Query 10

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course MTH 279

2/23 6pmI feel a little better about this section than I did with the query 4 section, but I do feel that I struggled with this section more than I should have. Between this section and that section I'm a little worried at this point about my abilities in this class. I think I'm okay with the techniques I've learned (separating variables, exact equations, Bernoulli equations etc), but when I get into these modeling problems I really start to struggle. Could you give me your honest opinion on how you feel I am performing in this class? Do you think I am performing on an adequate level based on my work? Or am I behind where I should be?

Query 10 Differential Equations

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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:

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

x = A -1, where A = e^-c

Changing the equation to x as the independent variable:

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

Solving using separation of variables:

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

Solving for x (which I assumed was the right thing to do. I didn't think I should integrate again to find position since x was already there).

Ae^(-mv/k) - 1 = x

The object would come to rest when v=0, so x(0) = A-1

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The text problem indicated that v(0) = v_0, which allows you to evaluate A.

Your solution appears to be consistent with the following, with the difference that the initial condition is applied:

dv/dt = dv/dx * dx/dt = v dv so the equation becomes

m v dv/dt = - k v / ( 1 + x ),

which rearranges to

dv = -k / m ( 1 + x ).

Integrating we get

v = -k / m ln | 1 + x | + c.

With the condition v(0) = v_0 we get

v_0 = - k / m ln | 1 + 0 | + c.

Since ln(1) = 0 we have c = v_0, giving us solution

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

To find where the object stops, solve for v(x) = 0.

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

when

-k/m ln | 1 + x | = -v_0,

which occurs when

ln | 1 + x | = m / k * v_0

| 1 + x | = e^(m / k * v_0)

If x > -1 then | 1 + x | = 1 + x and we get

x = e^(m / k * v_0) - 1.

If x < -1 then | 1 + x | = -1 - x and we get

x = -e^(m / k * v_0) - 1.
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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): I think I solved the first part right, but I'm not so sure if I found the x position correctly.

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Self-critique rating: 3

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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:

y_max= -m/(2k) ln[ (gm/k)/(v_0^2 + gm/k) ]

k = ????

Original equation:

mv dv/dy = -mg - kv^2

Solved as a Bernoulli Equation. Imposed initial condition v(0) = v_0

v^2 = -gm/k + (v_0^2 +gm/k)e^(-2ky/m)

Solved eqution for y.

y = -m/(2k) ln[ (v^2 + gm/k)/(v_0^2 + gm/k) ]

y_max will occur when v = 0, which leads to the above solution.

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If we know x_max, m and v_0 then k is given implicitly by this result. However we can't explicitly solve this equation for k, so we need to use graphical or numerical techniques.
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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I think I solved the equation for y_max corrctly, but I couldn't figure out how to isolate k.

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Self-critique rating: 3

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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:

x = m/(3P) (v_2^3 - v_1^3)

Starting equation:

m dv/dt = P/v

Solved using separation of variables:

v = sqrt(2Pt/m + C_1)

At this point, I solved for t at v_1 and v_2 and got:

t_1 = m/(2P) (v_1^2 - C_1)

t_2 = m/(2P) (v_2^2 - C_1)

Integrating v to find x,

x = m/(3P) (2Pt/m + C_1)^(3/2) + C_2

the distance the object traveled would be x(t_2 - t_1)....and after much algebra I arrived at the above solution.

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The following parallels your solution:

F= P/v

Fnet = P/v = m dv/dt

so dv/dt = P / v

dv/dt = dv/dx * dx/dt = v dv/dx

m v dv/dx = P / v

and

m v^2 dv/dx = P,

m v^2 dv = P dx

P x +c = m v^3 / 3 = m/3 * v^3

This implies that

m / 3 * v_2 ^ 3 - m / 3 * v_1 ^ 3 = P x_2 + c - (P x_1 + c) = P (x_2 - x_1)

x_2 - x_1 is the distance traveled, which is therefore

dist traveled = m / (3 P) (v_2^3 - v_1^3)
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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): I feel ok about my approach, but the process was a little messy.

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Self-critique rating: OK

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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:

v= sqrt( 2GM(2 - 2/r +4kr/m - 4k/m + Ce^(2kr/m) )

My work is extremely messy, so I'll instead give you my thought process...

Original equation:

mv dv/dr = -GMm/(r^2) +kv^2

I solved as a Bernoulli equation and had to use integration by parts 3 times to arrive at a solution.

confidence rating #$&*:

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Given Solution: Pretty sure I started out the problem right, I think I went through the right process to find a solution, but I'm not so sure if the actual answer I got was right.

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Self-critique (if necessary):

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Self-critique rating:

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Question:

Query 10

@& If we know x_max, m and v_0 then k is given implicitly by this result. However we can't explicitly solve this equation for k, so we need to use graphical or numerical techniques. *@

&#This looks good. See my notes. Let me know if you have any questions. &#

@&

If we know x_max, m and v_0 then k is given implicitly by this result. However we can't explicitly solve this equation for k, so we need to use graphical or numerical techniques.
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