Answer:
[tex]\dfrac{dy}{dt}=-0.059\ m/s[/tex]
Explanation:
It is given that,
Distance between the spotlight and the wall, y = 24 m
Height of the woman, h = 2 m
The woman walks toward the building at the rate of 0.6 m/s, [tex]\dfrac{dx}{dt}=0.6\ m/s[/tex]
In the attached figure, triangle ABC and MNC are similar. So,
[tex]\dfrac{2}{y}=\dfrac{x}{24}[/tex]............(1)
[tex]y=\dfrac{48}{x}[/tex]
When she is 2 meters from the building. So x = 24-2 = 22 m
[tex]y=\dfrac{48}{22}=2.18\ m[/tex]
Differentiating equation (1) i.e.
[tex]xy=48[/tex]
[tex]x.\dfrac{dy}{dt}+y.\dfrac{dx}{dt}=0[/tex]
[tex]22.\dfrac{dy}{dt}+2.18\times 0.6=0[/tex]
[tex]\dfrac{dy}{dt}=-0.059\ m/s[/tex]
So, her shadow is decreasing at the rate of 0.059 m/s. Hence, this is the required solution.
