A person of height 2 meters is walking away from an 8 meter tall street light at a speed of .5 meters per second. At what rate is the person's shadow

Thank you Melody ! :)

Hi Solveit and Guest.   :)

Solveit asked me to look at this question too :)  

Thanks Solveit, it is a good question  :)

A person of height 2 meters is walking away from an 8 meter tall street light at a speed of .5 meters per second. At what rate is the person's shadow increasing in length??

$\frac{dx}{dt}=0.5 \qquad find \quad \frac{dy}{dt}$

$tan\theta = \frac{8}{x+y}=\frac{2}{y}\\ 8y=2x+2y\\ 6y=2x\\ y=\frac{x}{3}\\ \frac{dy}{dx}=\frac{1}{3}$

$\frac{dy}{dt}=\frac{dy}{dx}\times \frac{dx}{dt}\\ \frac{dy}{dt}=\frac{1}{3}\times0.5\\ \frac{dy}{dt}=\frac{1}{6} \\ \mbox{So the shadow increases at at the rate of } \;\frac{1}{6 }\;\;m/s$

You are most welcome Solveit :)