#$&*
course Mth 279
April 12 around 1 pm.
Query 10 Differential Equations*********************************************
Question: 3.7.4. Solve the equation m dv/dt = - k v / (1 + x), where x is position as a function of t and v is velocity as a function of t.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
m dv/dt = - k v / (1 + x)
m dv/dt = F(x, v)
F(x,v) = - k v / (1 + x)
V(t) = v0-(k/m)*x
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
@&
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.
NOTE ON ERRONEOUS ANSWER:
Erroneous answer from careless algebra in last step (k/m instead of m/k, a very easy mistake to make since the quantity k / m occurs frequently and we're used to writing it):
x = e^(k / m * v_0)-1
This answer is not dimensionally consistent. The exponent must be unitless.
The correct answer is
x = e^(m / k * v_0) - 1.
Dimensional analysis would be useful here:
k has units of force * position / velocity, or mass * acceleration * position / velocity.
m / k therefore has units of
velocity * mass / (mass * acceleration * position),
which you can verify come out in units of
position / time * mass / (mass * position / time^2 * position) = time / position
so when m / k is multiplied by v_0 in units of position / time (to get m / k * v_0) we get a unitless exponent for v.
*@
*********************************************
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?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Velocity = v0
F_drag = kv^2
V0 = 80 m/s
Mass = 0.12 g
Y = 40 m
Find k.
If mv(dv/dx) + kv = 0
(dv/dx) = -(k/m)
(dv/dx)*m = -k
-(dv/dx)m = k
(-80m/s)*(0.12g) = k
-9.6 = k
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
@& Question:*********************************************Self-critique rating: ------------------------------------------------ Self-critique (if necessary):&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
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.
NOTE ON ERRONEOUS ANSWER:
Erroneous answer from careless algebra in last step (k/m instead of m/k, a very easy mistake to make since the quantity k / m occurs frequently and we're used to writing it):
x = e^(k / m * v_0)-1
This answer is not dimensionally consistent. The exponent must be unitless.
The correct answer is
x = e^(m / k * v_0) - 1.
Dimensional analysis would be useful here:
k has units of force * position / velocity, or mass * acceleration * position / velocity.
m / k therefore has units of
velocity * mass / (mass * acceleration * position),
which you can verify come out in units of
position / time * mass / (mass * position / time^2 * position) = time / position
so when m / k is multiplied by v_0 in units of position / time (to get m / k * v_0) we get a unitless exponent for v.
*@
*********************************************
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?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Fnet = ma
Power = (m*a)*(v)
mv(dv/dx) = F(x,v)
I’m not sure how to solve this, I have an idea, but can’t get it.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
@&
F= P/v
Fnet = P/v = m dv/dt
dv/dx = v dx/dt
P dx = mv^2 dv
P x +c = 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)
*@
*********************************************
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:
Height = h
Fdrag = v^2
Fgrav = -GMm/r^2
Impact velocity = v^2/2 = (GM/r) + c
V^2/2 = GM[(1/R)-(1/R+h)]
V = - sqrt (2GM[(1/R) - (1/R+h)])
@&
m dv/dt = -G M m / r^2 + k v^2
dv/dt = dv/dr*dr/dt = v dv/dr
mv dv/dr = -GMm/r^2 + kv^2
m v v ' = -GMm/r^2 + kv^2
where v ' means dv / dr.
v ' = - G M / r^2 * ( 1 / v ) + k / m * v
v ' - k / m * v = - G M / r^2 * (1 / v)
This is a Bernoulli equation
v ' - k / m * v = - G M / r^2 * v^(-1),
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
with integrating factor e^(-k / m * r), giving us
(u e^(-k / m * r) ) ' = - G M / r^2 * e^(-k / m * r).
Integrating the right-hand side, we find that we can't integrate the right-hand side.
If we could integrate, we would get an integration constant, which we could use to account for the initial position, which would be h + r_Earth.
Another comment:
This model might not be very realistic, because if h is great the atmosphere will thin. However the thinning is exponential in nature, and this could to an extent balance the fact that the air resistance at high speeds involves turbulence and an effective proportionality to a power of v greater than the power 2 assumed here. This might be interesting to investigate, but in the interest of time we'll probably have to leave it at that.
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
*********************************************
Question:
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
*********************************************
Question:
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#*&!
&#Your work looks good. See my notes. Let me know if you have any questions. &#