 Namaste, Myself, Mr. Birajdar Bala Sahib, Assistant Professor, Department of Humanities and Sciences, Walchand Institute of Technology, SolarPur. In this video lecture, we will discuss Applications of First Order and First Degree Ordinary Differential Equation, Part 3. At the end of this session, students will be able to find orthogonal trajectories of the family of polar curve. Let us start with general equation of polar curve. In general, the polar curve are represented by a functional relation between r and theta that is of the type f of r, theta, c is equal to 0, where r is called radius factor at point p of r, theta and theta is angle made by radius factor with initial line as shown in figure. This is the initial line and it is the perpendicular line. Its equation is theta equal to pi by 2 and this is the equation of a curve. p of r, theta is the point on this curve and when we join point p with origin, we get o p and that o p is nothing but radius factor denoted by r and theta is the angle made by this radius factor with the initial line in positive direction. So, the functional relation between r and theta at any point on the curve is equation of a curve in polar form. Now, in this video session, we are discussing how to find orthogonal trajectories of given polar curve. Let us start with the working rule to find orthogonal trajectory of a polar curve. Let us denote the given curve is f of r, theta, c equal to 0 by equation 1. In step 1, first we differentiate the given curve f of r, theta, c equal to 0 with respect to theta and eliminate parameter c from the equation of the curve. We get an ordinary differential equation of given curve as f of r, theta, c by d theta is equal to 0. Denote this equation by 2. Now, in step 2, replace d r by d theta by minus r square d theta upon d r in above equation 2. We get a differential equation f of r, theta, c minus r square d theta by d r equal to 0. Denote this equation by 3. So, equation 3 is an ordinary differential equation of orthogonal trajectory. Now, in step 3, we have to solve this ordinary differential equation denoted by 3 in step 2. We get a required a family of a polar curve, which is orthogonal to given family of a polar curve. Let us see the illustration. Example 1, find the orthogonal trajectories of the family of polar curve r is equal to a in bracket sec theta plus tan theta, where a is a parameter solution. Let us denote the given curve r is equal to a in bracket sec theta plus tan theta by equation 1. Now, by first step, differentiate the given curve with respect to 2 theta. We get d r by d theta equal to constant a in bracket. Derivative of sec theta is sec theta into tan theta plus derivative of tan theta is sec square theta bracket close. Therefore, d r by d theta equal to a constant as it is and in bracket, when we take sec theta is common, we get in bracket tan theta plus sec theta and into sec theta. Denote this equation by 2. Now, we have to eliminate parameter a between equation 1 and 2. For that, from equation 1, we know that the value of a in bracket sec theta plus tan theta is equal to r. Put it in the right hand side of equation 2. We get d r by d theta is equal to r into sec theta. Denote this equation by 3. This is the differential equation of given polar curve. Now, in order to find orthogonal trajectory, first of all, we have to replace d r by d theta is equal to minus r square d theta upon d r in equation 3. We get minus r square d theta upon d r equal to r into sec theta and 1 r gets cancelled from both the side and attach minus sign to right side. We get r into d theta upon d r equal to minus sec theta. Denote this equation by 4. This is the ordinary differential equation of orthogonal trajectory. Now, we have to solve this equation 4 using method of separating the variable. For that, we can write equation 4 as dividing sec theta and r theta on cross division. We get 1 by sec theta d theta equal to minus 1 by r d r which is equal to 1 by sec theta is cos theta into d theta equal to minus 1 by r into d r. This is the invariable separable form. Now, integrating both the side, integration of cos theta d theta equal to minus integration of 1 by r into d r. Therefore, integration of cos theta is sin theta which is equal to minus integration of 1 by r with respect to r is log r plus k constant of integration. This is the family of polar core which is orthogonal to given family of a polar core. Let us pause the video for a while and write the answer to the given question. Question is find the orthogonal trajectories of the family of core r is equal to a into e raised to theta where a is a parameter. Come back, I hope you have written answer to this question. Here, I will going to explain the solution. Solution is let us denote the given core r is equal to a into e raised to theta by equation 1. Now, first of all differentiate this core with respect to 2 theta, we get d r by d theta equal to constant a as it is into derivative of e raised to theta is e raised to theta denote this equation by 2. Now, we have to eliminate parameter a between equation 192 for that from equation 1 we know that value of a into e raised to theta is r put it in the right hand side of equation 2. We get d r by d theta equal to r denote this equation by 3. Now, in this equation 3 we replace d r by d theta by minus r square d theta by d r we get minus r square d theta by d r equal to r. Now, 1 r gets cancelled from both the side we get minus r d theta by d r equal to 1 denote this equation by 4. Now, we have to solve this equation 4 by separating the variable for that we can write equation 4 as d theta equal to minus 1 by r into d r this is in variable separable form. Now, integrating both the side integration of d theta equal to minus integration of 1 by r into d r which gives integration of d theta is theta is equal to minus integration of 1 by r is log r plus constant k this is the family of a core which is orthogonal to given family of a core. Let us consider another example example 2 find the orthogonal trajectories of the family of core r raise to n equal to a raise to n cos of n theta where a is a parameter and n is a constant. Now, solution let us denote the given core r raise to n equal to a raise to n into cos n theta by equation 1. Now, instead of differentiating directly we can take log of r raise to n log of a raise to n into cos n theta. Now, using properties of logarithm we can write log of r raise to n as n into log of is equal to in the right hand side log of a raise to n into cos n theta we can write as log of a raise to n plus log of cos of n theta. Now, differentiating both the side with respect to theta we get n as it is and derivative of log r is 1 by r into d r by d theta is equal to now log of a raise to n is constant and its derivative is 0 plus derivative of log of cos n theta is 1 upon cos n theta into derivative of cos n theta is minus sin n theta into n. Now, 1 n gets cancelled from both the side and we get 1 by r d r by d theta equal to minus sin as it is sin n theta upon cos n theta is tan n theta. Therefore, we get 1 by r into d r by d theta equal to minus tan n theta denote this equation by 2 this is the differential equation of given curve. Now, to find orthogonal trajectory we have to replace d r by d theta by minus r square d theta upon d r in equation 2. Therefore, we get 1 by r as it is and d r by d theta by minus r square d theta by d r which is equal to minus tan n theta as it is. Now, in the left hand side 1 r gets cancelled from numerator and denominator and multiply minus sign on both the side which gives totally r into d theta by d r is equal to tan n theta denote this equation by 3. This is the differential equation of orthogonal trajectory. Now, we have to solve this equation for that we can use the method of separating the variable. Therefore, divide tan n theta to left side and r to right side we get 1 upon tan n theta d theta equal to 1 upon r into d r, but 1 upon tan n theta is cot n theta d theta equal to 1 by r into d r which is in a variable separable form. Now, integrating both the side integration of cot n theta d theta equal to integration of 1 by r d r which is equal to integration of cot n theta is log sin n theta upon n is equal to integration of 1 by r is log r plus log of k which is a constant of integration. Now, using logarithm property we can simplify right hand side as therefore, log of sin n theta upon n left side as it is and in right side log of r plus log of k we can write log of r into k by property log a plus log b equal to log of a b. Now, multiply n on both the side we get log of sin n theta equal to n into log of r k. Therefore, log of sin n theta equal to log of r k rest to n. Now, cancel logarithm on both the side we get sin n theta equal to r rest to n into k rest to n. This is the required family of a polar core which is orthogonal to given family. To prepare this video lecture I refer this book as references. Thank you.