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PAGE 1

2.3 (continued)

from last time:

\[ \frac{dQ}{dt} = 2 - \frac{1}{150}Q \quad Q(0) = 40 \]

Annotations for the differential equation:

  • \( 2 \) kg/min in
  • \( \frac{Q}{150} \) kg/min out

solution: \( Q(t) = 300 - 260 e^{-\frac{1}{150}t} \)

As \( t \to \infty \) (run this tank for a long time)

\[ \lim_{t \to \infty} Q(t) = 300 \]

300 kg of salt in the tank

concentration is \( \frac{300 \text{ kg}}{600 \text{ L}} = \frac{1}{2} \text{ kg/L} \)

which matches the concentration of salt water going in.

Graph of Q versus t showing curves approaching a horizontal asymptote at Q = 300.
Diagram of a vertical pipe representing a tank with salt water flowing through it.

same flow rates in and out

our tank

pipe salt water through

PAGE 2

What if flow rates in and out don't match?

for example, flow rate in is 4 L/min, \( \frac{1}{2} \) kg/L

flow rate out is 5 L/min

initial tank volume is 600 L

\[ \frac{dQ}{dt} = (\text{rate in}) - (\text{rate out}) \]

concentration of salt water = \( \frac{Q}{\text{volume}} \)

\[ = (4 \text{ L/min})(\frac{1}{2} \text{ kg/L}) - (5 \text{ L/min})(\frac{Q}{600 - t} \text{ kg/L}) \]

a net loss of 1 L/min

\[ Q' = 2 - \frac{5}{600 - t} Q \quad \text{linear} \]
\[ = \frac{2(600 - t) - 5Q}{600 - t} \quad \text{not separable} \]
PAGE 3

Comparing Differential Equations

Compare the following differential equations:

\[ \frac{dQ}{dt} = 2 - \frac{1}{150}Q \]

to

\[ \frac{dv}{dt} = g - \frac{\gamma}{m}v \quad \text{(free fall)} \]

structurally identical

free-fall

Graph of velocity v versus time t for free fall, showing curves approaching a horizontal asymptote labeled terminal velocity.

salt

Graph of quantity Q versus time t for salt concentration, showing curves approaching a horizontal asymptote at 300.

now let's solve the projectile problem more completely \(\rightarrow\) up and down.

assume vertical motion only: air resistance is opposite velocity

PAGE 4

on the way up:

Free body diagram for upward motion showing a particle with two downward force vectors: mg and gamma times v.

both air and weight point down

on the way down:

Free body diagram for downward motion showing a particle with an upward force vector gamma times v and a downward force vector mg.

define up as positive

outline of solution:

solve upward with initial condition \( v(0) = v_0 \)

\[ m \frac{dv}{dt} = -mg - \gamma v \]

up to the max height \(\rightarrow t = t_{max}\)

then solve downward problem from there

\[ m \frac{dv}{dt} = -mg + \gamma v \]
PAGE 5

Example: Projectile Motion with Resistance

As an example, let's use \( v(0) = 20 \text{ m/s} \), \( m = 1 \text{ kg} \), \( \gamma = 1 \).
Initial height: \( 10 \text{ m} \).

\[ m \frac{dv}{dt} = -mg - \gamma v \]
\[ \frac{dv}{dt} = -9.8 - v = -(9.8 + v) \]
\[ \frac{1}{9.8 + v} dv = -dt \]
\[ \ln |9.8 + v| = -t + C \]
\[ 9.8 + v = C e^{-t} \]
\[ v = C e^{-t} - 9.8 \]

Given \( v(0) = 20 \):

\[ 20 = C - 9.8 \quad \text{so} \quad C = 29.8 \]

So, \( v(t) = 29.8 e^{-t} - 9.8 \)

upward flight

Time at max height?

At maximum height, the velocity is zero.

A simple sketch of a vertical path ending at a point labeled v=0.
PAGE 6
\[ 0 = 29.8 e^{-t} - 9.8 \]
\[ e^{-t} = \frac{9.8}{29.8} \quad t = -\ln \left( \frac{9.8}{29.8} \right) \approx 1.11 = t_{\text{max}} \]

What is the height?

Height is the integral of velocity:

\[ \text{height: } x(t) = \int v(t) dt = \int (29.8 e^{-t} - 9.8) dt \]
\[ = -29.8 e^{-t} - 9.8t + C \]

Given \( x(0) = 10 \):

\[ 10 = -29.8 + C \quad C = 39.8 \]

\( x(t) = -29.8 e^{-t} - 9.8t + 39.8 \)

Max height is:

\( x(t_{\text{max}}) = x(1.11) \approx 19.1 \)

Now the downward portion:

\[ \frac{dv}{dt} = -g + \frac{\gamma}{m} v \]
\[ \frac{dv}{dt} = -9.8 + v \]
PAGE 7

Initial Conditions and Differential Equation Solving

Initial Condition:

  • x starts at 19.1
  • v starts at 0

Redefine \( t = 0 \) at the top. Makes solving constants a bit neater.

\( x(0) = 19.1 \)

\( v(0) = 0 \)

instead of

\( x(1.11) = 19.1 \)

\( v(1.11) = 0 \)

(add 1.11 to \( t \) after to track time since launch)

Solving the Differential Equation

\[ \frac{dv}{dt} = -9.8 + v \]\[ \frac{1}{v - 9.8} dv = dt \]\[ \ln |v - 9.8| = t + C \]\[ v - 9.8 = C e^t \]\[ v = 9.8 + C e^t \]

\( v(0) = 0 \rightarrow 0 = 9.8 + C \quad \text{so} \quad C = -9.8 \)

PAGE 8

\( v(t) = 9.8 - 9.8 e^t \)

\( t = 0 \rightarrow \text{start of drop} \)

"real" \( t \) is \( t + 1.11 \)

Height Calculation:

\[ x(t) = \int v(t) dt \]\[ = 9.8t - 9.8e^t + C \]

\( x(0) = 19.1 \)

\( 19.1 = -9.8 + C \quad \rightarrow \quad C = 28.9 \)

\( x(t) = 9.8t - 9.8e^t + 28.9 \)

When is the impact w/ ground?

not when \( v = 0 \) but when \( x = 0 \)

\[ 0 = 9.8t - 9.8e^t + 28.9 \]

\( t \approx 1.49 \) since the drop

(or \( t = 1.11 + 1.49 = 2.6 \) since the launch)