Chapter 1.5, Problem 3QR

Solve the initial value problem: y= 0.05y + 5 y(0) = 100 y(t) =

y=f'(x) 1 8 The function f is defined on the closed interval [0,8]. The graph of its derivative f' is shown above. How many relative minima are there for f(x)? O 2 6 4 00

60! 5!.7!.15!.33!

• • Let > be a potential for the vector field F = (−2 y³, −6 xy² − 4 z³, −12 yz² + 4 2). Then the value of sin((-1.63, 2.06, 0.57) – (0,0,0)) is - 0.336 -0.931 -0.587 0.440 0.902 0.607 -0.609 0.146

The value of cos(4M) where M is the magnitude of the vector field with potential ƒ = e² sin(лy) cos(π²) at x = 1, y = 1/4, z = 1/3 is 0.602 -0.323 0.712 -0.816 0.781 0.102 0.075 0.013

There is exactly number a and one number b such that the vector field F = conservative. For those values of a and b, the value of cos(a) + sin(b) is (3ay + z, 3ayz + 3x, −by² + x) is -0.961 -0.772 -1.645 0.057 -0.961 1.764 -0.457 0.201

A: Tan Latitude / Tan P A = Tan 04° 30'/ Tan 77° 50.3' A= 0.016960 803 S CA named opposite to latitude, except when hour angle between 090° and 270°) B: Tan Declination | Sin P B Tan 052° 42.1'/ Sin 77° 50.3' B = 1.34 2905601 SCB is alway named same as declination) C = A + B = 1.35 9866404 S CC correction, A+/- B: if A and B have same name - add, If different name- subtract) = Tan Azimuth 1/Ccx cos Latitude) Tan Azimuth = 0.737640253 Azimuth = S 36.4° E CAzimuth takes combined name of C correction and Hour Angle - If LHA is between 0° and 180°, it is named "west", if LHA is between 180° and 360° it is named "east" True Azimuth= 143.6° Compass Azimuth = 145.0° Compass Error = 1.4° West Variation 4.0 East Deviation: 5.4 West

ds 5. Find a solution to this initial value problem: 3t2, s(0) = 5. dt 6. Find a solution to this initial value problem: A' = 0.03A, A(0) = 100.

2) Drive the frequency responses of the following rotor system with Non-Symmetric Stator. The system contains both external and internal damping. Show that the system loses the reciprocity property.