#$&*
course Mth 279
4/9
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:
The Equation, has the variables (v,x,t)
m dv/dt = - k v / (1 + x)
Using dv/dt = (dv/dx)(dx/dt) = (dv/dx)v
Equation becomes: (with only variables v,x)
mv dv/dx = - k v / (1 + x)
dv/dx = -k/(m(1+x))
Integration of both sides yeilds:
v(x) = (-k/m)*ln(1+x) + c
v(0) = c = v_0 due to ln(1) = 0
letting v = 0, and solving for x:
0 = (-k/m)*ln(1+x) + v_0
v_0 = (k/m)*ln(1+x)
(m/k) v_0 = ln(1+x)
1 + x = e^(mv_0/k)
x_f = e^(mv_0/k) - 1
which is the final position of x at rest.
@&
Not bad, but you equation has the gravitational force and the air resistance in opposite directions. For an ascending object they would be in the same direction.
Your equation would apply to a descending projectile, and your use of integration by parts would be correct.
If the two forces are in the same direction, the denominator of the equation doesn't factor and you end up with v integral
dv / (F/m + k v^2) = m / F * dv / (1 + m k / F * v^2), which leads to an arcTangent. This can lead to a solution, but it gets to be a mess.
Since the question doesn't involve time, it will be easier to solve in terms of velocity and position.
The equation would be as follows:
F_net = m v ' = F_grav + F_drag,
where F_grav = - m g and F_drag = - k v^2.
This leads to
m v' = -k v^2 - m g
or, using x for position,
m v dv/dx = - k v^2 - m g.
Since the question involves position rather than time, we use the latter form, which yields the equation
v dv / (k v^2 + m g) = -1/m dx.
Factoring k out of the denomonator on the left-hand side, then multiplying both sides by k we get
v dv / (v^2 + m g / k) = -k/m dx.
Solving this equation we get
v = sqrt(A e^(- 2 k / m * x) - m g / k ).
If v(0) = v_0 we solve for A:
v_0 = sqrt( A e^(-2 k/m * 0) - m g / k)
v_0 = sqrt( A - m g / k)
A = v_0^2 + m g / k.
This gives us
v(x) = sqrt( (v_0^2 + m g / k) e^(-2 k/m * x) - m g / k).
The projectile will reach maximum height when v(x) = 0. Thus, if x_max is the maximum height,
sqrt( (v_0^2 + m g / k) e^(-2 k/m * x_max) - m g / k) = 0
so that
v_0^2 + m g / k) e^(-2 k/m * x_max) - m g / k = 0
and
e^(-2 k/m * x_max) = (m g / k) / (v_0^2 + m g / k) = 1 / (k / (m g) v_0^2 + 1)
Taking the log of both sides of the resulting equation we get
-2 k/m * x_max = -ln( k / (mg) v_0^2 + 1)
with solution
x_max = m / (2 k) * ln( k / (mg) v_0^2 + 1).
To solve the numerical problem we would need to solve this equation for k. However since k appears both inside and outside the natural log, this will not be possible. So approximate methods must be used.
Regarding k as variable, we could determine the value of k as accurately as desired by apploying Newton's Method to the function
m / (2 k) * ln( k / (mg) v_0^2 + 1) - x_max
The result, accurate to 3 significant figures, is k = 5.01 * 10^-6 kg / meter.
We could also resort to technology. Graphing the expression m / (2 k) * ln( k / (mg) v_0^2 + 1) vs. k, using the given values of m and v_0 and the known value of g, we find that the value of the expression is 40 meters when k is about 5 * 10^-6 kg / meter.
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confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
3
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Given Solution:
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Self-critique (if necessary):
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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:
Did not understand as well, but will attempt to explain.
mv' = F - kv^2
v' = F/m - kv^2/m
v'/(F/m - kv^2/m) = 1
dv/(F/m - kv^2/m) = dt
dv/((F/m)(1 - kv^2/F)) = dt
dv/(1 - kv^2/F) = (F/m) dt
dv/((1 - kv/F)(1 + kv/F)) = (F/m) dt
Integration of both sides yeilds:
(1/2)(sqrt(k/F))*ln((1+sqrt(k/F))/(1-sqrt(k/F))) = (F/m)t + c
(1+sqrt(k/F))/(1-sqrt(k/F)) = e^(2sqrt(Fk)t/m + c) = Ce^(2sqrt(Fk)t/m)
(1+av)/(1-av) = B
v = (1/a)((B-1)/(B+1))
v = sqrt(F/k)((Ce^(2sqrt(Fk)t/m) - 1)/(Ce^(2sqrt(Fk)t/m) + 1))
Divide by e^("" "")
v = sqrt(F/k)((C - e^(2sqrt(Fk)t/m))/(C + e^(2sqrt(Fk)t/m)))
----------------------------------------------------------------------------
mv' = mg - kv^2
Using the previous equation, F = mg:
v = sqrt(mg/k)((C - e^(2sqrt(mgk)t/m))/(C + e^(2sqrt(mgk)t/m)))
Max height would be at v = 0
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v_0 = 80
m = 0.12E-3
If v(x), v(40) = 0 is max height
k?
at t = 0
v(0) = sqrt(mg/k) (c/c) = 80
80 = sqrt(mg/k) = sqrt((0.12E-3)(9.81)/k)
k = (0.12E-3)(9.81)/80^2 = 1.84E-7
@&
Not bad, but you equation has the gravitational force and the air resistance in opposite directions. For an ascending object they would be in the same direction.
Your equation would apply to a descending projectile, and your use of integration by parts would be correct.
If the two forces are in the same direction, the denominator of the equation doesn't factor and you end up with v integral
dv / (F/m + k v^2) = m / F * dv / (1 + m k / F * v^2), which leads to an arcTangent. This can lead to a solution, but it gets to be a mess.
Since the question doesn't involve time, it will be easier to solve in terms of velocity and position.
The equation would be as follows:
F_net = m v ' = F_grav + F_drag,
where F_grav = - m g and F_drag = - k v^2.
This leads to
m v' = -k v^2 - m g
or, using x for position,
m v dv/dx = - k v^2 - m g.
Since the question involves position rather than time, we use the latter form, which yields the equation
v dv / (k v^2 + m g) = -1/m dx.
Factoring k out of the denomonator on the left-hand side, then multiplying both sides by k we get
v dv / (v^2 + m g / k) = -k/m dx.
Solving this equation we get
v = sqrt(A e^(- 2 k / m * x) - m g / k ).
If v(0) = v_0 we solve for A:
v_0 = sqrt( A e^(-2 k/m * 0) - m g / k)
v_0 = sqrt( A - m g / k)
A = v_0^2 + m g / k.
This gives us
v(x) = sqrt( (v_0^2 + m g / k) e^(-2 k/m * x) - m g / k).
The projectile will reach maximum height when v(x) = 0. Thus, if x_max is the maximum height,
sqrt( (v_0^2 + m g / k) e^(-2 k/m * x_max) - m g / k) = 0
so that
v_0^2 + m g / k) e^(-2 k/m * x_max) - m g / k = 0
and
e^(-2 k/m * x_max) = (m g / k) / (v_0^2 + m g / k) = 1 / (k / (m g) v_0^2 + 1)
Taking the log of both sides of the resulting equation we get
-2 k/m * x_max = -ln( k / (mg) v_0^2 + 1)
with solution
x_max = m / (2 k) * ln( k / (mg) v_0^2 + 1).
To solve the numerical problem we would need to solve this equation for k. However since k appears both inside and outside the natural log, this will not be possible. So approximate methods must be used.
Regarding k as variable, we could determine the value of k as accurately as desired by apploying Newton's Method to the function
m / (2 k) * ln( k / (mg) v_0^2 + 1) - x_max
The result, accurate to 3 significant figures, is k = 5.01 * 10^-6 kg / meter.
We could also resort to technology. Graphing the expression m / (2 k) * ln( k / (mg) v_0^2 + 1) vs. k, using the given values of m and v_0 and the known value of g, we find that the value of the expression is 40 meters when k is about 5 * 10^-6 kg / meter.
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
2
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Given Solution:
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Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
*********************************************
Question:
3.7.8. Mass m accelerates from velocity v_1 to velocity v_2, while constant power is exerted by the net force.
At any instant power = force * velocity (physics explanation: power = dw / dt, the rate at which work is being done; w = F * dx so dw/dt = F * dx/dt = F * v).
How far does the mass travel as it accelerates?
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Your solution:
This problem confused me a little more than the others.
P = Fv = mav
a = dv/dt = dv/dx * dx/dt = dv/dt * v
P = m(dv/dx)v^2
I suppose solving for v(x) could give us a solition here as well, if its in the right form,
but that I am unsure of at the moment.
@&
You should try to answer as much of this question as possible before looking at the given solution in my note below.
P = m (dv/dx) v^2
is a good equation.
You can separate the variables and integrate, then solve for x in terms of v. You should be able to do this easily enough.
Then see if you can use this result to answer the question of the problem. The way to do this is simple and obvious once you see it; seeing it is the challenge.
*@
@&
Given solution:
Your equation can be rearranged to give you
v^2 dv = P / m dx.
Integrating you get
v^3 / 3 = P / m * x + constant
which can be solved for x to obtain
x = m v^3 / (3 P) - constant.
When velocity changes from v1 to v2, then. x changes from
x1 = m v1^3 / (3 P) - constant
to
x2 = m v2^3 / (3 P) - constant,
a change of
x2 - x1 = m / (3 P) * (v2^3 - v1^3).
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
1
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Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
*********************************************
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?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
F_net = kv^2 - GMm/x^2
I supposed that mv(dv/dx) = kv^2 - GMm/x^2
could give an answer in the form v(x), but the kv^2 adds to the problem.
Impact velocity would be at v(0) for v(x)
The video I watched detailed one kind of solution for -GMm/x^2 using v(x)
and the other for -kv^2 did not use this but used the other,
with them combined I'm a bit confused.
Could you show me how this done?
@&
As you see your equation doesn't separate.
The form of the equation is
m v v' - k v^2 = - G M m / x^2
where ' indicates a derivative with respect to x.
Dividing through by m v the equation is
v ' - k/m v = -G M m / x^2 * (1/v).
The left-hand side is first-order linear, but the right-hand side contains the term 1 / v. So the equation is a Bernoulli equation, of form
v ' - p * v = q * v^n
with p = - k / m, q = - G M / r^2 and n = -1.
Letting u = v^m, with m = 2, we will get the form
u ' - k / m * u = - G M / r^2
The integrating factor e^(-k / m * r), gives us
(u e^(-k / m * r) ) ' = - G M / r^2 * e^(-k / m * r).
Unfortunately the right-hand side has basic form 1/r^2 * e^r and cannot be integrated in closed form.
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
1
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
*********************************************
Question:
3.7.11. An object falls to the surface of the Earth from a great height h, and as it falls experiences a drag force proportional to the square of its velocity.
Assume that the gravitational force is - G M m / r^2.
What will be its impact velocity?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
F_net = kv^2 - GMm/x^2
I supposed that mv(dv/dx) = kv^2 - GMm/x^2
could give an answer in the form v(x), but the kv^2 adds to the problem.
Impact velocity would be at v(0) for v(x)
The video I watched detailed one kind of solution for -GMm/x^2 using v(x)
and the other for -kv^2 did not use this but used the other,
with them combined I'm a bit confused.
Could you show me how this done?
@&
As you see your equation doesn't separate.
The form of the equation is
m v v' - k v^2 = - G M m / x^2
where ' indicates a derivative with respect to x.
Dividing through by m v the equation is
v ' - k/m v = -G M m / x^2 * (1/v).
The left-hand side is first-order linear, but the right-hand side contains the term 1 / v. So the equation is a Bernoulli equation, of form
v ' - p * v = q * v^n
with p = - k / m, q = - G M / r^2 and n = -1.
Letting u = v^m, with m = 2, we will get the form
u ' - k / m * u = - G M / r^2
The integrating factor e^(-k / m * r), gives us
(u e^(-k / m * r) ) ' = - G M / r^2 * e^(-k / m * r).
Unfortunately the right-hand side has basic form 1/r^2 * e^r and cannot be integrated in closed form.
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
1
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#*&!
*********************************************
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?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
F_net = kv^2 - GMm/x^2
I supposed that mv(dv/dx) = kv^2 - GMm/x^2
could give an answer in the form v(x), but the kv^2 adds to the problem.
Impact velocity would be at v(0) for v(x)
The video I watched detailed one kind of solution for -GMm/x^2 using v(x)
and the other for -kv^2 did not use this but used the other,
with them combined I'm a bit confused.
Could you show me how this done?
@&
As you see your equation doesn't separate.
The form of the equation is
m v v' - k v^2 = - G M m / x^2
where ' indicates a derivative with respect to x.
Dividing through by m v the equation is
v ' - k/m v = -G M m / x^2 * (1/v).
The left-hand side is first-order linear, but the right-hand side contains the term 1 / v. So the equation is a Bernoulli equation, of form
v ' - p * v = q * v^n
with p = - k / m, q = - G M / r^2 and n = -1.
Letting u = v^m, with m = 2, we will get the form
u ' - k / m * u = - G M / r^2
The integrating factor e^(-k / m * r), gives us
(u e^(-k / m * r) ) ' = - G M / r^2 * e^(-k / m * r).
Unfortunately the right-hand side has basic form 1/r^2 * e^r and cannot be integrated in closed form.
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
1
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#*&!#*&!
@&
Check my notes.
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