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
[tex]\dfrac{AD}{AB} = \dfrac{DE}{BC}[/tex]
[tex]\dfrac{8}{12} = \dfrac{2}{y}[/tex]
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
∴
[tex]\dfrac{12-x}{12}=\dfrac{2}{y}[/tex]
[tex]1- \dfrac{1}{12}x = 2* \dfrac{1}{y}[/tex]
To find the derivatives of both sides ;we have:
[tex]- \dfrac{1}{12}dx = 2* \dfrac{1}{y^2}dy[/tex]
[tex]- \dfrac{1}{12} \dfrac{dx}{dt} = 2* \dfrac{1}{y^2} \dfrac{dy}{dt}[/tex]
During that time ;
[tex]\dfrac{dx}{dt }= 1.4 \ m/s[/tex] and y = 3
So; replacing the value into above ; we have:
[tex]-\dfrac{1}{12}(1.4) = - \dfrac{2}{9} \dfrac{dy}{dt}[/tex]
[tex]\dfrac{dy}{dt} = \dfrac{\dfrac{ 1.4} {12 } }{ \dfrac{2}{9}}[/tex]
[tex]\dfrac{dy}{dt} = {\dfrac{ 1.4} {12 } }*{ \dfrac{9}{2}}[/tex]
[tex]\dfrac{dy}{dt} =0.525 \ m/s[/tex]
Thus; the length of his shadow on the building is decreasing at the rate of 0.525 m/s
