Word problem help?

Question

A balloon is at a height of 20 meters, and is rising at the constant rate of 5 m/sec. A bicyclist passes beneath it, traveling in a straight line at the constant speed of 10 m/sec. How fast is the distance between the bicyclist and the balloon increasing 2 seconds later?

Answer 1

related rates problem, under derivatives

a=20+5(2) = 30, a'=5= balloon speed, b=10(2) = 20, b' = 10= bicycle speed

d^2 = a^2 +b^2 = 30^2+20^2 = 900+400=1300, d= sqr1300 = 10sqr13= distance from balloon to bicycle

helps to draw a right triangle

take derivatives of the Pythagorean equation

2dd' = 2aa' +2bb'

divide by 2

dd' = aa' +bb'

plug in known values

10sqr13(d') =30(5)+20(10) = 150+200 = 350

d' = 35/sqr13 = 35(sqr13)/13 m/sec

= about 9.707 meters per second