, I believe that they are equations equivalent since sine function is odd, in other words, I’m confused about when whatever is moving down from being pulled being up or the opposite. Any direction on this is welcomed. ------------------------------------------------ Self-critique Rating: 1
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Given Solution: ** This equation is of form d = A cos(omega * t) with A = 5 and omega = pi/2, corresponding to motion on a reference circle of radius 5 at angular velocity pi/2 rad/sec. The period of motion is 2 pi / omega = 2 pi / (pi/2) = 4. So frequency is 1 / period = 1/4. The amplitude of motion is 5 units, the coefficient of the cosine function. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK **** query problem 7.4.14 b = 4, c = 1, alpha = 120 deg What is the area of the triangle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the formula for area of triangle where area = K = (1/2)bc * sin(alpha) we have: K = (1/2)(4)(1)sin(120 deg) K = 2sin(120 deg) or approx. 1.73 deg
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Given Solution: ½ b c sin alpha .5(4)(1)sin120 = 1.73 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK **** query problem 7.4.28 cone from circle 24 ft diameter, 100 deg sector removed Find the area of the cone obtained when you fold join the cut edges.
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00:56:16 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To answer this question, we need to find the area of the circle and subtract the area of the sector from the result. Circle area = A = pi * r^2 where r = radius so A = 144pi ft or approx. 452.4 ft Sector Area = A = (1/2) r^2 * theta radians where r = radius so A = (1/2)(144 ft)(100 deg * pi/180 deg) A = 72 (100pi ft/180) A = 7200pi ft /180 = 40pi ft or approx. 125.66 ft Cone area = A = 144pi ft - 40pi ft A = 104pi ft or approx. 326.73 ft confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 100deg = 1.745rad Area of Circle = pi*r^2 = pi*12^2 = 452.389 Area of Sector = 1/2(r^2)(`theta rad) = 1/2(12^2)(1.745) = 125.64 Area of Cone = 452.389 - 125.64 = 326.75
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!