# Orthogonal Trajectories- Differential Equations

Two families of curves such that every member of either family cuts each member of the other family at right angles are called orthogonal trajectories of each other.

The concept of the orthogonal trajectories is of wide use in applied mathematics, especially in field problems.

For instance, in an electric field, the paths along which the current flows are the orthogonal trajectories of equipotential curves and vice versa.

In fluid flow, the streamlines and the equipotential lines are orthogonal trajectories.

Example

Find the Orthogonal Trajectories of the family of curves xy=Constant

Solution: Family of curves  xy= c

Differentiate w.r.t  'x'

$$x.\frac{dy}{dx} +y.1=0$$

Now replace $$\frac{dy}{dx}$$ by $$-\frac{dx}{dy}$$

$$x\frac{dx}{dy}=y$$

By variable separable, $$\int xdx = \int ydy$$

$$\frac{x^{2}}{2}=\frac{y^{2}}{2}+k$$

x² -y² = k₁ is the Orthogonal Trajectories of the given family of curves.

Orthogonal Trajectories of Polar curves

Example

Find the Orthogonal Trajectories of the family of curve $$r^{n}=a^{n}sin\mathit{n}\Theta$$

Solution: Family of curves $$r^{n}=a^{n}sin\mathit{n}\Theta$$ ..... i

Differentiate w.r.t  'Ө' and eliminate 'a'

$$nr^{n-1}\frac{\mathrm{dr} }{\mathrm{d} \Theta }=a^{n}cos\mathit{n\Theta} \times n$$   .... ii

Divided equation (ii) by equation (i)

$$\frac{nr^{n-1}\frac{\mathrm{dr} }{\mathrm{d} \Theta }}{r^{n}} = \frac{a^{n}cos\mathit{n\Theta} \times n}{a^{n}sin\mathit{n}\Theta}$$

$$\frac{dr}{d\Theta }\frac{1}{r}=cot n\Theta$$

The differential equation represents the given family of curves.

Replace $$\frac{\mathrm{dr} }{\mathrm{d} \Theta } by -r^{2}\frac{\mathrm{d\Theta } }{\mathrm{d} r}$$

$$\frac{1}{r}\left(-r^{2} \frac{d \theta}{d r}\right)=\cot n \theta$$

$$-r \frac{d \theta}{d r} =\cot n \theta$$

$$\int \frac{1}{r} d r =-\int \tan n \theta d \theta$$

$$\log r =-\frac{\log \sec n \theta}{n}+\log c$$

$$\log r^{n} =\log \left[c^{n} \cos n \theta\right]$$

$$r^{n} =c^{n} \cos n \theta$$

This is required Orthogonal Trajectories

Newton's Law of Cooling

The temperature of a body changes at a rate that is proportional to the difference in temperature between that of the surrounding medium and that of the body itself.

The differential equation is $$\frac{d \theta}{d t} =-k\left(\theta-\theta_{s}\right)$$

by variable separable $$\int \frac{d \theta}{\theta-\theta_{s}} =\int-k d t$$

$$\log \left(\theta-\theta_{s}\right) =-k t+\log c$$

$$\theta-\theta_{s} =c e^{-k t}$$

This is the solution of Newton's Law of Cooling.

Example

A body originally at 80'C cools down to 60'C in 20 minutes, the temperature of the air being 40'C. What will be the temperature of the body after 40 minutes from the original?

Solution: According to Newton's law of cooling

$$\frac{d \theta}{d t} =-k(\theta-40)$$

$$\int \frac{d \theta}{\theta-40} =-\int k d t$$

$$\log (\theta-40) =-k t+\log c$$

$$\theta-40 =c e^{-k t}$$

Put$$t =0, \theta=80^{\circ} \text { in equation (i) }$$

We get,$$c =40$$

Put, $$t =20 \text { min, } \theta=60^{\circ}$$

Then, $$k =\frac{1}{20} \log 2$$

By equation(i)$$\theta =40+40 e^{\left(-\frac{1}{2} \log 2\right)t}$$

Put, $$t =40 \text { min, then } \theta=50^{\circ} \mathrm{C}$$

Law of Growth

The rate of the change amount of a substance with respect to time is directly proportional to the amount of substance present.

$$\frac{d x}{d t} \propto x$$

$$\frac{d x}{d t} =k x(k>0)$$

$$\int \frac{d x}{x} =\int k d t$$

$$\log x =k t+\log c$$

$$x =c e^{k t} \text { is solution of law of growth }$$

Law of Decay

The rate of change of amount of substances is directly proportional to the amount of substance present.

$$\frac{d x}{d t} \propto x$$

The differential equation is

$$\frac{d x}{d t} =-k x(k>0)$$

$$\int \frac{d x}{x} =-\int k d t$$

$$\log x =-k t+\log c$$

$$x =c e^{-k t} \text { is solution of law of decay. }$$