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course MTH 279
2/28 10 am
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:
Call:
D the density of the cylinder.
H the height of the cylinder
A the area of the cylindrical bases
Y the y distance which the cylinder is submerged at equilibrium
D_w the density of water.
F = W - B
W we know is mg, which is constant. B is DVg, which changes with V (volume of water displaced)
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Using your notation, B would be D_w V g.
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Also, we don't know m, but we can write it in terms of its volume and density, which is easy to calculate with the given information.
DAHy'' = DAHg - D_wA(Y + y)g
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You haven't defined y, but it appears that it is the vertical position of the cylinder relative to Y.
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To make some simplifications, distribute, and note that DAHg = D_wAYg (you can find this from looking at the way the forces balance when the object is at equilibrium)
DAHy'' = -D_wAgy
Rewriting the equation in standard form,
y'' + D_wg/(DH) y = 0
Subbing in known values,
y'' + 14y = 0
Which has a general solution of
y = C_1sin14t + C_2cos14t
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The general solution would be of this form, but instead of 14 you should have sqrt(14).
For example the second derivative of c1 sin(14 t) is -196 c1 sin(14 t), which will not result in 0 when added to 14 c1 sin(14 t).
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To solve for C_1 and C_2, we can plug y(0) = .1 and y'(0) = 0. We find C_1 = 0 and C_2 = .1
Our final equation is therefore:
y = .1cos14t
The second scenario is exactly the same, but with different initial conditions; that is y(0) = .3 and y'(0) = 0 (I'm ""starting the timer"" at the instant the cylinder is fully submerged)
This leads to the solution:
y = .3cos14t
and y' = -4.2sin14t
The instant the cylinder is struck, it would have maximum velocity. Looking at the y' equation, this would occur when the arguement of the sin function = pi/2, which gives us the velocity:
-4.2 m/s
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Very good. The only errors I see result from your use of 14 rather than sqrt(14) (see my preceding note).
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
On the second part of the question, I wanted to say that y(0) = 0 to say that it had moved 0 from its equilibrium condition, but that equation didn't really help me. That is why I decided to use the initial condition (and a little knowledge of simple harmonic motion) that I did.
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Self-critique rating: 3
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Question:
For each of the equations and initial conditions below, find the largest t interval in which a solution is known to exist:
y '' + y ' + 3 t y = tan(t), y(pi) = 1, y ' (pi) = -1
t y '' + sin(2 t) / (t^2 - 9) y ' + 2 y = 0, y(1) = 0, y ' (1) = 1.
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Your solution:
1) part of the equation is undefined when cos = 0, so the largest interval would occur after the initial condition until a time when cos = 0. This gives us the largest interval being: (pi , 3pi/2)
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The cosine is 1 when its argument is pi.
The interval would be (pi/2, 3 pi/2).
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2) I first rearranged the equation into standard form: y '' + sin(2 t) /[t (t^2 - 9)] y ' + 2/t y = 0
So, the equation has discontinuities at t = 0, -3, and 3. Therefore there are two equally ""large"" intervals on which the function is defined : (-inf, -3) and (3, +inf). This is omitting the tiny intervals of (-3, 0) and (0, 3). HOWEVER, initial condition states that t = 1, so the interval in question is: (0,3)
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary): OK
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Self-critique rating: OK
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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:
y '' + y = - 2 t, y(0) = 1, y ' (0) = -1 concave down and decreasing
y '' + y = - 2 t, y(0) = 1, y ' (0) = -1 same function and answer as above
y '' - y = t^2, y(0) = 1, y ' (0) = 1 concave up and increasing
y '' - y = - 2 cos(t), y (0) = 1, y ' (0) = 1. concave down and increasing
One can determine increasing/decreasing from the sign of y'. A negative y' means decreasing and a positive means increasing. Concave up/down is determined by the sign of y''. A negative means concave down and a positive means concave up. So it really just comes down to ""plugging in "" the given initial values.
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary): OK
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Self-critique rating: OK
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Very good. You did have a couple of easily-corrected errors, but your approach throughout is excellent.
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