Question

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 1.4 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?

Answer 1

Answer:

the length of his shadow on the building is decreasing at the rate of 0.525 m/s

Step-by-step explanation:

From the diagram attached below;

the man is standing at point D with his head at point E

During that time, his shadow on the wall is y = BC

ΔABC and Δ ADE are similar in nature; thus their corresponding sides have equal ratios; i.e

$\dfrac{AD}{AB} = \dfrac{DE}{BC}$

$\dfrac{8}{12} = \dfrac{2}{y}$

8y = 24

y = 24/8

y = 3 meters

Let take an integral look  at the distance of the man from the building as x, therefore the distance from the spotlight to the man is  12 - x

∴

$\dfrac{12-x}{12}=\dfrac{2}{y}$

$1- \dfrac{1}{12}x = 2* \dfrac{1}{y}$

To find the derivatives of both sides ;we have:

$- \dfrac{1}{12}dx = 2* \dfrac{1}{y^2}dy$

$- \dfrac{1}{12} \dfrac{dx}{dt} = 2* \dfrac{1}{y^2} \dfrac{dy}{dt}$

During that time ;

$\dfrac{dx}{dt }= 1.4 \ m/s$   and y = 3

So; replacing the value into above ; we have:

$-\dfrac{1}{12}(1.4) = - \dfrac{2}{9} \dfrac{dy}{dt}$

$\dfrac{dy}{dt} = \dfrac{\dfrac{ 1.4} {12 } }{ \dfrac{2}{9}}$

$\dfrac{dy}{dt} = {\dfrac{ 1.4} {12 } }*{ \dfrac{9}{2}}$

$\dfrac{dy}{dt} =0.525 \ m/s$

Thus; the length of his shadow on the building is decreasing at the rate of 0.525 m/s