A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks along the x-axis from the spotlight toward the building at a speed of 1.8 m/s, which is taken as the given dx/dt, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?

Answer:0.675 m/sStep-by-step explanation:Let height of shadow= y,CD=xHeight of man=2 mSpeed of man= [tex]\frac{dx}{dt}=1. 8 m/s[/tex][tex]\triangle ABD\sim\triangle ECD[/tex]Therefore, [tex]\frac{AB}{EC}=\frac{BD}{CD}[/tex][tex]\frac{y}{2}=\frac{12}{x}[/tex][tex]xy=24[/tex]Differentiate w.r.t t[tex]x\frac{dy}{dt}+y\frac{dx}{dt}=0[/tex][tex]x\frac{dy}{dt}=-y\frac{dx}{dt}[/tex][tex]\frac{dy}{dt}=-\frac{y}{x}\frac{dx}{dt}[/tex]When the man is 4 m from  the building Then, we have x=12-4=8 m[tex]\frac{dx}{dt}=1.8 m/s[/tex]Substitute the values in above equation then, we get[tex]8y=24[/tex][tex]y=\frac{24}{8}=3[/tex]Substitute the values then we get [tex]\frac{dy}{dt}=-\frac{3}{8}\times 1.8=-0.675 m/s[/tex]Hence, the length of his shadow on the building decreasing at the rate 0.675 m/s.