Question

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 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’s 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

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

Important hint

For the solution apply Euler’s formula.

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’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

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’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

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’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 \({\bf{v(t) = }}2.215\).