A calculus problem by A Former Brilliant Member

Calculus Level 2

A lamp post is 4 m 4\text{ m} high. A man with a height of 1.9 m 1.9\text{ m} walks away from this post at a speed of 1.5 m/s 1.5\text{ m/s} . How fast (in m/s \text{ m/s} ) does the end of his shadow move with respect to the lamp post?

Round your answer to 2 decimal places.


The answer is 2.86.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

By similar triangles,

S 4 = x 2.1 \dfrac{S}{4}=\dfrac{x}{2.1}

S = 4 2.1 x S=\dfrac{4}{2.1}x

Differentiate with respect to t t .

d S d t = 4 2.1 d x d t = 4 2.1 ( 1.5 ) = 2.86 m s \dfrac{dS}{dt}=\dfrac{4}{2.1}\dfrac{dx}{dt}=\dfrac{4}{2.1}(1.5)=2.86~\dfrac{m}{s}

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