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A point source of light along a straight road is at a height of ‘a’ metres. A boy ‘b’ metres in height is walking along the road. How fast is his shadow increasing if he is walking away from the light at the rate of c metres per minute?


Let AB = a metres be the lamp-post and PQ = b metres the boy, CP = y be his shadow at time t. Let AP = x.
Now ∆CAB and ∆CPQ are equiangular and hence similar

therefore space space space space space space space PC over AC space equals space PQ over AB
rightwards double arrow space space space space space space space fraction numerator straight y over denominator straight x plus straight y end fraction space equals space straight b over straight a
therefore space space space space space space space space space space space space space space space space space space space space space space space space cy space equals space bx plus by
rightwards double arrow space space space space space space space space space space space space space space space space left parenthesis straight a minus straight b right parenthesis space straight y space equals space bx
therefore space space space space space space space space space space space space space space space space space space space space space straight y space equals space fraction numerator straight b over denominator straight a minus straight b end fraction straight x
therefore space space space space space space space space space space space space space space space space space space dy over dx space equals space fraction numerator straight b over denominator straight a minus straight b end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
Also comma space space space space space space space space space space space dx over dt space equals space straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
Now space space space space space dy over dt space equals space dy over dx space. dx over dt space space equals space fraction numerator straight b over denominator straight a minus straight b end fraction cross times straight c space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space of space left parenthesis 1 right parenthesis comma space left parenthesis 2 right parenthesis close square brackets
therefore space space space space space dy over dt space space equals space fraction numerator bc over denominator straight a minus straight b end fraction comma space space space space space space which space gives space the space rate space at space which space boy apostrophe straight s space shadow space is space increasing. space space


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