#$&*
course Mth 279
April 28 around 6:45pm.
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:
We went over something similar to this in class. I understand it and have no further questions on it.
@& Good. For reference, in case it's useful to you:
We will assume the upward direction to be positive.
The net force on the system when its displacement from equilibrium is y is equal to the change in the buoyant force. If the cylinder rises above equilibrium the buoyant force is less than that at equilibrium and the net force is negative; at positions below equilibrium the buoyant force is greater than that at equilibrium and the net force is positive.
At position y relative to equilibrium the buoyant force differs from that at equilibrium by and amount equal in magnitude to the difference in the weights of displaced water. The difference in the volumes of displaced water is
volume difference = - cross-sectional area * position relative to equilibrium = - pi r^2 y,
so the difference in the mass of displaced water is - rho r^2 y and the net force is the product of this difference in mass and the acceleration of gravity:
F_net = - rho_water * pi r^2 y * g.
The mass of the entire cylinder is
mass = rho_cylinder * pi r^2 h,
where h is the altitude of the cylinder.
Since F_net = mass * y '' we have
rho_cyl * pi r^2 h * y '' = - rho_water * pi r^2 y * g
which we simplify to obtain
y '' = -rho_water * g / (h * rho_cyl) * y.
This is of the form y '' = - c * y with c = -rho_water * g / (h * rho_cyl) = -1000 kg / m^3 * 9.8 m/s^2 / (1.00 m * 700 kg / m^3) = -14 s^-2, approx..
The mass is that of the cylinder
Solving the equation
y '' = - c y
we obtain
y = B cos(sqrt(c) * t) + C sin(sqrt(c) * t),
which can be expressed in the form
y = A cos(sqrt(c) * t + phi).
A and phi are arbitrary constants, while sqrt(c) = sqrt( 14 s^-2) = 3.7 s^-1., approx..
If the cylinder is raised 10 cm and released then y(0) = .10 meters and y ' (0) = 0 so that
y(0) = A cos( sqrt(c) * 0 + phi) = .10 m and
y ' (0) = -sqrt(c) A sin(sqrt(c) * 0 + phi) = 0.
Thus
A cos(phi) = .10 m and
A sin(phi) = 0.
The latter indicates that phi = 0 or pi radians. The former dictates that phi = 0 radians rather than pi radians, with A = .10 m.
Thus our solution is
y(t) = 10 cm * cos(sqrt(c) * t ) or approximately
y(t) = 10 cm * cos( 3.7 t ).
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).
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Question: 4.1 Find the largest t interval in which a solution exists for each of the equations and initial conditions below:
• 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:
Need to make sure these are in standard form first:
y’’ + p(t)y’ + q(t)y = g(t)
The first one is.
The second one looks to be but would you divide the “t” out of the ty’’ or just leave it?!
After getting in standard form, you can then identify the coefficient functions in the hypothesis of theorem 4.1 (page 122).
There looks to be no restrictions on the first equation. (t0) = pi
Since there is a denominator in the second equation, one of the coefficient functions is not continuous at t=3, bc that would make the denominator 0, but no other points of discontinuity on the t-axis. (t0) = 1
How would I go about finding the interval from here?! I feel like I have the majority of it.
@& 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.
We rearrange the equation t y '' + sin(2 t) / (t^2 - 9) y ' + 2 y = 0 to the form
y '' + sin(2 t) / (t ( t^2 - 9) + 2 y / t = 0,
with initial conditions y(1) = 0, y ' (1) = 1:
The functions 1 / (t^2 - 9) and 1 / t are restricted in domain, the first being undefined at t = -3 and t = 3, the second at t = 0.
So the intervals on which solutions might be known to exist are
(-infinity, -3), (-3, 0), (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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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 = 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:
First: decreasing, concave down
Second: increasing, concave up
Third: increasing, concave down
@& Good. For reference, just in case:
For each situation, visualize a curve through the point (0, y(0)) with slope at that point equal to y ' (0). Knowing the value of y at t = 0, you can in each case calculate y '' (0).
For y '' + y = - 2 t, y(0) = 1, y ' (0) = -1, the initial point is (0, 1) and the slope -1.
So the function is decreasing. y '' = -y - 2 t, so y ''(0) = -y(0) - 2 * 0 = -1 - 2 * 0 = -1.
With the negative second derivative the function is concave down.
For y '' - y = t^2, y(0) = 1, y ' (0) = 1, the initial point is (0, 1) and the slope 1.
So the function is increasing. y '' = y + t^2, so y ''(0) = y(0) + 0^2 = 1 + 0 = 1.
With the positive second derivative the function is concave up.
For y '' - y = - 2 cos(t), y (0) = 1, y ' (0) = 1, the initial point is (0, 1) and the slope 1.
So the function is increasing. y '' = y - 2 cos(t), so y ''(0) = y(0) + 0^2 = 1 - 2 cos(0) = -1.
With the negative second derivative the function is concave down.
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Given Solution:
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&#Good responses. See my notes and let me know if you have questions. &#