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Chapter 3: Q20E (page 131)

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 鈥揵v. 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 rin (14),take \({\bf{m = 1,}}\,\,{\bf{g = 9}}{\bf{.81,}}\,\,{\bf{b = 2}}\) and \({{\bf{v}}_{\bf{o}}}{\bf{ = 0}}\).Then use the improved Euler鈥檚 method subroutine with \({\bf{h = 0}}{\bf{.2}}\) to approximate the solution to (14) on the interval \(0 \le {\bf{t}} \le 5\)for \({\bf{r = 1}}{\bf{.0,}}\,\,{\bf{1}}{\bf{.5}}\) and 2.0. What is the relationship between these solutions and the constant solution\({\bf{v(t) = }}{\left( {\frac{{{\bf{9}}{\bf{.81}}}}{{\bf{2}}}} \right)^{\frac{{\bf{1}}}{{\bf{r}}}}}\)?

Short Answer

Expert verified

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 \({\bf{v(t) = }}2.215\).

Step by step solution

01

Important hint

For the solution apply Euler鈥檚 formula.

02

Find the value of temperature when \({\bf{r = 1}}{\bf{.0}}\).

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鈥檚 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

03

Determine the value of temperature when\({\bf{r = 1}}{\bf{.5}}\).

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鈥檚 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

04

Evaluate the value of temperature when \({\bf{r = }}2.0\).

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鈥檚 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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