By Bhaskar Rao, P. B.; Sriramachary, S. K. V. S.; Bhujanga Rao, M.

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I) Engineering Mathematics - I 56 x log- =kl 50 x = 150, when 1= 1 .. 7 Example The rate of cooling of a body is proportional to the difference between the temperature of the body and the surrounding air. If the air temperature is 20°C and the body cools for 20 minutes from 140°C to 80°C, find when the temperature will be 35°C. Solution If 8 is the temperature of the body at time '1' then from Newton's law of cooling -d8 -a(8-20) dl = f~ 8-20 ~ de - =-k(8-20) dl k rldl + c J' log(8 -20) = - kt + c Initially when t = 0, 8 = 140 log(8 - 20) = 0 + c, ....

XJI, - :)=0 ... (3) is the differential equation of the system of orthogonal trajectories, and its solution is the fami Iy of orthogonal trajectories of (I) . 2 Orthogonal Trajectories - Polar Coordinates Suppose f(r,O,c) = 0 ..... (1) is the family of curves where c is the arbitrary constant. We can form a differential equation F(r,o, :~ )= 0 ..... (2) of the family (I), after elimination of the constant 'c'. Let ¢ be the angle between the radius vector and the tangent at any point (r, 0) .

In the flow of electricity in thin conducting sheets, the paths along which the current flows are the orthogonal trajectories of the equipotential curves and vice - versa. 1 Orthogonal Trajectories: Cartesian Coordinates f(x,y,c)=O Let ..... (1) be a family of curves, where c is a parameter. , ..... (2) is the differential equation whose general solution is (1 ) If the two curves are orthogonal (curves intersecting at right angles) the product of the slopes of the tangents at their point of intersection must be equal to -I.