Solve it quickly please Find the orthogonal trajectories ofa sin? 0 where a be the arbitrary constant. Find the orthogonal trajectories of r = a sin 0 where a be the arbitrary constant.

Answered: Find the orthogonal trajectories of r=…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Solve it quickly please

Find the orthogonal trajectories of
a sin? 0 where a be the arbitrary constant.

Find the orthogonal trajectories of r = a sin 0 where a be the arbitrary constant.

Expert Solution

Step 1

Given:

Given that $r = a \sin^{2} θ$ , where a is arbitrary constant

we want to find orthogonal trajectories of $r = a \sin^{2} θ$

Step 2

Solution:

Given equation $r = a \sin^{2} θ$ , where a is arbitrary constant

Differentiating given equation with respect to $θ$ gives

$r' = 2 a \sin θ \cos θ$ ..............*

Now , $r = a \sin^{2} θ \Rightarrow a = \frac{r}{\sin^{2} θ}$

Therefore substituting $a = \frac{r}{\sin^{2} θ}$ in * we get

$\begin{array}{rcl} r' & = & 2 (\frac{r}{\sin^{2} θ}) \sin θ \cos θ \\ = & 2 r (\frac{\cos θ}{\sin θ}) \\ r' & = & 2 r co t (θ) \end{array}$

Now replace $r'$ with $(\frac{- 1}{r'})$ to write the differential equation of the orthogonal curve

$- \begin{array}{rcl} \frac{1}{r'} \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} 2 \end{array} \begin{array}{rcl} r \end{array} \begin{array}{rcl} co \end{array} \begin{array}{rcl} t \end{array} \begin{array}{rcl} (θ) \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \Rightarrow \end{array} \begin{array}{rcl} r \end{array} \begin{array}{rcl} ' \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} - \end{array} \begin{array}{rcl} \frac{1}{2 r c o t (θ)} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \Rightarrow \end{array} \begin{array}{rcl} r \end{array} \begin{array}{rcl} ' \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} - \end{array} \begin{array}{rcl} \frac{\tan (θ)}{2 r} \end{array}$

Now we can integrate this differential equation

$r' = - \frac{\tan (θ)}{2 r} \Rightarrow \frac{d r}{d θ} = - \frac{\tan (θ)}{2 r} \Rightarrow 2 r d r = - \tan (θ) d θ \Rightarrow \int 2 r d r = - \int \tan (θ) d θ \Rightarrow 2 (\frac{r^{2}}{2}) = \ln |\cos θ| + \ln C$

$\begin{array}{rcl} \end{array} \begin{array}{rcl} \Rightarrow \end{array} \begin{array}{rcl} r \end{array} \begin{array}{rcl} ^{2} \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} \ln \end{array} \begin{array}{rcl} |\cos θ| \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} \ln \end{array} \begin{array}{rcl} C \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \Rightarrow \end{array} \begin{array}{rcl} r \end{array} \begin{array}{rcl} ^{2} \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} \ln \end{array} \begin{array}{rcl} (C |\cos θ|) \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \Rightarrow \end{array} \begin{array}{rcl} C \end{array} \begin{array}{rcl} |\cos θ| \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} e \end{array} \begin{array}{rcl} x \end{array} \begin{array}{rcl} p \end{array} \begin{array}{rcl} (r^{2}) \end{array} \begin{array}{rcl} \Rightarrow \end{array} \begin{array}{rcl} \cos \end{array} \begin{array}{rcl} θ \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} \pm \end{array} \begin{array}{rcl} \frac{1}{C} \end{array} \begin{array}{rcl} e \end{array} \begin{array}{rcl} x \end{array} \begin{array}{rcl} p \end{array} \begin{array}{rcl} (r^{2}) \end{array}$

by denoting $C_{1} = \pm \frac{1}{C}$ we obtain the final implicit equation of the orthogonal trajectories

$\begin{array}{rcl} \cos \end{array} \begin{array}{rcl} θ \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} \pm \end{array} \begin{array}{rcl} C_{1} \end{array} \begin{array}{rcl} e \end{array} \begin{array}{rcl} x \end{array} \begin{array}{rcl} p \end{array} \begin{array}{rcl} (r^{2}) \end{array}$

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