Question

g A spotlight on the ground is shining on a wall 20m away. If a woman 2m tall walks from the spotlight toward the building at a speed of 0.8m/s, how fast is the length of her shadow on the building decreasing when she is 4m from the building

Answer 1

Answer:

0.3 m/s

Step-by-step explanation:

$\dfrac{dx}{dt}=\text{Speed of person}=0.8\ \text{m/s}$

As the two triangles in the diagram are similar to each other we have

$\dfrac{y}{12}=\dfrac{2}{8}\\\Rightarrow y=12\times \dfrac{2}{8}\\\Rightarrow y=3\ \text{m}$

Again as the triangles are similar we have

$\dfrac{y}{2}=\dfrac{12}{12-x}\\\Rightarrow x=12-\dfrac{24}{y}$

Differentiating the above equation with respect to time we get

$\dfrac{dx}{dt}=\dfrac{-24\times-1}{y^2}\dfrac{dy}{dt}\\\Rightarrow \dfrac{dy}{dt}=\dfrac{dx}{dt}\times\dfrac{y^2}{24}\\\Rightarrow \dfrac{dy}{dt}=0.8\times \dfrac{3^2}{24}\\\Rightarrow \dfrac{dy}{dt}=0.3\ \text{m/s}$

The speed at which the shadow is changing is 0.3 m/s.