course Mth 164 SOLUTIONS/COMMENTARY ON QUERY 9
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**** Query problem 7.3.14 b = 4, c = 1, alpha = 120 deg
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00:36:10 ********************************************* Your solution: a^2 = 16+1-2(4)(1)cos(120) = 16+1-8(-.5) = 21 a = sq. rt. of 21 cos^-1 B .654 = 49.156 deg. 180 – 120 – 49.156 = 10.844 = C Confidence Assessment: 2 ********************************************* 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): Self-critique Rating: **** 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 ********************************************* Your solution: I first assumed that this was not a right triangle and tried to figure it like this: Sin 15 deg./500 = sinB/100 but this gave me angle B of 2.96 so I knew this wasn’t right, so I tried to figure it as a right triangle and for 500 sin(75) = a sin(15) I got a = 1866.014 so I knew this wasn’t right either, so I tried again with the Pythagorean thereom and got 500^2+ b^2 = 100^2 which gives b^2 = -240,000. Confidence Assessment: 0 ********************************************* 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): I did not understand the part about needing 2 triangles but I knew there was something wrong with the way I was setting the problem up. Self-critique Rating: **** 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 ********************************************* Your solution: D = 10sin(2pi/3t) 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). **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don’t understand why + pi is necessary on the end
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: **** query problem 7.4.14 b = 4, c = 1, alpha = 120 deg What is the area of the triangle? ********************************************* Your solution: I didn’t think we were assigned sec. 8.4, which covers area of triangles, and I didn’t read the question completely first, so I computed the 3rd side length and angle before discovering that I needed to compute the area, then I used Heron’s formula : S=(1/2)(1+sq. rt. of 21+4) = 4.79 K = Sq. rt. of(4.79*94.79-sq. rt. of 21)*(4.79-4)*(4.79-1) = 1.723 Given Solution: ½ b c sin alpha .5(4)(1)sin120 = 1.73 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: **** 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 ********************************************* Your solution: Radius = 12 ft. ½ the angle = 50 deg. ½(12)(12)sin(50) = 55.16 ft. I thought the folded sector formed a triangle so I found the area of the triangle(couldn’t find the formula for area of a cone) Confidence Assessment: ********************************************* 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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