calculus
posted by nesri .
A shark, looking for dinner, is swimming parallel to a straight beach and is 90 meters offshore. The shark is swimming at a constant speed of 30 meter per second. At time t = 0, the shark is directly opposite a lifeguard station. How fast is the shark moving away from the lifeguard station when the distance between them is 150 meters?

make a diagram, you will have a rightangled triangle
let the horizontal distance covered by the shark since t=0 be x m , and
let the distance between the shark and the lifeguard be y m
x^ + 90^2 = y^2
2x(dx/dt) + 0 = 2y(dy/dt)
when y = 150
x^2 + 90^2 = 150^2
x = √14400 = 120
dy/dt = 2x(dx/dt)/(2y)
= 2(120)(30)/300
= 24 m/sec