Query_12

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course Mth 279

3 / 20 13

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:

The volume of water displaced or delta_V is

delta_V = - A y

delta_V = - pi r^2 y

The net force of the volume displaced is

delta_V * - rho_w * g

The mass of the cylinder is

mass = rho_c pi r^2 h

From Newton’s law:

F_net = mass * acceleration = m a

mass = rho_c pi r^2 h

acceleration = y’’

Newton’s law becomes:

-rho_w pi r^2 y g = rho_c pi r^2 h y’’

-rho_w y g = rho_c h y’’

solving for y’’ we have

y’’ = (-rho_w y g ) / (rho_c h)

This is in the form of y’’ = - c y

with c equal to

c = (-rho_w g) / (rho_c h)

We can plug in the following given information to solve for c with

rho_w = - 1000 kg / m^3

g = 9.8 m / s^2

rho_c = 700 kg / m^3

h = 1 m

Solving for c we have c = - 14/ s^2

??? These units are odd to me but I will keep them to see if it cancels out ‘down the road’

y = k_2 cos(sqrt(c) t) + k_1 sin(sqrt(c) t)

This can be expressed as

y = A cos(sqrt(c) t + phi)

sqrt(c) = sqrt(14 / s^2) = 3.7 / s

Given conditions:

y(0) = 0.10 m

y’(0) = 0

y(0) = A cos(sqrt(c) (0) + phi) = 0.1

A cos(phi) = 0.1 so phi cannot be 0

y’(0) = -sqrt(c) A sin(sqrt(c) (0) + phi) = 0

A sin(phi) = 0 so phi is equal to 0 or pi

From the above two equations phi must be equal to pi

The solution becomes:

y(t) = (10 cm) cos(sqrt(c) t)

y’(t) = (10 cm) cos (3.7 t )

With the cylinder at the surface of the water we know that the displacement must be 30cm so

A = 30 cm

y(0) = A cos(sqrt(c) (0) + phi)

y’(0) < 0

with cos(phi) = 0 , phi = pi / 2 or 3 pi / 2

with sin(phi) < 0 , phi = 3 pi / 2

The solution is:

y(t) = 30 cm cos(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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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:

The first function:

tan(t) is continues except for multiples of pi / 2

The initial condition of t = pi falls between the interval of ( pi/ 2 , 3 pi / 2)

The second function:

divide each side of the function by t to isolate the y’ term and we have

y’’ + (sin( 2 t) / t (t^2 - 9)) y’ + ( 2 / t) y = 0

With t (t^2 - 9) we cannot have t = -3, 0 , 3

The intervals for the solution to exist are from

(-infinty, -3) (-3, 0) (0, 3) and (3 , infinity)

The initial condition is t = 1 so this falls under the interval of

(0, 3)

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Given Solution:

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Self-critique (if necessary):

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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:

• y '' + y = - 2 t, y(0) = 1, y ' (0) = -1

Given initial point is (0,1) with slope (y’) of -1 so function is decreasing

y’’(0) = - y - 2(0)

y’’(0) = - y , the function is concave down

• y '' - y = t^2, y(0) = 1, y ' (0) = 1

Given initial point is (0,1) with slope (y’) of 1 so function is increasing

y’’(0) = (0)^2 + y

y’’ = y , the function is positive so concave up

• y '' - y = - 2 cos(t), y (0) = 1, y ' (0) = 1.

Given initial point of (0, 1) with slope (y’) 1 the function is increasing.

y’’(0) = y - 2 cos(0)

y’’(0) = y - 2 = 1 - 2 = -1

With a negative y’’ the function is 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:

Query_12

&#Very good responses. Let me know if you have questions. &#