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A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.9 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?

Respuesta :

Answer:

v = 2.85 m/s

Explanation:

By the law of similarity of the triangle we can say

[tex]\frac{height\: of\: man}{distance\: of \: man\: from \: spotlight} = \frac{height \:of \:image}{distance\: of \:wall\: from\: spotlight}[/tex]

here we know that

height of man = 2 m

let the distance of man from spotlight = x

distance of wall from spotlight = 12 m

height of image is let say "y"

so we will have

[tex]\frac{2}{x} = \frac{y}{12}[/tex]

[tex]y = \frac{24}{x}[/tex]

now we have

[tex]\frac{dy}{dt} = \frac{24}{x^2}\frac{dx}{dt}[/tex]

here we know

[tex]\frac{dx}{dt} = speed = 1.9 m/s[/tex]

x = 4 m

now we have

[tex]\frac{dy}{dt} = \frac{24}{4^2}(1.9)[/tex]

[tex]v = 2.85 m/s[/tex]

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