Question

If air resistance is proportional to the square of the instantaneous velocity, then the velocityof a massdropped from a given height is determined from

Let$\mathit{v}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{,}\mathit{k}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{125}\mathbf{,}\mathit{m}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}$slugs, and $\mathit{g}\mathbf{=}\mathbf{32}\mathbf{\text{ft}}\mathbf{/}{\mathbf{\text{s}}}^{\mathbf{2}}$.

1. Use the RK4 method with h = 1to approximate the velocity v(5).
2. Use a numerical solver to graph the solution of the IVP on the interval \([0,6]\).
3. Use separation of variables to solve the IVP and then find the actual value v(5)

Short answer

1. the approximation $v_5=35.82054$is the functional value at t=5 , and thus $v\left(5\right)=35.82054$.
2. it can be concluded that velocity after a particular time remains constant.
3. The solution of the given initial value problem is $v\left(t\right)=\sqrt{1280\frac{e^{\frac{4}{\sqrt{5}}t}-1}{\frac{4}{{\sqrt{5}}_t^t}+1}}$and the actual value of the solution is$\underset{¯}{v\left(5\right)=32.9503}.$

Step-by-step solution

First and initial approximation method

Begin by making the first or initial approximation aswhen independent variable value is$t_0$.

The $(n+1{)}^{\text{th}}$approximation is obtained from the $n^{\text{th}}$approximation by

$v_{n+1}=v_n+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)$.

Let $F(t,v)$be the given function with the width of approximation h, then,

$k_1=hF\left(t_n,v_n\right)k_2=hF\left(t_n+\frac{h}{2},v_n+\frac{k_1}{2}\right)k_3=hF\left(t_n+\frac{h}{2},v_n+\frac{k_2}{2}\right)k_4=hF\left(t_n+h,v_n+k_3\right)$

Applying First and initial approximation method

Rewrite the given differential equation in the form of$v^{\text{'}}=F(t,v)$as shown below.

$$\begin{gathered} m\frac{dv}{dt}=mg-k{v}^2,k>0\frac{dv}{dt}=g-\frac{k}{m}{v}^2 \\ F(t,v)=g-\frac{k}{m}{v}^2 \end{gathered}$$

Substitute the values $v\left(0\right)=0,k=0.125,m=5,g=32$in the above formula and solve as shown below.

$\begin{array}{rcl}F(t,v)& =& g-\frac{k}{m}{v}^2\\ & =& 32-\frac{0.125}{5}{v}^2\\ & =& 32-0.025v^2\\ & & \end{array}$

Obtain the values of$k_1,k_2,k_3$and$k_4$.

$k_1=1\times F(0,0)=32-0.025\times (0{)}^2=32$

Use$k_1$to obtain$k_2$as shown below.

$\begin{array}{rcl}{k}_2& =& 1\times F\left(0+\frac{1}{2},0+\frac{32}{2}\right)\\ & =& F(0.5,16)\\ & =& 32-0.025\times (16{)}^2\\ & =& 25.6\\ & & \end{array}$

Use$k_2$to $k_3$obtainas follows.

$\begin{array}{rcl}{k}_3& =& 1\times F\left(0+\frac{1}{2},0+\frac{25.6}{2}\right)\\ & =& F(0.5,12.8)\\ & =& 32-0.025\times (12.8{)}^2\\ & =& 27.904\end{array}$

Use $k_3$to $k_4$obtain as shown below.

$\begin{array}{rcl}{k}_4& =& 1\times F(0+1,0+27.904)\\ & =& F(1,27.904)\\ & =& 32-0.025\times (27.904{)}^2\\ & =& 12.53417\\ & & \end{array}$

Now, evaluate the functional value v(1).

$\begin{array}{rcl}{v}_1& =& v_0+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0+\frac{1}{6}(32+(2\times 25.6)+(2\times 27.904)+12.5342)\\ & =& \frac{1}{6}\times (32+51.2+55.808+12.5342)\\ & =& 25.2570\\ & & \end{array}$

Calculate t1 using t0 and h as shown below.

$\begin{array}{rcl}{t}_1& =& t_0+h\\ & =& 0+1\\ & =& 1\end{array}$

Using the approximation $v_1=25.2570$, evaluate the approximation $v_2=v\left(2\right)$.

Obtain the values of $k_1,k_2,k_3$and $k_4$
.

$$\begin{gathered} k_1=1\times F(1,25.2570) \\ =32-0.025\times (25.2570{)}^2 \\ =15.9329 \end{gathered}$$

Use k₁ to obtain k₂ :

$$\begin{gathered} k_2=1\times F\left(1+\frac{1}{2},25.3512+\frac{15.9329}{2}\right) \\ =F(1.5,33.31765) \\ =32-0.025\times (33.31765{)}^2 \\ =4.24835 \end{gathered}$$

Use k₂ to obtain k₃ :

$\begin{array}{rcl}{k}_3& =& 1\times F\left(1+\frac{1}{2},25.3512+\frac{4.24835}{2}\right)\\ & =& F(0.5,27.47538)\\ & =& 32-0.025\times (27.47538{)}^2\\ & =& 13.12759\\ & & \end{array}$

Use k₃ to obtain k₄ :

$\begin{array}{rcl}{k}_4& =& 1\times F(1+1,25.3512+13.12759)\\ & =& F(1,38.47879)\\ & =& 32-0.025\times (38.47879{)}^2\\ & =& -5.01543\end{array}$

Now, evaluate the functional value v(2).

$\begin{array}{rcl}{v}_2& =& v_1+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 25.3512+\frac{1}{6}(15.9329+(2\times 4.24835)+(2\times 13.12759)-5.01543)\\ & =& 25.3512+\frac{1}{6}\times 45.66935=32.96276\end{array}$

Therefore, calculate t₂ using t₁ and h as follows.

$\begin{array}{rcl}{t}_2& =& t_1+h\\ & =& 1+1\\ & =& 2\end{array}$

Using the approximation v₂,= 32.96276 evaluate the approximation v₃ = v(3)

Obtain the values of k₁ , k_{2 ,} k₃ and k₄

$\begin{array}{rcl}{k}_1& =& 1\times F(2,32.96276)\\ & =& 32-0.025\times (32.96276{)}^2\\ & =& -4.83641\end{array}$

Use k₁ to obtain k₂ :

$\begin{array}{rcl}{k}_2& =& 1\times F\left(2+\frac{1}{2},32.96276+\frac{-4.83641}{2}\right)\\ & =& F(2.5,30.54456)\\ & =& 32-0.025\times (30.54456{)}^2\\ & =& 8.67575\\ & & \end{array}$

Use k₂ to obtain k₃ :

$$\begin{gathered} k_3=1\times F\left(2+\frac{1}{2},32.96276+\frac{8.67575}{2}\right) \\ =F(2.5,37.30064) \\ =32-0.025\times (37.30064{)}^2=-2.78344 \end{gathered}$$

Use k₃ to obtain k₄ :

$$\begin{gathered} k_4=1\times F(2+1,32.96276-2.78344) \\ =F(3,30.17932) \\ =32-0.025\times (30.17932{)}^2=9.23022 \end{gathered}$$

Now, evaluate the functional value v(3) .

$\begin{array}{rcl}{v}_3& =& v_2+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 32.96276+\frac{1}{6}(-4.83641+(2\times 8.67575)+(2\times -2.78344)+9.23022)\\ & =& 32.96276+\frac{1}{6}\times 16.17843=35.65917\end{array}$

Therefore, calculate using and h ,

$t_3=t_2+h=2+1=3$

Using the approximation v₃ = 35.65917 , evaluate the approximation v₄ = v(4) Obtain the values of k₁ , k_{2 ,} k₃ and k₄

$\begin{array}{rcl}{k}_1& =& 1\times F(3,35.65917)\\ & =& 32-0.025\times (35.65917{)}^2\\ & =& 0.21060\end{array}$

Use k₁ to obtain k₂ :

$\begin{array}{rcl}{k}_2& =& 1\times F\left(3+\frac{1}{2},35.65917+\frac{0.21060}{2}\right)\\ & =& F(3.5,35.76447)\\ & =& 32-0.025\times (35.76447{)}^2\\ & =& 0.02257\end{array}$

Use k₂ to obtain k₃ :

$\begin{array}{rcl}{k}_3& =& 1\times F\left(3+\frac{1}{2},35.65917+\frac{0.02257}{2}\right)\\ & =& F(3.5,35.67046)\\ & =& 32-0.025\times (35.67046{)}^2\\ & =& 0.19046\end{array}$

Use k₃ to obtain k₄ :

$\begin{array}{rcl}{k}_4& =& 1\times F(3+1,35.65917+0.19046)\\ & =& F(4,35.84963)\\ & =& 32-0.025\times (35.84963{)}^2\\ & =& -0.12990\\ & & \end{array}$

Now, evaluate the functional value v(4).

$$\begin{gathered} v_4=v_3+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =35.84963+\frac{1}{6}(0.21060+(2\times 0.02257)+(2\times 0.19046)-0.12990) \\ =35.84963+\frac{1}{6}\times 0.50676 \\ =35.93409 \end{gathered}$$

Therefore, calculate t₄ using t₃ and h $t_4=t_3+h$$=3+1=4$

Obtain the values of k₁, k₂, k₃ and k₄.

$k_1=1\times F(4,35.93409)=32-0.025\times (35.93409{)}^2=-0.28147$

$\begin{array}{rcl}{k}_2& =& 1\times F\left(4+\frac{1}{2},35.93409+\frac{-0.28147}{2}\right)\\ & =& F(4.5,35.79336)\\ & =& 32-0.025\times (35.79336{)}^2\\ & =& -0.02912\end{array}$

Use k₂to obtain k₃

$\begin{array}{l}{k}_3=1\times F\left(4+\frac{1}{2},35.93409+\frac{-0.02912}{2}\right)\\ =F(4.5,35.91953)\\ =32-0.025\times {\left(35.91953\right)}^2\\ =-0.25532\end{array}$

Use ${\mathrm{K}}_3$to obtain ${\mathrm{K}}_4$:

$k_4=1\times F(4+1,35.93409-0.25532)=F(5,35.77877)=32-0.025\times (30.17932{)}^2=-0.00301$

Now, evaluate the functional value v(5).

$\begin{array}{rcl}{v}_5& =& v_4+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 35.93409+\frac{1}{6}(-0.28147+(2\times -0.02912)+(2\times -0.25532)-0.00301)\\ & =& 32.96276-\frac{1}{6}\times 0.85336=35.82054\\ & & \end{array}$

Therefore, the approximation $v_5=35.82054$is the functional value at t = 5, and thus $v\left(5\right)=35.82054$.

Use a numerical solver to graph the solution of the IVP on the interval [0,6].

1. Prepare a table in MS-Excel which can be used to assign iterations as rows and the intermediate parameters in columns, relating to the method of numerical approximation using RK4.
2. Fill the cells on the table with initial values.
3. Plot the resultant solution function.

The table obtained from MS-Excel is as shown below in Table 1.

The first column represents the timestamps at which the solution is plotted and the second column represents functional values of the points.

Third column contains slope of the solution curve at the time stamps.

The numbers in columns fromthroughdenote the intermediate values for the RK4 calculation.

Now, graph the initial value problem using the values in the Table 1 as shown below.

From the Figure 1, it can be concluded that velocity after a particular time remains constant.

Compute the initial value problem using the variable separable method and also find the actual value of v(5)

Solutions of a differential equation,is obtained by separating the variables on two sides of the equation to get the form, whereandare homogeneous functions inandrespectively, followed with integration with respect to the corresponding variables.

Substitute the values of $m, g$ and k in the given differential equation as shown below.

$5\frac{dv}{dt}=160-0.125v^2$

Separate the variables such that L.H.S. contains only the variable and the R.H.S. contains the variable .

$$\begin{gathered} 5\frac{dv}{dt}=160-0.125v^2\frac{5dv}{160-0.125v^2} \\ =dt\frac{40dv}{1280-v^2}=dt \end{gathered}$$

Integrate both sides of the resultant equation, consideringas the integrating constant.

$\int \frac{40dv}{1280-v^2}=\int dt$

$40\times \frac{1}{2\sqrt{1280}}\log \left(\frac{\sqrt{1280}+v}{\sqrt{1280}-v}\right)=t+c$

$\frac{\sqrt{1280}+v}{\sqrt{1280}-v}=e^{\frac{4}{\sqrt{5}}(t+c)}$

On further simplifications,

$v\left(t\right)=\sqrt{1280}\frac{e^{\frac{4}{\sqrt{5}}(t+c)}-1}{e^{\frac{4}{\sqrt{5}}(t+c)}+1}$

Substitute the initial conditions t = 5 and v(0) = 0 in equation (1).

$$\begin{gathered} 0=\sqrt{1280}\frac{e^{\frac{4}{\sqrt{5}}(0+c)}-1}{e^{\frac{4}{\sqrt{5}}(0+c)}+1}\left.e^{\left(\frac{4}{\sqrt{5}}c\right.}\right)-1 \\ =0\left.\ln \left\{e^{\left(\frac{4}{\sqrt{5}}c\right.}\right)\right\} \\ =\ln \left\{1\right\}\frac{4}{\sqrt{5}}c=0c=0 \end{gathered}$$

Therefore, the solution of given IVP is obtained by substituting c = (0) in equation (1) as obtained in equation (2).

Substitute t = 5 in equation (2) to find the functional value v(5).

$$\begin{gathered} dv\left(5\right)=\sqrt{1280}\frac{e^{\frac{4}{\sqrt{5}}\times 5}-1}{\frac{4}{\sqrt{5}}\times 5}+1 \\ =\sqrt{1280}\times \frac{e^{4\sqrt{5}}-1}{e^{4\sqrt{5}+1}} \\ =35.77709\times \frac{24.31305-1}{24.31305+1} \\ =32.95032 \end{gathered}$$

Therefore, the actual value is v(5) = 32.95032.