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course Mth 279
7/22 Sorry it is taking so long to upload the assignments. They are completed it just takes a while to type in all the responses.
Query 12 Differential Equations*********************************************
Question: A cylinder floating vertically in the water has radius 50 cm, height 100 cm and uniform density 700 kg / m^3. It is raised 10 cm from its equilibrium position and released. Set up and solve a differential which describes its position as a function of clock time.
The same cylinder, originally stationary in its equilibrium position, is struck from above, hard enough to cause its top to come just to but not below the level of the water. Solve the differential equation for its motion with this condition, and use a particular solution to determine its velocity just after being struck.
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Your solution:
A)p_w = 1000kg/m^3; p_cyl = 700 kg/m^3; h= 100cm = 1m
Vol difference = -Area * position = -(pir^2)y
mass displaced = -pr^2y
F_net = -p_w(pir^2)y*g
m_cyl = p*pir^2h
F_net = my’’
so the plugging in everything we get
p_cyl*pir^2h*y’’ = -p_w*pir^2gy
y’’ = (-p_w*g)y/(h*p_cyl) = (-1000*9.8)y/(700*10) = -14y
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This is the equation of motion for the cylinder. The number 14 has units of seconds^-2.
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
I took problem to as far as I could work it then I didn’t know what to do to finish/answer the problem.
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Self-critique rating:
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The solution to the equation
y '' = - c y
is
y = B cos(sqrt(c) * t) + C sin(sqrt(c) * t)
as you can (and should) easily verify for yourself.
See how far you can get using this information.
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Question:
For each of the equations and initial conditions below, find the largest t interval in which a solution is known to exist:
y '' + y ' + 3 t y = tan(t), y(pi) = 1, y ' (pi) = -1
t y '' + sin(2 t) / (t^2 - 9) y ' + 2 y = 0, y(1) = 0, y ' (1) = 1.
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Your solution:
A) y’’ + y’ + 3ty = tan(t)
so p(t) = 1 ; q(t) = 3t and g(t) = tan(t)
p and q are continuous everywhere
g discontinuous at every odd multiple of pi/2
t_0 = pi
so now we need to find the interval where t_0 falls in-between.
we would have intervals of (-3pi/2, -pi/2) , (-pi/2, pi/2), (pi/2, 3pi/2) and so on.
But the only integral that t_0 falls between is
(pi/2, 3pi/2)
B) ty’’ + sin(2t)/(t^2 - 9) y’ + 2y = 0
divide by t
y’’ + sin(2t)/(t(t^2 -9)) y’ + 2y/t = 0
p = sin(2t)/(t(t^2 - 9)); q = 2y/t and g = 0
to find where p and q are discontinuous you find where the denominators are = 0
for p:
t(t^2 - 9) = 0
t = 0; t = 3, and t = -3
for q:
t= 0
so find the largest interval when t = 1
we would have the intervals of:
(-infinity, -3) (-3,0) (0,3) (3,infinity)
so t=1 falls between (0,3)
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
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Question: Decide whether the solution of each of the following equations is increasing or decreasing, and whether each is concave up or concave down in the vicinity of the given initial point:
y '' + y = - 2 t, y(0) = 1, y ' (0) = -1
y '' + y = - 2 t, y(0) = 1, y ' (0) = -1
y '' - y = t^2, y(0) = 1, y ' (0) = 1
y '' - y = - 2 cos(t), y (0) = 1, y ' (0) = 1.
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Your solution:
A) y’’ + y = -2t
initial point (0,1) with a slope of -1, so function is decreasing.
y’’ = -1 - 2(0) = -1 with a neg second derivative the function is concave down
B) y’’ -y = t^2
slope = 1 so function is increasing
y’’ = y + t^2 = 1 + 0^2 = 1 so function is concave up
C) y’’ - y = -2cos(t)
slope =1 so increasing
y’’ = 1 - 2cos(0) = 1 - 2(1) = -1 so concave down
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:"
Self-critique (if necessary):
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&#Your work looks good. See my notes. Let me know if you have any questions. &#