Question

Shan, who is 2 meters tall, is approaching a post that holds a lamp 6 meters above the ground. If he is walking at a speed of 1:5 m/s, how fast is the end of his shadow moving (with respect to the lamp post) when he is 6 meters away from the base of the lamp post? 

Answer 1

equation of a straight line with the end of the shadow from time to time. we know that it passes through two points depending on time. the first point is (6-$\frac{t}{5}$ , 2). Second (0, 6).$y=k(t)x+b(t). x=0,y=6\implies b(t)=6\\ x=6-\frac{t}{5},y=2=> 2=k(6-\frac{t}{5})+6;k=\frac{-4}{6-\frac{t}{5}}\\ y=0,0=k(t)x+b(t),x=-6*\frac{6-\frac{t}{5}}{-4}=9-\frac{3t}{10}\\ |x'(0)|=0.3$

Answer:0.3