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Application Of Derivatives

Question
CBSEENMA12035034
Wired Faculty App

A man 2 metres high walks at a uniform speed of 6 metre/sec away from a lamp-post 6 metres high. Find the rate at which the length of his shadow increases.

Solution

Let AB be the lamp-post and PQ the man, CP be his shadow at time t. Let AP = x, PC = y. Also AB = 6 m, PQ = 2 m.
Now ∆CAB and ∆CPQ are equiangular and hence similar.
therefore space space space PC over AC space equals space PQ over AB space space space space space space space space space space space space space space space space space space rightwards double arrow space space space space space space fraction numerator straight y over denominator straight x plus straight y end fraction space equals space 2 over 6 space space space or space space space fraction numerator straight y over denominator straight x plus straight y end fraction space equals space 1 third therefore space space space space space space 3 straight y space equals space straight x plus straight y comma space space space space space space space space space space space space space space space or space space space 2 straight y space equals space straight x
       therefore space space space space space straight y equals space 1 half straight x 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 therefore space space space space space space dy over dx space equals space 1 half 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 space space space space space Also comma space space space space space dx over dt space equals space 6 Now space space space dy over dt space equals space dy over dx space dx over dt space equals space 1 half cross times 6 space equals space 3 therefore space space space space length space of space shadow space increases space at space the space rate space of space 3 straight m divided by sec.

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

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