Assignment 12

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course Mth 279

6/28

Query 10 Differential Equations*********************************************

Question: 3.7.4. Solve the equation m dv/dt = - k v / (1 + x), where x is position as a function of t and v is velocity as a function of t.

How far does the object travel before coming to rest?

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Your solution:

m dv/dt = -kv/(1+x)

mv dv/dx = -kv(1+x)

m dt = -k/(1+x) dx integrate

mv = -k ln(1+x) +C

v = (-kln(1+x))/m + C

find C v(0) = v_0

v(0) = (-kln(1+ 0))/m + C

v_0 = C

v(x) = -k/m ln( 1+x) + v_0

find how far it travels?

v(x) = 0

0 = -k/m *ln(1+ x) + v_0

v_0 = k/m ln(1+x)

mv_0/k = ln(1+x)

e^(mv_0/k) -1 = x

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

checked with same problem in book and solution is correct

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Self-critique rating:

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Question:

3.7.6. A vertical projectile has initial velocity v_0, and experiences drag force k v^2 in the direction opposite its motion. Assume that acceleration g of gravity is constant. Find the maximum height to which the projectile rises.

If the initial velocity is 80 m/s and the projectile, whose mass is .12 grams, rises to a height of 40 meters (estimated quantities for a plastic BB), what is the value of k?

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Your solution:

mv dv/dx = -kv^2 -mg

v/(kv^2/m + g) dv = -dx

using u-sub integrate

u = kv^2/m + g

du = 2kv/m

m/(2k) int (1/u)du

m/(2k) ln|kv^2/m + g| = -x + C

solve for v

ln | kv^2/m + g| = -2kx/m + C

kv^2/m + g = Ae^(-2kx/m)

v=sqrt(mAe^(-2kx/m)/k - mg/k)

find and solve for A

v(0) = v_0

v_0 = sqrt(mAe^(-2k(0)/m)/k - mg/k)

v^2_0 = mA/k - mg/k

mA/k = v^2_0 + mg/k

A = kv^2_0/m + mg/k *(k/m) = kv^2_0/m + g

so

v(x) = sqrt((m(kv^2_0/m + g)e^(-2kx/m))/k - mg/k)

v(x) = sqrt((v^2_0 + g)e^(-2kx/m) - mg/k)

max height v(x) = 0

0 = sqrt((v^2_0 + g)e^(-2kx/m) - mg/k)

0 = (v^2_0 + g)e^(-2kx/m) - mg/k

mg/k = (v^2_0 + g)e^(-2kx/m)

gm/(k(v^2_0 + g) = -2kx/m

x_max = -m/(2x) * ln(mg/(k(v^2_0 + g)))

find k? plug in values given

40 = -0.00012/(2k) ln((9.8*.00012)/(k(80^2+9.8))

40 = 6*10^-6/k ln(5.45*10^6 * k)

40k = 6*10^-6 *ln(5.45*10^6 *k)

Seem to be stuck here solving for k???

@&

You could plot

40k - 6*10^-6 *ln(5.45*10^6 *k)

vs. k and estimate the value that gives you zero.

Combining this with Newton's Method would quickly provide a sufficiently accurate estimate.

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confidence rating #$&*:

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1

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Given Solution:

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Self-critique (if necessary):

Got stuck solving for k

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Self-critique rating:

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Question:

3.7.8. Mass m accelerates from velocity v_1 to velocity v_2, while constant power is exerted by the net force. At any instant power = force * velocity (physics explanation: power = dw / dt, the rate at which work is being done; w = F * dx so dw/dt = F * dx/dt = F * v).

How far does the mass travel as it accelerates?

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Your solution:

P = Fv therefore F=P/v

mv dv/dx = P/v

mv^2 dv/dx = P

integrate

mv^3/3 = Px

on the left you have limits from v_1 to v_2 and on the right side of equation limits x_1 to x_2

mv^3_2/3 - mv^3_1/3 = Px_2 - Px_1

x_2 - x_1 = m(v^3_2 - v^3_1)/(3P)

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Good.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Question:

3.7.11. An object falls to the surface of the Earth from a great height h, and as it falls experiences a drag force proportional to the square of its velocity. Assume that the gravitational force is - G M m / r^2.

What will be its impact velocity?

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Your solution:

mv dv/dt - kv^2 = -GMm/x^2

dv/dt -kv/m = -GM/(x^2*v)

v’ - kv/m = (-GM/(x^2))v^-1 use bernoullis

v’ + 2kv/m = -2GM/(x^2) use first order non homogenous)

p(t) = -2k/m

v(t) = Ce^(2kt/m) + e^(2kt/m)*int(e^(-2kt/m)*(-GM/x^2)dt

v(t) = Ce^(2kt/m) - (2GM)/(x^2)*e^(2kt/m)*(-m/2k)(e^(-2kt/m)

v(t) = Ce^(2kt/m) + (2GMm)/(kx^2)

Here I am starting to get lost in the math and the use of t’s and x’s are getting confusing.

@&

In your first steps m dv/dt becomes m v dv/dr.

So there is no t dependence in your equation. Your solution will give v as a function of r.

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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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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. &#