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Problem 23: Skydiver Dynamics

A skydiver weighs 180 lb and jumps from 5000 ft. They open their chute after 10 seconds of free fall.

Drag Coefficients:

  • Drag: \( 0.75 |v| \) when chute is closed
  • Drag: \( 12 |v| \) when chute is open

Define down as positive (because falling downward, so \( v \) is always positive).

Equation of Motion

\[ m \frac{dv}{dt} = \begin{cases} mg - 0.75 v & \text{closed (first 10 seconds)} \\ mg - 12 v & \text{open (after first 10 s)} \end{cases} \]

Mass Calculation

Given weight \( = 180 = mg \). Using \( g = 32 \text{ ft/s}^2 \):

\[ 180 = m(32) \implies m = \frac{180}{32} = 5.625 \]

Initial condition: \( v(0) = 0 \)

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a) Speed when chute opens

The differential equation for the first 10 seconds is:

\[ \frac{dv}{dt} = g - \frac{0.75}{m} v = 32 - \frac{2}{15} v \]

This is a separable/linear differential equation. Solving for \( v \):

\[ v = 240 + Ce^{-\frac{2}{15}t} \]

Using the initial condition \( v(0) = 0 \):

\[ 0 = 240 + C \implies C = -240 \]

The velocity function for \( 0 \le t \le 10 \) is:

\[ v(t) = 240 - 240 e^{-\frac{2}{15}t} \]

At \( t = 10 \):

\[ v(10) = 176.7 \text{ ft/s} \]

b) Distance fallen before chute opens

Let \( y \) be the distance fallen, with \( y(0) = 0 \).

\[ y(t) = \int v(t) dt = 240t + 1800 e^{-\frac{2}{15}t} + C \]

Using \( y(0) = 0 \):

\[ 0 = 1800 + C \implies C = -1800 \]

The distance function is:

\[ y(t) = 240t + 1800 e^{-2/15 t} - 1800 \]

At \( t = 10 \):

\[ y(10) = 1074.47 \text{ ft} \]
Vertical line diagram showing a drop from 5000 ft down to a point labeled 'open chute'.
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c) Terminal velocity after chute opens

\[ \frac{dv}{dt} = g - \frac{12}{m}v \]
\(\vdots\)
\[ v(t) = 15 + Ce^{-\frac{12}{m}t} \]
\[ \lim_{t \to \infty} v(t) = 15 \text{ ft/s} \]
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29. \( v_0 = \sqrt{2gR} \)

\( R \): radius of Earth

a).

Physics diagram showing a horizontal surface at x=0, an upward arrow for positive x, and a downward arrow for gravity.

Surface: \( x = 0 \)

gravity: \( -\frac{mK}{(R+x)^2} \) \( \downarrow \) constant

for Earth, at \( x = 0 \)

gravity is \( -mg \)

so \( +mg = +\frac{mK}{R^2} \)

\( K = gR^2 \)

DE: \( m \frac{dv}{dt} = -\frac{mgR^2}{(R+x)^2} \) two indep. variables

change to \( x \) (see example 4 in textbook)

\(\vdots\)
\[ v = \sqrt{\frac{2gR^2}{R+x}} \]
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Problem b): Time to Reach

Given parameters:
\( x = 240,000 \)
\( R = 4000 \)

\[ V = \sqrt{\frac{2gR^2}{R+x}} = \frac{dx}{dt} \]

Find \( x(t) \)

\[ \sqrt{R+x} \, dx = \sqrt{2gR^2} \, dt \]

Separable.

\[ \frac{2}{3}(R+x)^{3/2} = \sqrt{2gR^2} \, t + C \]

Initial Condition (IC): \( x(0) = 0 \) (surface)

\[ C = \frac{2}{3}R^{3/2} \]

Substituting \( x = 240,000 \):

\( t \approx 51 \) hours

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2.4 Differences Between Linear and Nonlinear Eqs

Linear Equations

Standard form: \( y' + p(t)y = g(t) \) with initial condition \( y(t_0) = y_0 \).

Integrating factor: \[ \mu(t) = e^{\int p(t) dt} \]

\[ \frac{d}{dt} [\mu(t)y] = \mu(t)g(t) \]

Solution: \[ \mu(t)y = \int \mu(t)g(t) dt + C \]

\[ y = \frac{1}{e^{\int p(t) dt}} \left[ \int e^{\int p(t) dt} g(t) dt + C \right] \]

A solution only exists if \( \int p(t) dt \) and \( \int e^{\int p(t) dt} g(t) dt \) exist.

This implies that \( p(t) \) and \( g(t) \) are both continuous.
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Solution exists and is unique on any interval on which \( p(t) \) and \( g(t) \) are continuous, AND containing \( y(t_0) = y_0 \).

Example

\[ y' + \frac{1}{t(t-3)}y = \sin t \]

\( y(1) = 2 \)

\( t = 1 \)

\( p(t) = \frac{1}{t(t-3)} \) Continuous where \( t \neq 0, t \neq 3 \)

\( g(t) = \sin t \) continuous for all \( t \)

\( p(t) \) and \( g(t) \) are both continuous on

\( (-\infty, 0) \),

\( (0, 3) \)

, \( (3, \infty) \)

\( ↑ \) \( y(1) = 2 \) is in here

so solution exists and is unique on \( (0, 3) \)

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  • unique: one solution only
  • → solutions cannot intersect
A coordinate graph with t and y axes showing a green solution curve passing through a point at t=1.

For nonlinear eqs,

\[ y' = f(t, y) \quad y(t_0) = y_0 \]

Solution exists and is unique in a rectangle in \( ty \)-plane where \( f \) and \( \frac{\partial f}{\partial y} \) are continuous and containing \( y(t_0) = y_0 \).

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Example: Existence and Uniqueness

\[ y' = \frac{4t}{y-1} \]

Let \( f(t,y) = \frac{4t}{y-1} \). This function is continuous for \( y \neq 1 \). Note that \( \frac{4t}{y-1} = 4t(y-1)^{-1} \).

The partial derivative with respect to \( y \) is:

\[ \frac{\partial f}{\partial y} = -\frac{4t}{(y-1)^2} \]

This partial derivative is also continuous for \( y \neq 1 \).

If the initial condition is \( y(0) = 0 \):

We look for a rectangle containing the initial condition (IC). We don't know how big the rectangle is, but it cannot grow past the line \( y = 1 \) because the function and its derivative are not continuous there.

Coordinate graph showing t and y axes with a dashed rectangle around the origin and a red boundary line at y=1.

Possible rectangles containing the initial condition (IC) are limited by the discontinuity at \( y = 1 \).

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Nonlinear Equations

Nonlinear equations are usually very hard to solve. Exceptions include some separable equations and the Bernoulli equation.

Bernoulli Equation

\[ y' + p(t)y = q(t)y^n, \quad n \neq 0, 1 \]
  • If \( n = 0 \), then \( y' + p(t)y = q(t) \), which is linear.
  • If \( n = 1 \), then \( y' + p(t)y = q(t)y \), which simplifies to:
    \[ y' + [p(t) - q(t)]y = 0 \]
    This is also linear.

Solution Strategy:

To solve a Bernoulli equation, use the substitution: \[ v = y^{1-n} \]

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Example: Bernoulli Equation Substitution

\[ y' - y = -y^2 \]

For this Bernoulli equation, we identify the following parameters:

  • \( p(t) = -1 \)
  • \( q(t) = -1 \)
  • \( n = 2 \)

Substitution Step

Let \( v = y^{1-n} = y^{1-2} = y^{-1} \).

Differentiating with respect to \( t \):

\[ v' = -y^{-2} y' \implies y' = -y^2 v' \]

Transformation to Linear Equation

Substitute \( y' \) back into the original equation:

\[ -y^2 v' - y = -y^2 \]

Divide the entire equation by \( -y^2 \):

\[ v' + \frac{1}{y} = 1 \]

Since \( v = y^{-1} = \frac{1}{y} \), the equation becomes:

\[ v' + v = 1 \]

This is now a linear equation in v.

Solving for y

Solving the linear equation for \( v \):

\[ v = 1 + Ce^{-t} \]

Substitute back \( v = \frac{1}{y} \) (which implies \( y = \frac{1}{v} \)):

\[ y = \frac{1}{1 + Ce^{-t}} \]