A 5 ft long man walks away from the foot of a 12(½) ft high lamp post at the rate of 3 mph. What will be the rate at which the shadow increases?(a) 0mph(b) 1mph(c) 2mph(d) 3mphI got this question at a job interview.My question is from Application of Derivative in chapter Application of Derivatives of Mathematics – Class 12

A 5 ft long man walks away from the foot of a 12(½) ft high lamp post

1.	A 5 ft long man walks away from the foot of a 12(½) ft high lamp post at the rate of 3 mph. What will be the rate at which the shadow increases?(a) 0mph(b) 1mph(c) 2mph(d) 3mphI got this question at a job interview.My question is from Application of Derivative in chapter Application of Derivatives of Mathematics – Class 12
Answer» The correct option is (C) 2mph Explanation: LET, AB be the lamp-post whose FOOT is A, and B is the source of light, and given (AB)’ = 12(½) ft. Let MN denote the position of the man at time t where (MN)’ = 5ft. Join BN and produce it to meet AM(produced) at P. Then the length of man’s shadow= (MP)’ Assume, (AM)’ = x and (MP)’ = y. Then, (PA)’ = (AM)’ + (MP)’ = x + y And dx/dt = velocity of the man = 3 Clearly, TRIANGLES APB and MPN are similar. Thus, (PM)’/(MN)’ = (PA)’/(AB)’ Or, y/5 = (x + y)/12(½) Or, (25/2)y = 5X + 5y Or, 3y = 2x Or, y = (2/3)x Thus, dy/dt = (2/3)(dx/dt) As, dx/dt = 2, = 2/3*3 = 2mph

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