#$&*
course Mth 279
5/2
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.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Using the Equation form: W = W_l
where W is the object's weight, and W_l is the weight of the displaced water.
pALg = p_1 AYg
Y (depth) = p/p_1 (L) = 700/1000 (1m) = 0.7m
Y + y(t) = depth
pAL(d^2/dt^2)(Y+y(t)) = pALg - p_l A(Y+y(t))g
simplified: pALy''(t) = -p_l Agy(t)
or y''(t) + w^2y(t) = 0
w^2 = (p_l g)/(pL) = (1000*9.81)/(700*1m) = 981/70 = 14.01 approx.
w = sqrt (981/70) = 3.74 approx.
Solutions:
y(t) = sin(wt) and y(t) = cos(wt)
General solution:
y(t) = Asin(wt) + Bcos(wt)
y(t) = Asin(3.74t) + Bcos(3.74t)
If y(0) = 0.1m
y'(0) = 0
y(0) = Asin(0) + Bcos(0) = 0.1
Bcos(0) = 0.1
B = 0.1
y'(0) = Awcos(0) - Bwsin(0) = 0
Awcos(0) = 0
Aw = 0
A = 0
Final solution becomes:
y(t) = (o.1)cos(3.74(t))
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The second part:
Struck and sinks the distance of L-Y = 30cm = 0.3m
y(0) = -0.3m
y''(t) + w^2y(t) = 0
Gen. Solution. -> y(t) = Asin(wt) + Bcos(wt)
y(0) = Asin(0) + Bcos(0) = -0.3
Bcos(0) = -0.3
B = -0.3
y'(0) = Awcos(0) - Bwsin(0) = 0
Awcos(0) = 0
Aw = 0
A = 0
Solution:
y(t) = -0.3cos(3.74(t))
@&
This is a good solution, if we assume that t = 0 when the cylinder's top reaches the level of the water.
If we assume that t = 0 at the instant the cylinder is struck, we get the following:
The cylinder is easily seen to float in equilibrium when 70 cm is below and 30 cm above the water line.
If the cylinder is struck from above at equilibrium and comes just to and not below the surface of the water, then we know that its maximum displacement from equilibrium has magnitude 30 cm. So we can let A = 30 cm.
The velocity of the cylinder at the initial instant is negative and its position is zero, so
y(0) = A cos(sqrt(c) * 0 + phi) = 0
and
y ' (0) < 0.
The first condition requires that cos(phi ) = 0, so that phi = pi/2 or 3 pi / 2.
The second condition requires that sin(phi) < 0, so that phi = 3 pi / 2.
The solution is thus
y(t) = 30 cm * cos(sqrt(c) * t + 3 pi / 2) or approximately
y(t) = 30 cm * cos( 3.7 t + 3 pi / 2).
The velocity function wasn't requects, but it is
v(t) = y ' (t) = -sqrt(c) * 30 cm sin (sqrt(c) t + 3 pi / 2) = -110 cm/s * sin(3.7 t + 3 pi/2).
*@
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:
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:
y '' + y ' + 3 t y = tan(t)
y(pi) = 1
y ' (pi) = -1
The equation should only be undefined at (pi/2)(2n+1) due to tan(t)
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t y '' + sin(2 t) / (t^2 - 9) y ' + 2 y = 0
y(1) = 0
y'(1) = 1
First we need the equation in the correct form:
t y '' + sin(2 t) / (t^2 - 9) y ' + 2 y = 0
y '' + sin(2 t) / (t(t^2 - 9)) y ' + (2 y)/t = 0
Should be undefined at zero because t cannot equal zero,
if it was you would have 2 points where you cannot divide by zero.
and if
t^2 -9 = 0
We get another point or 2 where you cannot divide by zero.
t^2 = 9
t = sqrt(9)
t = + or - 3
Therefore the equation would be undefined at these points.
Every place that the equation is undefined there should be asymptotes.
@&
Good, but you need to specify the intervals where the solutions are defined.
Compare with the following:
For the equation y '' + y ' + 3 t y = tan(t), y(pi) = 1, y ' (pi) = -1:
3 t y is defined for all values of t and y.
tan(t) is defined, continuous and differentiable except at odd multiples of t = pi / 2. So for example the function is continuous on the intervals ..., (-3 pi/2, -pi/2), (-pi/2, pi/2), (pi/2, 3 pi/2), ...
The initial condition is given at t = pi, which lies in the interval (pi/2, 3 pi/2). Thus the solution exists within this interval.
The equation t y '' + sin(2 t) / (t^2 - 9) y ' + 2 y = 0, y(1) = 0, y ' (1) = 1 is rearranged to the form y '' + sin(2 t) / (t ( t^2 - 9 ) ) t ; + 2 y / t = 0.
The only function that is restricted in domain is 1 / (t (t^2 - 9) ), which is undefined at t = -3 and t = 3, and a t = 0.
So the intervals on which solutions are known to exist are (-infinity, -3), (-3, 9), (0, 3) and (3, infinity).
The initial conditions are given at t = 1, which is in the interval (0, 3). So for these initial conditions, the solution is known to exist within the interval (0, 3).
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confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
2
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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 - sin(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:
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y '' + y = 2 - sin(t), y(0) = 1, y ' (0) = -1
y''(0) + y(0) = 2 - sin(0)
y''(0) + 1 = 2
y''(0) = 1
Because y'(0) is negative, it would be a decreasing slope.
@&
y ' negative means that the solution is decreasing, not that the slope is decreasing.
If the slope was decreasing the graph would be concave down.
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and y''(0) is positive, Concave up
Graph B
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y '' + y = - 2 t, y(0) = 1, y ' (0) = -1
y''(0) + y(0) = - 2(0)
y''(0) + 1 = 0
y''(0) = -1
Because y'(0) is negative, it would be a decreasing slope.
@&
The slope is decreasing but because y '' is negative, not because y' is negative.
*@
and y''(0) is negative, Concave down
Graph D
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y '' - y = t^2, y(0) = 1, y ' (0) = 1
y''(0) - y(0) = (0)^2
y''(0) - 1 = 0
y''(0) = 1
Because y'(0) is positive, it would be a increasing slope.
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Positive y ' doesn't tell you whether the slope is increasing or decreasing, only that it's positive.
That is, positive y ' tells you that the solution at this point is increasing (not that the slope is increasing).
*@
and y''(0) is positive, Concave up
@&
good
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Graph A
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y '' - y = - 2 cos(t), y (0) = 1, y ' (0) = 1
y''(0) - y(0) = -2cos(0)
y''(0) - 1 = -2
y''(0) = -1
Because y'(0) is positive, it would be a increasing slope.
@&
y ' positive tells you that the solution is increasing, not that the slope is increasing.
*@
and y''(0) is negative, Concave down
Graph C
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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&#Good responses. See my notes and let me know if you have questions. &#