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Q20E

Expert-verifiedFound in: Page 131

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**Falling Body. ****In Example 1 of Section 3.4, page 110, we modeled the velocity of a falling body by the initial value problem \({\bf{m}}\frac{{{\bf{dv}}}}{{{\bf{dt}}}}{\bf{ = mg - bv,v(0) = }}{{\bf{v}}_{\bf{o}}}{\bf{ = 0}}\)under the assumption that the force due to air resistance is –bv. However, in certain cases the force due to air resistance behaves more like\({\bf{ - b}}{{\bf{v}}^{\bf{r}}}\), where \({\bf{(r > 1)}}\) is some constant. This leads to the model \({\bf{m}}\frac{{{\bf{dv}}}}{{{\bf{dt}}}}{\bf{ = mg - b}}{{\bf{v}}^{\bf{r}}}{\bf{,v(0) = }}{{\bf{v}}_{\bf{o}}}\) (14).To study the effect of changing the parameter r in (14), take \({\bf{m = 1,}}\,\,{\bf{g = 9}}{\bf{.81,}}\,\,{\bf{b = 2}}\) and \({{\bf{v}}_{\bf{o}}}{\bf{ = 0}}\)**.

When \({\bf{r = 1}}{\bf{.0}}\), the constant solution is \({\bf{v(t) = 4}}{\bf{.905}}\).

When \({\bf{r = }}\,\,{\bf{1}}{\bf{.5}}\), the constant solution is \({\bf{v(t) = }}2.887\).

When \({\bf{r = }}\,\,{\bf{2}}{\bf{.0}}\), the constant solution is \({\bf{v(t) = }}2.215\).

For the solution apply Euler’s formula.

The given equation is

\(\begin{array}{l}{\bf{v(0) = 0}}\\\frac{{{\bf{dv}}}}{{{\bf{dt}}}}{\bf{ = 9}}{\bf{.81 - 2}}{{\bf{v}}^{\bf{r}}}\end{array}\)

Now apply improved Euler’s method subroutine with \({\bf{h = 0}}{\bf{.2}}\) and \({\bf{N = 25}}\).

\(\begin{array}{c}{\bf{f(t,T) = 9}}{\bf{.81 - 2v}}\\{\bf{G = f(t + h,v + hF)}}\\{\bf{ = 9}}{\bf{.81 - 2(v + 0}}{\bf{.2(9}}{\bf{.81 - 2v))}}\end{array}\)

Apply initial conditions \({\bf{t = }}{{\bf{t}}_{\bf{o}}}{\bf{ = 0,v = }}{{\bf{v}}_{\bf{o}}}{\bf{ = 0}}\).

\(\begin{array}{c}{\bf{F(0,0) = 0}}{\bf{.2}}\\{\bf{G(0,0) = 5}}{\bf{.886}}\\{\bf{t = }}{{\bf{t}}_{\bf{0}}}{\bf{ + h = 0}}{\bf{.2}}\\{\bf{T = }}{{\bf{T}}_{\bf{o}}}{\bf{ + h}}\frac{{{\bf{F + G}}}}{{\bf{2}}}{\bf{ = 1}}{\bf{.570}}\end{array}\)

Therefore, at \({\bf{x = 0}}{\bf{.2}}\), the solutions are approximately \({\bf{v = 1}}{\bf{.570}}\).

The other values are

t | v | t | v | t | v |

0.4 | 2.637 | 2 | 4.801 | 3.6 | 4.900 |

0.6 | 3.363 | 2.2 | 4.834 | 3.8 | 4.902 |

0.8 | 3.856 | 2.4 | 4.857 | 4 | 4.903 |

1 | 4.192 | 2.6 | 4.872 | 4.2 | 4.904 |

1.2 | 4.420 | 2.8 | 4.883 | 4.4 | 4.904 |

1.4 | 4.575 | 3 | 4.890 | 4.6 | 4.904 |

1.6 | 4.681 | 3.2 | 4.895 | 4.8 | 4.905 |

1.8 | 4.753 | 3.4 | 4.898 | 5 | 4.905 |

The given equation is

\(\begin{array}{c}{\bf{v(0) = 0}}\\\frac{{{\bf{dv}}}}{{{\bf{dt}}}}{\bf{ = 9}}{\bf{.81 - 2}}{{\bf{v}}^{\bf{r}}}\end{array}\)

Now apply improved Euler’s method subroutine with \({\bf{h = 0}}{\bf{.2}}\) and N=25.

\(\begin{array}{c}{\bf{f(t,T) = 9}}{\bf{.81 - 2}}{{\bf{v}}^{{\bf{1}}{\bf{.5}}}}\\{\bf{G = f(t + h,v + hF)}}\\{\bf{ = 9}}{\bf{.81 - 2(v + 0}}{\bf{.2(9}}{\bf{.81 - 2}}{{\bf{v}}^{{\bf{1}}{\bf{.5}}}}{\bf{)}}{{\bf{)}}^{{\bf{1}}{\bf{.5}}}}\end{array}\)

Apply initial conditions \({\bf{t = }}{{\bf{t}}_{\bf{o}}}{\bf{ = 0,v = }}{{\bf{v}}_{\bf{o}}}{\bf{ = 0}}\).

\(\begin{array}{c}{\bf{F(0,0) = 9}}{\bf{.81}}\\{\bf{G(0,0) = 4}}{\bf{.3136}}\\{\bf{t = }}{{\bf{t}}_{\bf{0}}}{\bf{ + h = 0}}{\bf{.2}}\\{\bf{T = }}{{\bf{T}}_{\bf{o}}}{\bf{ + h}}\frac{{{\bf{F + G}}}}{{\bf{2}}}\\{\bf{ = 1}}{\bf{.412}}\end{array}\)

Thus, at \({\bf{x = 0}}{\bf{.2}}\), the solutions are approximately \({\bf{v = 1}}.412\).

The other values are

t | v | t | v | t | v |

0.4 | 2.150 | 2 | 2.884 | 3.6 | 2.887 |

0.6 | 2.519 | 2.2 | 2.885 | 3.8 | 2.887 |

0.8 | 2.703 | 2.4 | 2.886 | 4 | 2.887 |

1 | 2.795 | 2.6 | 2.887 | 4.2 | 2.887 |

1.2 | 2.841 | 2.8 | 2.887 | 4.4 | 2.887 |

1.4 | 2.864 | 3 | 2.887 | 4.6 | 2.887 |

1.6 | 2.875 | 3.2 | 2.887 | 4.8 | 2.887 |

1.8 | 2.881 | 3.4 | 2.887 | 5 | 2.887 |

The given equation is

** **

\(\begin{array}{c}{\bf{v(0) = 0}}\\\frac{{{\bf{dv}}}}{{{\bf{dt}}}}{\bf{ = 9}}{\bf{.81 - 2}}{{\bf{v}}^{\bf{r}}}\end{array}\)

Now apply improved Euler’s method subroutine with \({\bf{h = 0}}{\bf{.2}}\) and \({\bf{N = 25}}\).

\(\begin{array}{c}{\bf{f(t,T) = 9}}{\bf{.81 - 2}}{{\bf{v}}^{\bf{2}}}\\{\bf{G = f(t + h,v + hF)}}\\{\bf{ = 9}}{\bf{.81 - 2(v + 0}}{\bf{.2(9}}{\bf{.81 - 2}}{{\bf{v}}^{\bf{2}}}{\bf{)}}{{\bf{)}}^{\bf{2}}}\end{array}\)

Apply initial conditions\({\bf{t = }}{{\bf{t}}_{\bf{o}}}{\bf{ = 0,v = }}{{\bf{v}}_{\bf{o}}}{\bf{ = 0}}\).

\(\begin{array}{c}{\bf{F(0,0) = 9}}{\bf{.81}}\\{\bf{G(0,0) = 2}}{\bf{.111}}\\{\bf{t = }}{{\bf{t}}_{\bf{0}}}{\bf{ + h = 0}}{\bf{.2}}\\{\bf{T = }}{{\bf{T}}_{\bf{o}}}{\bf{ + h}}\frac{{{\bf{F + G}}}}{{\bf{2}}}{\bf{ = 1}}{\bf{.192}}\end{array}\)

Hence, at \({\bf{x = 2}}{\bf{.0}}\), the solutions are approximately \({\bf{v = 1}}.192\).

The other value is

t | v | t | v | t | v |

0.4 | 1.533 | 2 | 2.218 | 3.6 | 2.201 |

0.6 | 1.719 | 2.2 | 2.146 | 3.8 | 2.204 |

0.8 | 1.841 | 2.4 | 2.160 | 4 | 2.206 |

1 | 1.927 | 2.6 | 2.171 | 4.2 | 2.208 |

1.2 | 1.991 | 2.8 | 2.180 | 4.4 | 2.209 |

1.4 | 2.039 | 3 | 2.187 | 4.6 | 2.210 |

1.6 | 2.077 | 3.2 | 2.193 | 4.8 | 2.211 |

1.8 | 2.105 | 3.4 | 2.197 | 5 | 2.211 |

Therefore, when \({\bf{r = 1}}{\bf{.0}}\), the constant solution is \({\bf{v(t) = 4}}{\bf{.905}}\).

When \({\bf{r = }}\,{\bf{1}}{\bf{.5}}\), the constant solution is \({\bf{v(t) = }}2.887\).

When \({\bf{r = }}\,{\bf{2}}{\bf{.0}}\), the constant solution is \({\bf{v(t) = }}2.215\).

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