Aspire Faculty ID #16326 · Topic: NIMCET 2010 · Just now
NIMCET 2010

The rate of increase of length of the shadow of a man $2$ meters high, due to a lamp at $10$ meters height, when he is moving away from it at $2 \text{ m/sec}$ is

Solution

Let $x =$ distance of man from lamp 
Let $y =$ length of shadow 
 By similar triangles: $\dfrac{10}{x+y} = \dfrac{2}{y}$  Cross-multiply: 
$10y = 2(x+y)$ 
$10y = 2x + 2y$ 
$8y = 2x$ 
$y = \dfrac{x}{4}$ 
 Differentiate w.r.t time $t$: 
 $\dfrac{dy}{dt} = \dfrac{1}{4}\dfrac{dx}{dt}$ 
 Given $\dfrac{dx}{dt} = 2$ m/sec
: $\dfrac{dy}{dt} = \dfrac{1}{4} \times 2 = \dfrac{1}{2}$
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