Mathematics: A child, 1 m tall, is walking directly under a street lamp that is 6 m above the ground. If the child walks away from the light at the rate of 20 m/min, how fast is the child s shadow lengthening?

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A child, 1 m tall, is walking directly under a street

hallee25

20.01.2022

The shadow is increasing with the rate of 4 m/min

Step-by-step explanation:

Consider x represents the length of the boy from the lamp and y represents the length of his shadow,

Given,

The height of boy = 1 m,

The height of lamp = 6 m,

By making the diagram of this situation,

We get two similar triangles one having sides 6 and x+y , and other having sides 1 and y,

Since in similar triangles corresponding sides are in same proportion,

Thus,

$\frac{6}{x+y}=\frac{1}{y}$

6y=x+y

6y-y=x

5y=x

$\implies y=\frac{x}{5}$

Differentiate with respect to t (time),

$\frac{dy}{dt}=\frac{1}{5}\frac{dx}{dt}$

We have $\frac{dx}{dt}=20\text{ m per min}$

$\frac{dy}{dt}=\frac{1}{5}(20)=4\text{ m per min}$

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